Holt Geometry. Complete Solutions Manual (2007)

Solutions KeyFoundations for Geometry1

CHAPTER

ARE YOU READY? PAGE 3

1. C 2. E

3. A 4. D

5. 7 1 _ 2 in. 6. 2 1 _

2 cm

7. 100 yd 8. 10 ft

9. 30 in. 10. 15.6 cm

11. 8y 12. -2x + 56

13. -x - 14 14. -2y + 31

15. x + 3x + 7x = 11x = 11(-5) = -55

16. 5p + 10 = 5(78) + 10 = 390 + 10 = 400

17. 2a - 8a = -6a = -6(12) = -72

18. 3n - 3 = 3(16) - 3 = 48 - 3 = 45

19. (0, 7) 20. (-5, 4)

21. (6, 3) 22. (-8, -2)

23. (3, -5) 24. (6, -4)

1-1 UNDERSTANDING POINTS, LINES, AND PLANES, PAGES 6–11

CHECK IT OUT! PAGES 6–8

1. Possible answer: plane � and plane ABC

2.

3. Possible answer: plane GHF

4.

THINK AND DISCUSS, PAGE 8

1. By Post. 1-1-1, through any 2 pts. there is a line. Therefore any 2 pts. are collinear.

2. Post. 1-1-4

3. Any 3 noncollinear pts. determine a plane.

4. � �� AB , −− AB ,

�� AB ,

�� BA ; 0 planes

5.

EXERCISES, PAGES 9–11GUIDED PRACTICE, PAGE 9

1. Possible answer: the intersection of 2 floor tiles

2. S 3. A, B, C, D, E

4. Possible answer: � �� AC , � �� BD

5. Possible answer: ABC and �

6. Possible answer: B, C, or D

7. 8.

9. Possible answer: � �� AB

10. Possible answer: plane ABD

11. 12.

PRACTICE AND PROBLEM SOLVING, PAGES 9–10

13. B, E, A

14. Possible answer: B, C, D, E

15. Possible answer: plane ABC

16. 17.

18. Possible answer: G, J, and � 19. Possible answer: planes and 20. 21.

22a. Possible answer: tip of a stake

b. Possible answer: string

c. Possible answer: grid formed by string

23.

24.

25. U 26. U

27. U

28. If 2 pts. lie in a plane, then the line containing those pts. lies in the plane.

29. If 2 lines intersect, then they intersect in exactly 1 pt.

30. It is not possible. By Post. 1-1-2, any 3 noncollinear pts. are contained in a unique plane. If the 3 pts. are collinear, they are contained in infinitely many planes. In either case, the 3 pts. will be coplanar.

Copyright © by Holt, Rinehart and Winston. 1 Holt GeometryAll rights reserved.

31. A;

32. N;

33. A;

34. S;

35. Post. 1-1-3

36. There are 4 outcomes: (A, B, C), (A, B, D), (A, C, D), (B, C, D); only collinear outcome is (A, B, C). So probability is 1 __

4 .

37. Post. 1-1-2

38. Lines may not intersect: 0 pts. of intersection.

All 3 lines may intersect in 1 pt.

Two of the lines may not intersect, but they might each intersect a third line.

Each line may intersect each of the other lines.

TEST PREP, PAGE 11

39. C; Other 3 sets are collinear.

40. F; Greatest number is when each pair of lines has separate intersection; there are 6 pairs of lines.

41. D; The 2 walls are planes, and they meet in a line.

42. 4; Greatest number is when each triple of pts. determines a separate plane; there are 4 triples of pts.

CHALLENGE AND EXTEND, PAGE 11

43. 6

44. Among 10 pts, there are 45 pairs of pts. Therefore maximum is 45 segs.

45. Maximum = number of pairs of pts. 1st pt. can be chosen in n ways. 2nd pt. can be chosen in n - 1 ways. In this way, each pair is counted twice. Therefore

maximum = n(n - 1)

_______ 2 .

46. Rescue teams can use the principles of Post. 1-1-1 and Post. 1-1-4. A distress signal is received by 2 rescue teams. By Post. 1-1-1, 2 pts. determine a line. So 2 lines are created by the 3 pts., the locations of the rescue teams and the distress signal. By Post. 1-1-4, the unique intersection of the 2 lines will be the location of the distress signal.

SPIRAL REVIEW, PAGE 11

47. Age of mother = aAge of each daughter = a - 25a + 2(a - 25) = 58

3a - 50 = 58 ______ + 50 ____ + 50

3a = 108

3a ___ 3 = 108 ____

3

a = 36Mother is 36. Daughters are 36 - 25 = 11.

48. Yes, each element in the range is assigned to exactly one element in the range.

49. No, since the x-value 10 is assigned to the y-value 6 and the y-value -6.

50. mean = Σx _ n

= 34 _ 8

= 4.25

median = 3 + 5 _ 2 = 4

mode: none

51. mean = Σx _ n

= 2.21 _ 5

= 0.442

median = 0.44

mode = 0.44

TECHNOLOGY LAB: EXPLORE PROPERTIES ASSOCIATED WITH POINTS, PAGE 12

TRY THIS, PAGE 12

1. Check students’ work.

2. No; D must be between A and C.

1-2 MEASURING AND CONSTRUCTING SEGMENTS, PAGES 13–19


1a. XY = 5 - 1 1 _ 2 b. XZ = -3 - 1 1 _

2

= 3 1 __ 2 = -4 1 _

2

= 3 1 _ 2 = 4 1 _

2


3a. XZ = XY + YZ

3 = 1 1 __ 3 + YZ

____

- 1 1 __ 3

_________ - 1 1 __

3

1 2 __ 3 = YZ

b. DF = DE + EF 6x = 3x - 1 + 13 6x = 3x + 12 ____ - 3x ________ - 3x 3x = 12

3x ___ 3 = 12 ___

3

x = 4 DF = 6x = 6(4) = 24

4. Let current location be X, and location of new drinks station be T.

XT = 1 _ 2 XY

= 1 _ 2 (1182.5) = 591.25 m


5. Step 1 Solve for x. RS = ST-2x = -3x - 2 ____ + 3x _______ + 3x x = -2Step 2 Find RS, ST, and RT. RS = -2x = -2(-2) = 4 ST = -3x - 2 = -3(-2) - 2 = 4 RT = RS + ST = 4 + 4 = 8


1. Since R is the mdpt. of −− ST , you know SR = RT.

Also, ST = SR + RT. By subst., ST = SR + SR = 2SR. So ST is twice SR.

2.


1. −− XM and

−− MY 2. distance

3. AB = 1 - (-2.5) = 3.5 = 3.5

4. BC = 3.5 - 1 = 2.5 = 2.5


6. AC = AB + BC 15.8 = 9.9 + BC ____ - 9.9 _________ - 9.9 5.9 = BC

7. MP = MN + NP 5y + 9 = 17 + 3y _______ - 3y _______ - 3y 2y + 9 = 17 ______ - 9 ___ - 9

2y

___ 2 = 8 __

2

y = 4MP = 5y + 9 = 5(4) + 9 = 29

8. Let road sign be R, Sacramento be S, Oakland be O, and location of picnic area be P.

RS + SO = RO 23 + SO = 110 _________ - 23 ____ - 23 SO = 87

RP = RS + SP = RS + 1 _

2 SO

= 23 + 1 _ 2 (87) = 66.5 mi

9. Step 1 Solve for x. JL = 2JK 4x - 2 = 2(7) 4x - 2 = 14 _____ + 2 ___ ___ + 2

4x ___ 4 = 16 ___

4

x = 4Step 2 Find KL and JL.KL = JK = 7JL = 4x - 2 = 4(4) - 2 = 14

10. Step 1 Solve for y. DE = EF 2y = 8y - 3 ____ - 8y _______ - 8y -6y = -3

-6y

____ -6 = -3 ___ -6

y = 0.5Step 2 Find DE, EF,

and DF.DE = 2y = 2(0.5) = 1EF = 8y - 3 = 8(0.5) - 3 = 1DF = DE + EF = 1 + 1 = 2


11. DB = 2 _ 3

- (-5 1 _ 4 ) 12. CD = -5 1 _

4 - (-2)

= 8 _ 12

+ 5 3 _ 12

= -3 1 _ 4

= 5 11 _ 12

= 3 1 _ 4

= 5 11 _ 12


14. CE = CD + DE 17.1 = CD + 8 ____ - 8 ______ - 8 9.1 = CD

15. MR = MN + NR 5x - 3 = 2.5x + x 5x - 3 = 3.5x _______ - 5x ____ - 5x -3 = -1.5x

-3 _____ -1.5 = -1.5x _____ -1.5

2 = x MN = 2.5x = 2.5(2) = 5

16. Total yards = pass + run = (24 - 9) + 1 _

2 (50 - 24) = 28 yd

17. Step 1 Solve for x. DE = EF

2x + 4 = 3x - 1 _______ - 2x _______ - 2x 4 = x - 1 ___ + 1 _____ + 1 5 = x

Step 2 Find DE, EF, and DF.

DE = 2x + 4 = 2(5) + 4 = 14EF = 3x - 1 = 3(5) - 1 = 14DF = DE + EF = 14 + 14 = 28

18. PQ = 1 _ 2 PR

3y = 1 __ 2 (42)

3y = 21

3y

___ 3 = 21 ___

3

y = 7

QR = 1 _ 2 PR

= 1 _ 2 (42) = 21

19a. C is the mdpt. of −− AE . b. EF = 2(AC) + 2

= 2(7) + 2 = 16 AB = 2(EF) - 16 = 2(16) - 16 = 16

20. GH = 9 1 __ 3


21. EF = 1 _ 2 (DF)

= 1 _ 2 (CD)

= 1 _ 2 (14.2) = 7.1

22. GH = 2(DH)4x - 1 = 2(8) 4x = 17 x = 4.25

23. CF = 2(CD)2y - 2 = 2(3y - 11)2y - 2 = 6y - 22 20 = 4y

y = 5CD = 3y - 11 = 3(5) - 11 = 4

24. A;

25. S;

26. A;

27. AM � MB is incorrect. The statement should be written as

−− AM � −− MB , not as two distances that are

�.

28. Let x be length of shorter piece. Let A and B be ends of dowel, and let C be cut point, nearest to A.AC + CB = 72 x + 5x = 72 6x = 72

6x ___ 6 = 72 ___

6

x = 12AC = x = 12CB = 5x = 5(12) = 60Dowel pieces are 12 cm long and 60 cm long.

29. MN = x N - x M 4 = x N - 2.5 ±4 = x N - 2.5 x N = 2.5 ± 4 = 6.5 or -1.5

30.

Possible answer: −− DE +

−− EF = −− DF

31. RS + ST = RT 7y - 4 + y + 5 = 28 8y + 1 = 28 8y = 27 y = 3.375

32. RS + ST = RT

3x + 1 + 1 __ 2 x + 3 = 18

7 __ 2 x + 4 = 18

7 __ 2 x = 14

2 __ 7 ( 7 __

2 x) = 2 __

7 (14)

x = 4

33. RS + ST = RT 2z + 6 + 4z - 3 = 5z + 12 6z + 3 = 5z + 12 z + 3 = 12 z = 9

34. B is not between A and C, because A, B, and C are not collinear.

35. Check students’ constructions.

TEST PREP, PAGE 19

36. D Order of pts. P, Q, S, R, T. PQ + QS + SR + RT = PT

1 __ 2 QR + QR + RT = PT

1 __ 2 (8) + 8 + RT = 34

12 + RT = 34 RT = 22

37. JAD = 2AC = 2(2BC) = 4BC = 4(12) = 48

38. BStatement must refer to segments

−− XY and −− YZ , and

use � symbol for congruence.

39. HThink: In AC = AB + BC, subst. BC for AB and CD for BC.AC = AB + BC = BC + CD = BD = 16AC + CE = AE 16 + CE = 34 CE = 18


40. JK = 1 _ 2 HJ

= 1 _ 2 (4x) = 2x

HJ + JK = HK 4x + 2x = 78 6x = 78 x = 13JK = 2x = 2(13) = 26

41.

42. Race distance is: 13 + 9(8.5) + x = 100 89.5 + x = 100 x = 10.5 m

43. Race distance is: 13.72 + 9(9.14) + x = 110 95.98 + x = 110 x = 14.02 m

44. JK cannot be equal to JL because JK + KL = JL and KL ≠ 0.



45. 20 - 8 = 12 = 12 46. -9 + 23 = 14 = 14

47. - 4 - 27 = - -23 = -23

48. 8a - 3(4 + a) - 10= 8a - 12 - 3a -10= 5a - 22

49. x + 2(5 - 2x) - (4 + 5x)= x + 10 - 4x - 4 - 5x= -8x + 6

50. � �� AB , � �� CB 51. −− AD ,

−− BD

52. A, B, D 53. �� CB

1-3 MEASURING AND CONSTRUCTING ANGLES, PAGES 20–27


1. ∠RTQ, ∠T, ∠STR, ∠1, ∠2

2a. m∠BOA = 40°∠BOA is acute.

b. m∠DOB = 165° - 40° = 125°

∠DOB is obtuse.

c. m∠EOC = 105°∠EOC is obtuse.

3. m∠XWZ = m∠XWY + m∠YWZ 121° = 59° + m∠YWZ 62° = m∠YWZ

4a. Step 1 Find y.

m∠PQS = 1 __ 2 m∠PQR

(5y - 1)° = 1 __ 2 (8y + 12)°

5y - 1 = 4y + 6 y - 1 = 6 y = 7

Step 2 Find m∠PQS.m∠PQS = 5y - 1 = 5(7) - 1

= 34°

b. Step 1 Find x. m∠LJK = m∠KJM (-10x + 3)° = (-x + 21)° 3 = 9x + 21 -18 = 9x x = -2

Step 2 Find m∠LJM.m∠LJM = 2m∠LJK = 2(-10x + 3) = 2 (-10(-2) + 3) = 46°


1. Two � with the same measure are �. All rt. � measure 90°, so any 2 rt. � are �.

2. m∠ABD = m∠DBC = 1 _ 2 m∠ABC

3.


1. ∠A, ∠R, ∠O 2. C; �� CB ,

�� CD

3. ∠AOB, ∠BOA, or ∠1; ∠BOC, ∠COB, or ∠2; ∠AOC or ∠COA

4. m∠VXW = 15°∠VXW is acute.

5. m∠TXW = 105°∠TXW is obtuse.

6. m∠RXU = 110°∠RXU is obtuse.

7. m∠JKM = m∠JKL + m∠LKM = 42° + 28° = 70°

8. m∠JKM = m∠JKL + m∠LKM 82.5° = 56.4° + m∠LKM m∠LKM = 26.1° 9. Step 1 Find x. m∠ABD = m∠DBC (6x + 4)° = (8x - 4)° 4 = 2x - 4 8 = 2x x = 4

Step 2 Find m∠ABD.m∠ABD = 6x + 4 = 6(4) + 4

= 28°

10. Step 1 Find y. m∠ABD = m∠DBC (5y - 3)° = (3y + 15)° 2y - 3 = 15 2y = 18 y = 9

Step 2 Find m∠ABC.m∠ABC = 2m∠ABD = 2(5y - 3) = 2 (5(9) - 3)

= 84°


11. ∠1 or ∠JMK; ∠2 or ∠LMK; ∠M or ∠JML

12. m∠CGE = 110 - 20 = 90°∠CGE is a rt. ∠.

13. m∠BGD = 113 - 20 = 93°∠BGD is obtuse.

14. m∠AGB = 20°∠AGB is acute.

15. m∠RSU = m∠RST + m∠TSU = 38° + 28.6° = 66.6°


16. m∠RSU = m∠RST + m∠TSU 83.5° = m∠RST + 46.7° m∠RST = 36.8° 17. Step 1 Find x. m∠RSP = m∠PST (3x - 2)° = (9x - 26)° -2 = 6x - 26 24 = 6x x = 4

Step 2 Find m∠RST.m∠RST = 2m∠RSP = 2(3x - 2) = 2 (3(4) - 2)

= 20°

18. Step 1 Find y.

m∠PST = 1 __ 2 m∠RST

(y + 5)° = 1 __ 2 ( 5 __

2 y) °

y + 5 = 5 __ 4 y

5 = 1 __ 4 y

y = 20

Step 2 Find m∠RSP.m∠RSP = m∠PST = y + 5 = 20 + 5

= 25°

19. acute 20. rt.

21. acute 22. obtuse

23–26. Check students’ drawings.

27. Let x be the measure of ∠BSC. m∠ASB + m∠BSC = m∠ASC 3x + x = 90 4x = 90

4x ___ 4 = 90 ___

4

x = 22.5m∠ASB = 3x = 3(22.5) = 67.5°m∠BSC = x = 22.5°

28. First construct the bisector of the given ∠. Then choose one of the smaller � that was constructed and construct its bisector. The resulting � will

have 1 _ 4 the measure of the original ∠.

29. m∠AOC + m∠DOC + m∠EOD = 180° 7x - 2 + 2x + 8 + 27 = 180 9x + 33 = 180 9x = 147

x = 147 ____ 9 = 16 1 _

3

30. m∠AOB + m∠BOC + m∠COD + m∠DOE = 180° 4x - 2 + 5x + 10 + 3x - 8 + 5x + 10 = 180 17x + 10 = 180 17x = 170 x = 10

31. m∠AOB + m∠BOC = m∠AOC 6x + 5 + 4x - 2 = 8x + 21 10x + 3 = 8x + 21 2x + 3 = 21 2x = 18 x = 9

32. Let m∠QRS = x. Then m∠PRQ = 4x.Step 1 Find x.

m∠PRQ + m∠QRS = m∠PRS 4x + x = 90 5x = 90 x = 18 m∠PRQ = 4 (m∠QRS) = 4(18) = 72°

33a. m∠LOK = 57° 3x + 2x + 12 = 57 5x + 12 = 57 5x = 45 x = 9

b. ∠LOJ � ∠JOK m∠LOJ = m∠JOK 3x = 2x + 12 x = 12

c. Think: m∠LOJ = 3x° > 0°; so x > 0. m∠LOK < 90° 5x + 12 < 90 5x < 78 x < 15.6 0 < x < 15.6

34. m∠AOB = (360) · 0.25 = 90°; rt.m∠BOC = (360) · 0.35 = 126°; obtusem∠COD = (360) · 0.10 = 36°; acutem∠DOA = (360) · 0.30 = 108°; obtuse

35. m∠COD = 2(36) = 72°m∠BOC = 126 - 36 = 90°

36. m∠ = 360 _ 5 = 72°

37. No; an obtuse ∠ measures greater than 90°, so it cannot be � to an acute ∠ (less than 90°).

38. 5x + 45 < 180 5x < 135 x < 27

39. m∠EFG = m∠EFH + m∠HFG = 2m∠EFH, so m∠EFH = 1 _

2 m∠EFG.

40. Check students’ constructions. Each ∠ should measure 35°.

TEST PREP, PAGE 26

41. Dm∠VOY = m∠VOW + m∠WOX + m∠XOY

= 1 _ 2 m∠UOW + m∠WOX + m∠XOY

= 1 __ 2 (50) + 90 + 40

= 155° 42. H

m∠UOX = m∠UOW + m∠WOX = 50 + 90 = 140°

43. C m∠ABC = m∠ABD + m∠DBC = 2m∠ABD

4x + 5 = 2(3x - 1) 4x + 5 = 6x - 2 5 = 2x - 2 7 = 2x x = 3.5

44. J

1 __ 2 m∠ + 30 = 90

1 __ 2 m∠ = 60

2 ( 1 __ 2 m∠) = 2(60)

m∠ = 120° 45. The � are acute. An obtuse ∠ measures between

90° and 180°. Since 1 _ 2 of 180 is 90, the resulting �

must measure less than 90°.



46. Each hour interval measures 360 ÷ 12 = 30°. The angle formed at 7:00 is the (lesser) angle between the 12 and the 7, which measures 5(30) = 150°.

47. m∠PQR = 2m∠PQS x 2 = 2(2x + 6) x 2 = 4x + 12 x 2 - 4x - 12 = 0 (x - 6)(x + 2) = 0

x - 6 = 0 or x + 2 = 0 x = 6 or x = -2 m∠PQR = x 2 = 36° or 4° 48. 81°24'15" - 42°30'10"

= (81 - 42)° + (24 - 30)' + (15 - 10)"= (80 - 42)° + (60 + 24 - 30)' + (15 - 10)"= 38°54'5"

49. 2.25° = (60 · 60 · 2.25)" = 8100

50. ∠ABC � ∠DBC m∠ABC = m∠DBC

3x ___ 2 + 4 = 2x - 27 1 __

4

31.25 = 1 __ 2 x

2(31.25) = 2 ( 1 __ 2 x)

x = 62.5No; x = 62.5, and substituting this value into the expressions for the ∠ measures gives a sum of 195.5.


51. 35 · 64% = 35 · 0.64 = 22.4

52. 33.6 _ 280

· 100% = 12%

53.

54.

55.

56. JK + KL = JL x + 3x = 2x + 4 2x = 4 x = 2 JK = x = 2

57. KL = 3x = 3(2) = 6

58. JL = 2x + 4 = 2(2) + 4 = 8

1-4 PAIRS OF ANGLES, PAGES 28–33


1a. ∠5 and ∠6 are adjacent angles. Their noncommon

sides, �� PQ and

�� PT , are opposite rays, so ∠5 and ∠6

also form a linear pair.

b. ∠7 and ∠SPU share −− SP but are on the same side of

it, so ∠7 and ∠SPU are not adjacent angles.

c. ∠7 and ∠8 share vertex P but do not have a common side, so ∠7 and ∠8 are not adjacent angles.

2a. (90 - y)°90° - (7x - 12)° = 90° - 7x + 12°

= (102 - 7x)° b. (180 - x)°

180° - 116.5° = 63.5° 3. Step 1 Let m∠A = x °. Then ∠B, its supplement,

measures (180 - x)°.Step 2 Write and solve an equation.

x = 1 __ 2 (180 - x) + 12

x = 90 - x __ 2 + 12

x = 102 - x __ 2

3 __ 2 x = 102

2 __ 3 ( 3 __

2 x) = 2 __

3 (102)

x = 68m∠A = x ° = 68°

4. 1 Understand the ProblemAnswers are measures of ∠1, ∠2, and ∠4.List important information:· ∠1 � ∠2· ∠1 and ∠3 are comp., and ∠2 and ∠4 are comp.· m∠3 = 27.6°2 Make a PlanIf ∠1 � ∠2, then m∠1 = m∠2.If ∠1 and ∠3 are comp., then m∠1 = (90 - 27.6)°.If ∠2 and ∠4 are comp., then m∠4 = (90 - 27.6)°.3 Solvem∠1 = m∠2 = (90 - 27.6)° = 62.4°m∠3 = m∠4 = (90 - 62.4)° = 27.6°4 Look BackAnswer makes sense because 27.6° + 62.4° = 90°, so ∠1 and ∠3 are comp., and ∠2 and ∠4 are comp. Thus m∠1 = 62.4°, m∠2 = 62.4°, and m∠4 = 27.6°.

5. Possible answer: ∠EDG and ∠FDH are vert. angles and appear to have the same measure.∠EDG ≈ ∠FDH ≈ 45°.



1. All rt. � measure 90°, so the sum of the measures of any 2 rt. � is 180°. Therefore any 2 rt. � are supp.

2. Vert. � cannot be adj. � because the def. of vert. � states that they are nonadj. � formed by intersecting lines.

3.


1. (90 - x)°; (180 - x)° 2. �� BC

3. ∠1 and ∠2 are adj. angles. Their noncommon sides, �� EG and

�� EJ , are opposite rays, so ∠1 and ∠2 also

form a lin. pair.

4. ∠1 and ∠3 share vertex E but do not have a common side, so ∠1 and ∠3 are not adj. angles.

5. ∠2 and ∠4 share vertex E but do not have a common side, so ∠2 and ∠4 are not adj. angles.

6. ∠2 and ∠3 are adj. angles. Their noncommon sides, �� EF and

�� EH , are not opposite rays, so ∠1 and ∠2

are only adj. angles.

7. (180 - x)°180° - 81.2° = 98.8°

8. (90 - x)°90° - 81.2° = 8.8°

9. (180 - y)°180° - (6x - 5)° = 180° - 6x + 5 = (185 - 6x)°

10. (90 - y)°90° - (6x - 5)° = 90° - 6x + 5 = (95 - 6x)°

11. Step 1 Let m∠A = x°. Then ∠B, its comp., measures (90 - x)°.Step 2 Write and solve an equation.

x = 3(90 - x) + 6 x = 270 - 3x + 6 x = 276 - 3x 4x = 276 x = 69

m∠A = x = 69°

12. 1 Understand the ProblemAnswers are measures of ∠2, ∠3, and ∠4.List Important information:· ∠1 � ∠2· ∠1 and ∠3 are comp., and ∠2 and ∠4 are comp.· m∠1 = 18.5°2 Make a PlanIf ∠1 � ∠2, then m∠1 = m∠2.If ∠1 and ∠3 are comp., then m∠3 = (90 - 18.5)°.If ∠2 and ∠4 are comp., then m∠4 = (90 - 18.5)°.3 Solvem∠1 = m∠2 = 18.5°m∠3 = m∠4 = (90 - 18.5)° = 71.5°4 Look BackAnswer makes sense because 18.5° + 71.5° = 90°, so ∠1 and ∠3 are comp., and ∠2 and ∠4 are comp. Thus m∠2 = 18.5°, m∠3 = 71.5°, and m∠4 = 71.5°.

13. ∠ABE and ∠CBD are vert. �; ∠ABC and ∠EBD are vert. �.


14. ∠1 and ∠4 are adj. angles. Their noncommon sides are opposite rays, so ∠1 and ∠4 also form a lin. pair.


16. ∠3 and ∠4 are adj. angles. Their noncommon sides are not opposite rays, so ∠3 and ∠4 are only adj. angles.

17. ∠3 and ∠1 share a vertex but do not have a common side, so ∠3 and ∠1 are not adj. angles.

18. (180 - x)° 180° - 56.4° = 123.6°

19. (90 - x)° 90° - 56.4° = 33.6°

20. (180 - y)° 180° - (2x - 4)° = 180° - 2x + 4 = (184 - 2x)° 21. (90 - y)° 90° - (2x - 4)° = 90° - 2x + 4 = (94 - 2x)° 22. Step 1 Let m∠A = x°, so m∠B = (90 - x)°.

Step 2 Write and solve an equation. x = 3(90 - x) x = 270 - 3x 4x = 270 x = 67.5

m∠A = x = 67.5°m∠B = 90 - 67.5° = 22.5°

23. ∠1 � ∠2m∠1 = m∠2 = 22.3°m∠4 = m∠3 = 90° - 22.3° = 67.7°

24. ∠PTU, ∠VTR; ∠UTQ, ∠STV; ∠QTR, ∠PTS; ∠PTQ, ∠STR; ∠UTR, ∠PTV; ∠QTV, ∠UTS


25. Possible outcomes are (30°, 60°), (30°, 120°), (30°, 150°), (60°, 120°), (60°, 150°), (120°, 150°). Supp. outcomes are (30°, 150°) and (60°, 120°). So P(supp.) = 2 _

6 = 1 _

3

26. Step 1 Find x.m∠ABD + m∠BDE = 180°

5x + 17x - 18 = 180 22x - 18 = 180 22x = 198 x = 9

Step 2 Find ∠ measures.m∠ABD = 5x = 5(9) = 45°m∠BDE = 17x - 18 = 17(9) - 18 = 135°

27. Step 1 Find x.m∠ABD + m∠BDE = 180°

3x + 12 + 7x - 32 = 180 10x - 20 = 180 10x = 200 x = 20

Step 2 Find ∠ measures.m∠ABD = 3x + 12 = 3(20) + 12 = 72°m∠BDE = 7x - 32 = 7(20) - 32 = 108°

28. Step 1 Find x. m∠ABD + m∠BDE = 180°

12x - 12 + 3x + 48 = 180 15x + 36 = 180 15x = 144 x = 9.6

Step 2 Find ∠ measures.m∠ABD = 12x - 12 = 12(9.6) - 12 = 103.2°m∠BDE = 3x + 48 = 3(9.6) + 48 = 76.8°

29. Step 1 Find y.m∠ABD + m∠BDC = 90°

5y + 1 + 3y - 7 = 90 10y - 6 = 90 8y = 96 y = 12

Step 2 Find ∠ measures.m∠ABD = 5y + 1 = 5(12) + 1 = 61°m∠BDC = 3y - 7 = 3(12) - 7 = 29°


4y + 5 + 4y + 8 = 90 8y + 13 = 90 8y = 77 y = 9.625

Step 2 Find ∠ measures.m∠ABD = 4y + 5 = 4(9.625) + 5 = 43.5°m∠BDC = 4y + 8 = 4(9.625) + 8 = 46.5°


y - 30 + 2y = 90 3y - 30 = 90 3y = 120 y = 40

Step 2 Find ∠ measures.m∠ABD = y - 30 = 40 - 30 = 10°m∠BDC = 2y = 2(40) = 80°

32. The measure of an acute ∠ is less than 90°. Therefore the measure of its supp. must be between 90° and 180°, which means the supp. is an obtuse ∠.

33a.

Step 1 Find x. m∠JAH + m∠KAH = 90° 3x - 8 + x + 2 = 90 4x - 6 = 90 4x = 96 x = 24 Step 2 Find ∠ measures. m∠JAH = 3x - 8 = 3(24) - 8 = 64° m∠KAH = x + 2 = 24 + 2 = 26° b.

Step 1 Find x. m∠JAH + m∠KAH = 180° 3x - 8 + x + 2 = 180 4x - 6 = 180 4x = 186 x = 46.5 Step 2 Find ∠ measures. m∠JAH = 3x - 8 = 3(46.5) - 8 = 131.5° m∠KAH = x + 2 = 46.5 + 2 = 48.5° c.

Step 1 Find x. m∠JAH = m∠KAH 3x - 8 = x + 2 2x = 10 x = 5 Step 2 Find ∠ measures. m∠JAH = 3x - 8 = 3(5) - 8 = 7° m∠KAH = x + 2 = 5 + 2 = 7° 34. F; the supp. must be greater than the comp.

35. F; vert. � cannot be adj. �, so they cannot form a lin. pair.

36. T 37. T

38. The 2 � must both measure 45°. 45° + 45° = 90°, so the � are comp. and �.

TEST PREP, PAGE 33

39. C x + 90 + x = 180 2x + 90 = 180 2x = 90 x = 45

40. H x + 2x = 90 3x = 90 x = 30

m∠2 = 2(30) = 60°


41. C m∠A + m∠B = 180° 3y + 2(3y) = 180 9y = 180

9y

___ 9 = 180 ____

9

y = 20

42. H 7x + 5x = 180 12x = 180

12x ____ 12

= 180 ____ 12

x = 15 m∠2 = 5(15) = 75


43. 4 pairs of 45° � + 4 pairs of 90° � + 4 pairs of 135° � = 12 pairs of vert. �

44. Let ∠ measure be x °. Then180 - x = 2(90 - x) + 4180 - x = 180 - 2x + 4

180 - x = 184 - 2xx = 4

m∠ = x = 4° 45. m∠1 + m∠2 = 90°

2m∠2 + m∠ 2 = 90° 3m∠2 = 90°

m∠2 = 30°m∠1 - m∠2 = 2m∠2 - m∠2 = m∠2 = 30°

46. Let ∠ measure be x °. Then180 - x = 2 (180 - (90 - x)) - 36

180 - x = 2(180 - 90 + x) - 36 180 - x = 144 + 2x

36 = 3x x = 12(180 - x)° 180° - 12° = 168°


47. 4x + 10 = 42 4x = 32 x = 8

48. 5m - 9 = m + 4 4m - 9 = 4 4m = 13 m = 3.25

49. 2(y + 3) = 12 2y + 6 = 12 2y = 6 y = 3

50. -(d + 4) = 18 -d - 4 = 18 -d = 22 d = -22

51. XY + YZ = XZ 3x + 1 + 2x - 2 = 84 5x - 1 = 84 5x = 85 x = 17

52. XY = 3x + 1 = 3(17) + 1 = 52

53. YZ = 2x - 2 = 2(17) - 2 = 32

54. m∠XYZ = m∠WYX = 26°

55. m∠WYZ = m∠WYX + m∠XYZ = 26° + 26° = 52°

MULTI-STEP TEST PREP, PAGE 34

1. collinear 2. X is the mdpt. of −− RT .

RX = XT10x - 6 = 3x + 8 7x = 14 x = 2RT = RX + XT = 10x - 6 + 3x + 8 = 13x + 2 = 13(2) + 2 = 28 ft

3. m∠RXM + m∠MXW + m∠WXB = m∠RXBm∠RXM + 44° + m∠RXM = 90°

2m∠RXM + 44 = 90 2m∠RXM = 46 m∠RXM = 23°; acutem∠RXW = m∠RXM + m∠MXW = 23° + 44° = 67°; acutem∠WXB = 23°; acutem∠MXW = 44°; acutem∠RXT = 180°; straightm∠MXT = m∠RXT - m∠MXR = 180° - 23° = 157°; obtusem∠WXT = m∠WXB + m∠BXT = 23° + 90° = 113°; obtuse

4. adj.; adj. and a lin. pair; supp.; comp.; vert.

READY TO GO ON? PAGE 35

1. 2.

3. 4.

5. Possible answer: T, V, W

6. � �� XZ and � �� WY

7. plane TVX 8. �

9. −− SV = 5 - (-1.5) = 6.5 = 6.5

10. −− TR = 2 - (-4) = 6 = 6

11. −− ST = 2 - (-1.5) = 3.5 = 3.5

12. HJ + JK = HK4x + 6 + 9 = 39

4x + 15 = 39 4x = 24 x = 6HJ = 4x + 6 = 4(6) + 6 = 30


14. PR = 2PQ 8z - 12 = 2(2z)

8z - 12 = 4z -12 = -4z z = 3

PQ = 2z = 2(3) = 6PR = 8z - 12 = 8(3) - 12 = 12

15. ∠LMN, ∠NML, or ∠1; ∠NMP, ∠PMN, or ∠2; ∠LMP or ∠PML


16. acute 17. obtuse

18. obtuse

19. m∠QRS = m∠SRT 3x + 8 = 9x - 4

12 = 6x x = 2

m∠SRT = 9x - 4 = 9(2) - 4 = 14°



22. ∠4 and ∠5 are adj. angles. Their noncommon sides are not opposite rays, so ∠4 and ∠5 are only adj. �.

23. ∠3 and ∠4 share a vertex but do not have a common side, so ∠3 and ∠4 are not adj. angles.

24. (180 - y)° 180° - (5x - 10)° = 180° - 5x + 10 = (190 - 5x)° 25. (90 - y)° 90° - (5x - 10)° = 90° - 5x + 10 = (100 - 5x)°

1-5 USING FORMULAS IN GEOMETRY, PAGES 36–41


1. P = 4s = 4(3.5) = 14 in.

A = s 2 = (3.5) 2 = 12.25 in 2

2. The area of one rectangle isA = �w = (6.5)(2.5) = 16.25 in 2 .The total area of the 4 rectangles is4(16.25) = 65 in 2 .

3. C = 2πr = 2π(14) = 28π

≈ 88.0 m

A = π r 2 = π(14 ) 2 = 196π

≈ 615.8 m 2


1. Possible answer: A rect. with length 8 in. and width 2 in.; a square with sides 4 in. long; a � with base 4 in. and height 8 in.

2.


1. Both terms refer to the dist. around a figure.

2. base and height

3. P = 2� + 2w = 2(11) + 2(4) = 22 + 8 = 30 mm

A = �w = (11)(4) = 44 m m 2

4. P = 4s = 4(y - 3) = 4y - 12

A = s 2 = (y - 3 ) 2 = y 2 - 6y + 9

5. P = a + b + c = 5 + (x + 3) + 13 = (x + 21) m

A = 1 _ 2 bh

= 1 _ 2 (x + 3)(4)

= (2x + 6) m 2

6. Area of one � is A = 1 _ 2 bh = 1 _

2 (3)(4) = 6 i n 2 .

There are 80(20) = 1600 � in total. The total area of the 1600 � is 1600(6) = 9600 i n 2 .

7. C = 2πr = 2π(2.1) = 4.2π

≈ 13.2 m

A = π r 2 = π(2.1 ) 2 = 4.41π

≈ 13.9 m 2

8. C = 2πr = 2π(7) = 14π

≈ 44.0 in.

A = π r 2 = π(7 ) 2 = 49π

≈ 153.9 in 2

9. r = d _ 2 = 16 ___

2 = 8 cm

C = 2πr = 2π(8) = 16π

≈ 50.3 cm

A = π r 2 = π(8 ) 2 = 64π

≈ 201.1 cm 2


10. P = 4s = 4(7.4) = 29.6 m

A = s 2 = (7.4 ) 2 = 54.76 m 2

11. P = 2� + 2w = 2(x + 6) + 2x = 4x + 12

A = �w = (x + 6)x = x 2 + 6x

12. P = a + b + c = 5x + 8 + 4x = 9x + 8

A = 1 __ 2 bh

= 1 __ 2 (8)(3x)

= 12x

13. Area of one � is 1 __ 2 bh = 1 __

2 (3)(1.5) = 2.25 in 2 .

Area of 32 � is 32(2.25) = 72 in 2 .

14. C = 2πr = 2π(12) = 24π

≈ 75.4 m

A = π r 2 = π(12 ) 2 = 144π

≈ 452.4 m 2

15. r = d _ 2 = 6.25 ft

C = 2πr = 2π(6.25) = 12.5π

≈ 39.3 ftA = π r 2 = π(6.25 ) 2 = 39.0625π

≈ 122.7 ft 2


16. r = d _ 2 = 1 _

4 mi

C = 2πr

= 2π ( 1 _ 4 ) = 1 _

4 π

≈ 1.6 mi

A = π r 2

= π ( 1 _ 4 )

2 = 1 _

16 π

≈ 0.2 mi 2

17. A = s 2 = (9.1 ) 2 = 82.81 yd 2

18. A = s 2 = (x + 1 ) 2 = x 2 + 2x + 1

19. A = 1 __ 2 bh

= 1 __ 2 (5.5)(2.25) = 6.1875 in 2

20. A = 1 __ 2 bh

6.75 = 1 __ 2 (3)h

6.75 = 3 __ 2 h

2 __ 3 (6.75) = 2 __

3 ( 3 __

2 h)

h = 4.5 m

21. A = �w 347.13 = 20.3w w = 17.1 cm

22. Possible answer: A = π r 2

64π = π r 2 64 = r 2 r = √ �� 64 = 8

23. A = π(8 ) 2 is incorrect. The radius is 4 cm, not 8 cm.A = π r 2 = π(4 ) 2 = 16π cm 2

24. r = d _ 2 = 14 m

A = π r 2 = π(14 ) 2 = 196π m 2

25. A = π r 2 = π(3y ) 2 = 9 y 2 π

26. Dist. around Earth at equator is the circumference.C = 2πr = 2π(3964) = 7928π

≈ 24,907 mi

27. For a square, the length and width are both s, soP = 2� + 2w = 2s + 2s = 4s and A = �w = s(s) = s 2 .

28. P = 2� + 2w = 2(x + 1) + 2(x - 3) = 4x - 4

A = �w = (x + 1)(x - 3) = x 2 - 2x - 3

29. Step 1 Write two equations.h = b - 3; A = 1 _

2 bh = 19b

Step 2 Find h.

1 _ 2 bh = 19b

1 _ 2 h = 19

h = 38 in.Step 3 Find b. h = b - 3

38 = b - 3 b = 41 in.

30a. P = π(4) + 2(3) + 2 √ �� 3 2 + 4 2 ≈ 28.6 ft = 343.2 in. number of strips = 342.2 in. ÷ 24 in. = 14.3

someone cannot buy part of a strip, thereforecost of material = (15)($1.39) = $20.85

b. charge for labor = total cost - material cost = $120.30 - $20.85 = $99.45

c. A = 1 __ 2 π r 2

= 1 __ 2 π(4 ) 2 = 8π

≈ 25.1 ft 2

d. A = 1 __ 2 bh

= 1 __ 2 (4)(3) = 6 ft 2

e. Area of garden ≈ 25.1 + 2(6) ≈ 37 ft 2

31a. Areas of small rects. are ac, ad, bc, and bd. Sum of areas = ac + ad + bc + bd. This must be equal to (a + b)(c + d), because sum

of areas of 4 small rects. equals area of large rect.

b. (a + 1)(c + 1) Areas of small rects. are ac, a, c, and 1. Sum of areas = ac + a + c + 1. This must be equal to (a + 1)(c + 1), because sum

of areas of 4 small rects. equals area of large rect.

c. (a + 1) 2 Areas of small rects. are a 2 , a, a, and 1. Sum of areas = a 2 + 2a + 1. This must be equal to the product (a + 1) 2

because the sum of the areas of the 4 small rects. equals the area of the large rect.

32. max. area - min. area = (110)(75) - (100)(64)= 8250 - 6400 = 1850 m 2

33. A = 1 __ 2 bh

28 = 1 __ 2 (2)h

h = 28 ft

34. A = 1 __ 2 bh

282.5 = 1 __ 2 (22.6)h

282.5 = 11.3h h = 25 yd

35. A = bh = (9.8)(2.7) = 26.46 ft 2

36. A = bh = (4(5280) + 960) (440) = 9,715,200 ft 2

= 9,715,200

_________ 5280 2

= 0.3 −− 48 mi 2

37. A = bh = (3(3) + 12) (11) = 231 ft 2

= 231 ____ 3

2 = 25 2 __

3 yd 2

38. P = 2b + 2h = 2(21.4) + 2(7.8) = 58.4 in.

39. b = 4 ft 6 in. = 4.5 ft; h = 6 in. = 0.5 ftP = 2b + 2h = 2(4.5) + 2(0.5) = 10 ft

40. b = 2 yd 8 ft = (2(3) + 8) ft = 14 ftP = 2b + 2h

= 2(14) + 2(6) = 40 ft = 13 yd 1 ft


41. C = 2πr 14 = 2πr

7 __ π

= r

d = 2r = 14 ___ π

42. A = π r 2 100π = π r 2 100 = r 2 10 2 = r 2 r = √ �� 100 = 10 d = 2r = 20

43. C = 2πr 50π = 2πr 25 = r d = 2r = 50

44. Missing outer dimensions: 17 - 4 = 13 yd; 9 - 4 = 5 ydP = 17 + 5 + 4 + 4 + 13 + 9 = 52 ydA = sum of areas of two rects. = (13)(9) + (4)(5) = 137 yd 2

45. Measure any side as the base. Then measure the ht. of the � at a rt. ∠ to the base.

46. The method works because adding the length and width together and doubling the result is 2(� + w), which equals 2� + 2w.

TEST PREP, PAGES 40–41

47. B A = π r 2 452 = π r 2

452 ____ π

= r 2

r = √ �� 452 ____ π

≈ 12.0 in.

48. G � = 2w P = 2� + 2w P = 2(2w) + 2w P = 6w 48 = 6w w = 8 m � = 2w

= 2(8) = 16 m

49. AA = 1 __

2 bh

= 1 __ 2 (4x)(x + 2)

= 2 x 2 + 4x

50. JP = 4s = 4(90) = 360 in. = 30 ft


51. A metal = A rect. - A circle

= �w - π r 2 = (14)(8) - π(3 ) 2 = 112 - 9π

≈ 83.7 in 2

52a. P = 2� + 2w P - 2� = 2w

P - 2�

______ 2 = w

b. w = P - 2�

______ 2

= 9 -2(3)

_______ 2

= 1.5 ft

53. Think: assume that � ≥ w. Values for (�, w) satisfy 12 = 2� + 2w or � + w = 6: (5, 1), (4, 2), or (3, 3). So possible areas are A = �w

= (5)(1) = 5; or (4)(2) = 8; or (3)(3) = 9

54. A actual = π r 2 = π (4.5) 2 = 20.25π

A estimate = s 2 = 8 2 = 64

Percent error = A estimate - A actual ______________

A actual · 100%

= 64 - 20.25π

___________ 20.25π

· 100%

≈ 0.6% (overestimate)

55. Step 1 Write an equation relating � and w.w = 4 _

5 �

Step 2 Substitute in formula for area of a rectangle and solve to find �.

A = �w

320 = � ( 4 __ 5 �)

320 = 4 __ 5 � 2

5 __ 4 (320) = 5 __

4 ( 4 __

5 � 2 )

400 = � 2 √ �� 400 = � � = 20 in.

Step 3 Find w.w = 4 _

5 � = 4 __

5 (20) = 16 in.


56. D: {2, -5, -3}R: {4, 8}

57. D: {4, -2, 16}R: {-2, 8, 0}

58. plane 59. line or segment

60. Let a and b be the lengths. Write 2 equations. a = 4b 10 = a + b 10 = 4b + b 10 = 5b

10 ___ 5 = 5b ___

5

b = 2 yd a = 4b = 4(2) = 8 yd

61. −− AB �

−− BC AB = BC -2.5 - (-8) = C - (-2.5) 5.5 = C - (-2.5) Think: B > A, so C > B. ThereforeC - (-2.5) = 5.5 C = -2.5 + 5.5 = 3

62. m∠ = 2(180 - m∠) + 9 m∠ = 360 - 2m∠ + 9 3m∠ = 369 m∠ = 123°


CONNECTING GEOMETRY TO ALGEBRA: GRAPHING IN THE COORDINATE PLANE, PAGE 42

TRY THIS, PAGE 42

1. (0, 0) 2. (0, 4)

3. (3, 2) 4. (-1, 0)

5. Spruce and Beech 6. Spruce and Hickory

7. Maple and Elm 8. Pine and Birch

1-6 MIDPOINT AND DISTANCE IN THE COORDINATE PLANE, PAGES 43–49


1. M ( x 1 + x 2

_ 2 ,

y 1 + y 2 _

2 )

( -2 + 5 _ 2 ,

3 + (-3) _

2 ) = ( 3 _

2 , 0 _

2 )

= ( 3 _ 2 , 0)

2. Step 1 Let coords. of T equal (x, y).Step 2 Use Mdpt. Formula.

(-1, 1) = ( -6 + x _ 2 ,

-1 + y _

2 )

Step 3 Find x-coord.

-1 = -6 + x _ 2

2(-1) = 2 ( -6 + x _ 2 )

-2 = -6 + x x = 4

Find y-coord.

1 = -1 + y

_ 2

2(1) = 2 ( -1 + y

_ 2 )

2 = -1 + y y = 3

The coordinates of T are (4, 3).

3. Step 1 Find coords. of each point.E(-2, 1), F(-5, 5), G(-1, -2), and H(3, 1)Step 2 Use Dist. Formula.

d = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

EF = √ �� (-5 - (-2) ) 2 + (5 - 1 ) 2

= √ �� (-3 ) 2 + 4 2 = √ �� 9 + 16 = 5

GH = √ �� (3 - (-1) ) 2 + (1 - (-2) ) 2

= √ �� 4 2 + 3 2

= √ �� 16 + 9 = 5Since EF = GH,

−− EF � −− GH .

4a. Method 1 Use Dist. Formula. Subst. values for coords. of R and S into Dist. Formula.

RS = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (-3 - 3 ) 2 + (-1 - 2 ) 2

= √ �� (-6 ) 2 + (-3 ) 2

= √ �� 36 + 9 = √ �� 45 ≈ 6.7Method 2 Use Pyth. Thm. Count the units for the legs of the rt. � formed by R and S.a = 6 and b = 3 c 2 = a 2 + b 2

= 6 2 + 3 2 = 36 + 9 = 45

c = √ �� 45 ≈ 6.7

b. Method 1 Use Dist. Formula. Subst. values for coords. of R and S into Dist. Formula.

RS = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (2 - (-4) ) 2 + (-1 - 5 ) 2

= √ �� (-6 ) 2 + (-6 ) 2

= √ �� 36 + 36 = √ �� 72 ≈ 8.5Method 2 Use Pyth. Thm. Count the units for the legs of the rt. � formed by R and S.a = 6 and b = 6 c 2 = a 2 + b 2

= 6 2 + 3 2 = 36 + 36 = 72

c = √ �� 72 ≈ 8.5

5.

H (0,0) F(90,0)

S(90,90)T(0,90)

Let the pitching mound be M(42.8, 42.8). The dist. MH from center of mound to home plate is the length of hyp. of a rt. �.

MH = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (42.8 - 0 ) 2 + (42.8 - 0 ) 2

= √ �� 42. 8 2 + 42. 8 2

= √ �� 1831.84 + 1831.84 = √ �� 3663.68 ≈ 60.5 ft



1. yes; x 1 + x 2

_ 2 =

x 2 + x 1 _

2 and

y 1 + y 2 _

2 =

y 2 + y 1 _

2

2. s and t represent lengths of legs. r represents length of hyp.

3. Yes; you can use either method to find dist. between 2 pts.

4. Possible answer: to make locating addresses easier

5.


1. hypotenuse

2. M ( x 1 + x 2

_ 2 ,

y 1 + y 2 _

2 )

( 4 + (-4)

_ 2 , -6 + 2 _

2 ) = ( 0 _

2 , -4 _

2 )

= (0, -2)

3. M ( x 1 + x 2

_ 2 ,

y 1 + y 2 _

2 )

( 0 + 3 _ 2 , -8 + 0 _

2 ) = ( 3 _

2 , -8 _

2 )

= (1 1 _ 2 , -4)

4. Step 1 Let coords. of N equal (x, y).Step 2 Use Mdpt. Formula:

(0, 1) = ( -3 + x _ 2 ,

-1 + y _

2 )


0 = -3 + x _ 2

2(0) = 2 ( -3 + x _ 2 )

0 = -3 + x x = 3

Find y-coord.

1 = -1 + y

_ 2

2(1) = 2 ( -1 + y

_ 2 )

2 = -1 + y y = 3

The coordinates of N are (3, 3).

5. Step 1 Let coords. of C equal (x, y).Step 2 Use Mdpt. Formula.

(-1 1 _ 2 , 1) = ( -3 + x ______

2 ,

4 + y _____

2 )


-1 1 _ 2 = -3 + x _

2

2 ( 3 _ 2 ) = 2 ( -3 + x _

2 )

-3 = -3 + x x = 0

Find y-coord.

1 = 4 + y

_ 2

2(1) = 2 ( 4 + y

_ 2

)

2 = 4 + y y = -2

The coordinates of C are (0, -2).

6. Step 1 Find coords. of each point.F(5, 4), G(3, -1), J(-4, 0), and K(1, -2)Step 2 Use Dist. Formula.

d = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

JK = √ �� (1 - (-4) ) 2 + (-2 - 0 ) 2

= √ �� 5 2 + (-2 ) 2

= √ �� 25 + 4 = √ �� 29

FG = √ �� (3 - 5 ) 2 + (-1 - 4 ) 2

= √ �� (-2 ) 2 + (-5 ) 2

= √ �� 4 + 25 = √ �� 29 Since JK = FG,

−− JK � −− FG .

7. Step 1 Find coords. of each point.J(-4, 0), K(1, -2), R(-3, -2), and S(3, -5)Step 2 Use Dist. Formula.

d = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

JK = √ �� (1 - (-4) ) 2 + (-2 - 0 ) 2

= √ �� 5 2 + (-2 ) 2

= √ �� 25 + 4 = √ �� 29

RS = √ �� (3 - (-3) ) 2 + (-5 - (-2) ) 2

= √ �� 6 2 + (-3 ) 2

= √ �� 36 + 9 = √ �� 45 = 3 √ � 5 Since JK ≠ RS,

−− JK � −− RS .

8. Method 1 Use Dist. Formula. Subst. values for coords. of A and B into Dist. Formula.

AB = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (-4 - 1 ) 2 + (-4 - (-2) ) 2

= √ �� (-5 ) 2 + (-2 ) 2

= √ �� 25 + 4 = √ �� 29 ≈ 5.4Method 2 Use Pyth. Thm. Count the units for the legs of the rt. � formed by A and B.a = 5 and b = 2 c 2 = a 2 + b 2

= 5 2 + 2 2 = 25 + 4 = 29

c = √ �� 29 ≈ 5.4


9. Method 1 Use Dist. Formula. Subst. values for coords. of X and Y into Dist. Formula.

XY = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (-2 - (-2) ) 2 + (-8 - 7 ) 2

= √ �� (0 ) 2 + (-15 ) 2

= √ �� 0 + 225 = 15.0Method 2 Use Pyth. Thm. Count the units for the legs of the rt. � formed by X and Y.a = 0 and b = 15 c 2 = a 2 + b 2

= 0 2 + 15 2 = 225

c = √ �� 225 = 15.0

10. Method 1 Use Dist. Formula. Subst. values for coords. of V and W into Dist. Formula.

VW = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (-4 - 2 ) 2 + (8 - (-1) ) 2

= √ �� (-6 ) 2 + 9 2

= √ �� 36 + 81 = √ �� 117 ≈ 10.8Method 2 Use Pyth. Thm. Count the units for the legs of the rt. � formed by V and W.a = 6 and b = 9 c 2 = a 2 + b 2

= 6 2 + 9 2 = 36 + 81 = 117

c = √ �� 117 ≈ 10.8

11. Use Pyth. Thm. From the plan, a = 22 andb = 16. c 2 = a 2 + b 2 = 22 2 + 16 2

= 484 + 256 = 740 c = √ �� 740 ≈ 27.2 ft


12. M ( x 1 + x 2

_ 2 ,

y 1 + y 2 _

2 )

( -3 + (-1) _

2 , -7 + 1 _

2 ) = ( -4 _

2 , -6 _

2 )

= (-2, -3)

13. M ( x 1 + x 2

_ 2 ,

y 1 + y 2 _

2 )

( 12 + (-5)

_ 2 ,

-7 + (-2) _

2 ) = ( 7 _

2 , -9 _

2 )

= (3 1 _ 2 , -4 1 _

2 )

14. Step 1 Let coords. of R equal (x, y).Step 2 Use Mdpt. Formula.

(7, -9) = ( -3 + x _ 2 ,

5 + y _

2 )


7 = -3 + x _ 2

14 = -3 + x x = 17

Find y-coord.

-9 = 5 + y

_ 2

-18 = 5 + y y = -23

The coordinates of R are (17, -23).

15. Step 1 Let coords. of C equal (x, y).Step 2 Use Mdpt. Formula.

(2 1 _ 2 , 1) = (

x + (-3) _

2 ,

y + (-2) _

2 )


2 1 _ 2 =

x + (-3) _

2

5 = x - 3 x = 8

Find y-coord.

1 = y + (-2)

_ 2

2 = y - 2 y = 4

The coordinates of C are (8, 4).

16. Step 1 Find coords. of each point.D(-4, 0), E(0, -2), F(2, 3), and G(4, -1).Step 2 Use Dist. Formula.

DE = √ �� (0 - (-4) ) 2 + (-2 - 0 ) 2

= √ �� 4 2 + (-2 ) 2

= √ �� 16 + 4 = √ �� 20 = 2 √ � 5

FG = √ �� (4 - 2 ) 2 + (-1 - 3 ) 2

= √ �� 2 2 + (-4 ) 2

= √ �� 4 + 16 = √ �� 20 = 2 √ � 5 Since DE = FG,

−− DE � −− FG .

17. Step 1 Find coords. of each point.D(-4, 0), E(0, -2), R(-3, -4), and S(2, -2).Step 2 Use Dist. Formula.

DE = √ �� (0 - (-4) ) 2 + (-2 - 0 ) 2

= √ �� 4 2 + (-2 ) 2

= √ �� 16 + 4 = √ �� 20 = 2 √ � 5

RS = √ �� (2 - (-3) ) 2 + (-2 - (-4) ) 2

= √ �� 5 2 + 2 2

= √ �� 25 + 4 = √ �� 29 Since DE ≠ RS,

−− DE � −− RS .

18. Method 1 Dist. Formula.

UV = √ �� (-3 - 0 ) 2 + (-9 - 1 ) 2

= √ �� (-3 ) 2 + (-10 ) 2

= √ �� 9 + 100 = √ �� 109 ≈ 10.4Method 2 Pyth. Thm. Count the units for the legs of the rt. � formed by U and V.a = 3 and b = 10 c 2 = a 2 + b 2

= 3 2 + 10 2 = 100 + 9 = 109

c = √ �� 109 ≈ 10.4


MN = √ �� (2 - 10 ) 2 + (-5 - (-1) ) 2

= √ �� (-8 ) 2 + (-4 ) 2

= √ �� 64 + 16 = √ �� 80 ≈ 8.9Method 2 Pyth. Thm. Count the units for the legs of the rt. � formed by M and N.a = 8 and b = 4 c 2 = a 2 + b 2

= 8 2 + 4 2 = 64 + 16 = 80

c = √ �� 80 ≈ 8.9



PQ = √ �� (5 - (-10) ) 2 + (5 - 1 ) 2

= √ �� 15 2 + 4 2

= √ �� 225 + 16 = √ �� 241 ≈ 15.5Method 2 Pyth. Thm. Count the units for the legs of the rt. � formed by P and Q.a = 15 and b = 4 c 2 = a 2 + b 2

= 15 2 + 4 2 = 225 + 16 = 241

c = √ �� 241 ≈ 15.5

21. Use Pyth. Thm.a = 11 and b = 14 c 2 = a 2 + b 2

= 11 2 + 14 2 = 121 + 196 = 317

c = √ �� 317 ≈ 18 in.

22. Step 1 Find coords. of each point.A(-4, 2), B(1, 4), C(2, 5), D(4, 1), E(-2, -2), and F(3, -1).Step 2 Use Dist. Formula.

AB = √ �� (1 - (-4) ) 2 + (4 - 2 ) 2

= √ �� 5 2 + 2 2

= √ �� 25 + 4 = √ �� 29

CD = √ �� (4 - 2 ) 2 + (1 - 5 ) 2

= √ �� 2

2 + (-4 ) 2

= √ �� 4 + 16 = √ �� 20

EF = √ �� (3 - (-2) )

2 + (-1 - (-2) )

2

= √ �� 5

2 + 1

2

= √ �� 25 + 1 = √ �� 26 −− CD ,

−− EF , −− AB

23. a = 2 and b = 4 c 2 = 2 2 + 4 2

= 20 c = √ �� 20 ≈ 4.47

24. ( a + (-5a)

_ 2 , 3a + 0 _

2 ) = ( -4a _

2 , 3a _

2 )

= (-2a, 3 _ 2 a)

25. Divide each coord. by 2.

26. Coords. of Cedar City are (2, 3).Coords. of Milltown are (3, -3).

Dist. = √ �� (3 - 2 ) 2 + (-3 - 3 ) 2

= √ �� 1 2 + (-6 ) 2 = √ �� 37 ≈ 6.1 mi

27. Coords. of Jefferson are (-2, -2).Dist. = 1 __

2 (dist. from Jefferson to Milltown)

= 1 __ 2 √ �� (3 - (-2) ) 2 + (-3 - (-2) ) 2

= 1 __ 2 √ �� 5 2 + (-1 ) 2

= 1 __ 2 √ �� 26 ≈ 2.5 mi

28. Use Pyth. Thm.a = 960 and b = 750 c 2 = a 2 + b 2

= 960 2 + 750 2 = 1,484,100

c = √ �� 1,484,100 ≈ 1218 m

29. Possible answer: seg. with endpts. (1, 2) and (-1, -2)

30. Step 1 Find AB, BC, and AC.

AB = √ �� (-2 - 1 ) 2 + (-1 - 4 ) 2

= √ �� (-3 ) 2 + (-5 ) 2

= √ �� 9 + 25 = √ �� 34

BC = √ �� (-3 - (-2) ) 2 + (-2 - (-1) ) 2

= √ �� (-1 ) 2 + (-1 ) 2

= √ �� 1 + 1 = √ � 2

AC = √ �� (-3 - 1 ) 2 + (-2 - 4 ) 2

= √ �� (-4 ) 2 + (-6 ) 2

= √ �� 16 + 36 = √ �� 52 Step 2 Find perimeter.P = AB + BC + AC

= √ �� 34 + √ � 2 + √ �� 52 ≈ 14.5

31. A = 1 _ 2 bh

= 1 _ 2 (BC) ( √ � 2 )

= 1 _ 2 ( √ � 2 ) ( √ � 2 )

= 1 _ 2 (2) = 1 square unit

32. When 2 pts. lie on a horiz. or vert. line, they share a common y-coord. or x-coord. To find the dist. between the pts., find the difference of the other coords.

33. Let M be the mdpt. of −− AC .

AM = MC = 1 _ 2 (10) = 5.0 ft

MB = MD = √ �� 5 2 + 4 2 = √ �� 41 ≈ 6.4 ft

TEST PREP, PAGE 49

34. BGH =

√ �� 3 2 + 3 2 = √ �� 18 ≈ 4.2

35. GCoords. of mdpt. of

−− LM are (2.5, 1); coords. of mdpt. of

−− JK are (1.5, -2.5).

Dist. = √ �� (2.5 - 1.5 ) 2 + (1 - (-2.5) ) 2

= √ �� 1 2 + 3.5 2

= √ �� 13.25 ≈ 3.6

36. D

( 7 + (-5)

_ 2 , -3 + 6 _

2 ) = ( 2 _

2 , 3 _

2 )

= (1, 1 1 _ 2 )

37. JDist. = √ �� (3 - (-5) ) 2 + (5 - 1 ) 2

= √ �� 8 2 + 4 2

= √ �� 80 ≈ 8.9



38a. x-coord. of Q = x-coord. of P = 1 + 4 _ 2 = 2.5;

y-coord. of Q = y-coord. of R = 1 + 3 _ 2 = 2;

coords. of Q are (2.5, 2).

b. Area of PBRQ = �w = (1.5)(1) = 1.5 square units

c. DB = √ �� 3 2 + 2 2 = √ �� 13 ≈ 3.6

39. XY 2 = (a + 1 - (a - 5) ) 2 + (2a - (-2a) ) 2 10 2 = 6 2 + (4a ) 2 100 - 36 = (4a ) 2 4a = ± √ �� 64 = ±8 a = ±2

40. Let coords. of pt. on y-axis be (0, y). 5 2 = (4 - 0 ) 2 + (2 - y ) 2 25 = 16 + 4 - 4y + y 2 0 = y 2 - 4y - 5 0 = (y - 5)(y + 1) y = 5 or -1Coords. of 2 pts. are (0, 5) and (0, -1).

41. By Pyth. Thm., AB 2 = AC 2 + BC 2 = x 2 + y 2

AB = √ �� x 2 + y 2


42. y 3x - 1

4 3(-1) -1

4 -4 �no

43. f(x) 5 - x 2

4 5 - (-1 ) 2

4 4 �yes

44. g(x) x 2 - x + 2

4 (-1 ) 2 - (-1) + 2

4 4 �yes

45. m∠ABD = 1 _ 2 (180) = 90°; rt.

46. m∠CBE = 1 _ 2 m∠CBD = 1 _

2 (180 - 90) = 45°; acute

47. m∠ABE = 180 - m∠CBE = 135°; obtuse

48. P = 4s20 = 4s s = 5 in. A = s 2

= 5 2 = 25 in 2

49. b = 2h = 2(2) = 4 ft

A = 1 _ 2 bh

= 1 _ 2 (4)(2) = 4 ft 2

50. A = �w= (x)(4x + 5)

= 4 x 2 + 5x

1-7 TRANSFORMATIONS IN THE COORDINATE PLANE, PAGES 50–55


1a. The transformation is a translation. MNOP → M′N′O′P′

b. The transformation is a 90° rotation. �XYZ → �X′Y′Z′

2.

The transformation is a rotation of 90° because each pt. has been rotated 90° counterclockwise about the origin.

3. Step 1 Find the coordinates of rect. JKLM.The vertices of rect. JKLM are J(1, 1), K(3, 1), L(3, -4), and M(1, -4).Step 2 Apply the rule to find the vertices of the image. J ′(1 - 2, 1 + 4) = J ′(-1, 5) K ′(3 - 2, 1 + 4) = K ′(1, 5) L′(3 - 2, -4 + 4) = L′(1, 0)M ′(1 - 2, -4 + 4) = M ′(-1, 0)Step 3

4. Step 1 Choose 2 pts.Choose a pt. A on the preimage and a corr. pt. A� on the image. A has coords. (3, 1) and A� has coords. (-1, -3).Step 2 Translate.To translate A to A�, 4 units are subtracted from both the x-coord. and the y-coord. Therefore, the translation rule is (x, y) → (x - 4, y - 4).



1. Possible answer: The preimage and image will be mirror images of each other.

2.


1. Preimage is �XYZ; image is �X ′Y ′Z ′. 2. translation; reflection; rotation

3. transformation is a reflection; �ABC → �A′B ′C ′ 4. transformation is a translation; PQRS → P ′Q ′R ′S ′ 5.

The transformation is a reflection across the y-axis because each pt. and its image are the same dist. from the y-axis.

6. Step 1 State the coordinates of �DEF.The vertices of �DEF are D(2, 3), E(1, 1), and F(4, 0).Step 2 Apply the rule to find the vertices of the image. D ′(2 - 3, 3 - 2) = D ′(-1, 1) E ′(1 - 3, 1 - 2) = E ′(-2, -1) F ′(4 - 3, 0 - 2) = F ′(1, -2)Step 3

7. Step 1 Choose 2 pts.Choose a pt. A on the preimage and a corr. pt. A′ on the image. A has coords. (0, 0) and A′ has coords. (4, 4).Step 2 Translate.To translate A to A′, 4 units are added to both the x-coord. and the y-coord. Therefore, the translation rule is (x, y) → (x + 4, y + 4).


8. rotation: DEFG → D ′E ′F ′G ′ 9. reflection: WXYZ → W ′X ′Y ′Z ′ 10.

translation; each pt. moves the same dist. right and the same dist. down.

11. Step 1 Apply the rule to find the vertices of the image. A′(-4 + 3, 1 - 2) = A′(-1, -1) B ′(1 + 3, 1 - 2) = B ′(4, -1) C ′(1 + 3, -2 - 2) = C ′(4, -4)D ′(-4 + 3, -2 - 2) = D ′(-1, -4)Step 2

12. Step 1 Choose 2 pts.Choose a pt. A on the preimage and a corr. pt. A′ on the image. A has coords. (-5, 1) and A′ has coords. (6, -3).Step 2 Translate.To translate A to A′, 11 units are added to the x-coord. and 4 units are subtracted from the y-coord. Therefore, the translation rule is (x, y) → (x + 11, y - 4).

13. reflection 14. translation

15. reflection

16. Vertices of image are F ′(3, -5), G ′(-1, -4), and H ′(5, 0).


17. Vertices of image are F �(-3, 5), G �(1, 4), and H �(-5, 0).

18. Vertices of �1 are (1, 1), (3, 1), and (2, 3).�1 to �2: (x, y) → (x, -y)�2 to �3: (x, y) → (-x, y)�3 to �4: (x, y) → (x, -y)

19. B 20. A

21. D 22. C

23. R ′(1 - 2, -4 - 8) = R ′(-1, -12)S ′(-1 - 2, -1 - 8) = S ′(-3, -9) T ′(-5 - 2, 1 - 8) = T ′(-7, -7)

24. Both translations move the preimage right and up. The second translation moves the preimage farther in each direction.

25. M′(2 + 2, 8 - 5) = M′(4, 3)N′(-3 + 2, 4 - 5) = N′(-1, -1)

26. K ′(-1 - 4, 1 + 3) = K ′(-5, 4) L′(3 - 4, -4 + 3) = L′(-1, -1)

27. Find the coords. of the preimage. Then, find the coords. of the image after translation. Plot the vertices of image pts. and use a straightedge to draw the image �.

28. Possible answer: 2 reflections (across the y-axis and across � �� EC )

TEST PREP, PAGE 55

29. A

30. HE ′(-3 - 2, -3 + 1) = E ′(-5, -2)

31. A1 + (-3) = -2 �

32. H-7 + 6 = -1 �


33a. R �((-2 - 1) + 4, (-2 + 3) - 1) = R �(1, 0) S �((-3 - 1) + 4, (1 + 3) - 1) = S �(0, 3) T �((1 - 1) + 4, (1 + 3) - 1) = T �(4, 3)

b. (x, y) → ((x - 1) + 4, (y + 3) - 1) = (x + 3, y + 2)

34. m∠ = 12 ___ 60

(360) = 72°

35. Transformation is (x, y) → (-y, x). So vertices of image are A ′(0, 1), B ′(0, 5), and C ′(-2, 2).

36. (x, y) → (x, -y) 37. (x, y) → (-x, y)


38. 0 = x 2 + 12x + 350 = (x + 7)(x + 5)x = -7 or -5

39. 0 = x 2 + 3x - 180 = (x + 6)(x - 3)x = -6 or 3

40. 0 = x 2 - 18x + 810 = (x - 9) 2 x = 9

41. 0 = x 2 - 3x + 20 = (x - 2)(x - 1)x = 2 or 1

42. m(supp. of ∠A) = 180 - m∠A= 180 - 76.1 = 103.9°

43. m(comp. of ∠A) = 90 - m∠A= 90 - 76.1 = 13.9°

44. Method 1 Use Dist. Formula. Subst. values for coords. of 2 pts. into Dist. Formula.

AB = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (4 - 2 ) 2 + (6 - 3 ) 2

= √ �� 2 2 + 3 2

= √ �� 4 + 9 = √ �� 13 ≈ 3.6Method 2 Use Pyth. Thm. Count the units for the legs of the rt. � formed by 2 pts.a = 3 and b = 2 c 2 = a 2 + b 2

= 3 2 + 2 2 = 9 + 4 = 13

c = √ �� 13 ≈ 3.6



AB = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (0 - (-1) ) 2 + (8 - 4 ) 2

= √ �� 1 2 + 4 2


= 1 2 + 4 2 = 1 + 16 = 17

c = √ �� 17 ≈ 4.1


AB = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (-6 - (-3) ) 2 + (-2 - 7 ) 2

= √ �� (-3 ) 2 + (-9 ) 2


= 3 2 + 9 2 = 9 + 81 = 90

c = √ �� 90 ≈ 9.5


AB = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (-1 - 5 ) 2 + (3 - 1 ) 2

= √ �� (-6 ) 2 + 2 2


= 6 2 + 2 2 = 36 + 4 = 40

c = √ �� 40 ≈ 6.3

TECHNOLOGY LAB: EXPLORE TRANSFORMATIONS, PAGES 56–57

ACTIVITY 1, PAGE 56

4. They appear to be �.

5. The �s move together and stay the same size and shape.

TRY THIS, PAGE 56

1. The image of the � moves the same dist. in the same direction as the endpt.

2. The � move together and remain a fixed dist. apart in a fixed direction.

3. Each measurement is the same for the preimage and image �. The � appear to be �.

ACTIVITY 2, PAGE 57

6. The � and its image rotate by the same ∠ measure and remain the same size and shape.

TRY THIS, PAGE 57

4. The image rotates by the same ∠ measure as the marked ∠.

5. The � rotates by the same ∠ measure. When P is inside, the image overlaps the �. When P coincides with a vertex, the image also coincides with the vertex.

6. The � rotates by an ∠ of 30°, not by the measure of the marked ∠.

7. 360°



1. A = (6 ) 2 + (12)(6) = 36 + 72 = 108 ft 2

P = 4(6) + 2(12) = 24 + 24 = 48 ft

She would need 144 stones. Total cost is $324.00. Explanation: 2 stones make a 12 in. by 18 in. rect.(12)(18) = 216 in 2 = 1.5 ft 2 . 108 _

1.5 = 72.

Since 2 stones make up each rect., 72(2) = 144 stones, and 144($2.25) = $324.

2. Let G be position of fountain.

BG = √ �� 6 2 + 6 2 = √ �� 72 ≈ 8.5 ft

CG = 6.0 ft

EG = √ �� 12 2 + 6 2 = √ �� 180 ≈ 13.4 ft

FG = 6.0 ft

3. She used a reflection across −− AB . Check students’

drawings; possible answer: reflection across −− AF ;

rotation about B; translation from D to F; rotation about E; translation from F to E.


1. P = 2� + 2w= 2(20) + 2(8)= 40 + 16 = 56 in.

A = �w= (20)(8) = 160 in 2

2. P = a + b + c= 13 + 2x + 20 + 3x - 11= 5x + 22

A = 1 __ 2 bh

= 1 __ 2 (2x + 20)(13)

= 1 __ 2 (26x + 260)

= 13x + 130

3. P = 2� + 2w= 2(6x) + 2(3x + 2)= 12x + 6x + 4= 18x + 4

A = �w= (6x)(3x + 2)=18 x 2 + 12x

4. P = a + b + c= 10 + 5x + 14 + 14x - 2= 19x + 22

A = 1 __ 2 bh

= 1 __ 2 (5x + 14)(10)

= 1 __ 2 (50x + 140)

= 25x + 70

5. C = 2πr= 2π(6) ≈ 37.7 m

A = π r 2 = π(6 ) 2 = 113.1 m 2

6. ( -4 + 3 _ 2 , 6 + 8 _

2 ) = ( -1 _

2 , 14 _

2 ) = (-0.5, 7)

7. Step 1 Let coords. of K equal (x, y).Step 2 Use Mdpt. Formula.

(9, 3) = ( 6 + x _ 2 ,

-2 + y _

2 )


9 = 6 + x _ 2

18 = 6 + x12 = x

Find y-coord.

3 = -2 + y

_ 2

6 = -2 + y8 = y

The coordinates of K are (12, 8).

8. Step 1 Find coords. of each point.Q(4, 3), R(-3, 1), S(-2, -4), and T(5, -2).Step 2 Use Dist. Formula.

QR = √ �� (-3 - 4 ) 2 + (1 - 3 ) 2

= √ �� (-7 ) 2 + (-2 ) 2 = √ �� 49 + 4 = √ �� 53 ≈ 7.3

ST = √ �� (5 - (-2) ) 2 + (-2 - (-4) ) 2

= √ �� 7 2 + 2 2

= √ �� 49 + 4 = √ �� 53 ≈ 7.3Since QR = ST,

−− QR � −− ST .


FG = √ �� (-3 - 4 ) 2 + (-2 - 3 ) 2

= √ �� (-7 ) 2 + (-5 ) 2

= √ �� 49 + 25 = √ �� 74 ≈ 8.6Method 2 Pyth. Thm. Count the units for the legs of the rt. � formed by F and G.a = 7 and b = 5 c 2 = a 2 + b 2

= 7 2 + 5 2 = 74

c = √ �� 74 ≈ 8.6

10. reflection; �ABC → �A′B′C′ 11. translation; PQRS → P′Q′R′S′ 12. Vertices of figure are H(2, 1), J(5, 1), K(5, -2), and

L(2, -2).Vertices of image are: H ′(2 - 3, 1 + 2) = H ′(-1, 3)

J ′(5 - 3, 1 + 2) = J ′(2, 3)K ′(5 - 3, -2 + 2) = K ′(2, 0) L′(2 - 3, -2 + 2) = L′(-1, 0)


13.

From graph, transformation is a rotation of 180° about the origin.

STUDY GUIDE: REVIEW, PAGES 60–63

1. angle bisector 2. complementary angles

3. hypotenuse

LESSON 1-1, PAGES 60–61

4. A, F, E, G or C, G, D, B

5. Possible answer: � �� GC

6. Possible answer: plane AEG

7. 8.

9.

LESSON 1-2, PAGE 61

10. JL = 2 - (-1.5) = 3.5 = 3.5

11. HK = 1 - (-4) = 5 = 5

12. Use Seg. Add. Post. XY + YZ = XZ13.8 + YZ = 21.4 YZ = 21.4 - 13.8 = 7.6

13. Step 1 Find x. Use Seg. Add. Post.PQ + QR = PR3x + 6x + 4 = 14x - 6 10 = 5x x = 2Step 2 Find PR.PR = 14x - 6

= 14(2) - 6 = 22

14. Step 1 Find x. TU = UV3x + 4 = 5x - 2 6 = 2x x = 3Step 2 Find TU, UV, and TV.TU = 3x + 4

= 3(3) + 4 = 13UV = TU = 13TV = TU + UV

= 13 + 13 = 26

15. Step 1 Find x.DE = EF 9x = 4x + 10 5x = 10 x = 2Step 2 Find DE, EF, and DF.DE = 9x

= 9(2) = 18EF = DE = 18DF = DE + EF

= 18 + 18 = 36

LESSON 1-3, PAGE 61

16. ∠VYX rt.; ∠XYZ acute; ∠ZYW acute; ∠VYZ obtuse; ∠XYW rt; ∠VYW straight.

17. Step 1 Find x.Use ∠ Add. Post. m∠HJK + m∠KJL = m∠HJL13x + 20 + 10x + 27 = 116 23x = 69 x = 3Step 2 Find m∠HJK.m∠HJK = 13x + 20

= 13(3) + 20 = 59° 18. Step 1 Find x.

m∠MNP = m∠PNQ 6x - 12 = 4x + 8 2x = 20 x = 10Step 2 Find m∠MNQ.m∠MNQ = 6x - 12 + 4x + 8

= 10x - 4= 10(10) - 4 = 96°

LESSON 1-4, PAGE 62

19. only adj. 20. adj. and a lin. pair

21. not adj.

22. 90 - m∠ = 90 - 74.6 = 15.4°180 - m∠ = 180 - 74.6 = 105.4°

23. 90 - m∠ = 90 - (2x - 4) = (94 - 2x)°

180 - m∠ = 180 - (2x - 4) = (184 - 2x)°

24. m∠ = 4(90 - m∠) + 5 m∠ = 365 - 4m∠5m∠ = 365 m∠ = 73°

LESSON 1-5, PAGE 62

25. P = 2� + 2w= 2(4x - 1) + 2(3x)= 14x - 2

A = �w= (4x - 1)(3x)= 12 x 2 - 3x

26. P = 4s= 4(x + 4)= 4x + 16

A = s 2 = (x + 4 ) 2 = x 2 + 8x + 16


27. P = a + b + c= 8 + x - 5 + 12= x + 15

A = 1 __ 2 bh

= 1 __ 2 (x - 5)(8)

= 4x - 20

28. P = 2� + 2w= 2(5x + 7) + 2(20)= 10x + 54

A = �w= (5x + 7)(20)= 100x + 140

29. C = 2πr= 2π(21)= 42π ≈ 131.9 m

A = π r 2 = π(21 ) 2 = 441π

≈ 1385.4 m 2

30. r = 14 _ 2 = 7 ft

C = 2πr= 2π(7) = 14π ≈ 44.0 ft

A = π r 2 = π(7 ) 2 = 49π ≈ 153.9 ft 2

31. A = 1 __ 2 bh

102 = 1 __ 2 (17)h

h = 2 ___ 17

(102) = 12 m

LESSON 1-6, PAGE 63

32. Y ( 3 + (-1)

________ 2 , 2 + 4 _____

2 ) = ( 2 __

2 , 6 __

2 ) = (1, 3)

33. Step 1 Let coords. of B equal (x, y).Step 2 Use Mdpt. Formula.

(-2, 3) = ( 5 + x _ 2 ,

0 + y _

2 )


-2 = 5 + x _ 2

-4 = 5 + x-9 = x

Find y-coord.

3 = 0 + y

_ 2

6 = 0 + y6 = y

The coordinates of B are (-9, 6).

34. Step 1 Let coords. of A equal (x, y).Step 2 Use Mdpt. Formula.

(-2, 3) = ( x + (-4)

_ 2 ,

y + 4 _

2 )


-2 = x + (-4)

_ 2

-4 = x - 4 0 = x

Find y-coord.

3 = y + 4

_ 2

6 = y + 42 = y

The coordinates of A are (0, 2).

35. Method 1 Use Dist. Formula. Subst. values for coords. of X and Y into Dist. Formula.

XY = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (6 - (-2) ) 2 + (1 - 4 ) 2

= √ �� 8 2 + (-3 ) 2

= √ �� 64 + 9 = √ �� 73 ≈ 8.5Method 2 Use Pyth. Thm. Count the units for the legs of the rt. � formed by X and Y.a = 8 and b = 3 c 2 = a 2 + b 2

= 8 2 + 3 2 = 64 + 9 = 73

c = √ �� 73 ≈ 8.5

36. Method 1 Use Dist. Formula. Subst. values for coords. of H and K into Dist. Formula.

HK = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (-2 - 0 ) 2 + (-4 - 3 ) 2

= √ �� (-2 ) 2 + (-7 ) 2

= √ �� 4 + 49 = √ �� 53 ≈ 7.3Method 2 Use Pyth. Thm. Count the units for the legs of the rt. � formed by H and K.a = 2 and b = 7 c 2 = a 2 + b 2

= 2 2 + 7 2 = 4 + 49 = 53

c = √ �� 53 ≈ 7.3

37. Method 1 Use Dist. Formula. Subst. values for coords. of L and M into Dist. Formula.

LM = √ �� ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

= √ �� (3 - (-4) ) 2 + (-2 - 2 ) 2

= √ �� 7 2 + (-4 ) 2

= √ �� 49 + 16 = √ �� 65 ≈ 8.1Method 2 Use Pyth. Thm. Count the units for the legs of the rt. � formed by L and M.a = 7 and b = 4 c 2 = a 2 + b 2

= 7 2 + 4 2 = 49 + 16 = 65

c = √ �� 65 ≈ 8.1

LESSON 1-7, PAGE 63

38. 90° rotation; DEFG → D′E′F′G′ 39. translation; PQRS → P′Q′R′S′ 40. X ′(-5 + 4, -4 + 5) = X ′(-1, 1)

Y ′(-3 + 4, -1 + 5) = Y ′(1, 4) Z ′(-2 + 4, -2 + 5) = Z ′(2, 3)


CHAPTER TEST, PAGE 64

1.

2. Possible answer: D, E, C, A

3. Possible answer: � �� BE 4. AB = 0.5 - (-3) = 3.5 = 3.5

5. Step 1 Find x.Use Seg. Add. Post. EF + FG = EG6x - 4 + 3x = 5x + 8 4x = 12 x = 3Step 2 Find EF.EF = 6x - 4

= 6(3) - 4 = 14

6. Step 1 Find x. HJ = JK3x + 5 = 9x - 3 8 = 6x x = 4 _

3

Step 2 Find HJ, JK, and HK.HJ = 3x + 5

= 3 ( 4 _ 3 ) + 5 = 9

JK = HJ = 9HK = HJ + JK

= 9 + 9 = 18

7. acute 8. rt.

9. obtuse

10. Step 1 Find x.m∠RTV = m∠VTS 16x - 6 = 13x + 9 3x = 15 x = 5Step 2 Find m∠RTV.m∠RTV = 16x - 6

= 16(5) - 6 = 74° 11. m∠ = 3(180 - m∠) - 5

m∠ = 540 - 3m∠ - 54m∠ = 535 m∠ = 133.75°m(supp. of ∠) = 180 - m∠

= 180 - 133.75 = 46.25° 12. only adj. 13. adj. and a lin. pair

14. not adj.

15. P = 2b + 2h= 2(8) + 2(4) = 16 + 8 = 24 ft

A = bh= (8)(4) = 32 ft 2

16. C = 2πr= 2π(15)= 30π ≈ 94.2 m

A = π r 2 = π(15 ) 2 = 225π ≈ 706.9 m 2

17. r = d _ 2 = 12.5 ft

C = 2πr= 2π(12.5)= 25π ≈ 78.5 ft

A = π r 2 = π(12.5 ) 2 = 156.25π ≈ 490.9 ft 2

18. r = d _ 2 = 1.4 cm

C = 2πr= 2π(1.4)= 2.8π ≈ 8.8 cm

A = π r 2 = π(1.4 ) 2 = 1.96π ≈ 6.2 cm 2

19. ( -4 + 3 _ 2 , 6 + 2 _

2 ) = ( -1 _

2 , 8 _

2 ) = (-0.5, 4)

20. Step 1 Let coords. of N equal (x, y).Step 2 Use Mdpt. Formula.

(-5, 1) = ( 2 + x _ 2 ,

4 + y _

2 )


-5 = 2 + x _ 2

-10 = 2 + x-12 = x

Find y-coord.

1 = 4 + y

_ 2

2 = 4 + y-2 = y

The coordinates of N are (-12, -2).

21. Use Dist. Formula.

AB = √ �� (-1 - (-5) ) 2 + (3 - 1 ) 2

= √ �� 4 2 + 2 2

= √ �� 16 + 4 = √ �� 20

CD = √ �� (4 - 1 ) 2 + (1 - 4 ) 2

= √ �� 3 2 + (-3 ) 2 = √ �� 9 + 9 = √ �� 18 Since AB ≠ CD,

−− AB � −− CD

22. 180° rotation; QRS → Q ′R ′S ′ 23. reflection; WXYZ → W ′X ′Y ′Z ′ 24. Step 1 Find the coordinates of �ABC.

Vertices of �ABC are A(-5, 1), B(-2, 4), and C(-1, 1).Step 2 Use (x, y) → (x + 3, y - 3) to find vertices of image.A′(-5 + 3, 1 - 3) = A′(-2, -2)B′(-2 + 3, 4 - 3) = B′(1, 1)C′(-1 + 3, 1 - 3) = C′(2, -2)Step 3


COLLEGE ENTRANCE EXAM PRACTICE, PAGE 65

1. EUse Seg. Add. Post.DF = DE + EF 6 = 2 + EFEF = 4EG = EF + FG = 4 + 5 = 9DG = DF + FG = 6 + 5 = 11EG(DG) = 9(11) = 99

2. CUse ∠ Add. Post.m∠PQR = 2m∠SQR4x + 2 = 2(3x - 6)4x + 2 = 6x - 12 14 = 2x x = 7

3. E P = 2� + 2w42 = 2(2w) + 2w = 6w w = 7 m

A = �w= (2(7))(7) = 98 m 2

4. Cs =

√ �� 4 2 + 4 2 = √ �� 32 A = ( √ �� 32 ) 2 = 32

5. CUse Seg. Add. Post.m∠BFC + m∠CFD = m∠BFD m∠BFC + 72 = 90 m∠BFC = 18°m∠AFB + m∠BFC = m∠AFC x + 18 = 90 x = 72


Solutions KeyGeometric Reasoning2

CHAPTER


1. B 2. A

3. F 4. C

5. D 6. lin. pair

7. vert. ! 8. comp. !

9. natural, whole, integer, rational

10. rational

11. integer, rational 12. rational

13. rational 14. whole, integer, rational

15. Possible answer: B 16. Possible answer: " #$ BD

17. Possible answer: ##$ CA 18. Possible answer:

%% CD

19. Possible answer: plane F

20. 8 + x = 5 ______ &8 ___ &8 x = &3

21. 6y = &12

6y

_ 6

= &12 _ 6

y = &2

22. 9 = 6s

9 _ 6 = 6s _

6

1.5 = s

23. p & 7 = 9 _____ + 7 ___ +7 p = 16

24. z _ 5 = 5

5 ( z _ 5 ) = 5(5)

z = 25

25. 8.4 = &1.2r

8.4 _ &1.2

= &1.2r _ &1.2

&7 = r

2-1 USING INDUCTIVE REASONING TO MAKE CONJECTURES, PAGES 74–79

CHECK IT OUT! PAGES 74–76 1. 0.0004 2. odd

3. Female whales are longer than male whales.

4a. Possible answer: x = 1 _ 2

b. Possible answer:

c. Jupiter or Saturn

THINK AND DISCUSS, PAGE 76 1. No; possible answer: a conjecture cannot be proven

true just by giving examples, no matter how many.

2.


1. Possible answer: A conjecture is based on observation and is not true until proven true in every case.

2. September 3. 4 _ 6

4. 5. even

6. 1 = 11 + 3 = 41 + 3 + 5 = 91 + 3 + 5 + 7 = 16Rule is n 2 .

7. The number of bacteria doubles every 20 min.

8. Roosevelt was inaugurated at age 42.

9. The 3 pts. are collinear.

10. Possible answer: x = &3


11. 5 P.M. 12. 42

13. 14. 2 = 1(2)2 + 4 = 6 = 2(3)2 + 4 + 6 = 12 = 3(4)Rule is n(n + 1).

15. n & 1

16. About 5%(526) ' 26 students will participate.

17. Possible answer: y = &1

18. Possible Answer: x = &1

19. m(1 = m(2 = 90°

20. Each term is the square of the previous term; 16 2 = 256, 256 2 = 65,536

21. Possible answer: each term is the previous term multiplied by 1 _

2 ; 1 _

16 , 1 _

32 .

22. The terms are multiples of 3 with alternating signs; &15, 18

23. 2n + 1 24. T

25. F; possible answer: n = 2

26. F; possible answer: 27. T

28. Amount increases by about $50 per day. Therefore, about $300 + 2($50) = $400 is raised during the 6th day.


29. 1 _ 11

= 0. %%

09 , 2 _ 11

= 0. %%

18 , 3 _ 11

= 0. %%

27 ,…; fraction pattern

is multiples of 1 _ 11

, decimal pattern is repeating

multiples of 0.09.

30. 6 = 3 + 3; 8 = 3 + 5; 10 = 3 + 7 or 5 + 5; 12 = 5 + 7; 14 = 3 + 11 or 7 + 7

31. 13 + 21 = 34; 21 + 34 = 55; 34 + 55 = 89; each term is the sum of the 2 previous terms.

32. The middle number is the mean of the other 2 numbers.

33. 2n & 1 is odd

34. Feb. 19; possible answer: the weather or the whales’ health

35. Possible answer: Even numbers are divisible by 2, but odd numbers are not. So the conjecture is true for even numbers but not necessarily for all numbers.

36a. 8 b. tenth day

TEST PREP, PAGE 79

37. CFor example, 2 & 4 is negative.

38. J2 and 2 + 1 = 3 are both prime.

39. DIn 2010, 75 & 3(15) = 30 students predicted.


40. x x 2 + x + 111 132 173 234 315 416 537 678 83

Possible answer: prime numbers. x = 10

41. Seats are up for election 6 years and 12 years later. 6 is not divisible by 4, but 12 is; so seats are next up for election during a presidential election 12 years later.

42a. Week Sit-ups1 152 353 554 755 956 1157 1358 1559 17510 195

b. Week 8. c. Rob does 20(n & 1) + 15 or 20n & 5 sit-ups during week n.

43.

m(CAB = m(CBA; AC = CB; possible answer: if a pt. is equidistant from the endpts. of a seg., then the 2 ! formed by connecting the pt. to the endpts. of the seg. are ).


44. y = 3x & 5 8 3(1) & 5 8 3 & 5 8 &2 ! no

45. y = 3x & 5 &11 3(&2) & 5 &11 &6 & 5 &11 &11 " yes

46. y = 3x & 5 4 3(3) & 5 4 9 & 5 4 4 " yes

47. y = 3x & 5 0.5 3(&3.5) & 5 0.5 &10.5 & 5 0.5 &15.5 ! no

48. A = s 2 = x 2 s = xP = 4s = 4x

49. P = 2! + 2w= 2x + 2(4x & 3)= 10x & 6

50. P = 3s= 3(x + 2)= 3x + 6

51. A = " r 2 = 9" x 2 r 2 = 9 x 2 r = 3xC = 2"r

= 2"(3x) = 6"x

52. (&1, &1) * (&1, &1 + 2) = (&1, 1)(0, 1) * (0, 1 + 2) = (0, 3)(4, 0) * (4, 0 + 2) = (4, 2)

53. (&1, &1) * (&1 + 4, &1 & 1) = (3, &2)(0, 1) * (0 + 4, 1 & 1) = (4, 0)(4, 0) * (4 + 4, 0 & 1) = (8, &1)

CONNECTING GEOMETRY TO NUMBER THEORY: VENN DIAGRAMS, PAGE 80

TRY THIS, PAGE 80 1.


2.

3.

2-2 CONDITIONAL STATEMENTS, PAGES 81–87

CHECK IT OUT! PAGES 81–83 1. Hypothesis: A number is divisible by 6.

Conclusion: The number is divisible by 3.

2. If 2 ! are comp., then they are acute.

3. F; possible answer: 7

4. Converse: If an animal has 4 paws, then it is a cat; F. Inverse: If an animal is not a cat, then it does not have 4 paws; F.Contrapositive: If an animal does not have 4 paws, then it is not a cat; T.

THINK AND DISCUSS, PAGE 84 1. T; F 2. T

3. Yes; possible answer: “If x = 3, then 2x = 6” is true, and so is the conv. “If 2x = 6, then x = 3.”

4.


1. converse 2. logically equivalent

3. Hypothesis: A person is at least 16 years old.Conclusion: The person can drive a car.

4. Hypothesis: A figure is a rectangle.Conclusion: A figure is a parallelogram.

5. Hypothesis: a & b < a.Conclusion: b is a positive number.

6. If a person is 18 years old, then that person is eligible to vote.

7. If 0 < a < b, then ( a _ b ) 2 < a _

b .

8. If something is a rotation, then it is a transformation.

9. T

10. F; possible answer: x = 2 and y = &4

11. F; possible answer: April

12. Converse: If Brielle travels 10 mi in 20 min, then she drives at exactly 30 mi/h; F.Inverse: If Brielle does not drive at exactly 30 mi/h, then she does not travel 10 mi in 20 min; F.Contrapositive: If Brielle does not travel 10 mi in 20 min, then she does not drive at exactly 30 mi/h; T.


13. Hypothesis: An animal is a tabby.Conclusion: The animal is a cat.

14. Hypothesis: Two lines intersect.Conclusion: Four angles are formed.

15. Hypothesis: 8 oz of cereal cost $2.99.Conclusion: 16 oz of cereal cost $5.98.

16. If a patient is ill, then you should monitor the patient’s heart rate.

17. If the batter makes 3 strikes, then the batter is out.

18. If segs. are ), then they have equal measures.

19. T

20. F; by Post. 1-1-5, if 2 planes intersect, they do so in exactly 1 line.

21. T

22. Converse: If an event is unlikely to occur, then the probability of the event is 0.1; F.Inverse: If the probability of an event is not 0.1, then the event is likely to occur; F.Contrapositive: If an event is likely to occur, then the probability of the event is not 0.1; T.

23. Converse: If the air temperature is 32°F or less, then freezing rain is falling; F.Inverse: If freezing rain is not falling, then the air temperature is greater than 32°F; F.Contrapositive: If the air temperature is greater than 32°F, then freezing rain is not falling; T

24. T 25. T

26. T 27. F

28. T 29. F

30.


31.

32.

33.

34. If an animal is a dolphin, then it is a mammal.

35. If a person is a Texan, then the person is an American.

36. If x < &4, then x < &1.

37a. Hypothesis: Only you can find it.Conclusion: Everything’s got a moral.

b. If only you can find it, then everything’s got a moral.

38. x = 5

39. Possible answer: 40. Possible answer:

41. Possible answer: You did not go out in the sun.

42. If a mineral is calcite, then it has a hardness of 3; T.

43. If a mineral has a hardness less than 5, then it is not apatite; T.

44. If a mineral is not apatite, then it has a hardness of less than 5; F.

45. If a mineral is not apatite, then it is calcite; F.

46. If a mineral has a hardness of 3, then it is not apatite; T.

47. If a mineral is calcite, then it has a hardness less than 5; T.

48. Converse: If 2 ! have the same measure, then they are ); T.Inverse: If 2 ! are not ), then they do not have the same measure; T.Contrapositive: If 2 ! do not have the same measure, then they are not ); T.

49. Possible answer: A conditional statement is false when the hypothesis is true and the conclusion is false. A conditional statement with a false hypothesis is always true because nothing has been guaranteed by the hypothesis.

TEST PREP, PAGE 86

50. C 51. H

52. D 53. J


54. No lines are pts. No pts. are lines.

55. Some students are adults. Some adults are students.

56a.

Possible answer: Figure A is not a rect., so it belongs outside the larger oval in the Venn diag. It cannot be inside the smaller oval, so it cannot be a square.

b. If a figure is not a rect., then it is not a square. By the contrapositve, since the figure is not a rect., it is not a square.

57. 3 true conditionals: r * q, q * p, and r * p


58. y = x + 3 59. y = 2x + 1

60. y = 5 _ 2 x & 4 61. T

62. F; possible answer: acute ! measure less than 90°, so the sum of the measures of 2 acute ! must be less than 180°. Therefore, 2 acute ! cannot be supp.

63. T 64. 13,131

65. 2 _ 81

66. 5 x 5


2-3 USING DEDUCTIVE REASONING TO VERIFY CONJECTURES, PAGES 88–93

CHECK IT OUT! PAGES 88–90 1. The myth rests on a false premise, that eelskin

wallets are made from electric eels. The conclusion is a result of deductive reasoning.

2. Hypothesis: A student passes his classes.Conclusion: The student is eligible to play sports.The given statement “Ramon passed his classes” matches the hypothesis of the given conditional. By the Law of Detachment, Ramon is eligible to play sports. The conjecture is valid.

3. Let p, q, and r represent the following:p: An animal is a mammal.q: An animal has hair.r : An animal is a dog.You are given that p * q and r * p. Since p is the conclusion of the 2nd conditional and the hypothesis of the 1st conditional, you can conclude that r * q. The conjecture is valid by the Law of Syllogism.

4. Conclusion: Polygon P is not a quad.

THINK AND DISCUSS, PAGE 90 1. Yes; the given information is false.

2. Possible answer: Using symbols instead of words forces you to look at the validity of the argument itself, without being distracted by the truth values of the individual statements.

3.


1. Possible answer: Inductive reasoning is based on a pattern of specific cases. Deductive reasoning is based on logical reasoning.

2. The conclusion is based on logical reasoning. It is a result of deductive reasoning.

3. The conclusion is based on logical reasoning. It is a result of deductive reasoning.

4. Hypothesis: You want to go on a field trip.Conclusion: You must have a signed permission slip.The given statement “Zola has a signed permission slip” matches the conclusion of a true conditional. But this does not mean the hypothesis is true. Zola could have a permission slip for another reason. The conclusion is not valid.

5. Hypothesis: The side lengths of a rect. are 3 ft and 4 ft.Conclusion: The rect.’s area is 12 ft 2 .The given statement “A rect. has side lengths 3 ft and 4 ft” matches the hypothesis of the given conditional. By the Law of Detachment, the rect. has area 12 ft 2 . The conjecture is valid.

6. Let p, q, and r represent the following:p: You fly from Texas to California.q: You travel from the central to the Pacific time zone.r : You gain two hours.You are given that p * q and q * r. Since q is the conclusion of the 1st conditional and the hypothesis of the 2nd conditional, you can conclude that p * r. The conjecture is valid by the Law of Syllogism.

7. Let p, q, and r represent the following:p: A figure is a square.q: A figure is a rectangle.r : A figure is a parallelogram.You are given that p * q and p * r. The Law of Syllogism cannot be used to draw a conclusion since p is the hypothesis of both conditionals. Even though the conjecture r * q is true, the logic used to draw the conclusion is not valid.

8. Conclusion: Alex’s car might not start.


9. The conclusion is based on mathematical calculation. So it is the result of deductive reasoning.

10. Since the conclusion is based on a pattern of observation, it is a result of inductive reasoning.

11. Hypothesis: One integer is odd and another integer is even.Conclusion: The product of 2 integers is even.Statement “The product of 2 integers is 24” matches the conclusion of a true conditional. However, hypothesis is not necessarily true. For example, 6(4) = 24. Conclusion is not valid.

12. Let p, q, and r represent the following:p: An element is an alkali metal.q: An element reacts with water.r : An element is in the 1st column of the periodic table.You are given that p * q and r * p. Since p is the conclusion of the 2nd conditional and the hypothesis of the 1st conditional, you can conclude that r * q. The conjecture is valid by the Law of Syllogism.

13. Conclusion: Dakota gets better grades in Social Studies.

14. If Cheetah-Net is 75 times as fast as dial-up, then dial-up is 75 times as slow as Cheetah-Net. Let c and d be the download times with Cheetah-Net and dial-up.

d = 75c 18 = 75c

18 _ 75

= 75c _ 75

c = 0.24 min or 14.4 s

15. valid 16. invalid

17. valid 18. invalid


19. yes (p * q and q * r, so p * r); no (counterexample: x = 1); because the 1st conditional is false (same counterexample)

20. A: comp. ! are not necessarily adj., so they may not form a rt. (.

21. Possible answers: If Mary goes to the store, then I will go with her. Mary goes to the store. The conclusion “I will go with her” is valid by Law of Detachment. If Jon goes to the movies, then he will eat popcorn. If Jon eats popcorn, then he needs a drink. The conclusion “If Jon goes to the movies, then he will need a drink” is valid by Law of Syllogism.

22a. If a creature is a serpent, then it eats eggs.

b. No; possible answer: the Pigeon did not correctly apply Law of Detachment; “Alice eats eggs” matches conclusion of conditional, not hypothesis.

TEST PREP, PAGE 93

23. D 24. H

25. 196 HzA G note is 3 octaves above low G. So its frequency is the frequency of low G, doubled 3 times.2 (2 (2(24.50)) ) = 196 Hz


26. Either Andre is less than 35 years old, or he is not a natural-born citizen. Possible answer: Since there are 3 criteria and he meets 1, he must not meet 1 of the other 2.

27a. If you live in San Diego then you live in the United States.

b. If you do not live in the United States, then you do not live in California. If you do not live in California, then you do not live in San Diego.

c. If you do not live in the United States, then you do not live in San Diego.

d. They are contrapositives of each other.

28. If Cassie goes, at least 3 people will go, contradicting the hypothesis. If either Hanna or Amy goes, then Marc will go, so by Law of Syllogism, Dallas will also go: again, at least 3 people. Therefore neither Cassie, Amy, or Hanna will go. By elimination, Marc and Dallas will go.


29. 2(x + 5) = 2x + 10

30. (4y + 6) & (3y & 5)= 4y + 6 & 3y + 5= y + 11

31. (3c + 4c) + 2(&7c + 7) = 7c & 14c + 14= &7c + 14

32. ( 1 + 4 _ 2 , 2 + 5 _

2 ) = ( 5 _

2 , 7 _

2 ) = (2.5, 3.5)

33. ( &3 + 0 _ 2 , 6 + 1 _

2 ) = (& 3 _

2 , 7 _

2 ) = (&1.5, 3.5)

34. ( &2.5 + 2.5 _ 2 ,

9 + (&3) _

2 ) = ( 0 _

2 , 6 _

2 ) = (0, 3)

35. Hypothesis: The fire alarm rings.Conclusion: Everyone should exit the building.

36. Hypothesis: Two diffferent lines intersect.Conclusion: Two lines intersect at exactly 1 pt.

37. Hypothesis: %%

AB ) %%

CD Conclusion: AB = CD

GEOMETRY LAB: SOLVE LOGIC PUZZLES, PAGES 94–95

TRY THIS, PAGE 94 1. Because no one else can own the bird and Fiona

owns only 1 pet.

2. Danny Frank Jude Kian

Ally + + + ,

Emily + + , +

Misha , + + +

Tracy + , + +

TRY THIS, PAGE 95 3. WG and GC; because the wolf will eat the goat and

the goat will eat the cabbage.

4. 2 solutions; 7 steps each

5. Possible answer: You can see all solutions instead of just 1 possible solution.

6.

Possible answer: Fill 1-c cup. Fill 3 _ 4 -c cup from

1-c cup. Empty 3 _ 4 -c cup; have 1 _

4 c in the 1-c cup.

Transfer this to 3 _ 4 -c cup, and fill 1-c cup. Fill 3 _

4 -c cup

from 1-c cup, leaving 1 _ 2 c in the 1-c cup.


2-4 BICONDITIONAL STATEMENTS AND DEFINITIONS, PAGES 96–101

CHECK IT OUT! PAGES 96–98 1a. Let p and q represent the following:

p: An ( is acute.q: An (’s measure is greater than 0° and less than 90°.2 parts of biconditional p - q are p * q and q * p.Conditional: If an ( is acute, then its measure is greater than 0° and less than 90°.Converse: If an (’s measure is greater than 0° and less than 90°, then the ( is acute.

b. Let x and y represent the following:x: Cho is a member.y: Cho has paid the $5 dues.2 parts of biconditional x - y are x * y and y * x.Conditional: If Cho is a member, then he has paid the $5 dues.Converse: If Cho has paid the $5 dues, then he is a member.

2a. Converse: If it is Independence Day, then the date is July 4.Biconditional: It is July 4th if and only if it is Independence Day.

b. Converse: If pts. are collinear, then they lie on the same line.Biconditional: Pts. lie on the same line if and only if they are collinear.

3a. Conditional: If an ( is a rt. (, then its measure is 90°. (T)Converse: If an (’s measure is 90°, then it is

a rt. (. (T)Since conditional and converse are true, biconditional is true.

b. Conditional: y = &5 * y 2 = 25; TConverse: y 2 = 25 * y = &5; FIf y = 5, then y 2 = 25, so converse is false. Therefore biconditional is false.

4a. A figure is a quad. if and only if it is a 4-sided polygon.

b. An ( is a straight ( if and only if its measure is 180°.

THINK AND DISCUSS, PAGE 98 1. Possible answer: Find truth values of conditional

and converse that biconditional contains. If both are true, then biconditional is true.

2. A . has 3 sides and 3 vertices. A quad. has 4 sides and 4 vertices.

3.


1. Possible answer: A biconditional statement contains the conditional and its converse. A conditional is not reversible, but a biconditional is.

2. Let p and q represent the following:p: Perry can paint the entire living room.q: Perry has enough paint.2 parts of biconditional p - q are p * q and q * p.Conditional: If Perry can paint the entire living room, then he has enough paint.Converse: If Perry has enough paint, then he can paint the entire living room.

3. Let p and q represent the following:p: Your medicine will be ready by 5 P.M.q: You drop your prescription off by 8 A.M.2 parts of biconditional p - q are p * q and q * p.Conditional: If your medicine is ready by 5 P.M., then you dropped your prescription off by 8 A.M.Converse: If you drop your prescription off by 8 A.M., then your medicine will be ready by 5 P.M.

4. Converse: If a student is in the tenth grade, then the student is a sophomore.Biconditional: A student is a sophomore if and only if the student is in the tenth grade.

5. Converse: If 2 segs. are ), then they have the same length.Biconditional: 2 segs. have the same length if and only if they are ).

6. Conditional: xy = 0 * x = 0 or y = 0; TConverse: x = 0 or y = 0 * xy = 0; TSince conditional and converse are true, biconditional is true.

7. Conditional: If a figure is a quad., then it is a polygon; TConverse: If a figure is a polygon, then it is a quad.; FA . is a polygon but not a quad., so converse is false. Therefore, biconditional is false.


8. 2 lines are / if and only if they are coplanar and never intersect.

9. An animal is a hummingbird if and only if it is a tiny, brightly colored bird with narrow wings, a slender bill, and a long tongue.


10. Conditional: If 3 pts. are coplanar, then they lie in the same plane.Converse: If 3 pts. lie in the same plane, then they are coplanar.

11. Conditional: If a 0 is a rect., then it has 4 rt. !.Converse: If a 0 has 4 rt. !, then it is a rect.

12. Conditional: If a lunar eclipse occurs, then Earth is between the Sun and the Moon.Converse: If Earth is between the Sun and the Moon, then a lunar eclipse occurs.

13. Converse: If it is the weekend, then today is Saturday or Sunday.Biconditional: Today is Saturday or Sunday if and only if it is the weekend.

14. Converse: If Greg wins the race, then he has the fastest time.Biconditional: Greg has the fastest time if and only if he wins the race.

15. Converse: If a . is a rt. ., then it contains a rt. (.Biconditional: A . contains a rt. ( if and only if it is a rt. .

16. Conditional is true because a swimmer is an athlete. Converse is false because an athlete might not be a swimmer. Therefore, biconditional is false. Possible counterexample: Felipe could be a runner.

17. Conditional is true because if 2n is even, it is

divisible by 2, so 2n _ 2 = n is an integer. Converse is

true because if n is an integer, then 2n is an integer divisible by 2, and so is even. Therefore, biconditional is true.

18. A figure is a 1 if and only if it is the set of all pts. that are a fixed dist. from a given pt.

19. A player is a catcher if and only if the player is positioned behind home plate and catches throws from the pitcher.

20. no; possible answer: a = 3, b = &3

21. yes (x = 5 is the solution of both equations)

22. no; possible answer: y = &8

23. no; possible answer: x = &2

24. An equil. . is a . with 3 ) sides.

25. A square is a quad. with 4 ) sides and 4 rt. !.

26. A cell is a white blood cell if and only if it defends the body against invading organisms by engulfing them or releasing antibodies.

27. Possible answer: A bicycle is a vehicle that moves along the ground but is not an automobile.

28. Possible answer: A computer is a machine that performs calculations but is not a calculator.

29. Possible answer: Definition does not say that the rays have a common endpt.

30. no 31. no

32. no 33. 5 (since 2"(5) = 10")

34. quad.

35. A statement is a biconditional if and only if it can be written in the form “p if and only if q.”Conditional: If a statement is a biconditional, then it can be written in the form “p if and only if q.”Converse: If a statement can be written in the form “p if and only if q,” then it is a biconditional.Since conditional and converse are true, biconditional is true.

36. Possible answer: If you write the def. as a biconditional, “A ray is an ( bisector iff it divides the ( into 2 ) !,” then you can use it either forward or backward. If you know the ray is an ( bisector, then you can conclude that the 2 ! formed are ). If you know that 2 adj. ! formed by a ray are ), then you can conclude that the ray is an ( bisector.

37a. If I say it, then I mean it. If I mean it, then I say it.

b. Possible answer: The biconditional Alice implies is “I say it if and only if I mean it.” This biconditional is not true. People often mean things without saying them or say things they don’t mean.

TEST PREP, PAGE 101

38. Am(S 2 80° but (S is acute.

39. G 40. BConverse is also true.

41. Conditional: If you get a traffic ticket, then you are speeding.Converse: If you are speeding, then you will get a traffic ticket.

The biconditional is false because both statements are false. It is possible to get a traffic ticket for running a red light while not speeding. Also, it is possible to speed without getting a ticket.


42. The two ovals within Venn diagram will exactly overlap each other. If one condition is met then the other is necessarily met, which is true of the conditions in a good def.

43a. If an ( does not measure 105°, then the ( is not obtuse.

b. If an ( is not obtuse, then it does not measure 105°.

c. It is the contrapositive of the original.

d. F; the inverse is false, so the biconditonal formed is false.

44. T; It is given that conditional is true. Converse “If D is in int. of (ABC, then m(ABD + m(DBC = m(ABC ” is true by ( Add. Post. Since conditional and its converse are true, biconditional is true.


45. Possible answer: n = 2 (since n is not divisible by 4, but n 2 = 4 is even.)


46. The graph is shifted 5 units up and is wider than graph of parent function.

47. The graph is reflected across x-axis and shifted 1 unit down, and is narrower than graph of parent function.

48. y = (x + 2)(x & 2) = x 2 & 4

The graph is shifted 4 units down.

49. T 50. Y

51. S 52. F; poss. answer: n = 0

53. F; poss. answer: x = 2 54. T


1. Alice is mad.

2. If you only walk long enough, then you’re sure to get somewhere.

3. No; no hypothesis is known to be true, so Law of Detachment cannot be applied. No conclusion matches another hypothesis, so Law of Syllogism cannot be applied.

4. I breathe if and only if I sleep. This biconditional is made of 2 conditionals: If I breathe, then I sleep, and if I sleep, then I breathe. 2nd is true, but 1st is not. So biconditional is false.


1. 31 2. January

3. &1 4.

5. Possible answer: A male lion weighs about 412.4 lb.

6. negative 7. Possible answer: 6

8. Hypothesis: An (’s measure is 107°.Conclusion: An ( is obtuse.

9. If a number is a whole number, then it is an integer.

10. If a figure is a square, then it is a rect.

11. If a figure is a square, then its diags. are ).

12. F; possible answer: an ( that measures 60°

13. T

14. Converse: If a number is divisible by 4, then it is even; TInverse: If a number is not even, then it is not divisible by 4; TContrapositive: If a number is not divisible by 4, then it is not even; F

15. Hypothesis: Sue finishes her science project.Conclusion: Sue can go to the movie.The given statement “Sue goes to the movie” matches the conclusion of a true conditional. But this does not mean the hypothesis is true. Sue could have gone to the movie on another night. The conclusion is not valid.

16. Let p, q, and r represent the following.p: 1 ( of a . is 90°.q: A . is a rt. ..r : A .’s acute ( measures are comp.You are given that p * q and q * r. Since q is the conclusion of the 1st conditional and the hypothesis of the 2nd conditional, you can conclude that p * r, or that if 1 ( of a . is 90°, then its acute ( measures are comp.

17. Converse: If the sum of 2 ( measures is 180°, then the ! are supp.Biconditional: 2 ! are supp. if and only if the sum of their measures is 180°.

18. T

2-5 ALGEBRAIC PROOF, PAGES 104–109

CHECK IT OUT! PAGES 104–106 1. 1 _

2 t = &7 Given equation

2 ( 1 _ 2 t) = 2(&7) Mult. Prop. of =

t = &14 Simplify.

2. 1 Understand the ProblemAnswer will be temperature in °C.Important information:

• C = 5 _ 9 (F & 32) • F = 86°F

2 Make a PlanSubst. given information into formula and solve.3 Solve

C = 5 _ 9 (F & 32) Given equation

= 5 _ 9 (86 & 32) Subst.

= 5 _ 9 (54) Simplify.

= 30°C Simplify.4 Look BackCheck answer by substituting it back into original formula. C = 5 _

9 (F & 32)

30 3 5 _ 9 (86 & 32)

9(30) 3 5(54) 270 = 270 "

3. ( Add. Post.Subst.Simplify.Subtr. Prop. of =Add. Prop. of =

4a. Sym. Prop. of = b. Reflex. Prop. of =

c. Trans. Prop. of = d. Sym. Prop. of )


THINK AND DISCUSS, PAGE 107 1. Mult. Prop. of =

2. Use a ) symbol for geometric figures. Use an = sign for numbers.

3.


1. Possible answer: A proof is an argument that uses logic, definitions, and previously proven statements to show that a statement is always true.

2. y + 1 = 5 Given equation& 1 & 1 Subtr. Prop. of =

y = 4 Simplify.

3. t & 3.2 = &8.3 Given equation+ 3.2 + 3.2 Add. Prop. of =

t = &5.1 Simplify.

4. 2p & 30 = &4p + 6 Given equation +4p +4p Add. Prop. of = 6p & 30 = 6 Simplify. _______ + 30 ____ +30 Add. Prop. of = 6p = 36 Simplify.

6p

_ 6 = 36 _

6 Div. Prop. of =

p = 6 Simplify.

5. x + 3 _ &2

. = 8 Given equation

(&2) ( x + 3 _ &2

) = &2(8) Mult. Prop. of =

x + 3 = &16 Simplify.& 3 & 3 Subtr. Prop. of =

x = &19 Simplify.

6. 1 _ 2

n = 3 _ 4 Given equation

2 ( 1 _ 2 n) = 2 ( 3 _

4 ) Mult. Prop. of =

n = 3 _ 2 Simplify.

7. 0 = 2(r & 3) + 4 Given equation0 = 2r & 2 Distribute.

+ 2 + 2 Add. Prop. of =2 = 2r Simplify.

2 _ 2 = 2r _

2 Div. Prop. of =

1 = r Simplify.r = 1 Sym. Prop. of =

8. 1 Understand the ProblemAnswer will be amount of fat in g.Important information:• C = 9f + 90 • C = 102 calories2 Make a PlanSubst. given information into formula and solve.3 Solve

C = 9f + 90 Given equation102 = 9f + 90 Subst.& 90 & 90 Subtr. Prop. of =

12 = 9f Simplify.

12 _ 9 = 9f _

9 Div. Prop. of =

4 _ 3 = f Simplify.

Cereal contains 4 _ 3 g of fat.

4 Look BackCheck answer by substituting it back into original formula. C = 9f + 90

102 3 9 ( 4 _ 3 ) + 90

90 3 12 + 90102 = 102 "

9. 1 Understand the ProblemAnswer will be number of movie rentals.Important information:• C = $5.75 + $0.89m • C = $11.982 Make a PlanSubst. given information into formula and solve.3 Solve

C = 5.75 + 0.89m Given equation11.98 = 5.75 + 0.89m Subst.& 5.75 & 5.75 Subtr. Prop. of =

6.23 = 0.89m Simplify.

6.23 _ 0.89

= m Div. Prop. of =

7 = m Simplify.

Elias rented 7 movies.

4 Look BackCheck answer by substituting it back into original formula. C = 5.75 + 0.89m11.98 3 5.75 + 0.89(7)11.98 3 5.75 + 6.2311.98 = 11.98 "

10. Def. of ) segs.Subst.Subtr. Prop. of =Subtr. Prop. of =Div. Prop. of =

11. Seg. Add. Post.Subst.Subtr. Prop. of =Add. Prop. of =Div. Prop. of =

12. Reflex. Prop. of ) 13. Trans. Prop. of =

14. Sym. Prop. of = 15. Trans. Prop. of )


16. 5x & 3 = 4(x + 2) Given equation5x & 3 = 4x + 8 Distrib. Prop.x & 3 = 8 Subtr. Prop. of =

x = 11 Add. Prop. of =


17. 1.6 = 2n Given equation0.5 = n Div. Prop. of =

18. z _ 3 & 2 = &10 Given equation

z _ 3

= &8 Add. Prop. of =

z = &24 Mult. Prop. of =

19. &(h + 3) = 72 Given equation&h & 3 = 72 Distrib. Prop.

&h = 75 Add. Prop. of =h = &75 Mult. Prop. of =

20. 9y + 17 = &19 Given equation9y = &36 Subtr. Prop. of =y = &4 Div. Prop. of =

21. 1 _ 2 (p & 16) = 13 Given equation

p & 16 = 26 Mult. Prop. of =p = 42 Add. Prop. of =

22. T = 0.03c + 0.05b Given equation147 = 0.03c + 0.05(150) Subst.147 = 0.03c + 7.5 Simplify.

139.5 = 0.03c Subtr. Prop. of =4,650 = c Div. Prop. of =4,650 bottles were collected.

23. ( Add. Post.Subst.Simplify.Subtr. Prop. of =Add. Prop. of =Div. Prop. of =

24. ( Add. Post.Subst.Distrib. Prop.Simplify.Subtr. Prop. of =Div. Prop. of =

25. Sym. Prop. of ) 26. Reflex. Prop. of =

27. Trans. Prop. of = 28. Reflex. Prop. of )

29. Estimate: 2(3x & 1) = 94 3x & 1 = 47 3x = 48 x = 16Equation:2(3.1x & 0.87) = 94.36 Given equation

3.1x & 0.87 = 47.18 Div. Prop. of =3.1x = 48.05 Add. Prop. of =

x = 15.5 Div. Prop. of =Possible answer: The exact solution rounds to the estimate.

30. 3x & 1 31. (A ) (T

32. %%

NP ) %%

BC

33. ( 1 + x _ 2 ,

y + 1 _

2 ) = (3, 5)

1 + x _ 2 = 3 Midpt. formula

1 + x = 6 Mult. Prop. of = x = 5 Subtr. Prop. of =

y + 1

_ 2 = 5 Midpt. formula

y + 1 = 10 Mult. Prop. of = y = 9 Subtr. Prop. of =

34. C = 35 + 21h + 1.1p Given equation169.5 = 35 + 21(3) + 1.1p Subst.169.5 = 98 + 1.1p Simplify.71.5 = 1.1p Subtr. Prop. of =

65 = p Div. Prop. of =Cost of parts was $65.

35a. C = 92.5 + 79.96 + 983 + 10,820x

Given equation

1,733.65 = 92.5 + 79.96 + 983 + 10,820x

Subst.

1,733.65 = 1,155.46 + 10,820x Simplify.578.19 = 10,820x Subtr. Prop. of =0.0534 ' x Div. Prop. of =

Average cost of gas per mile is ' $0.05.

b. 1 gal costs ' 1 gal · 32 mi _ 1 gal

· $0.05 _ 1 mi

' $1.71

36. Given %%

PR , you know from Reflex. Prop. of = that PR = PR. By def. of ) segs.,

%% PR ) %%

PR . Given that

%% PR )

%% ST , you know from def. of ) segs.

that PR = ST. By Sym. Prop. of =, ST = PR. By def. of ) segs.,

%% ST ) %%

PR .Given that

%% AB ) %%

CD and %%

CD ) %%

EF , you know from def. of ) segs. that AB = CD and CD = EF. By Trans. Prop. of =, AB = EF. Therefore,

%% AB )

%% EF by

def. of ) segs.

37a. x + 15 4 63 Given inequal.x 4 48 Subtr. Prop. of Inequal.

b. &2x > 36 Given inequal.x < &18 Div. Prop. of Inequal.

38. Possible answer: The conclusion of a deductive proof has been proven true in all cases, but a conjecture is based on observation and is not proven to be true.

TEST PREP, PAGE 109

39. B 40. H

41. D

42. 90° m(1 + m(2 + m(3 = 180°m(1 + m(1 + 2m(1 = 180° 4m(1 = 180° m(1 = 45°m(3 = 2m(1

= 2(45) = 90°


43. PR = PA + RA Seg. Add. Post. PA = QB, QB = RA Given PA = RA Trans. Prop. of = PR = PA + PA Subst. PA = 18 Given PR = 18 + 18 Subst. PR = 36 in. Simplify.

44. Possible answer: You cannot add geometric figures.

45. 7 & 3x > 19 Given &3x > 12 Subtr. Prop. of Inequal. x < &4 Div. Prop. of Inequal.



46. the interest rate the account earns

47. Check students’ contructions.

48. Check students’ contructions.

49. deductive reasoning 50. inductive reasoning

2-6 GEOMETRIC PROOF, PAGES 110–116

CHECK IT OUT! PAGES 110–112 1. 1. Given

2. Def. of mdpt.3. Given4. Trans. Prop. of )

2a. (1 and (2 are supp., and (2 and (3 are supp.

b. m(1 + m(2 = m(2 + m(3

c. Subtr. Prop. of = d. (1 ) (3

3. Statements Reasons

1. (1 and (2 are comp., (2 and (3 are comp.

1. Given

2. m(1 + m(2 = 90°, m(2 + m(3 = 90°

2. Def. of comp. !

3. m(1 + m(2 = m(2 + m(3 3. Subst.4. m(2 = m(2 4. Reflex. Prop.

of =5. m(1 = m(3 5. Subtr. Prop.

of =6. (1 ) (3 6. Def. of ) !

THINK AND DISCUSS, PAGE 113 1. the last step

2. Possible answer: so another person can follow your proof and you can verify that your logical reasoning is correct.

3. postulate; theorem; definition; property

4.


1. statements; reasons 2. theorem

3. 1. Given2. Subst.3. Simplify.4. Add. Prop. of =5. Simplify.6. Def. of supp. !

4a. Def. of ) ! b. (1 and (2 are supp.

c. Subst. d. (1 and (3 are supp.


1. X is the mdpt. of %%

AY , Y is the mdpt. of

%% XB .

1. Given

2. %%

AX ) %%

XY , %%

XY ) %%

YB 2. Def. of mdpt.3. %%

AX ) %%

YB 3. Trans. Prop. of )


6. 1. Given2. Def. of ( bisector3. Def. of ) !4. Given5. Subst.6. ( Add. Post.7. Subst.8. Simplify.9. Def. of rt. (

7a. m(1 + m(2 = 180°,m(3 + m(4 = 180°

b. Subst.

c. m(1 = m(4

d. Def. of ) !

8a. Def. of rt. ( b. m(1 + m(2 = m(BAC

c. m(2 = m(3 d. Subst.

e. (1 and (3 are comp.


1. %%

BE ) %%

CE , %%

DE ) %%

AE 1. Given2. BE = CE, DE = AE 2. Def. of ) segs.3. AE + BE = AB,

CE + DE = CD3. Seg. Add. Post.

4. DE + CE = AB 4. Subst.5. AB = CD 5. Subst.6. %%

AB ) %%

CD 6. Def. of ) segs.


1. (1 and (3 are comp., (2 and (4 are comp.

1. Given

2. m(1 + m(3 = 90°,m(2 + m(4 = 90°

2. Def. of comp. !

3. m(1 + m(3 = m(2 + m(4 3. Subst.4. ( 3 ) (4 4. Given5. m( 3 = m(4 5. Def. of ) !6. m( 1 = m(2 6. Subtr. Prop.

of =7. (1 ) (2 7. Def. of ) !

11. m(1 + 48° = 180° m(1 = 132°

12. m(2 + 63° = 90° m(2 = 27°

13. m(3 + 31° = 90° m(3 = 59°

14. ) Supps. Thm.

15. Possible answer: because the ! can be supp. or comp. to the same ( or to 2 ) !.

16. sometimes 17. sometimes

18. sometimes 19. never

20. 4n + 5 + 8n & 5 = 180 12n = 180 n = 15

21. 9x & 6 = 8.5x + 2 9x = 8.5x + 8 0.5x = 8 x = 16

22. 4z + 3z + 6 = 90 7z + 6 = 90 7z = 84 z = 12


23. Possible answer: A thm. and a post. are both true statements of geometric facts. They are different because a post. is assumed to be true, while a thm. must be proven to be true.

24a. Given: Y is mdpt. of %%

AC . X is the mdpt. of %%

AB .

Prove: XY = 1 _ 2

BC

b. Given: (C is a rt. (.Prove: (A and (B are comp.

c. Given: (C is a rt. (.Prove: (AB) 2 = (AC) 2 + (BC) 2

TEST PREP, PAGE 116

25. C 26. G

27. D



1. m(LAN = 30° 1. Given2. m(1 + m(2 = m(LAN 2. ( Add. Post.3. m(1 + m(2 = 30° 3. Subst.4. m(1 = 15° 4. Given5. 15° + m(2 = 30° 5. Subst.6. m(2 = 15° 6. Subtr. Prop. of =7. m(1 = m(2 7. Trans. Prop. of =8. (1 ) (2 8. Def. of ) !9.

##$ AM bisects (LAN. 9. Def. of ( bisector

29. Step 1 Find a.2a + 3.5 = 2.5a & 52a + 8.5 = 2.5a 8.5 = 0.5a 17 = aStep 2 Find ( measures.2a + 3.5 = 2(17) + 3.5 = 37.5°3a + 1.5 = 3(17) + 1.5 = 52.5°2.5a & 5 = 2.5(17) & 5 = 37.5°

30. Step 1 Find x.4 x 2 & 6 + (&2 x 2 + 19x) = 180 2 x 2 + 19x & 6 = 180 2 x 2 + 19x & 186 = 0 (x & 6)(2x + 31) = 0 x = 6 (since &2 x 2 + 19x > 0)Step 2 Find ( measures.4 x 2 & 6 = 4(6 ) 2 & 6 = 138°&2 x 2 + 19x = &2(6 ) 2 + 19(6) = 42°


31. 60 _ 250

= 0.24 = 24% 32. 2(14) = 28 tires

33. Possible answer:


35. Sym. Prop. of ) 36. Trans. Prop. of =

GEOMETRY LAB: DESIGN PLANS FOR PROOFS, PAGE 117

ACTIVITY, PAGE 117


1. ( AXB ) ( CXD 1. Given 2. m(AXB = m(CXD 2. Def. of ) ! 3. m(BXC = m(BXC 3. Reflex. Prop. of = 4. m(AXB + m(BXC =

m(CXD + m(BXC 4. Add. Prop. of =

5. m(AXB + m(BXC = m(AXC m(BXC + m(CXD = m(BXD

5. ( Add. Post.

6. m(AXC = m(BXD 6. Subst. 7. (AXC ) (BXD 7. Def. of ) !

TRY THIS, PAGE 117 1. Possible answer: A plan for a proof is less formal

than a proof. A formal proof presents every logical step in detail, but a plan describes only the key logical steps.

2. Plan: Use the def. of ( bisector to show that m(1 = m(2. Then use ( Add. Post. and subst. to show that 2m(1 = m(ABC.2-column proof:

Statements Reasons

1. ##$ BD bisects (ABC. 1. Given

2. (1 ) (2 2. Def. of ( bis.3. m(1 = m(2 3. Def. of ) !4. m(1 + m(2 = m(ABC 4. ( Add. Post.5. m(1 + m(1 = m(ABC 5. Subst.6. 2m(1 = m(ABC 6. Simplify.

3. Plan: Since (LXN is a rt. (, its measure is 90°. By the ( Add. Post., m(1 + m(2 = m(LXN. So by subst., m(1 + m(2 = 90°, which means that (1 and (2 are comp.2-column proof:

Statements Reasons

1. (LXN is a rt. (. 1. Given2. m(LXN = 90° 2. Def. of rt. (3. m(1 + m(2 = m(LXN 3. ( Add. Post.4. m(1 + m(2 = 90° 4. Subst.5. (1 and (2 are comp. 5. Def. of comp. !

2-7 FLOWCHART AND PARAGRAPH PROOFS, PAGES 118–125



1. RS = UV, ST = TU 1. Given2. RS + ST = TU + UV 2. Add. Prop. of =3. RS + ST = RT,

TU + UV = TV3. Seg. Add. Post.

4. RT = TV 4. Subst.5. %%

RT ) %%

TV 5. Def. of ) segs.


2.


1. (WXY is a rt. (. 1. Given2. m(WXY = 90° 2. Def. of rt. (3. m(2 + m(3 = m(WXY 3. ( Add. Post.4. m(2 + m(3 = 90° 4. Subst.5. (1 ) (3 5. Given6. m(1 = m(3 6. Def. of ) !7. m(2 + m(1 = 90° 7. Subst.8. (1 and (2 are comp. 8. Def. of comp. !

4. It is given that (1 ) (4. By Vert. ! Thm., (1 ) (2 and (3 ) (4. By Trans. Prop. of ) (twice), (2 ) (4, and (2 ) (3.

THINK AND DISCUSS, PAGE 122 1. Possible answer: There may be more than one thm.

that you can apply to a proof, and the steps in a proof may sometimes be written in a different order.

2. Answers will vary.

3.

EXERCISES, PAGES 122–125GUIDED PRACTICE, PAGES 122–123

1. flowchart 2. paragraph


1. (1 ) (2 1. Given2. (1 and (2 are supp. 2. Lin. Pair Thm.3. (1 and (2 are rt. ! 3. ) ! supp. * rt. !

4.


1. (2 ) (4 1. Given2. (1 ) (2, (3 ) (4 2. Vert. ! Thm.3. (1 ) (4 3. Trans. Prop. of )4. (1 ) (3 4. Trans. Prop. of )

6. It is given that ##$ BD bisects (ABC, so (1 ) (2 by def.

of ( bis. By Vert. ! Thm., (1 ) (4 and (2 ) (3. By Trans. Prop. of ), (4 ) (2, and thus (4 ) (3.

Therefore ##$ BG bisects (FBH by def. of ( bis.



1. B is mdpt. of %%

AC . 1. Given2. %%

AB ) %%

BC 2. Def. of mdpt.3. AB = BC 3. Def. of ) segs.4. AD + DB = AB,

BE + EC = BC4. Seg. Add. Post.

5. AD + DB = BE + EC 5. Subst.6. AD = EC 6. Given7. DB = BE 7. Subtr. Prop. of =

8.


1. (1 ) (4 1. Given2. (1 ) (2 2. Vert. ! Thm.3. (4 ) (2 3. Trans. Prop. of )4. m(4 = m(2 4. Def. of ) !5. (3 and (4 are supp. 5. Lin. Pair Thm.6. m(3 + m(4 = 180° 6. Def. of supp. !7. m(3 + m(2 = 180° 7. Subst.8. (2 and (3 are supp. 8. Def. of supp. !

10. Since (1 and (2 are comp., m(1 + m(2 = 90°. (1 ) (3 by Vert. ! Thm. Thus m(1 = m(3. By subst., m(2 + m(3 = 90°, so (2 and (3 are comp.

11. 13 cm; by conv. of the Common Segs. Thm.

12. 90°; ) ! supp. * rt. !

13. 37°; By Vert. ! Thm.

14. By the Common Segs. Thm., 2x + 4 = 5x & 2 6 = 3x x = 2

15. By the Vert. ! Thm.,11y = 121 y = 11


16. By the Vert. ! Thm.,2x + 40 = 5x + 16 24 = 3x x = 8

17. A; diagram is marked with Prove information instead of Given information.

18.

19. Possible answer: Both ! adj. to given rt. ( must be rt. ! because they form lin. pairs with the given (. The fourth ( is a vert. ( of given (, so it, too, is a rt. (. Since all 4 ! are rt. !, they are all ) by Rt. ! ) Thm.

20. Answers will vary.

TEST PREP, PAGE 125

21. C(5 and (8 are vert. !.

22. Hm(2 = 90 + 38 = 128°

23. D


24.


1. (AOC ) (BOD 1. Given2. m(AOC = m(BOD 2. Def. of ) !3. m(AOB + m(BOC = m(AOC,

m(BOC + m(COD = m(BOD3. ( Add. Post.

4. m(AOB + m(BOC = m(BOC + m(COD

4. Subst.

5. m(BOC = m(BOC 5. Reflex. Prop. of =

6. m(AOB = m(COD 6. Subtr. Prop. of =

7. (AOB ) (COD 7. Def. of ) !

26. It is given that (2 and (5 are rt. !. By Rt. ( ) Thm., (2 ) (5. By def. of ) !, m(2 = m(5. It is also given that m(1 + m(2 + m(3 = m(4 + m(5 + m(6. By Subtr. Prop. of =, m(1 + m(3 = m(4 + m(6. (3 ) (6 by Vert. ! Thm. By def. of ) !, m(3 = m(6. By Subtr. Prop. of =, m(1 = m(4. So by def. of ) !, (1 ) (4.

27. Step 1 Find x and y.By Vert. ! Thm.,3x + 1 = 6y + x & 6 2x = 6y & 7 x = 3y & 3.5

By def. of supp. !,3x + 1 + 2x + 2y + 1 = 180 5x + 2y = 178 5(3y & 3.5) + 2y = 178 15y & 17.5 + 2y = 178 17y = 195.5 y = 11.5

x = 3y & 3.5 = 3(11.5) & 3.5 = 31Step 2 Find ( measures.3x + 1 = 3(31) + 1 = 94°2x + 2y + 1 = 2(31) + 2(11.5) + 1 = 86°6y + x & 6 = 6(11.5) + 31 & 6 = 94°m(4th () = m(2nd () = 86°


28. y = 2x + 14 &6x + 18 = 2x + 14 18 = 8x + 14 4 = 8x

1 _ 2 = x

y = 2x + 14

= 2 ( 1 _ 2 ) + 14 = 15

( 1 _ 2 , 15)

29. 7x & y = &33 3x + y = &7 10x = &40 x = &4

3x + y = &7 3(&4) + y = &7 &12 + y = &7 y = 5 (&4, 5)

30. 2(&x + 3y = 10) + 2x + y = 8

&2x + 6y = 20 + 2x + y = 8 7y = 28 y = 4

&x + 3y = 10 &x + 3(4) = 10 &x + 12 = 10 &x = &2 x = 2 (2, 4)


35. Converse: If a positive integer is a composite number, then it has more than 2 factors.Biconditional: A positive integer has more than 2 factors if and only if it is a composite number.

36. Converse: If a quad. has exactly 1 pair of / sides, then it is a trapezoid.Biconditional: A quad. is a trapezoid if and only if it has exactly 1 pair of / sides.



1. 75° 4 x 4 90°

2. 75 4 x 4 90180 & 75 5 180 & x 5 180 & 90 105 5 y 5 90 90° 4 y 4 105°


1. m(2 = 145° 1. Given2. (1 and (2 are supp. 2. Lin. Pair Thm.3. m(1 + m(2 = 180° 3. Def. of supp. !4. m(1 + 145° = 180° 4. Subst.5. m(1 = 35° 5. Subtr. Prop. of =6. (1 ) (3 6. Vert. ! Thm.7. m(1 = m(3 7. Def. of ) !8. m(3 = 35° 8. Trans. Prop. of =9. m(1 < 75° and m(3 < 75° 9. Def. of Inequal.

4. A surveyor found that m(2 = 145°. Since (1 and (2 are supp. by Lin. Pair Thm., m(1 + m(2 = 180° by def. of supp. !. By using subst. and Subtr. Prop. of =, we can conclude that m(1 = 35°. By Vert. ! Thm., (1 ) (3, so their measures are =. Therefore, m(3 = 35° by Trans. Prop. of =. Since 35° < 75°, we can conclude that m(1 < 75° and m(3 < 75°. Thus ! in the intersection do not meet safety guidelines of U.S. Department of Transportation, and intersection should be reconstructed.


1. m & 8 = 13 Given equationm = 21 Add. Prop. of =

2. 4y & 1 = 27 Given equation4y = 28 Add. Prop. of =y = 7 Div. Prop. of =

3. & x _ 3 = 2 Given equation

&x = 6 Mult. Prop. of =x = &6 Div. Prop. of =

4. Sym. Prop. of = 5. Reflex. Prop. of )

6. Trans. Prop. of ) 7. Trans. Prop. of =

8a. Given(given information)

b. (1 and (3 are supp.(deduce from line 1)

c. Reflex. Prop. of )(reason why (3 ) (3)

d. (1 ) (4(apply ) Supps. Thm. to lines 2 and 3)


1. %%

AB ) %%

EF 1. Given2. AB = EF 2. Def. of ) segs.3. EF = AB 3. Sym. Prop. of =4. %%

EF ) %%

AB 4. Def. of ) segs.

10.

11. It is given that (1 ) (3. By Vert. ! Thm., (1 ) (2 and (3 ) (4. By Trans. Prop. of ), (2 ) (3 and thus (2 ) (4.


1. theorem 2. deductive reasoning

3. counterexample 4. conjecture

LESSON 2-1, PAGE 130 5. The rightmost . is duplicated, rotated 180°, and

shifted right. The next two items are and

6. Each term is 1 _ 6 greater than previous one. The next

two terms are 5 _ 6 and 1.

7. The white section is halved. If the white section is a rect. but not a square, it is halved horiz. and the upper portion is colored yellow. If the white section is a square, it is halved vert. and the left portion is colored yellow. The next 2 items are and

8. odd 9. positive

10. F; 0 is a whole number but not a natural number

11. T 12. T

13. F; during a leap year, there are 29 days in February.

14. Check students’ constructions. Possible answer: The 3 ( bisector of a . intersect at the int. of ..

LESSON 2-2, PAGE 131 15. If it is Monday, then it is a weekday.

16. If something is a lichen, then it is a fungus.

17. T

18. F; possible answer: 6 7 2 and 6 7 2

19. Converse: If m(X = 90°, then (X is a rt. (; TInverse: If (X is not a rt. (, then m(X 2 90°; TContrapositive: If m(X 2 90°, then (X is not a rt. (; T

20. Converse: If x = 2, then x is a whole number; TInverse: If x is not a whole number, then x 2 2; TContrapositive: If x 2 2, then x is not a whole number; F


LESSON 2-3, PAGE 131 21. Let p, q, r, and s be the following;

p: The team practices, beginning at 8 A.M. on weekdays and at 12 noon on Saturday.q: Sue swims, beginning at 8 A.M. on weekdays and at 12 noon on Saturday.r : The pool opens at 8 A.M. on weekdays and at 12 noon on Saturday.Using symbols, given information is p * q, r * p, and r. By Law of Detachment, p is valid, so conjecture is not valid.

22. Using symbols, given information is p * q, r * p, and r. By Law of Syllogism, r * q, and by Law of Detachment, q is valid; so conjecture is valid.

23. Using symbols, given information is p * q, r * p, and r. By Law of Detachment, p is valid, so conjecture is not valid.

24. Let p be hypothesis: Cost of Sara’s call is $2.57Let q be conclusion: Sara’s call lasted 7 min. 2.57 = 2.15 + 0.07x 0.42 = 0.07x x = 6x + 1 = 7So p * q. Since statement “Cost of Sara’s call is $2.57.” matches hypothesis, can conclude that Sara’s call lasted 7 min.

25. Let p be hypothesis: Paolo’s call lasts 10 min.Let q be conclusion: Cost of Paolo’s call is $2.78.2.15 + 0.07(10 & 1) = 2.78So p * q. Since statement “Paolo’s call lasts 10 min.” matches hypothesis, can conclude that cost of Paolo’s call is $2.78.

26. No conclusion; the number and lengths of calls are unknown.

LESSON 2-4, PAGE 132 27. yes 28. no; x = 2

29. no; seg. with endpts. (3, 7) and (&5, &1)

30. yes

31. comp. 32. positive

33. greater than 50 mi/h 34. 4s

LESSON 2-5, PAGE 132 35. m _

&5 + 3 = &4.5 Given equation

m _ &5

= &7.5 Subtr. Prop. of =

m = 37.5 Mult. Prop. of =

36. &47 = 3x & 59 Given equation12 = 3x Add. Prop. of =4 = x Div. Prop. of =

37. Reflex. Prop. of = 38. Sym. Prop. of )

39. Trans. Prop. of = 40. figure ABCD

41. m(5 = m(2 42. %%

CD ) %%

EF

43. I = Prt Given equation4200 = P(0.06)(4) Subst.4200 = P(0.24) Simplify.

$17,500 = P Div. Prop. of =

LESSON 2-6, PAGE 133 44. 1. Given

2. Def. of comp. !3. Given4. Def. of ) !5. Subst.6. Def. of comp. !

45a. Given

b. TU = UV

c. SU + UV = SV

d. Subst.

46. Think: Use def. of supp. !.z & 2 + 2 + 7z = 180 8z = 180 z = 22.5

47. Think: Use def. of comp. !.3x + 2x + 5 = 90 5x = 85 x = 17

LESSON 2-7, PAGE 133

48.

49. It is given that (ADE and (DAE are comp. and (ADE and (BAC are comp. By ) Comps. Thm., (DAE ) (BAC. By Reflex. Prop. of ), (CAE ) (CAE. By Common ! Thm., (DAC ) (BAE.

50. w = 45; by Vert. ! Thm.

51. x = 45; since ) ! supp. * rt. !


1. 2. 5

3. even

4. Possible answer: (1 and (2 are comp. but not adj.

5. Hypothesis: It rains.Conclusion: The show is cancelled.

6. If 2 lines are /, then they do not intersect.

7. F; 8. T

9. Converse: If you live in Kentucky, then you live in the United States; T


10. Inverse: If you do not live in the United States, then you do not live in Kentucky; T

11. Contrapositive: If you do not live in Kentucky, then you do not live in the United States; F

12. Let p, q, and r be the following:p: it is colder than 50°F.q: Tom wears a sweater.r : it is 46°F today.You are given that p * q and that r is true. You also know that r * p, By Law of Detachment, q is true, so conjecture is valid.

13. Let p, q, and r be the following:p: A figure is a square.q: A figure is a quad.r : A figure is a polygon.You are given that p * q and q * r. By Law of Syllogism, p * r. The statement “Figure ABCD is a square” matches p, so by Law of Detachment, r is true for figure ABCD. Therefore, figure ABCD is a polygon.

14. Conditional: If Chad works on Saturday, then he gets paid overtime.Converse: If Chad gets paid overtime, he will work on Saturday.

15. F; A, B, and C with B not between A and C

16. 8 & 5s = 1 Given equation&5s = &7 Subtr. Prop. of =

s = 1.4 Div. Prop. of =

17. 0.4t + 3 = 1.6 Given equation0.4t = &1.4 Subtr. Prop. of =

t = &3.5 Div. Prop. of =

18. 38 = &3w + 2 Given equation36 = &3w Subtr. Prop. of =

&12 = w Div. Prop. of =

19. Trans. Prop. of = 20. Reflex. Prop. of =

21. Trans. Prop. of ) 22. Sym. Prop. of )


1. (AFB ) (EFD 1. Given2. (EFD ) (BFC 2. Vert. ! Thm.3. (AFB ) (BFC 3. Trans. Prop. of )4.

##$ FB bisects (AFC. 4. Def. of ( bisector

24. It is given that (AFB ) (EFD. By Vert. ! Thm., (EFD ) (BFC. Therefore, (AFB ) (BFC by

Trans. Prop. of ). So ##$ FB bisects (AFC. by Def. of

( bisector.

25.


1. CThink: Use (def. of ) ! and) Vert. ! Thm. to get 2nd equation.m(1 = m(2 60 = 3x x = 20

m(1 = m(5 60 = x + y 60 = 20 + y y = 40

2. EI and III contained in biconditional; II is contrapositive of I

3. DContrapositive of “If p, then q” is “If 8q, then 8p.”

4. CUse the Pyth. Thm. DE 2 = 3 2 + 2 2

= 9 + 4 = 13 DE = 6 77 13

5. BA is never true, C, D, and E are only sometimes true.


CHAPTER 2, PAGES 140–141

THE MYRTLE BEACH MARATHON , PAGE 140 1. Both given rates are equivalent to 7.8 mi/h; time to

complete marathon: (7.8)(26) = 202.8 min ! 3 1 _ 3 h.

2. There are 5 pts. with medical station and portable toilets: at 6 mi, 12 mi, 18 mi, 24 mi, and 26 mi.

3. Let x and y be distances from HQ to viewing stand and from viewing stand to 29th Ave. N. Given information: x + y = 3.25y, so x = 2.25y. From map,1.7 + x + y = 4.3 3.25y = 2.6 y = 0.8x = 2.25y

= 2.25(0.8) = 1.8 mi

SOUTH CAROLINA’S WATERFALLS , PAGE 141 1. Waterfalls < 100 ft with 1-way trail length " 1.5 mi:

Mill Creek Falls or Yellow Branch Falls

2a. F; round-trip hike to Mill Creek Falls is > 4 mi, but falls are < 400 ft tall

b. F; If you hike to Raven Cliff, then you have seen a waterfall that is " 200 ft tall.

c. T

3. Let height of middle falls be x.x + x + (x + 15) = 120 3x = 105 x = 35Heights are 35 ft, 35 ft, and 35 + 15 = 50 ft.


THE QUEEN’S CUP, PAGE 294 1. Think: Calculate new bearing at each change of

direction.At A: 50° + 43° = 93°, so new bearing is S 43° E.At C: 43° + 62° = 105°, so new bearing is N 62° E.At E: 62° + 20° = 82°, so new bearing is S 20° E.

2. Speed over first 49 mi is about 10 mi/h. So race distance (about 80 mi) should take about 8 h.

3. Yes; there is enough information to find m#MXY (101°). MX and MY are know, so a unique $MXY is determined by SAS.

THE AIR ZOO, PAGE 295 1. Think: 7-month data will give most reliable mean

painting rate. Use proportions.

n _ 28,800

= 7 _ 18,327

n = 7 ______ 18,327

(28,800) ! 11 mo

2. m#DGF = m#EFG = 29° (Alt. Int. #)m#EGF = m#DGF = 29° (bisected #)m#FEG = 180 % (29 + 29) = 180 % 58 = 122°m#AEG = 180 % m#FEG = 58°

3. Think: Solve a Simpler Problem. From diagram, d + 150 = 1000, so d = 850 ft.


HANDMADE TILES, PAGE 448 1. Height of tile is h = 1 _

2 (4) = 2 in.; base is b = 6 in.;

overlap width is x = 2 & ' 3 in.Can cut mn tiles, for greatest m and n such that mb + x ( 40 and nh ( 12. So m ( 1 _

6 (40 % 2 & ' 3 )

! 6.1 and n ( 12 _ 2 = 6. Therefore m = n = 6, so

(6)(6) = 36 tiles can be cut.

2. Inside boundary of rect. must be 25 in. by 49 in.Shorter bases of tiles meet at corners; so if 2m + 1 tiles fit along 25-in. side, (m + 1)(1) + m(3) = 25 4m + 1 = 25 4m = 24 m = 6So 2(6) + 1 = 13 tiles fit along each 25-in. side. Similarly, if 2n + 1 tiles fit along 49-in. side, (n + 1)(1) + n(3) = 49 4n + 1 = 49 4n = 48 n = 12So 2(12) + 1 = 25 tiles fit along each 49-in. side. Total number of tiles = 2(13) + 2(25) = 76 tiles.

3. Let a and b be shorter and longer half-diagonal lengths, so 2a = b. Each $ formed by diags. is a rt. $ with sides a, 2a, and 7, such that a 2 + (2a ) 2 = 7 2 5 a 2 = 49 a 2 = 9.8 a = & '' 9.8 Diag. lengths are 2 & '' 9.8 ! 6.26 cm and

4 & '' 9.8 ! 12.52 cm.

THE MILLENNIUM FORCE ROLLER COASTER, PAGE 449 1. ! = 310 & ' 2 ! 438.4 ft 2. ! = vt

438.4 ! 20t t ! 22 s

Solutions KeyProblem Solving On Location


Solutions KeyParallel and Perpendicular Lines3

CHAPTER


1. F 2. D

3. B 4. E

5. A

6. Hypothesis: E is on ! "# AC .Conclusion: E lies in plane P.

7. Hypothesis: A is not in plane Q.Conclusion: A is not on ! "# BD .

8. Hypothesis: Plane P and plane Q intersect.Conclusion: Plane P and plane Q intersect in a line.

9. Possible answer: $GHJ; acute

10. Possible answer: $KLM; obtuse

11. Possible answer: $QPN; right

12. Possible answer: $RST; straight

13. Possible answer: $AGB and $EGD

14. Possible answer: $AGB and $BGC

15. Possible answer: $BGC and $CGD

16. Possible answer: $AGC and $CGD

17. 4x + 9= 4(31) + 9= 133

18. 6x % 16= 6(43) % 16= 242

19. 97 % 3x= 97 % 3(20)= 37

20. 5x + 3x + 12= 8x + 12= 8(17) + 12= 148

21. 4x + 8 = 24 4x = 16 x = 4

22. 2 = 2x % 810 = 2x 5 = x

23. 4x + 3x + 6 = 90 7x + 6 = 90 7x = 84 x = 12

24. 21x + 13 + 14x % 8 = 180 35x + 5 = 180 35x = 175 x = 5

3-1 LINES AND ANGLES, PAGES 146–151


1a. Possible answer: &&

BF ' &&

EJ

b. Possible answer: &&

BF and &&

DE are skew.

c. Possible answer: &&

BF ( &&

FJ

d. Possible answer: plane FJH ' plane BCD

2a. Possible answer: $1 and $3

b. Possible answer: $2 and $7

c. Possible answer: $1 and $8

d. Possible answer: $2 and $3

3. transv.: n; same-side int. )

THINK AND DISCUSS, PAGE 148 1. Intersecting lines can intersect at any $. ( lines

intersect at 90° ).

2. The ) are outside lines m and n, on opposite sides of line p.

3.


1. alternate interior angles

2. Possible answer: &&

EH ( &&

DH


AB and &&

DH are skew.


AB ' &&

CD

5. Possible answer: plane ABC ' plane EFG

6. Possible answer: $2 and $4




10. transv.: n; corr. ) 11. transv.: m; alt. ext. )

12. transv.: n; alt. int. )

13. transv.: p; same-side. int. )



AB ' &&

DE


AB and &&

CF are skew.


BD ( &&

DF

17. Possible answer: plane ABC ' plane DEF





22. transv.: p; corr. ) 23. transv.: q; alt. int. )

24. transv.: !; alt. ext. )

25. transv.: p; same-side. int. )

26. The 30-yard line and goal line are ', and the path of the runner is the transv.

27. Possible answer: corr. )

28. Possible answer: alt. int. )


29. Possible answer: same-side int. )


CD ' &&

GH


CD and &&&

FG


DH ( &&

GH

33a. plane MNR ' plane KLP; plane LMQ ' plane KNP; plane PQR ' plane KLM

b. same-side int. )

34. parallel or skew;




38. transv.: !; corr. ) 39. transv.: n; alt. int. )

40. transv.: m; alt. ext. ) 41. The lines are skew.

42. m ' n

43. Possible answer: In a room, the intersection of the front wall and the ceiling forms part of one line, and the intersection of the right wall and the floor forms part of another line. The two lines are skew.


44. B

45. Gcorr. $ pairs: $1 and $8, $2 and $5, $3 and $6, $4 and $7

46. C 47. F

48. D


49. transv. m: $1 and $3, $2 and $4, $5 and $7, $6 and $8;transv. n: $9 and $11, $10 and $12, $13 and $15, $14 and $16;transv. p: $1 and $9, $2 and $10, $5 and $13, $6 and $14;transv. q: $3 and $11, $4 and $12, $7 and $15, $8 and $16

50. transv. m: $2 and $7, $3 and $6;transv. n: $10 and $15, $11 and $14;transv. p: $5 and $10, $6 and $9;transv. q: $7 and $12, $8 and $11



53. corr. );

54. the red and orange faces, the blue and purple faces, the yellow and green faces


55. 4(%1 ) 2 % 7 = 4 % 7 = %3;4(0 ) 2 % 7 = %7;4(1 ) 2 % 7 = 4 % 7 = %3;4(2 ) 2 % 7 = 16 % 7 = 9;4(3 ) 2 % 7 = 36 % 7 = 29

56. %2(%1 ) 2 + 5 = %2 + 5 = 3;%2(0 ) 2 + 5 = 5;%2(1 ) 2 + 5 = %2 + 5 = 3;%2(2 ) 2 + 5 = %8 + 5 = %3;%2(3 ) 2 + 5 = %18 + 5 = %13

57. (%1 + 3)(%1 % 3) = (2)(%4) = %8;(0 + 3)(0 % 3) = (3)(%3) = %9;(1 + 3)(1 % 3) = (4)(%2) = %8;(2 + 3)(2 % 3) = (5)(%1) = %5;(3 + 3)(3 % 3) = (6)(0) = 0

58. C = 2"r= 2"(80) = 160*+ 502.7 cm

A = " r 2 = "(80 ) 2 = 1600*+ 20,106.2 cm 2

59. r = 3.8 ÷ 2 = 1.9 mC = 2"r

= 2"(1.9) = 3.8*+ 11.9 m

A = " r 2 = "(1.9 ) 2 = 3.61*+ 11.3 m 2

60. Rt. $ , Thm. or Vert. ) Thm.

61. Lin. Pair Thm. 62. Vert. ) Thm.


CONNECTING GEOMETRY TO ALGEBRA: SYSTEMS OF EQUATIONS, PAGES 152–153

TRY THIS, PAGE 153 1. 10x + 4y = 90

_____________ 26x % 4y = 90 36x + 0 = 180 36x = 180 x = 5

10x + 4y = 9010(5) + 4y = 90 50 + 4y = 90 4y = 40 y = 10

2. 3x + 3y = 45 ___________________ %3x + 17y + 45 = 180 20y + 45 = 45 + 180 20y = 180 y = 9

3x + 3y = 45 3x + 3(9) = 45 3x + 27 = 45 3x = 18 x = 6

3. 6x + 10y + 36 = 180 6x + 10y = 144 6x + 10y = 144 - % 3(6x + 10y = 144) -

%18x % 30y = %432

%18x % 30y = %432 _________________ 18x + 6y = 144 %24y = %288 y = 12

6x + 10y = 144 6x + 10(12) = 144 6x + 120 = 144 6x = 24 x = 4

4. 32x + 2y = 90 - %2(3 2x + 2y = 90) - %64x % 4y = %180

%64x % 4y = %180 ________________ 19x + 4y = 90 %45x = %90 x = 2

32x + 2y = 90 32(2) + 2y = 90 64 + 2y = 90 2y = 26 y = 13

TECHNOLOGY LAB: EXPLORE PARALLEL LINES AND TRANSVERSALS, PAGE 154

ACTIVITY, PAGE 154

4. Angle $AGE $BGE $AGH $BGHMeasure 100° 80° 80° 100°Measure 72° 108° 108° 72°

Angle $CHG $DHG $CHF $DHFMeasure 100° 80° 80° 100°Measure 72° 108° 108° 72°

Possible measures are given in the tables. Possible answer: All acute ) are ,. All obtuse ) are ,. Any acute $ is supp. to any obtuse $.

TRY THIS, PAGE 154 1. The corr. ) are the pairs $AGE and $CHG, $BGE

and $DHG, $AGH and $CHF, and $BGH and $DHF. The ) in each pair have = measures.

2. The alt. int. ) are the pairs $CHG and $BGH, and $AGH and $DHG. The ) in each pair have = measures.The alt. ext. ) are the pairs $AGE and $DHF, and $BGE and $CHF. The ) in each pair have = measures.The same-side int. ) are the pairs $CHG and $AGH, and $BGH and $DHG. The ) in each pair have measures that add up to 180°.

3. Possible answer: If the ' lines are dragged farther apart or closer together, there is no change in the $ measures. Since the lines remain ', the amount of “tilt” of the line remains the same, so the $ measures remain the same.

3-2 ANGLES FORMED BY PARALLEL LINES AND TRANSVERSALS, PAGES 155–161

CHECK IT OUT! PAGES 155–157 1. x = 118

x + m$QRS = 180118 + m$QRS = 180 m$QRS = 62°

2. (2x + 10)° = (3x % 15)° 10 = x % 15 25 = x m$ABD = 2x + 10 = 2(25) + 10 = 60°

3. Acute ) measure 180 % 120 = 60° and 180 % 125 = 55°


THINK AND DISCUSS, PAGE 157 1. If the transv. is (, all the ) formed are rt. ), and all

rt. ) are ,.

2.


1. x = 127m$JKL = x = 127°

2. (7x % 14)° = (4x + 19)° 3x % 14 = 19 3x = 33 x = 11 m$BEF = 4x + 19 = 4(11) + 19 = 63°

3. m$1 = 90°

4. (6x)° + (3x + 9)° = 180° 9x + 9 = 180 9x = 171 x = 19 m$CBY = 3x + 9 = 3(19) + 9 = 66°

5. 5x + 6y = 94 ___________ 4x + 6y = 86 x = 8

4x + 6y = 86 4(8) + 6y = 86 32 + 6y = 86 6y = 54 y = 9


6. m$KLM = y = 115° 7. 4a = 2a + 502a = 50 a = 25m$VYX = 4(25) = 100°

8. m$ABC = x = 116° 9. 13x + 17x = 180 30x = 180 x = 6m$EFG = 17(6) = 102°

10. 3n % 45 = 2n + 15 n % 45 = 15 n = 60

m$PQR + (2n + 15) = 180 m$PQR + 2(60) + 15 = 180 m$PQR + 135 = 180 m$PQR = 45°

11. 4x % 14 = 3x + 12 x % 14 = 12 x = 26

m$STU + (3x + 12) = 180 m$STU + 3(26) + 12 = 180 m$STU + 90 = 180 m$STU = 90°

12. m$1 = 60 and m$2 + 60 = 180 2x % 3y = 60 _________________ x + 3y + 60 = 180 3x + 60 = 240 3x = 180 x = 60

2x % 3y = 60 2(60) % 3y = 60 120 % 3y = 60 %3y = %60 y = 20

13. m$1 = 120°Corr. ) Post.

14. m$1 + m$2 = 180 120 + m$2 = 180 m$2 = 60°Lin. Pair Thm.

15. 120 + m$3 = 180 m$3 = 60°Same-Side. Int. ) Thm.

16. m$4 = 120°Alt. Int. ) Thm.

17. 120 + m$5 = 180 m$5 = 60°Lin. Pair Thm.

18. 120 + m$6 = 180 m$6 = 60°Lin. Pair Thm.

19. m$7 = 120°Vert. ) Thm.

20. Alt. Ext. ) Thm.; m$1 = m$2 7x + 15 = 10x % 9 15 = 3x % 9 24 = 3x 8 = x

m$2 = 10x % 9 = 10(8) % 9 = 71° m$1 = m$2 = 71°

21. Same-Side Int. ) Thm.; m$3 + m$4 = 180 (23x + 11) + (14x + 21) = 180 37x + 32 = 180 37x = 148 x = 4

m$3 = 23x + 11 = 23(4) + 11 = 103°

m$4 = 14x + 21 = 14(4) + 21 = 77°

22. Alt. Int. ) Thm.; m$4 = m$5 37x % 15 = 44x % 29 %15 = 7x % 29 14 = 7x 2 = x

m$5 = 44x % 29 = 44(2) % 29 = 59° m$4 = m$5 = 59°


23. Corr. ) Post.; m$1 = m$4 6x + 24 = 17x % 9 24 = 11x % 9 33 = 11x 3 = x

m$4 = 17x % 9 = 17(3) % 9 = 42° m$1 = m$4 = 42°

24. Corr. ) Post.

25a. $1 , $3 b. Corr. ) Post.

c. $1 , $2 d. Trans. Prop. of ,

26. It is given that r ' s. By the Corr. ) Post., $1 , $3; so m$1 = m$3 by def. of , ). By the Lin. Pair Thm., m$3 + m$2 = 180°. By subst., m$1 + m$2 = 180°.

27.

28. The situation is impossible because when ' lines are intersected by a transverse, same-side int. ) are supp.

29a. same-side int. )

b. By the Same-Side Int. ) Thm., m$QRT + m$STR = 180 25 + 90 + m$STR = 180 115 + m$STR = 180 m$STR = 65°

30. same-side int. ); m$1 + m$2 = 180 (2x + 6) + (3x + 9) = 180 5x + 15 = 180 5x = 165 x = 33

m$1 = 2x + 6 = 2(33) + 6 = 72°

m$2 = 3x + 9 = 3(33) + 9 = 108°

31. A is incorrect because the ) are supp., not ,.

32. By the Alt. Int. ) Thm., x = y, so x _ y = 1.

33.

If all ) formed by m and p are ,, then m ( p. If the $ formed by ! and p is , to the ) formed by m and p, it must be a rt. $, so ! ( p.


34. C m$RST = m$STU x + 50 = 3x + 20 50 = 2x + 20 30 = 2x 15 = x

m$STU = 3x + 20 = 3(15) + 20 = 45° m$RVT = m$STU = 45°

35. Jm(comp. $) = 7°; this is smaller than other $ measures.

36. By the Lin. Pair Thm., m$1 + m$2 = 180°. By the Alt. Int. ) Thm., $2 , $3, so m$2 = m$3. By subst., m$1 + m$3 = 180°, so $1 and $3 are supp.


37. m$1 = 40° + (180 % 145)° = 40 + 35 = 75°

38. m$1 = 180° % (105 % 80)° = 180 % 25 = 155°

39. By the Same-Side Int. ) Thm., 10x + 5y + 80 = 180 and 15x + 4y + 72 = 180 10x + 5y = 100 and 15x + 4y = 108 10x + 5y = 100 - %1.5( 10x + 5y = 100) -

%15x % 7.5y = %150

%15x % 7.5y = %150 _________________ _______________ 15x + 4y = 108 3.5y = 42 y = 12

10x + 5y = 100 10x + 5(12) = 100 10x + 60 = 100 10x = 40 x = 4

40. a + b = 180 and a = 2b 2b + b = 180 3b = 180 b = 60

a = 2b = 2(60) = 120


41. increase 42. decrease

43. Let p, q, and r be the following:p: $1 and $2 form a lin. pair.q: $1 and $2 are supp.r : m$1 + m$2 = 180°It is given that p - q and q - r, and also that p is true. By the Law of Syllogism, p - r. So by the Law of Detachment r is true: m$1 + m$2 = 180°.


44. Let p, q, and r be the following:p: A figure is a square.q: A figure is a rect.r : A figure’s sides are (.It is given that p - q and q - r, and also that p is true for figure ABCD. By the Law of Syllogism, p - r. So by the Law of Detachment, r is true for figure ABCD: The sides of ABCD are (.




3-3 PROVING LINES PARALLEL, PAGES 162–169

CHECK IT OUT! PAGES 162–165 1a. m$1 = m$3

$1 , $3So ! ' m by the Conv. of the Corr. ) Post.

b. m$7 = (4x + 25)° = 4(13) + 25 = 77°m$5 = (5x + 12)° = 5(13) + 12 = 77°So $7 , $5. ! ' m by the Conv. of the Corr. ) Post.

2a. m$4 = m$8 $4 , $8So r ' s by the Conv. of the Alt. Ext. ) Thm.

b. m$3 = 2x° = 2(50) = 100°m$7 = (x + 50)° = 50 + 50 = 100°So $7 , $3. r ' s by the Conv. of the Alt. Int. ) Thm.


1. $1 , $4 1. Given2. m$1 = m$4 2. Def. , )3. $3 and $4 are supp. 3. Given4. m$3 + m$4 = 180° 4. Def. supp. )5. m$3 + m$1 = 180° 5. Subst.6. m$2 = m$3 6. Vert. ) Thm.7. m$2 + m$1 = 180° 7. Subst.8. ! ' m 8. Conv. of Same-Side

Int. ) Thm.

4. 4y % 2 = 4(8) % 2 = 30°3y + 6 = 3(8) + 6 = 30°The ) are ,, so the oars are ' by the Conv. of the Corr. ) Post.

THINK AND DISCUSS, PAGE 165 1. Prove 2 corr. ) are ,, prove 2 same-side int. ) are

supp., or prove 2 alt. int. ) are ,.

2. If m$5 = m$1, then m ' n by the Conv. of the Corr. ) Post. If m$7 = m$1, then m ' n by the Conv. of the Alt. Ext. ) Thm. $6 and $8 each form a lin. pair with $5, so you could use the Lin. Pair Thm. and the Conv. of the Corr. ) Post.

3.


1. $4 , $5, so p ' q by the Conv. of the Corr. ) Post.

2. m$1 = (4x + 16)° = 4(28) + 16 = 128°m$8 = (5x % 12)° = 5(28) % 12 = 128°So $1 , $8. p ' q by the Conv. of the Corr. ) Post.

3. m$4 = (6x % 19)° = 6(11) % 19 = 47°m$5 = (3x + 14)° = 3(11) + 14= 47°So $4 , $5. p ' q by the Conv. of the Corr. ) Post.

4. $1 , $5, so r ' s by the Conv. of the Alt. Ext. ) Thm.

5. $3 and $4 are supp., so r ' s by the Conv. of the Same-Side Int. ) Thm.

6. $3 , $7, so r ' s by the Conv. of the Alt. Int. ) Thm.

7. m$4 = (13x % 4)° = 13(5) % 4 = 61°m$8 = (9x + 16)° = 9(5) + 16 = 61°So $4 , $8. r ' s by the Conv. of the Alt. Int. ) Thm.

8. m$8 = (17x + 37)° = 17(6) + 37 = 139°m$7 = (9x % 13)° = 9(6) % 13 = 41°m$8 + m$7 = 139° + 41° = 180°So m$8 and m$7 are supp. r ' s by the Conv. of the Same-Side Int. ) Thm.

9. m$2 = (25x + 7)° = 25(5) + 7 = 132°m$6 = (24x + 12)° = 24(5) + 12 = 132°So $2 , $6. r ' s by the Conv. of the Alt. Ext. ) Thm.

10a. Trans. Prop. of ! b. &&

XY ' &&&

WV

10c. Conv. of the Alt. Int. ) Thm.

11. m$1 = (17x + 9)° = 17(3) + 9 = 60°m$2 = (14x + 18)° = 14(3) + 18 = 60°So $1 , $2. By the Conv. of the Alt. Int. ) Thm., landings are '.


12. $3 , $7, so ! ' m by the Conv. of the Corr. ) Post.

13. m$4 = 54°m$8 = (7x + 5)° = 7(7) + 5 = 54°So $4 , $8. ! ' m by the Conv. of the Corr. ) Post.

14. m$2 = (8x + 4)° = 8(15) + 4 = 124°m$6 = (11x % 41)° = 11(15) % 41 = 124°So $2 , $6. ! ' m by the Conv. of the Corr. ) Post.

15. m$1 = (3x + 19)° = 3(12) + 19 = 55°m$5 = (4x + 7)° = 4(12) + 7 = 55°So $1 , $5. ! ' m by the Conv. of the Corr. ) Post.


16. $3 , $6, so n ' p by the Conv. of the Alt. Int. ) Thm.

17. $2 , $7, so n ' p by the Conv. of the Alt. Ext. ) Thm.

18. $4 and $6 are supp., so n ' p by the Conv. of the Same-Side Int. ) Thm.

19. m$1 = (8x % 7)° = 8(14) % 7 = 105°m$8 = (6x + 21)° = 6(14) + 21 = 105°So $1 , $8. n ' p by the Conv. of the Alt. Ext. ) Thm.

20. m$4 = (4x + 3)° = 4(25) + 3 = 103°m$5 = (5x % 22)° = 5(25) % 22 = 103°So $4 , $5. n ' p by the Conv. of the Alt. Int. ) Thm.

21. m$3 = (2x + 15)° = 2(30) + 15 = 75°m$5 = (3x + 15)° = 3(30) + 15 = 105°m$3 + m$5 = 75° + 105° = 180°So m$3 and m$5 are supp. n ' p by the Conv. of the Same-Side Int. ) Thm.

22a. Corr. ) Post. b. Given

c. Trans. Prop. of , d. &&

BC ' &&

DE

e. Conv. of the Corr. ) Post.

23. m$1 = (3x + 2)° = 3(6) + 2 = 20°m$2 = (5x % 10)° = 5(6) % 10 = 20°So $1 , $2.

&& DJ ' &&

EK by the Conv. of the Corr. ) Post.

24. Conv. of the Corr. ) Post.

25. Conv. of the Alt. Ext. ) Thm.

26. Conv. of the Alt. Int. ) Thm.

27. Conv. of the Corr. ) Post.


29. Conv. of the Same-Side Int. ) Thm.

30. ! ' m; Conv. of the Alt. Int. ) Thm.

31. m ' n; Conv. of the Same-Side Int. ) Thm.

32. ! ' n; Conv. of the Alt. Ext. ) Thm.

33. m ' n; Conv. of the Alt. Ext. ) Thm.

34. ! ' n; Conv. of the Alt. Int. ) Thm.

35. ! ' n; Conv. of the Same-Side Int. ) Thm.


37a. $URT; m$URT = m$URS + m$SRT by the $ Add. Post. It is given that m$SRT = 25° and m$URS = 90°, so m$URT = 25° + 90° = 115°.

b. It is given that m$SUR = 65°. From part a,

m$URT = 115°. 65° + 115° = 180°, so ! "# SU ' ! "# RT by the Conv. of the Same-Side Int. ) Thm.

38a. $1 , $2 b. Trans. Prop. of ,

c. ! ' m d. Conv. of the Corr. ) Post.

39. It is given that $1 and $2 are supp., so m$1 + m$2 = 180°. By the Lin. Pair Thm., m$2 + m$3 = 180°. By the Trans. Prop. of =, m$1 + m$2 = m$2 + m$3. By the Subtr. Prop. of =, m$1 = m$3. By the Conv. of the Corr. ) Post., ! ' m.

40. The $ formed by the wall and the roof and the $ formed by the plumb line and the roof are corr. ). If they have the same measure, then they are ,, so the wall is ' to the plumb line, by the Conv. of the Corr. ) Post. Since the plumb line is perfectly vertical, the wall must also be perfectly vertical.

41. The Reflex. Prop. is not true for ' lines, because a line is not ' to itself. The Sym. Prop. is true, because if ! ' m, then ! and m are coplanar and do not intersect. So m ' !. The Trans. Prop. is not true for ' lines, because if ! ' m and m ' n, ! and n could be the same line. So they would not be '.

42. Yes; by the Vert. ) Thm.; the $ that forms a same-side int. $ with the 55° $ measures 125°. 125° + 55° = 180°, so the same-side int. ) are supp. By the Conv. of the Same-Side Int. ) Thm., a ' b.


43. C 44. D

45. 15 (5x % 10) + (8x % 5) = 180 13x % 15 = 180 13x = 195 x = 15


46. q ' r by the Conv. of the Alt. Ext. ) Thm.

47. No lines can be proven '.

48. s ' t by the Conv. of the Corr. ) Post.

49. q ' r by the Conv. of the Alt. Int. ) Thm.


51. s ' t by the Conv. of the Alt. Ext. ) Thm.

52. s ' t by the Conv. of the Same-Side Int. ) Thm.


54. It is given that m$E = 60° and m$BDE = 120°, so m$E + m$BDE = 180°. So $E and $BDE are supp., so

&& AE ' &&

BD by the Conv. of the Same-Side Int. ) Thm.

55. By the Vert. ) Thm., $6 , $3, so m$6 = m$3. It is given that m$2 + m$3 = 180°. By subst., m$2 + m$6 = 180°. By the Conv. of the Same-Side Int. ) Thm., ! ' m.

56. It is given that m$2 + m$5 = 180°. By the Lin. Pair Thm., m$4 + m$5 = 180°. By the Trans. Prop. of =, m$2 + m$5 = m$4 + m$5. By the Subtr. Prop of =, m$2 = m$4. By the Conv. of the Corr. ) Post., ! ' n.



57. a % b = %ca = b % c

58. y = 1 _ 2 x % 10

2y = x % 202y + 20 = x x = 2y + 20

59. 4y + 6x = 12 4y = 12 % 6x

y = % 3 _ 2

x + 3

60. Converse: If an animal has wings, then it is a bat; FInverse: If an animal is not a bat, then it has no wings; FContrapositive: If an animal has no wings, then it is not a bat; T

61. Converse: If a polygon has exactly 3 sides, then it is a triangle; TInverse: If a polygon is not a triangle, then it does not have exactly 3 sides; TContrapositive: If a polygon does not have exactly 3 sides, then it is not a triangle; T

62. Converse: If a whole number is even, then the digit in the ones place of the number is 2; FInverse: If the digit in the ones place of a whole number is not 2, then the number is not even; FContrapositive: If a whole number is not even, then the digit in the ones place of the number is not 2; T

63. &&

AD ' &&

BC


AB and &&

DE are skew.

65. &&

AB ( &&

AD

GEOMETRY LAB: CONSTRUCT PARALLEL LINES, PAGES 170–171

TRY THIS, PAGE 170 1. Yes; lines are still '. 2. Check students’ work.

3. ' Post.

4. If you draw quadrilateral PQRS in the diagram, then it is a rhombus, because the same compass setting was used to construct all 4 side lengths.

TRY THIS, PAGE 171 5. Yes; lines are still '.

6. The corr. ) measure 90°. By the Conv. of the Corr. ) Post., the lines must be '.

7. Check students’ work. 8. the lines are '

3-4 PERPENDICULAR LINES, PAGES 172–178


1a. &&

AB b. AB < ACx % 5 < 12

+ 5 + 5x < 17


1. $EHF , $HFG 1. Given

2. ! "# EH ' ! "# FG 2. Conv. of Alt. Int. ) Thm.

3. ! "# FG ( ! "# GH 3. Given

4. ! "# EF ( ! "# GH 4. ( Transv. Thm.

3. The shoreline and the path of the swimmer should both be ( to the current, so they should be ' to each other.


1. If two intersecting lines form a lin. pair of , ), then the ) in the lin. pair have the same measure. By the Lin. Pair Thm., they are also supp., so their measures add to 180°. This means the measure of each $ must be 90°, so the lines must be (.

2. If the transv. is ( to ' the lines, all pairs of corr. ) must be rt. ). Since all rt. ) are ,, the transv. and the ' lines form 8 , ).

3.


1. &&

AB and ! "# CD are (. &&

AC and &&

BC are ,.

2. &&

EB 3. ED > EBx + 12 > 7% 12 % 12

x > %5

4a. 2 intersecting lines form a lin. pair of , ) - lines (.

b. ! "# DE ( ! "# AF

c. 2 lines ( to same line - 2 lines '.

5. The service lines are coplanar lines that are ( to the same line (the center line), so they must be ' to each other.


6. &&&

WY 7. WY < WZx + 8 < 19% 8 % 8

x < 11

8a. Given b. ! "# AB ' ! "# CD

c. ( Transv. Thm.

9. The frets are lines that are ( to the same line (the string), so the frets must be ' to each other.


10. x < 2x % 5 %x < %5 x > 5

11. 6x + 5 < 9x % 3 %3x + 5 < %3 %3x < %8

x > 8 _ 3

12. Step 1 Find x. Use the ( Transv. Thm.2x = 90 x = 45Step 2 Find y. Use the fact that transv. is (.3y % 2x = 903y % 90 = 90 3y = 180 y = 60

13. Step 1 Find y. Use the fact that the transv. is (.6y = 90 y = 15Step 2 Find x. Use the ( Transv. Thm.

5x + 4y = 90 5x + 4(15) = 90 5x + 60 = 90 5x = 30 x = 6

14. Step 1 Write 2 equations for x and y. 2x + y = 9010x % 4y = 90Step 2 Solve equations.

8x + 4y = 360 ______________ 10x % 4y = 90 18x = 450 x = 25

2x + y = 90 2(25) + y = 90 50 + y = 90 y = 40

15. Step 1 Write 2 equations for x and y.x + y + x = 180 x + y = 2yStep 2 Solve equations.

2x + y = 180 ___________ x % y = 0 3x = 180 x = 60

x % y = 0 60 % y = 0 60 = y

16. yes 17. no

18. no 19. no

20. yes 21. yes

22. The Reflex. Prop. is not true for ( lines because a line is not ( to itself. The Sym. Prop. is true, because if ! ( m - ! and m intersect to form a 90° angle, so m ( !. The Trans. Prop. is not true, because ! ( m and m ( n, then ! ' n.

23a. It is given that &&

QR ( &&

PQ and &&

PQ ' &&

RS , so &&

QR ( &&

RS by the ( Transv. Thm. It is given that

&& PS ' &&

QR . Since

&& QR (

&& RS , &&

PS ( &&

RS by the ( Transv. Thm.

b. It is given that &&

PS ' &&

QR and &&

QR ( &&

PQ . So &&

PQ ( &&

PS by the ( Transv. Thm.

24. By the ( Transv. Thm., all given ) are rt. ).16x % 6 = 90 16x = 96 x = 6

3x + 12y = 903(6) + 12y = 90 18 + 12y = 90 12y = 72 y = 6

or 24x % 9y = 9024(6) % 9y = 90 144 % 9y = 90 %9y = %54 y = 6

25. Possible answer: 1.6 cm

26. Possible answer: The two edges of cube that are skew are ( to a third edge, but they are not '.


28. The rungs of the ladder are lines that are all ( to the same line, a side of the ladder, so the rungs must be parallel.

29. Check students’ work. 30. Check students’ work.


31. C 32. F 22x + 10y = 180 ______________ 4x + 10y = 90 18x = 90 x = 5

4x + 10y = 90 4(5) + 10y = 90 20 + 10y = 90 10y = 70 y = 7

33. D 34. C

35a. n ( p

b. AB; AB; the shortest distance from a point to a line is measured along a ( segment.

c. The distance between two ' lines is the length of a segment that is ( to both lines and has one endpoint on each line.


36. m$1 = 180° % 1 _ 2 (90°)

= 180 % 45 = 135°

37. Label the , ) $1 and $2. By def. of , ), m$1 = m$2. By the Lin. Pair Thm., m$1 + m$2 = 180°. By subst., 2(m$1) = 180°. By the Div. Prop. of =, m$1 = 90°, so the lines are ( by the def. of ( lines.

38. Label a pair of corr. rt. ) $1 and $2. By the Rt. $ , Thm., $1 , $2. So r ' s by the Conv. of the Corr. ) Post.



39. 2(5 + 4 + 3 + 2 + 1) = 2(15) = 30 games

40. m$ + m$DJE = 180 m$ + 28 = 180 m$ = 152°

41. m$ + m$FJG = 90 m$ + 65 = 90 m$ = 25°

42. m$ + m$GJH = 180 and m$DJG + m$GJH = 180, so m$ = m$DJG

= m$DJF + m$FJG= 90 + 65 = 155°

43. Conv. of the Alt. Ext. ) Thm.


45. Conv. of the Same-Side Int. ) Thm.

GEOMETRY LAB: CONSTRUCT PERPENDICULAR LINES, PAGE 179

TRY THIS, PAGE 179 1. Check students’ work. 2. Check students’ work.

3. Check students’ work. The lines are ' because two lines that are ( to the same line are ' to each other.


1. The table top is parallel to the floor and ceiling of the room, and perpendicular to the walls.

2. The table top is parallel to the floor, which forms a 25° $ with the ground. Thus the table top also forms a 25° $ relative to the ground. If a ball were placed on the table, it would roll down the table top. To a person in the room, the table would appear to be level and the ball would appear to roll of its own power.

3. &&

RS forms a transversal relative to the board and the ground. The board is parallel to the ground by the Converse of the Alternate Interior Angles Theorem. From the point of view of a person in the room, the person on one end of the board would appear higher than the person on other end of the board.

4. The lamp is hanging straight down, so it would form a 25° $ with the walls. To a person in the room, the walls would appear straight, so the lamp would appear to be hanging at a 25° tilt.



AE ( &&

AB


AB and &&

FG are skew.


AE ' &&

FB

4. Possible answer: plane AEF ' plane DHG





9. m$ = x ° = 135° 10. 15x % 7 = 19x % 15 %7 = 4x % 15 2 = x

m$ = 15x % 7 = 15(2) % 7 = 23°

11. 54x + 14 = 43x + 36 11x + 14 = 36 11x = 22 x = 2

m$ = 54x + 14 = 54(2) + 14 = 122°

12. m$8 = (13x + 20)° = 13(3) + 20 = 59°m$6 = (7x + 38)° = 7(3) + 38 = 59°So $8 , $6. a ' b by the Conv. of the Corr. ) Post.

13. $1 , $5, so a ' b by the Conv. of the Alt. Ext. ) Thm.

14. $8 and $7 are supp., so a ' b by the Conv. of the Same-Side Int. ) Thm.

15. $8 , $4, so a ' b by the Conv. of the Alt. int. ) Thm.

16. m$1 = (3x + 12)° = 3(14) + 12 = 54°m$2 = (4x % 2)° = 4(14) % 2 = 54°So $1 , $2. The guy wires are ' by the Conv. of the Corr. ) Post.


1. $1 , $2, ! ( n 1. Given2. p ' n 2. Conv. of Alt. Int. )

Thm.3. ! ( p 3. ( Transv. Thm.

3-5 SLOPES OF LINES, PAGES 182–187

CHECK IT OUT! PAGES 183–184 1. Substitute (3, 1) for ( x 1 , y 1 ) and (2, %1) for ( x 2 , y 2 )

in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = %1 % 1 _ 2 % 3

= %2 _ %1

= 2

2. From the graph, Tony will have traveled approximately 390 mi.

3a. ! ""# WX is vert. and ! "# YZ is horiz., so the lines are perpendicular.

b. slope of ! "# KL = %3 % 4 _ %2 % (%4)

= %7 _ 2 = % 7 _

2

slope of ! "# MN = %1 % 1 _ %5 % 3

= %2 _ %8

= 1 _ 4

The slopes are not the same, so the lines are not parallel. The product of the slopes is not %1, so the lines are not perpendicular.


c. slope of ! "# BC = 5 % 1 _ 3 % 1

= 4 _ 2 = 2

slope of ! "# DE = 4 % (%6)

_ 3 % (%2)

= 10 _ 5 = 2

The lines have the same slope, so they are parallel.

THINK AND DISCUSS, PAGE 185 1. Subtract the first y-value from the second y-value.

Subtract the first x-value from the second x-value. Divide the difference of the y-values by the difference of the x-values.

2. Any two points on a horiz. line have the same y-value, so the numerator of the slope is 0. Thus the slope of a horiz. line is 0. Any two points on a vert. line have the same x-value, so the denominator of the slope is 0. Thus the slope of a vert. line is undefined.

3.


1. rise; run

2. Substitute (5, 7) for ( x 1 , y 1 ) and (%2, 1) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 1 % 7 _ %2 % 5

= %6 _ %7

= 6 _ 7

3. Substitute (%5, 3) for ( x 1 , y 1 ) and (4, %2) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = %2 % 3 _ 4 % (%5)

= %5 _ 9 = % 5 _

9

4. Substitute (0, 1) for ( x 1 , y 1 ) and (%5, 1) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 1 % 1 _ %5 % 0

= 0 _ %5

= 0

5. Substitute (4, %2) for ( x 1 , y 1 ) and (6, 3) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 3 % (%2)

_ 6 % 4

= 5 _ 2

6. Use the points (8, 80) and (11, 200) to graph the line and find the slope.

m = y 2 % y 1

_ x 2 % x 1 = 200 % 80 _ 11 % 8

= 120 _ 3 = 40

The slope is 40, which means the bird is flying at an average speed of 40 mi/h.

7.

slope of ! "# HJ = 1 % 2 _ 4 % 3

= %1 _ 1 = %1

slope of ! "# KM = %5 % (%4)

_ %1 % (%2)

= %1 _ 1 = %1

The slopes are the same, so the lines are parallel.

8.

slope of ! "# LM = 5 % 2 _ 2 % (%2)

= 3 _ 4

slope of ! "# NP = %2 % 2 _ 3 % 0

= %4 _ 3

= % 4 _ 3

The product of the slopes is ( 3 _ 4 ) (% 4 _

3 ) = %1, so

lines are perpendicular.

9.

slope of ! "# QR = 4 % 1 _ %2 % 6

= 3 _ %8

= % 3 _ 8

slope of ! "# ST = %1 % 3 _ %3 % 5

= %4 _ %8

= 1 _ 2

The slopes are not the same, so the lines are not parallel. The product of the slopes is not %1, so lines are not perpendicular.

PRACTICE AND PROBLEM SOLVING, PAGE 186

10. Substitute (0, 7) for ( x 1 , y 1 ) and (0, 3) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 3 % 7 _ 0 % 0

= %4 _ 0

The slope is undefined.

11. Substitute (5, %2) for ( x 1 , y 1 ) and (3, %2) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = %2 % (%2)

_ 3 % 5

= 0 _ %2

= 0


m = y 2 % y 1

_ x 2 % x 1 = 3 % 4 _ 4 % 3

= %1 _ 1 = %1

13. Substitute (0, 4) for ( x 1 , y 1 ) and (3, %3) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = %3 % 4 _ 3 % 0

= %7 _ 3 = % 7 _

3


14. Use the points (2.5, 100) and (5, 475) to graph the line and find the slope.

m = y 2 % y 1

_ x 2 % x 1 = 475 % 100 _ 5 % 2.5

= 375 _ 2.5

= 150

The slope is 150, which means that the plane is flying at an average speed of 150 mi/h.

15.

slope of ! "# AB = 2 % (%1)

_ 7 % 2

= 3 _ 5

slope of ! "# CD = %6 % (%3)

_ %3 % 2

= %3 _ %5

= 3 _ 5


16.

slope of ! "# XY = %2 % 5 _ 6 % (%2)

= %7 _ 8 = % 7 _

8

slope of ! "# ZW = 0 % 6 _ 4 % (%3)

= %6 _ 7

= % 6 _ 7

The slopes are not the same, so the lines are not parallel. The product of the slopes is not %1, so lines are not perpendicular.

17.

! "# JK is horiz. and ! "# JL is vert, so the lines are perpendicular.

18. m = 1150 _ 2400

+ 0.5; the average change in the

elevation of the river is about 0.5 m per km of length.

19. Substitute (7, 6) for ( x 1 , y 1 ) and (5, %3) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 5 % 6 _ %3 % 7

= %1 _ %10

= 1 _ 10

20. Substitute (%3, 5) for ( x 1 , y 1 ) and (4, %2) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = %2 % 5 _ 4 % (%3)

= %7 _ 7

= %1

21. Substitute (%2, %3) for ( x 1 , y 1 ) and (6, 1) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 1 % (%3)

_ 6 % (%2)

= 4 _ 8

= 1 _ 2

22. Substitute (%3, 5) for ( x 1 , y 1 ) and (6, 1) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 1 % 5 _ 6 % (%3)

= %4 _ 9 = % 4 _

9

23. slope of ! "# AB > 0 - m < 0

.slope of ! "# AB / < 1 - .m/ > 1Therefore the inequal. is m < %1.

24. The lines have the same slope. They are either ' or they are the same line.

25a. speed = 330 % 132 _ 5 % 2

= 198 _ 3 = 66 ft/s

b. speed = 66 ft/s · 15 mi/h _ 22 ft/s

= 45 mi/h

TEST PREP, PAGE 187

26. A

slope of ! "# AB = %2 % 3 _ 4 % 1

= %5 _ 3 = % 5 _

3

slope of ! "# CD = y % 1

_ x % 6

= 3 _ 5

Since %2 % 1 _ 1 % 6

= 3 _ 5 , x = 1 and y = %2 are possible.

27. F

slope of ! "# MN = 3 % 1 _ 1 % (%3)

= 2 _ 4 = 1 _

2

slope of ! "# PQ = 1 % 4 _ 2 % 8

= %3 _ %6

= 1 _ 2

28. Cslope of C = 200 _

4.5 + 45 mi/h


29. ! "# JK is a vert. line.

30. ! "# JK is a horiz. line.

31a. slope of ! "# AB = 4 % (%2)

_ 6 % 0

= 6 _ 6 = 1

slope of ! "# CD = 4 % 10 _ %6 % 0

= %6 _ %6

= 1

slope of ! "# BC = 10 % 4 _ 0 % 6

= 6 _ %6

= %1

slope of ! "# DA = %2 % 4 _ 0 % (%6)

= %6 _ 6 = %1

The opp. sides &&

AB and &&

CD both have slope 1, so they are '. The opp. sides

&& BC and

&& DA both have

slope %1, so they are '.

b. The slopes of any two consecutive sides are opp. reciprocals, so the consecutive sides are (.


c. By the Dist. Formula,

AB = 0 111111111 (6 % 0) 2 + (4 % (%2)) 2

= 0 1111 6 2 + 6 2 = 6 0 1 2

BC = 0 11111111 (0 % 6) 2 + (10 % 4) 2

= 0 11111 (%6 ) 2 + 6 2

= 6 0 1 2

CD = 0 111111111 (%6 % 0) 2 + (4 % 10) 2

= 0 111111 (%6 ) 2 + (%6 ) 2

= 6 0 1 2

DA = 0 1111111111 [(0 % (%6) ] 2 + (%2 % 4) 2

= 0 11111 6 2 + (%6 ) 2

= 6 0 1 2 All 4 sides have the same length, so they are ,.

32. slope of ! "# ST = %1 % 5 _ 1 % (%3)

= %6 _ 4 = % 3 _

2

slope of ! ""# VW = y % (%3)

_ 1 % x

= % 3 _ 2

2(y + 3) = %3(1 % x) 2y + 6 = 3x % 3 2y = 3x % 9

y = 3 _ 2 x % 9 _

2

Possible answer: x = 3, y = 0

33. slope of ! "# MN = 0 % 1 _ %3 % 2

= %1 _ %5

= 1 _ 5

slope of ! "# PQ = y % 4

_ 3 % x

= %5

y % 4 = %5(3 % x)y % 4 = 5x % 15 y = 5x % 11Possible answer: x = 1, y = %6


34. x-int.: %5; y-int.: %5

35. The y-int. is 1.

m = y 2 % y 1

_ x 2 % x 1 = %7 % 1 _ 2 % 0

= %8 _ 2 = %4

y = mx + b y = %4x + 1 Find the x-int. y = %4x + 1 0 = %4x + 1 4x = 1 x = 0.25 The x-int. is 0.25.

36. m = y 2 % y 1

_ x 2 % x 1 = 3 % (%3)

_ 3 % 1

= 6 _ 2 = 3

y % y 1 = m(x % x 1 ) y % 3 = 3(x % 3) y % 3 = 3x % 9 y = 3x % 6 The y-int is %6. Find the x-int. y = 3x % 6 0 = 3x % 6 6 = 3x 2 = x The x-int. is 2.


1. $1 is supp. to $3 1. Given2. $1 and $2 are supp. 2. Lin. Pair Thm.3. $2 , $3 3. , Supps. Thm.

38. T; Alt. Ext. ) Thm. 39. T; Corr. ) Post.

40. F; Same-Side Int. ) Thm.

TECHNOLOGY LAB: EXPLORE PARALLEL AND PERPENDICULAR LINES, PAGES 188–189

ACTIVITY 1, PAGE 188 1. y = 3x % 4 and y = 3x + 1 appear to be '. The

slopes of the lines are the same.

2. Possible answer: y = 2x + 1; the slope of the new line is 2.

3. Possible answer: y = % 1 _ 2 x + 1; the slope of the

new line is % 1 _ 2 .

TRY THIS, PAGE 188 1. Possible answer: y = x and y = x + 1

2. Possible answer: Yes; the lines are still ' if the window settings changed; both lines appear steeper.

3. Changing the y-intercept of the lines does not change whether they are '.

ACTIVITY 2, PAGE 189 1. yes

2. Possible answer: y = % 1 _ 3 x + 1; the slope of the

new line is % 1 _ 3 .

3. Possible answer: y = % 3 _ 2 x; the slope of the new line

is % 3 _ 2 .


TRY THIS, PAGE 189 4. The students’ equations should have slopes that are

opp. reciprocals of each other. The product of the two slopes should be %1.

5. Possible answer: No; the lines still intersect, but the $ does not look like a rt. $.

6. Changing the y-intercept of the lines does not change whether they are (.

3-6 LINES IN THE COORDINATE PLANE, PAGES 190–197

CHECK IT OUT! PAGES 191–193 1a. y = mx + b

6 = 0(4) + b6 = by = 0x + 6y = 6

b. m = 2 % 2 _ 1 % (%3)

= 0

y % y 1 = m(x % x 1 ) y % 2 = 0 (x % (%3)) y % 2 = 0 y = 2

2a. The equation is given in slope-intercept form, with a slope of 2 and a y-intercept of %3. Plot the point (0, %3) and then rise 2 and run 1 to find another point. Draw the line containing the two points.

b. The equation is given in point-slope form, with a

slope of % 2 _ 3 through the point (%2, 1). Plot the point

(%2, 1) and then rise %2 and run 3 to find another point. Draw the line containing the two points.

c. The equation is given in the form for a horizontal line with a y-intercept of %4. The equation tells you that the y-coordinate of every point on the line is %4. Draw the horizontal line through (0, %4).

3. Solve both equations for y to find the slope-intercept form.3x + 5y = 2 5y = %3x + 2 y = % 3 _

5 x + 2 _

5

3x + 6 = %5y 5y = %3x % 6 y = % 3 _

5 x % 6 _

5

Both lines have a slope of % 3 _ 5 , and the y-intercepts

are different. So the lines are parallel.

4. The equation for Plan B becomes y = 35x + 60. The lines would have the same slope, so they would be parallel.

THINK AND DISCUSS, PAGE 193 1. If the slopes are the same and the y-intercepts are

different, then the lines are '.

2. If the slopes of the two ( lines are multiplied, the product is %1. Each slope is the opp. reciprocal of the other slope. However, if the lines are horiz. and vert., one slope is 0 and the other is undefined.

3.


1. The slope-intercept form of an equation is solved for y. The x-term is the first, and constant term is the second.

2. m = 1 % 7 _ %2 % 4

= %6 _ %6

=1

y = mx + b7 = 1(4) + b3 = by = x + 3

3. y % y 1 = m(x % x 1 ) y % 2 = 3 _

4 (x % (%4))

y % 2 = 3 _ 4

(x + 4)

4. m = %2 % 0 _ 0 % 4

= %2 _ %4

= 1 _ 2

y = mx + b

y = 1 _ 2 x % 2

5. The equation is given in slope-intercept form, with a slope of %3 and a y-intercept of 4. Plot the point (0, 4) and then rise %3 and run 1 to find another point. Draw the line containing the two points.


6. The equation is given in point-slope form, with a

slope of 2 _ 3 through the point (6, %4). Plot the point

(6, %4) and then rise 2 and run 3 to find another point. Draw the line containing the two points.

7. The equation is given in the form for a vertical line with an x-intercept of 5. The equation tells you that the x-coordinate of every point on the line is 5. Draw the vertical line through (5, 0).

8. Both lines have a slope of %3, and the y-intercepts are different. So the lines are parallel.

9. Solve both equations for y to find the slope-intercept form.6x % 12y = %24 6x + 24 = 12y y = 1 _

2 x + 2

3y = 2x + 18

y = 2 _ 3 x + 6

The lines have different slopes, so they intersect.

10. Solve the second equation for y to find the slope-intercept form.3y = x + 2 y = 1 _

3 x + 2 _

3

Both lines have a slope of 1 _ 3 and y-intercept of 2 _

3 ,

so they coincide.

11. Solve the first equation for y to find the slope-intercept form.4x + 2y = 10 2y = %4x + 10 y = %2x + 5Both lines have a slope of %2, and the y-intercepts are different. So the lines are parallel.

12. Write and slove the system of equations for the ticket costs.

Conroe: y = x + 115 Lakeville: ____________ y = 10x + 50 0 = %9x + 65 9x = 65 x + 7 For 55 + 10 + 7 = 72 mi/h, tickets would cost

approximately the same.


13. m = 6 % (%2)

_ 4 % 0

= 8 _ 4 = 2

y % y 1 = m(x % x 1 ) y % (%2) = 2(x % 0) y + 2 = 2x

14. m = 2 % 2 _ %2 % 5

= 0

y = mx + b 2 = 0(5) + b 2 = b y = 0x + 2 y = 2

15. y % y 1 = m(x % x 1 )

y % (%4) = 2 _ 3 (x % 6)

y + 4 = 2 _ 3 (x % 6)

16. The equation is given in point-slope form, with a slope of 1 through the point (%4, 7). Plot the point (%4, 7) and then rise 1 and run 1 to find another point. Draw the line containing the two points.

17. The equation is given in slope-intercept form, with a

slope of 1 _ 2 and a y-intercept of %2. Plot the point


18. The equation is given in the form for a horizontal line with a y-intercept of 2. The equation tells you that the y-coordinate of every point on the line is 2. Draw the horizontal line through (0, 2).

19. The lines have different slopes, so they intersect.

20. Solve the second equation for y to find the slope-intercept form.2y = 5x % 4 y = 5 _

2 x % 2

Both lines have a slope of 5 _ 2 , and the y-intercepts



21. Solve the first equation for y to find the slope-intercept form.x + 2y = 6 2y = %x + 6 y = % 1 _

2 x + 3

Both line have a slope of % 1 _ 2 and a y-intercept of 3,

so they coincide.

22. Solve both equations for y to find the slope-intercept form.7x + 2y = 10 2y = %7x + 10 y = % 7 _

2 x + 5

3y = 4x % 5 y = 4 _

3 x % 5 _

3


23. Job 1: y = 0.2x + 375Job 2: _______________ y = 0.25x + 325 0 = %0.05x + 50 0.05x = 50 x = 1000Chris must make $1000 in sales per week.

24. m = 6 % 2 _ 3 % (%6)

= 4 _ 9

y % y 1 = m(x % x 1 )

y % 6 = 4 _ 9

(x % 3)

y % 6 = 4 _ 9

x % 4 _ 3

y = 4 _ 9

x + 14 _ 3

The equation in slope-intercept form has a slope

of 4 _ 9

and a y-intercept of 14 _ 3 . Plot the point (0, 14 _

3 )

and then rise 4 and run 9 to find another point. Draw the line containing the two points.

25. y = 3The equation in the form for a horizontal line has a y-intercept of 3. The equation tells you that the y-coordinate of every point on the line is 3. Draw the horizontal line through (0, 3).

26. y % y 1 = m(x % x 1 )

y % (%2) = 2 _ 3 (x % 5)

y + 2 = 2 _ 3 x % 10 _

3

y = 2 _ 3 x % 16 _

3


of 2 _ 3 and a y-intercept of % 16 _

3 . Plot the point

(0, % 16 _ 3 ) and then rise 2 and run 3 to find another

point. Draw the line containing the two points.

27. m = %3 % 0 _ 0 % 4

= 3 _ 4

y = mx + b

y = 3 _ 4 x % 3


of 3 _ 4 and a y-intercept of %3. Plot the point (0, %3)

and then rise 3 and run 4 to find another point. Draw the line containing the two points.

28. y % y 1 = m(x % x 1 )

y % 2 = % 1 _ 2 (x % 0)

y % 2 = % 1 _ 2 x

The equation in point-slope form has a slope of % 1 _ 2

through the point (0, 2). Plot the point (0, 2) and then rise %1 and run 2 to find another point. Draw the line containing the two points.


29. y % y 1 = m(x % x 1 )

y % 0 = 3 _ 4 (x % (%2))

y = 3 _ 4 (x + 2)

The equation in point-slope form has a slope of 3 _ 4

through the point (%2, 0). Plot the point (%2, 0) and then rise 3 and run 4 to find another point. Draw the line containing the two points.

30. y % y 1 = m(x % x 1 ) y % (%1) = %1(x % 5) y + 1 = %(x % 5)

The equation in point-slope form has a slope of %1 through the point (5, %1). Plot the point (5, %1) and then rise %1 and run 1 to find another point. Draw the line containing the two points.

31. m = %5 % 6 _ %2 % 4

= 11 _ 6

y % y 1 = m(x % x 1 )

y % 6 = 11 _ 6

(x % 4)

The equation in point-slope form has a slope of 11 _ 6

through the point (4, 6). Plot the point (4, 6) and then rise 11 and run 6 to find another point. Draw the line containing the two points.

32. B is incorrect. In B, the x- and y-values of the pt. used to find the point-slope form are interchanged.

33. The product of the slopes is (3)(%3) = %9; no

34. The product of the slopes is (%1)(1) = %1; yes

35. The product of the slopes is (% 2 _ 3 ) ( 3 _

2 ) = %1; yes

36. The product of the slopes is (%2) (% 1 _ 2 ) = 1; no

37. Step 1 Find the slope.m = 3Step 2 Find the equation of the ' line through P.y % y 1 = m(x % x 1 ) y % 3 = 3(x % 2) y % 3 = 3x % 6 y = 3x % 3Step 3 Find the equation of the ( line through P.y % y 1 = m(x % x 1 )

y % 3 = % 1 _ 3 (x % 2)

y % 3 = % 1 _ 3 x + 2 _

3

y = % 1 _ 3

x + 11 _ 3

38. Step 1 Find the slope.m = %2Step 2 Find the equation of the ' line through P.y % y 1 = m(x % x 1 ) y % 4 = %2[(x % (%1)] y % 4 = %2x % 2 y = %2x + 2Step 3 Find the equation of the ( line through P.y % y 1 = m(x % x 1 )

y % 4 = 1 _ 2 [(x % (%1)]

y % 4 = 1 _ 2 x + 1 _

2

y = 1 _ 2 x + 9 _

2

39. Step 1 Find the slope.4x + 3y = 8 3y = %4x + 8 y = % 4 _

3 x + 8 _

3

m = % 4 _ 3

Step 2 Find the equation of the ' line through P. y % y 1 = m(x % x 1 )

y % (%2) = % 4 _ 3

(x % 4)

y + 2 = % 4 _ 3

x + 16 _ 3

y = % 4 _ 3 x + 10 _

3

Step 3 Find the equation of the ( line through P. y % y 1 = m(x % x 1 )

y % (%2) = 3 _ 4 (x % 4)

y + 2 = 3 _ 4 x % 3

y = 3 _ 4 x % 5


40. Step 1 Find the slope.2x % 5y = 7 2x % 7 = 5y

y = 2 _ 5 x % 7 _

5

m = 2 _ 5

Step 2 Find the equation of the ' line through P.y % y 1 = m(x % x 1 )

y % 4 = 2 _ 5

[(x % (%2)]

y % 4 = 2 _ 5

x + 4 _ 5

y = 2 _ 5

x + 24 _ 5

Step 3 Find the equation of the ( line through P.y % y 1 = m(x % x 1 )

y % 4 = % 5 _ 2 [(x % (%2)]

y % 4 = % 5 _ 2 x % 5

y = % 5 _ 2 x % 1

41. slope of &&

AB = %2 % 3 _ 0 % (%5)

= %5 _ 5 = %1

slope of &&

BC = 3 % (%2)

_ 5 % 0

= 5 _ 5 = 1

&&

AB ( &&

BC : yes; $B is a rt. $.

42. slope of &&

DE = 7 % 0 _ 2 % 1

= 7 _ 1 = 7

slope of &&

EF = 1 % 7 _ 5 % 2

= %6 _ 3 = %2

slope of &&

DF = 1 % 0 _ 5 % 1

= 1 _ 4

no

43. slope of &&

GH = 4 % 4 _ %3 % 3

= 0

slope of &&

HJ = %2 % 4 _ 1 % (%3)

= %6 _ 4 = % 3 _

2

slope of &&

GJ = %2 % 4 _ 1 % 3

= %6 _ %2

= 3no

44. slope of &&

KL = 1 % 4 _ 2 % (%2)

= % 3 _ 4

slope of &&

LM = 8 % 1 _ 1 % 2

= %7

slope of &&

KM = 8 % 4 _ 1 % (%2)

= 4 _ 3

&&

KL ( &&

KM : yes; $K is a rt. $.

45. Write and solve the system of equations for prices. y = 1.5x + 8

_____________ y = 0.75x + 11 0 = 0.75x % 33 = 0.75x4 = xy = 1.50(4) + 8 = 14For 4 toppings, both pizzas will cost $14.

46. Possible answer: x = 1.2 and y = 3.7

47. slope = 9 % 5 _ 4 % 2

= 2, mdpt. = ( 2 + 4 _ 2 , 5 + 9 _

2 ) = (3, 7)

y % y 1 = m(x % x 1 )

y % 7 = % 1 _ 2 (x % 3)

y % 7 = % 1 _ 2 x + 3 _

2

y = % 1 _ 2 x + 17 _

2

48. The segment is a horizontal line with a midpoint of (2, 1). The perpendicular bisector is a vertical line, so its equation is x = 2.

49. slope = 4 % 3 _ %1 % 1

= % 1 _ 2 , mdpt. = ( 1 + (%1)

_ 2 , 3 + 4 _

2 )

= (0, 7 _ 2 )

y % y 1 = m(x % x 1 )

y % 7 _ 2 = 2(x % 0)

y % 7 _ 2 = 2x

y = 2x + 7 _ 2

50. The segment is a vertical line with a midpoint of (%3, %4). The perpendicular bisector is a horizontal line, so its equation is y = %4.

51a. y % y 1 = m(x % x 1 ) y % 5 = 2(x % 3) y % 5 = 2x % 6 y = 2x % 1

b. y = 2x % 1

__________

__________

__________

y = % 1 _ 2 x + 4

0 = 5 _ 2 x % 5

5 = 5 _ 2 x

2 = x

y = 2x % 1 = 2(2) % 1 = 3 ! and m intercept at (2, 3).

c. D = 0 11111111 (3 % 2) 2 + (5 % 3) 2

= 0 1111 1 2 + 2 2 = 0 111 1 + 4 = 0 1 5 units

52a. Possible answer: y = %x + 1

b. Possible answer: Intersection of p and r: y = x + 3 __________ y = %x + 1 0 = 2x + 2 %2x = 2 x = %1

y = x + 3 = %1 + 3 = 2 (%1, 2)

Intersection of q and r: y = x % 1 __________ y = %x + 1 0 = 2x % 2 %2x = %2 x = 1

y = x % 1 = 1 % 1 = 0 (1, 0)


c. D = 0 111111111 (%1 % 1) 2 + (2 % 0) 2

= 0 11111 (%2) 2 + 2 2

= 0 111 4 + 4 = 0 1 8 = 2 0 1 2 units

53a–b.

b. the time when the car has traveled 300 ft

c. Possible answer: 3.5 s

54. It is given that the eqn. of the line through ( x 1 , y 1 )with slope m is y % y 1 = m(x % x 1 ). Let (0, b) be a pt. on the line. Then 0 is a possible value for x 1 , and b is a possible value for y 1 . Substitute these values into the eqn. y % y 1 = m(x % x 1 ) to get y % b = m(x % 0). Simplify to get y % b = mx. By the Add. Prop. of =, y = mx + b. Thus the equation of the line through (0, b) with slope m is y = mx + b.


56. The slope of the line is m = 2 % 6 _ 2 % (%4)

= % 2 _ 3 . The

pt.-slope form of the line is y % 6 = % 2 _ 3 (x + 4). To

see if the line crosses the x-axis at (5, 0), substitute

5 for x and 0 for y:

0 % 6 = % 2 _ 3 (5 + 4)

%6 = % 2 _ 3 (9)

%6 = %6 2These values make the equation true, so (5, 0) is on the line.

57. The top line passes through (%4, 0) and (0, 3), so

its slope is m = 3 % 0 _ 0 % (%4)

= 3 _ 4 . The bottom line

passes through (0, %2) and (3, 0), so its slope is

m = 0 % (%2)

_ 3 % 0

= 2 _ 3

. The lines do not have same

slope, so they are not parallel.


58. DFind the slope-intercept forms:%3x + y = 7 y = 3x + 7

2x + y = %3 y = %2x % 3

59. JFind the slope intercept-form of J:x + 1 _

2 y = 1

1 _ 2 y = %x + 1

y = %2x + 2

60. Dslope is % 2 _

3 , y-intercept is 3

61. J2 = % 1 _

2 (%4) 2 and %3 = % 1 _

2 (6) 2


62. The vertices of the hypotenuse are at intercepts (0,

5) and ( 5 _ 2 , 0) . By the Pyth. Thm.,

length of hyp. = 3 1111

( 5 _ 2 ) 2 + 5 2

= 3 1111 25 _ 4

+ 25

= 3 11 125 _ 4 = 5 0 1 5 _

2 units

63. Possible answer: let the leg lengths be 8 and 15,

so the intercepts are (0, 8) and (15, 0). The slope is

% 8 _ 15

, so the equation is y = % 8 _ 15

x + 8.

64. Possible answer: let the vertices be (0, 0), (0, 5), and (12, 0). The equations of the lines containing

the legs are x = 0 and y = 0. The slope of the line

containing the hyp. is % 5 _ 12

, so the equation is

y = % 5 _ 12

x + 5.

65. Possible answer: I found the equation of the line

through the first 2 pts., which is y = 2 _ 7 x % 24 _

7 . Then

I substituted the x- and y-values for the third pt. to see if it lies on the line. The values did not make the equation true, so the pts. are not colinear.

66a. d 2 = (x % 3) 2 + (y % 2) 2 = (x % 3) 2 + (x + 1 % 2) 2 = (x % 3) 2 + (x % 1) 2 = x 2 % 6x + 9 + x 2 % 2x + 1 = 2 x 2 % 8x + 10

b. Complete the square: d 2 = 2( x 2 % 4x + 4) + 2

= 2(x % 2 ) 2 + 2 The shortest distance is 0 1 2 (when x = 2). The

perpendicular line through P is y % 2 = %(x % 3) or y = %x + 5. They intersect at x + 1 = %x + 5 or x = 2, y = 2 + 1 = 3. The distance between

P(3, 2) and (2, 3) is 3 1111 1 2 + 1 2 = 0 1 2 , so the distances are the same.



67. c = 2.5d + 5

20 = 2.5d + 515 = 2.5d 6 = dIf his bill was $20.00, Sean rented 6 DVDs.

68. ( !3 + 2 _ 2 , 1 + 3 _

2 ) = (! 1 _

2 , 2)

69. ( 2 + 0 _ 2 ,

3 + (!3) _

2 ) = (1, 0)

70. ( !3 + 0 _ 2 ,

1 + (!3) _

2 ) = (! 3 _

2 , !1)

71. Substitute (!3, 1) for ( x 1 , y 1 ) and (2, 3) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 ! y 1

_ x 2 ! x 1 = 3 ! 1 _ 2 ! (!3)

= 2 _ 5

72. Substitute (2, 3) for ( x 1 , y 1 ) and (0, !3) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 ! y 1

_ x 2 ! x 1 = !3 ! 3 _ 0 ! 2

= !6 _ !2

= 3

73. Substitute (!3, 1) for ( x 1 , y 1 ) and (0, !3) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 ! y 1

_ x 2 ! x 1 = !3 ! 1 _ 0 ! (!3)

= !4 _ 3 = ! 4 _

3

CONNECTING GEOMETRY TO DATA ANALYSIS: SCATTER PLOTS AND LINES OF BEST FIT, PAGES 198–199

TRY THIS, PAGE 199

1. Possible answer: y = ! 4 _ 3 x + 34 _

3

2. Possible answer: y = 90 ! x



1. speed limit = slope

= 264 ! 88 _ 6 ! 2

= 44 ft/s

44 ft/s · 15 mi/h _ 22 ft/s

= 30 mi/h

2. d = 22 _ 15

(45)t

d = 66t

Yes, the lines intersect at (0, 0). The line for Porter Street is steeper because the slope of the line is greater.



m = y 2 ! y 1

_ x 2 ! x 1 = !3 ! 5 _ 6 ! (!2)

= !8 _ 8 = !1

2. Substitute (6, !3) for ( x 1 , y 1 ) and (!3, !2) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 ! y 1

_ x 2 ! x 1 = !2 ! (!3)

_ !3 ! (6)

= 1 _ !9

= ! 1 _ 9

3. Substitute (!2, 5) for ( x 1 , y 1 ) and (4, 1) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 ! y 1

_ x 2 ! x 1 = 1 ! 5 _ 4 ! (!2)

= !4 _ 6 = ! 2 _

3

4. Substitute (4, 1) for ( x 1 , y 1 ) and (!3, !2) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 ! y 1

_ x 2 ! x 1 = !2 ! 1 _ !3 ! 4

= !3 _ !7

= 3 _ 7


m = y 2 ! y 1

_ x 2 ! x 1 = 7 ! 3 _ 0 ! 2

= 4 _ !2

= !2


m = y 2 ! y 1

_ x 2 ! x 1 = !1 ! 4 _ 5 ! (!1)

= !5 _ 6 = ! 5 _

6

7. Substitute (4, 0) for ( x 1 , y 1 ) and (1, !3) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 ! y 1

_ x 2 ! x 1 = !3 ! 0 _ 1 ! 4

= !3 _ !3

= 1

8. Substitute (4, 2) for ( x 1 , y 1 ) and (!3, 2) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 ! y 1

_ x 2 ! x 1 = 2 ! 2 _ !3 ! 4

= 0 _ !7

= 0

9. Use the points (4, 0) and (4.75, 2.5) to graph the line and find the slope.

m = y 2 ! y 1

_ x 2 ! x 1 = 2.5 ! 0 _ 4.75 ! 4

= 2.5 _ 0.75

" 3.3

The slope is about 3.3, which means Sonia’s average speed was about 3.3 mi/h.


ge07_SOLKEY_C03_045-068.indd 64ge07_SOLKEY_C03_045-068.indd 64 10/2/06 4:27:42 PM10/2/06 4:27:42 PM

10.

slope of ! "# EF = 1 % 3 _ 6 % (%2)

= %2 _ 8

= % 1 _ 4

slope of ! "# GH = 5 % 4 _ 2 % 6

= % 1 _ 4


11.

slope of ! "# JK = %1 % 3 _ 5 % 4

= %4 _ 1 = %4

slope of ! "# LM = %5 % 4 _ 3 % (%2)

= %9 _ 5

= % 9 _ 5


12.

slope of ! "# NP = 4 % (%3)

_ 0 % 5

= 7 _ %5

= % 7 _ 5

slope of ! "# QR = 3 % (%2)

_ 4 % (%3)

= 5 _ 7

The product of the slopes is (% 5 _ 7 ) ( 5 _

7 ) = %1, so the


13.

! "# ST is vert. and ! ""# VW is horiz, so the lines are perpendicular.

14. m = 4 % 8 _ %3 % 3

= 4 _ %6

= 2 _ 3

y % y 1 = m(x % x 1 )

y % 8 = 2 _ 3 (x % 3)

y % 8 = 2 _ 3 x % 2

y = 2 _ 3 x + 6

15. y % y 1 = m(x % x 1 )

y % 4 = 2 _ 3

(x % (%5))

y % 4 = 2 _ 3

(x + 5)

16. m = 1 % 2 _ 4 % 0

= %1 _ 4 = % 1 _

4

y = mx + b

y = % 1 _ 4

x + 2

17. The equation is given in slope-intercept form, with a slope of %2 and a y-intercept of 5. Plot the point (0, 5) and then rise %3 and run 5 to find another point. Draw the line containing the two points.

18. The equation is given in point-slope form, with a

slope of 1 _ 4 through the point (4, %3). Plot the point


19. The equation is given in the form for a vertical line with an x-intercept of 3. The equation tells you that the x-coordinate of every point on the line is 3. Draw the vertical line through (3, 0).

20. horiz. line: y = 3 21. slope = rise _ run = 2 _ 1 = 2

y-intercept = 3y = 2x + 3

22. vert. line: x = %1

23. Both lines have a slope of %2, and the y-interceptsare different. So the lines are parallel.

24. Solve the first equation for y to find the slope-intercept form.3x + 2y = 8 2y = %3x + 8

y = % 3 _ 2 x + 4

Both lines have a slope of % 3 _ 2 and a y-intercept of

4, so they coincide.


25. Solve the second equation for y to find the slope-intercept form.3x + 4y = 7 4y = %3x + 7

y = % 3 _ 4 x + 7 _

4



1. alternate interior angles

2. skew lines 3. transversal

4. point-slope form 5. rise; run



DE and &&

BC are skew.


AB ' &&

DE


AD ( &&

DE


10. !; alt. int. ) 11. n; corr. )

12. !; same-side int. ) 13. m; alt. ext. )

LESSON 3-2, PAGE 203 14. x + 90 = 180

x = 90m$WYZ = x° = 90°

15. 26x + 22 = 38x % 14 22 = 12x % 14 36 = 12x 3 = xm$KLM = 38x % 14

= 38(3) % 14 = 100°

16. 33x + 35 = 26x + 49 7x + 35 = 49 7x = 14 x = 2 m$DEF + (26x + 49) = 180 m$DEF + 26(2) + 49 = 180 m$DEF + 101 = 180 m$DEF = 79°

17. 17x + 8 = 13x + 24 4x + 8 = 24 4x = 16 x = 4 m$QRS = 13x + 24 = 13(4) + 24 = 76°

LESSON 3-3, PAGE 204 18. $4 , $6, so c ' d by the Conv. of the Alt. Int. )

Thm.

19. m$1 = (23x + 38)° = 23(3) + 38 = 107°m$5 = (17x + 56)° = 17(3) + 56 = 107° $1 , $5, so c ' d by the Conv. of the Corr. ) Post.

20. m$6 = (12x + 6)° = 12(5) + 6 = 66°m$3 = (21x + 9)° = 21(5) + 9 = 114°m$6 + m$3 = 66° + 114° = 180°$6 and $3 are supp., so c ' d by the Conv. of the Same-Side Int. ) Thm.

21. m$1 = 99°m$7 = (13x + 8)° = 13(7) + 8 = 99°$1 , $7, so c ' d by the Conv. of the Alt. Ext. ) Thm.


22. &&

KM 23. KM < KLx % 5 < 8 x < 13


1. &&

AD ' &&

BC , &&

AD ( &&

AB, &&

DC ( &&

BC 1. Given

2. &&

AB ( &&

BC 2. ( Transv. Thm.3.

&& AB ' &&

CD 3. 2 lines ( to the same line - the two lines are '

LESSON 3-5, PAGE 205 25. Substitute (%3, 2) for ( x 1 , y 1 ) and (4, 1) for ( x 2 , y 2 )

in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 1 % 2 _ 4 % (%3)

= %1 _ 7 = % 1 _

7

26. Substitute (1, 4) for ( x 1 , y 1 ) and (%2, %1) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = %1 % 4 _ %2 % 1

= %5 _ %3

= 5 _ 3

27. slope of ! "# EF = 4 % 2 _ %3 % 8

= 2 _ %11

= % 2 _ 11

slope of ! "# GH = 3 % 1 _ %4 % 6

= 2 _ %10

= % 1 _ 5


28. slope of ! "# JK = %2 % 3 _ %4 % 4

= %5 _ %8

= 5 _ 8

slope of ! "# LM = 1 % 6 _ %3 % 5

= %5 _ %8

= 5 _ 8


29. slope of ! "# ST = 3 % 5 _ 2 % (%4)

= %2 _ 6

= % 1 _ 3

slope of ! "# UV = 4 % 1 _ 4 % 3

= 3

The product of the slopes is (% 1 _ 3 ) (3) = %1, so the




30. m = 5 % 1 _ %3 % 6

= % 4 _ 9

y % y 1 = m(x % x 1 )

y % 1 = % 4 _ 9 (x % 6)

y % 1 = % 4 _ 9 x + 8 _

3

y = % 4 _ 9 x + 11 _

3

31. y % y 1 = m(x % x 1 )

y % (%4) = 2 _ 3 (x % (%3))

y + 4 = 2 _ 3 (x + 3)

y + 4 = 2 _ 3 x + 2

y = 2 _ 3 x % 2

32. m = %2 % 0 _ 0 % 1

= %2 _ %1

= 2

y % y 1 = m(x % x 1 ) y % 0 = 2(x % 1)

33. Solve both equations for y to find the slope-intercept form.%3x + 2y = 5 2y = 3x + 5

y = 3 _ 2 x + 5 _

2

6x % 4y = 8 6x % 8 = 4y

y = 3 _ 2 x % 2

Both lines have a slope of 3 _ 2 , and the y-intercepts


34. Solve the second equation for y to find the slope-intercept form.5x + 2y = 1 2y = %5x + 1

y = % 5 _ 2 x + 1 _

2


35. Solve the second equation for y to find the slope-intercept form.2x % y = %12x + 1 = y y = 2x + 1Both lines have a slope of 2 and y-intercept of 1, so they coincide.




AC ' &&

DF


AB and &&

CF are skew.

4. 3x + 21 = 4x + 9 21 = x + 9 12 = x3(12) + 21 = 574(12) + 9 = 57Both labeled ) measure 57°.

5. 26x % 7 = 20x + 17 6x % 7 = 17 6x = 24 x = 426(4) % 7 = 9720(4) + 17 = 97Both labeled ) measure 97°.

6. 42x % 9 = 35x + 12 7x % 9 = 12 7x = 21 x = 342(3) % 9 = 11735(3) + 12 = 117Both labeled ) measure 117°.

7. m$4 = (16x + 20)° = 16(3) + 20 = 68°m$5 = (12x + 32)° = 12(3) + 32 = 68°$4 , $5, so f ' g by the Conv. of the Alt. Int. ) Thm.

8. m$3 = (18x + 6)° = 18(4) + 6 = 78°m$5 = (21x + 18)° = 21(4) + 18 = 102°m$3 + m$5 = 78° + 102° = 180°$3 and $5 are supp., so f ' g by the Conv. of the Same-Side Int. ) Thm.


1. $1 , $2, n ( ! 1. Given2. ! ' m 2. Conv. of the Corr. )

Post.3. n ( m 3. ( Transv. Thm.

10. Substitute (%3, %4) for ( x 1 , y 1 ) and (%1, 3) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 3 % (%4)

_ %1 % (%3)

= 7 _ 2

11. Substitute (%1, %3) for ( x 1 , y 1 ) and (2, %1) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = %1 % (%1)

_ 2 % (%3)

= 0 _ 5 = 0

12. Substitute (0, %3) for ( x 1 , y 1 ) and (5, 1) for ( x 2 , y 2 ) in the slope formula and then simplify.

m = y 2 % y 1

_ x 2 % x 1 = 1 % (%3)

_ 5 % 0

= 4 _ 5

13. Use the points (9.5, 0) and (14, 32) to graph the line and find the slope.

m = 32 % 0 _ 14 % 9.5

= 32 _ 4.5

+ 7.1

The slope is about 7.1, which means Greg’s average speed was about 7.1 mi/h.

14.

slope of ! "# QR = %5 % 3 _ 6 % 3

= %8 _ 3 = % 8 _

3

slope of ! "# ST = %2 % 6 _ %1 % (%4)

= %8 _ 3 = % 8 _

3


15. y % y 1 = m(x % x 1 )

y % (%5) = % 3 _ 4

(x % (%2))

y + 5 = % 3 _ 4 (x + 2)

16. Solve both equations for y to find the slope-intercept form.6x + y = 3 y = %6x + 3

2x + 3y = 1 3y = %2x + 1

y = % 2 _ 3

x + 1 _ 3




1. ASlope of line A = 4 _

5 = slope of given line;

%4(2) + 5(%3) = %8 % 15 = %23

2. KI is false; II is true by the Vert. ) Thm.; III is true by the Alt. Ext. ) Thm.

3. Bslope of the first line =

5 % (%7) _

%8 % 1 = 12 _

%9 = % 4 _

3

slope of the second line = b % 6 _ %1 % 3

= %3 _ %4

= 3 _ 4

4(b % 6) = 3(%4) 4b % 24 = %12 4b = 12 b = 3

4. HBy the Conv. of the Corr. ) Post., $2 , $5 - m ' n. m$2 = m$5x + 18 = 2x % 28 18 = x % 28 46 = x

5. Bslope of

&& EF =

%2 % (%2) _

7 % 1 = 0

The equation of the line through E and F is y = %2. The line through G must be vert., so its equation is x = 4. The point of intersection is at (4, %2); the distance between this point and G(4, 2) is 4.


Solutions KeyTriangle Congruence4

CHAPTER


1. F 2. D

3. B 4. A

5. E 6. 35°

7. 90°


12. 9 _ 2 x + 7 = 25

______ ! 7 ___ ! 7

9 _ 2

x = 18

x = 2(18)

_ 9 = 4

13. 3x ! 2 _ 3 = 4 _

3

______

+ 2 _ 3

____ + 2 _

3

3x = 2 x = 2 _

3

14. x ! 1 _ 5

= 12 _ 5

_____

+ 1 _ 5

____

+ 1 _ 5

x = 13 _ 5 = 2 3 _

5

15. 2y = 5y ! 21 _ 2

_____ ! 5y _________ ! 5y

!3y = ! 21 ___ 2

y = 7 _ 2 = 3 1 _

2

16. t is 3 times m.t = 3m

17. Twice x is 9 ft.2x = 9

18. 53° + twice y is 90°.53 + 2y = 90

19. Price r is price p less 25. r = p ! 25

20. Half j is b plus 5 oz.

1 _ 2

j = b + 5

4-1 CLASSIFYING TRIANGLES, PAGES 216–221

CHECK IT OUT! PAGES 216–218 1. "FHG and "EHF are complementary.

m"FHG + m"EHF = 90° m"FHG + 30° = 90° m"FHG = 60°All # are equal. So $FHG is equiangular by definition.

2. AC = AB = 15No sides are congruent. So $ACD is scalene.

3. Step 1 Find the value of y.

%% FG &

%% GH

FG = GH3y ! 4 = 2y + 3 3y = 2y + 7 y = 7Step 2 Substitute 7 for y.FG = 3y ! 4 = 3(7) ! 4 = 17

GH = 2y + 3 = 2(7) + 3 = 17

FH = 5y ! 18 = 5(7) ! 18 = 17

4a. P = 3(7) = 21 in.

100 ÷ 21 = 4 16 _ 21

4 triangles

b. P = 3(10) = 30 in.

100 ÷ 30 = 3 1 _ 3

3 triangles

THINK AND DISCUSS, PAGE 218 1.

%% DE , %%

EF , "E; %%

EF , %%

FD , "F; %%

FD , %%

DE , "D

2. Possible answer:

3. No; all 3 # in an acute $ must be acute, but they do not have to have the same measure;

possible answer:

4. In an equil. rt. $, all 3 sides have the same length. By the Pyth. Thm., the 3 side lengths are related by the formula c 2 = a 2 + b 2 , making the hyp. c greater than either a or b. So the 3 sides cannot have the same length.

5.


1. An equilateral triangle has three congruent sides.

2. One angle is obtuse and the other two angles are acute.

3. "DBC is a rt. ". So $DBC is a rt. $.

4. "ABD and "DBC are supp.

"ABD + "DBC = 180° "ABD + 90 = 180 "ABD = 90°"ABD is a rt. ". So $ABD is a rt. $.

5. m"ADC = m"ADB + m"BDC = 31 + 70 = 101°"ADC is obtuse. So $ADC is an obtuse $.


6. EG = 3 + 3 = 6, EH = 8, GH = 8 %%

EH & %%

GH Exactly two sides are &, so $EGH is isosc.

7. EF = 3, EH = 8, FH = 7.4No sides are congruent, so $EFH is scalene.

8. GF = 3, GH = 8, FH = 7.4No sides are congruent, so $HFG is scalene.

9. Step 1 Find y.6y = 4y + 122y = 12 y = 6Step 2 Find side lengths.$ is equilateral, so all three side lengths = 6y = 36.

10. Step 1 Find x.2x + 1.7 = x + 2.4 2x = x + 0.7 x = 0.7Step 2 Find side lengths.x + 2.4 = 0.7 + 2.4 = 3.12x + 1.7 = 2(0.7) + 1.7 = 1.4 + 1.7 = 3.14x + 0.5 = 4(0.7) + 0.5 = 2.8 + 0.5 = 3.3

11. Perimeter is P = 3 + 3 + 1.5 = 7.5 cm

50 ÷ 7.5 = 6 2 _ 3

earrings

The jeweler can make 6 earrings.


12. m"BEA = 90°; rt. $

13. m"BCD = 60 + 60 = 120°; obtuse

14. m"ABC = 30 + 30 = 60°m"ABC = m"ACB = m"BAC; equiangular

15. %%

PS & %%

ST & %%

PT ; equilateral

16. %%

PS & %%

RS , so PS = RS = 10; RP = 17; isosc.

17. RT = 10 + 10 = 20, RP = 17, PT = 10; scalene

18. Step 1 Find z.3z ! 1 = z + 5 3z = z + 6 2z = 6 z = 3Step 2 Find side lengths.z + 5 = 3 + 5 = 83z ! 1 = 3(3) ! 1 = 84z ! 4 = 4(3) ! 4 = 8

19. Step 1 Find x.8x + 1.4 = 2x + 6.8 8x = 2x + 5.4 6x = 5.4 x = 0.9Step 2 Find side lengths.8x + 1.4 = 8(0.9) + 1.4 = 7.2 + 1.4 = 8.62x + 6.8 = 2(0.9) + 6.8 = 1.8 + 6.8 = 8.6

20a. Check students’ drawings.

%%

XY , %%

YZ , %%

XZ , "X, "Y, "Z

b. Possible answer: scalene obtuse

21. PQ + PR + QR = 60

PQ + PQ + 4 _ 3 PQ = 60

10 _ 3 PQ = 60

PQ = 3 _ 10

(60) = 18 ft

PR = PQ = 18 ft

QR = 4 _ 3 PQ = 4 _

3 (18) = 24 ft

22. 150 ÷ 60 = 2 1 _ 2 ; 2 complete trusses

23. 24. Not possible: an equiangular $ has only acute #.

25. 26.

27. Not possible: an equiangular $ must also be equilateral.

28.

29. Let x represent each side length.x + x + x = 105 3x = 105 x = 35 in.

30. %%

AB & %%

AC , so $ is isosc."BAC and "CAD are supp., and "CAD is acute; so "BAC is obtuse.$ABC is isosc. obtuse.

31. %%

AC & %%

CD and m"ACD = 90°.$ACD is isosc. rt.

32. (4x ! 1) + (4x ! 1) + x = 34 9x ! 2 = 34 9x = 36 x = 4

33a. E 22nd Street side = 1 _ 2 (Broadway side) ! 8

= 1 _ 2 (190) ! 8 = 87 ft

5th Avenue side = 2(E 22nd Street side) ! 1 = 2(87) ! 1 = 173 ft

b. All sides are different, so $ is scalene.

34. No; yes; not every isosc. $ is equil. because only 2 of the 3 sides must be &. Every equil. $ has 3 & sides, and the def. of an isosc. $ requires that at least 2 sides be &.

35. S; equil, acute

36. S; scalene, acute


37. A; 3 congruent sides, so always satisfies isosceles $ classification

38. s = P _ 3 . The perimeter of an equil. $ is 3 times the

length of any 1 side, or P = 3s. Solve this formula for s by dividing both sides by 3.


40a. DE 2 = AD 2 + AE 2

= 5 2 + ( 10 _ 2 ) 2

= 25 + 25 = 50DE = ' (( 50 = 5 ' ( 2 cmThink:

%% CE &

%% DE .

CE = DE = 5 ' ( 2 cm

b. Think: DE bisects "AEF.

m"DEF = 1 _ 2 m"AEF

= 1 _ 2 (90) = 45°

Think: "CEF & "DEF, so m"CEF = 45°.m"DEC = m"DEF + m"CEF = 45 + 45 = 90°

c. CE = DE and m"DEC = 90° isosc. $; rt. $

TEST PREP, PAGE 221

41. D3s = P3s = 36 2 _

3

s = 1 _ 3 (36 + 2 _

3 )

= 12 2 _ 9 in.

42. FBy graphing, By graphing, RTRT && RSRS )) ST, ST, so so $RST is isosc.

43. D$LMN has no rt. ".

44. 3P = AB + BC + AC = 1 _

2 x + 1 _

4 + 5 _

2 ! x + 1 _

2 x + 1 _

4

= ( 1 _ 2 ! 1 + 1 _

2 ) x + 1 _

4 + 5 _

2 + 1 _

4

= 3


45. It is an isosc. $ since 2 sides of the $ have length a. It is also a rt. $ since 2 sides of the $ lie on the coord. axes and form a rt. ".


1. $ABC is equiangular. 1. Given

2. "A & "B & "C 2. Def. of equiangular $

3. %%

EF * %%

AC 3. Given

4. "BEF & "A, "BFE & "C

4. Corr. # Post.

5. "BEF & "B, "BFE & "B

5. Trans. Prop. of &

6. "BEF & "BFE 6. # & to the same " are &.

7. $EFB is equiangular. 7. Def. of equiangular $

47. Think: Each side has the same measure. Use the expression y + 10 for this measure.3(y + 10) = 21 3y + 30 = 21 3y = !9 y = !3

48. Step 1 Find x. Think: Average of x + 12, 3x + 4, and 8x ! 16 is 24.

1 _ 3 (x + 12 + 3x + 4 + 8x ! 16) = 24

1 _ 3 (12x) = 24

4x = 24 x = 6Step 2 Find side lengths. x + 12 = 6 + 12 = 18 3x + 4 = 3(6) + 4 = 228x !16 = 8(6) ! 16 = 32 longest side ! average = 32 ! 24 = 8


49. y = x 2 50. y = x

51. y = x 2

52. F; skew lines do not intersect and are not parallel.

53. T

54. F; Possible answer: 30 has a 0 in the ones place, but 30 is not a multiple of 20.

55. y = 4x + 2 has slope 4. Line is * to y = 4x.

56. 4y = !x + 8

y = ! 1 __ 4 x + 2

Slope is neg. reciprocal of 4. Line is + to y = 4x.

57. 1 _ 2 y = 2x

y = 4xLine coincides with y = 4x.

58. !2y = 1 _ 2 x

y = ! 1 __ 4 x

Slope is neg. reciprocal of 4. Line is + to y = 4x.


GEOMETRY LAB: DEVELOP THE TRIANGLE SUM THEOREM, PAGE 222

TRY THIS, PAGE 222 1. When placed together, the three # form a line.

2. yes

3. m"A + m"B + m"C = 180°

4. The sum of the " measures in a $ is 180°.

4-2 ANGLE RELATIONSHIPS IN TRIANGLES, PAGES 223–230

CHECK IT OUT! PAGES 224–226 1. Step 1 Find m"NKM.

m"KMN + m"MNK + m"NKM = 180° 88 + 48 + m"NKM = 180 136 + m"NKM = 180 m"NKM = 44°Step 2 Find m"MJK.m"JMK + m"JKM + m"MJK = 180° 44 + 104 + m"MJK = 180 148 + m"MJK = 180 m"MJK = 32°

2a. Let acute # be "A, "B, with m"A = 63.7°.m"A + m"B = 90° 63.7 + m"B = 90 m"B = 26.3°

b. Let acute # be "C, "D, with m"C = x°.m"C + m"D = 90° x + m"D = 90 m"D = (90 ! x)°

c. Let acute # be "E, "F, with m"E = 48 2 __ 5 ° .

m"E + m"F = 90 48 2 __

5 + m"F = 90

m"F = 41 3 _ 5

°

3. m"ACD = m"ABC + m"BAC 6z ! 9 = 90 + 2z + 1 4z = 100 z = 25m"ACD = 6z – 9 = 6(25) – 9 = 141°

4. "P & "Tm"P = m"T 2 x 2 = 4 x 2 ! 32!2 x 2 = !32 x 2 = 16So m"P = 2 x 2 = 32°.Since m"T = m"P, m"T = 32°.

THINK AND DISCUSS, PAGE 226 1.

Since "3 and "4 are supp. #, m"3 + m"4 = 180° by def. "1 + "2 + "3 = 180° by the $ Sum Thm. By the trans. Prop. of =,m"3 + m"4 = m"1 + m"2 + m"3. Subtract m"3 from both sides. Then m"4 = m"1 + m"2.

2. 2; 6

3.


1. Possible answers: think “out of the way”

2. Exterior " is next to "E. So the remote interior # are "D and "F.

3. auxiliary lines

4. Think: Use $ " Sum Thm.180 = 3y + 13 + 2y + 2 + 5y ! 5180 = 10y + 10170 = 10y y = 17

5. Deneb: 3y + 13 = 3(17) + 13 = 64°Altair: 2y + 2 = 2(17) + 2 = 36°Vega: 5y ! 5 = 5(17) ! 5 = 80°

6. 20.8 + m" = 90 m" = 69.2°

7. y + m" = 90 m" = (90 ! y)°

8. 24 2 _ 3 + m" = 90

m" = 65 1 _ 3 °


9. m"M + m"N = m"NPQ3y + 1 + 2y + 2 = 48 5y + 3 = 48 5y = 45 y = 9m"M = 3y + 1 = 3(9) + 1 = 28°

10. m"K + m"L = m"HJL 7x + 6x ! 1 = 90 13x = 91 x = 7m"L = 6x ! 1 = 6(7) ! 1 = 41°

11. m"A + m"B = 117 65 + m"B = 117 m"B = 52°m"A + m"B + m"BCA = 180 117 + m"BCA = 180 m"BCA = 63°

12. "C & "Fm"C = m"F 4 x 2 = 3 x 2 + 25 x 2 = 25m"C = 4 x 2 = 100° m"F = m"C = 100°

13. "S & "U m"S = m"U5x ! 11 = 4x + 9 x = 20m"S = 5x ! 11 = 5(20) ! 11 = 89°m"U = m"S = 89°

14. "C & "Z m"C = m"Z4x + 7 = 3(x + 5)4x + 7 = 3x + 15 x = 8m"C = 4x + 7 = 4(8) + 7 = 39°m"Z = m"C = 39°


15. m"A + m"B + m"P = 180 39 + 57 + m"P = 180 96 + m"P = 180 m"P = 84°

16. 76 1 __ 4

+ m" = 90

m" = 13 3 _ 4 °

17. 2x + m" = 90 m" = (90 ! 2x)°

18. 56.8 + m" = 90 m" = 33.2°

19. Think: Use Ext. " Thm. m"W + m"X = m"XYZ5x + 2 + 8x + 4 = 15x ! 18 13x + 6 = 15x ! 18 24 = 2x x = 12m"XYZ = 15x ! 18 = 15(12) ! 18 = 162°

20. Think: Use Ext. " Thm and subst. m"C = m"D.m"C + m"D = m"ABD 2m"D = m"ABD 2(6x ! 5) = 11x + 1 12x ! 10 = 11x + 1 x = 11m"C = m"D = 6x ! 5 = 6(11) ! 5 = 61°

21. Think: Use Third # Thm. "N & "Pm"N = m"P 3 y 2 = 12 y 2 ! 144!9 y 2 = !144 y 2 = 16m"N = 3 y 2 = 3(16) = 48°m"P = m"N = 48°

22. Think: Use Third # Thm. "Q & "Sm"Q = m"S 2 x 2 = 3 x 2 ! 64 64 = x 2 m"Q = 2 x 2 = 2(64) = 128°m"S = m"Q = 128°

23. Think: Use $ " Sum Thm.m"1 + m"2 + m"3 = 180 x + 4x + 7x = 180 12x = 180 x = 15m"1 = x = 15°m"2 = 4x = 60°m"3 = 7x = 105°


1. $DEF with rt. "F 1. Given

2. m"F = 90° 2. Def. of rt. "

3. m"D + m"E + m"F = 180°

3. $ Sum Thm.

4. m"D + m"E + 90° = 180°

4. Subst.

5. m"D + m"E = 90° 5. Subtr. Prop.

6. "D and "E are comp. 6. Def. of comp. #

25. Proof 1:

Statements Reasons

1. $ABC is equiangular 1. Given

2. m"A = m"B = m"C 2. Def. of equilangular

3. m"A + m"B + m"C = 180°

3. $ Sum Thm.

4. m"A + m"A + m"A = 180°m"B + m"B + m"B = 180°m"C + m"C + m"C = 180°

4. Subst. prop

5. 3m"A = 180°, 3m"B = 180°, 3m"C = 180°

5. Simplify.

6. m"A = 60°, m"B = 60°, m"C = 60°

6. Div. Prop. of =

Proof 2:"A, "B, and "C are all congruent, so their measures are equal. The sum of the three " measures is 180°, by $ Sum Thm. Therefore, 3 · (common " measure) = 180°. So the common " measure is 60°. That is, m"A = m"B = m"C = 60°.


26. Step 1 Write an equation.m"1 = 1 1 _

4 m"2

Step 2 Since the acute # of a rt. $ are comp. write and solve another equation. m"1 + m"2 = 90

1 1 _ 4

m"2 + m"2 = 90

9 _ 4 m"2 = 90

m"2 = 4 _ 9

(90) = 40°

Step 3 Find the larger acute ", m"1.

m"1 = 1 1 _ 4

m"2 = 5 _ 4 (40) = 50°

27.

Statements Reasons

1. $ABC, $DEF, "A & "D, "B & "E

1. Given

2. m"A + m"B + m"C = 180° 2. $ Sum Thm.

3. m"C = 180° ! m"A ! m"B 3. Subtr. Prop. of =

4. m"D + m"E + m"F = 180° 4. $ Sum Thm.

5. m"F = 180° ! m"D ! m"E 5. Subtr. Prop. of =

6. m"A = m"D, m"B = m"E 6. Def. of & #

7. m"F = 180° ! m"A ! m"B 7. Subst.

8. m"F = m"C 8. Trans. Prop. of =

9. "F & "C 9. Def. of & #


1. $ABC with ext. "ACD 1. Given

2. m"A + m"B + m"ACB = 180°

2. $ Sum Thm.

3. m"ACB + m"ACD = 180° 3. Lin. Pair Thm.

4. m"ACD = 180° ! m"ACB 4. Subtr. Prop. of =

5. m"ACD = (m"A + m"B + m"ACB) ! m"ACB

5. Subst.

6.m"ACD = m"A + m"B 7. Simplify.

29. Think: Use Alt. Int. # Thm.m"WUX + m"UXZ = 180 m"WUX + 90 = 180 m"WUX = 90°So $UWX is a rt. $.m"UXW + m"XWU = 90 m"UXW + 54 = 90 m"UXW = 36°

30. "XWU, "UWY, and "YWV are supp. #.m"XWU + m"UWY + m"YWV = 180 54 + m"UWY + 78 = 180 m"UWY + 132 = 180 m"UWY = 48°

31. Think: Use Third # Thm. "WUY & "ZXY "UYW & "XYZ "WZX & "UWYm"WZX = m"UWY = 48°

32. "XYZ and "WZX are acute # in a rt. $.m"XYZ + m"WZX = 90 m"XYZ + 48 = 90 m"XYZ = 42°

33. Let "1, "2, and "3 be internal #. Let "4, "5, and "6 be external #.Think: Use Ext. " Thm.m"4 = m"1 + m"2m"1 = m"2 = 60°So m"4 = 60 + 60 = 120°.Likewise, m"5 = m"6 = 120°.Ext. " sum = m"4 + m"5 + m"6 = 360°

34. Think: Use Third # Thm. "SRQ & "RSTm"SRQ = m"RST = 37.5°

35. Let acute " measures be x° and 4x°.x + 4x = 90 5x = 90 x = 18Smallest " measure is x° = 18°.

36a. hypotenuse b. x° + y° + 90° = 180°

c. x° + y° = 90 x and y are comp. "

measures.

d. z° = x° + 90°

e. x + y = 9037 + y = 90

y = 53°

z = x + 90z = 37 + 90 z = 127°

37.

The ext. # at the same vertex of a $ are vert. #. Since vert. # are &, the 2 ext. # have the same measure.


1. %%

AB + %%

BD , %%

BD + %%

CD , "A & "C

1. Given

2. "ABD and "CDB are rt. # 2. Def. of + lines

3. m"ABD = m"CBD 3. Def. of rt. #

4. "ABD & "CDB 4. Rt. " & Thm.

5. "ADB & "CBD 5. Third # Thm.

6. %%

AD * %%

CB 6. Conv. of Alt. Int. # Thm.

39. Check students’ sketches. Ext. " measures = sums of remote int. " measures: 155°, 65°, and 140°.


40a. m"FCE = 1 __ 2

m"DCE

= 1 __ 2

(90) = 45°

m"FCB = 1 __ 2

m"FCE

= 1 __ 2

(45) = 22.5°

b. m"CBE + m"BEC + m"BCE = 180 m"CBE + 90 + 22.5 = 180 m"CBE + 112.5 = 180 m"CBE = 67.5°

TEST PREP, PAGE 230

41. C128 = 71 + x x = 57

42. F(2s + 10) + 58 + 66 = 180 2s + 134 = 180 2s = 46 s = 23

43. Dm"A + m"B = m"BCD m"B = m"BCD ! m"A

44. Let 2x, 3x, and 4x represent the " measures. The sum of the " measures of a $ is 180°, so 2x + 3x + 4x = 180°. Solving the eqn. for the value of x, yields x = 20. Find each measure by subsituting 20 for x in each expression. 2x = 2(20) = 40; 3x = 3(20) = 60; 4x = 4(20) = 80. Since all of the # measure less than 90°, they are all acute # by def. Thus the $ is acute.


45. 117 = (2 y 2 + 7) + (61 ! y 2 )117 = y 2 + 68 49 = y 2 y = 7 or !7

46. A rt. $ is formed. The 2 same-side int. # are supp., so the 2 # formed by their bisectors must be comp. That means the remaining " of the $ must measure 90°.

47. Since an ext. " is = to a sum of 2 remote int. #, it must be greater than either ". Therefore, it cannot be & to a remote int. ".

48. Possible sets of " measures:(30, 30, 120), (30, 60, 90), (60, 60, 60)

Probability = 2 _ 3

49. Let m"A = x°.m"B = 1 1 _

2 (x) ! 5

m"C = 2 1 _ 2 (x) ! 5

m"A + m"B + m"C = 180x + 1 1 _

2 (x) ! 5 + 2 1 _

2 (x) ! 5 = 180

5x ! 10 = 180 5x = 190 x = 38m"A = x° = 38°


50.

51.

52.

53. Use Seg. Add. Post. MN + NP = MP 4 + NP = 6 NP = 2 in. NP + PQ = NQ 2 + 4 = NQ NQ = 6 in.

54. %%

AD & %%

CD ) %%

AC Isosc.

55. %%

BD , %%

CD , %%

BC are )Scalene

56. %%

AB , %%

AD , %%

BD are )Scalene

57. %%

AD & %%

CD & %%

AC Equilateral

4-3 CONGRUENT TRIANGLES, PAGES 231–237

CHECK IT OUT! PAGES 231–233 1. Angles: "L & "E, "M & "F, "N & "G, "P & "H

Sides: %%

LM & %%

EF , %%%

MN & %%

FG , %%

NP & %%

GH , %%

LP & %%

EH

2a. %%

AB & %%

DE 2x ! 2 = 6 2x = 8 x = 4

b. Since the acute # of a rt. $ are comp.m"B + m"C = 90 53 + m"C = 90 m"C = 37° "F & "Cm"F = m"C = 37°



1. "A & "D 1. Given

2. "BCA & "ECD 2. Vert. # are &.

3. "ABC & "DEC 3. Third # Thm.

4. %%

AB & %%

DE 4. Given

5. %%

AD bisects %%

BE , and %%

BE bisects

%% AD .

5. Given

6. %%

BC & %%

EC , %%

AC & %%

BC 6. Def. of bisector

7. $ABC & $DEC 7. Def. of & ,


1. %%

JK * %%

ML 1. Given

2. "KJN & "MLN, "JKN & "LMN

2. Alt. Int. # Thm.

3. " JNK & "LNM 3. Vert. # Thm.

4. %%

JK & %%

ML 4. Given

5. %%

MK bisects %%

JL , and

%% JL bisects

%% MK .

5. Given

6. %%

JN & %%

LN , %%%

MN & %%

KN 6. Def. of bisector

7. $JKN & $MLN 7. Def. of & ,

THINK AND DISCUSS, PAGE 233 1. Measure all the sides and all the #. The trusses are

the same size if all the corr. sides and #are &.

2.

EXERCISES, PAGES 234–237GUIDED PRATICE, PAGE 234

1. You find the # and sides that are in the same, or matching, places in the 2 ,.

2. "B

3. %%

LM 4. %%

RT

5. "M 6. %%%

NM

7. "R 8. "T

9. %%

JK & %%

FG JK = FG3y ! 15 = 12 3y = 27 y = 9 KL = y = 9

10. "G & "K m"G = m"K4x ! 20 = 108 4x = 128 x = 32


1. %%

AB * %%

CD 1. Given

2. "ABE & "CDE, "BAE & "DCE

2. Alt. Int. " Thm.

3. %%

AB & %%

CD 3. Given

4. E is the mdpt. of %%

AC and

%% BD

4. Given

5. %%

AE & %%

CE , %%

BE & %%

DE 5. Def. of mdpt.

6. "AEB & "CED 6. Vert. " Thm

. 7. $ABE & $CDE 7. Def. of & ,



1. "UST & "RST, "U & "R 1. Given

2. "STU & "STR 2. Third # Thm.

3. %%

SU & %%

SR 3. Given

4. %%

ST & %%

ST 4. Reflex. Prop. of &

5. %%

TU & %%

TR 5. Given

6. $RTS & $UTS 6. Def. of & ,

13. %%

LM 14. %%

CF

15. "N 16. "D

17. "ADB & "CDBm"ADB = m"CDB 4x + 10 = 90 4x = 80 x = 20 m"C = x + 11 = 31°

18. %%

AB & %%

CB AB = CBy ! 7 = 12 y = 19


1. "N & "R 1. Given

2. %%

MP bisects "NMR 2. Given

3. "NMP & "RMP 3. Def. of " bisector

4. "NPM & "RPM 4. Third # Thm.

5. P is the mdpt. of %%

NR 5. Given

6. %%

PN & %%

PR 6. Def. of mdpt.

7. %%%

MN & %%%

MR 7. Given

8. %%

MP & %%

MP 8. Reflex. Prop. of &

9. $MNP & $MRP 9. Def. of & ,


1. "ADC and "BCD are rt. #

1. Given

2. "ADC & "BCD 2. Rt. " & Thm.

3. "DAC & "CBD 3. Given

4. "ACD & "BDC 4. Third # Thm.

5. %%

AC & %%

BD , %%

AD & %%

BC 5. Given

6. %%

DC & %%

DC 6. Reflex. Prop. of &

7. $ADC & $BCD 7. Def. of & ,


21. $GSR & $KPH, $SRG & $PHK$RSG & $HPK,

22. RVUTS & VWXZY

23. %%

AB & %%

DE AB = DE2x ! 10 = x + 20 x = 30AB = 2x ! 10 = 2(30) ! 10 = 50

24. "L & "P m"L = m"P x 2 + 10 = 2 x 2 + 1 9 = x 2 m"L = x 2 + 10 = 9 + 10 = 19°

25. %%

BC & %%

QR BC = QR6x + 5 = 5x + 7 x = 2BC = 6x + 5 = 6(2) + 5 = 17

26a. %%

KL & %%

ML by the def. of a square.

b. Statements Reasons

1. JKLM is a square. 1. Given

2. %%

KL & %%

ML 2. Def. of a square

3. %%

JL and %%

MK are + bisectors of each other.

3. Given

4. %%%

MN & %%

KN 4. Def. of bisector

5. %%

NL & %%

NL 5. Reflex. Prop. of &

6. "MNL and "KNL are rt. #.

6. Def. of +

7. "MNL & "KNL 7. Rt. " & Thm.

8. "NML & "NKL 8. Given

9. "NLM & "NLK 9. Third # Thm.

10. $NML & $NKL 10. Def. of & ,

27.

Statements Reasons

1. %%

BD + %%

AC 1. Given

2. "ADB and "CDB are rt. #.

2. Def. of +

3. "ADB & "CDB 3. Rt. " & Thm.

4. %%

BD bisects "ABC. 4. Given

5. "ABD & "CBD 5. Def. of bisector

6. "A & "C 6. Third # Thm.

7. %%

AB & %%

CB 7. Given

8. %%

BD & %%

BD 8. Reflex. Prop. of &

9. D is the mdpt. of %%

AC . 9. Given

10. %%

AD & %%

CD 10. Def. of mdpt.

11. $ABD & $CBD 12. Def. of & ,


29. Solution A is incorrect. "E & "M, so m"E = 46°.

30. Yes; by the Third # Thm., "K & "W, so all 6 pairs of corr. parts are &. Therefore, the , are &.

TEST PREP, PAGE 236

31. BMatching up #, $ABC & $FDE.

32. G "N & "Sm"N = m"S 62 = 2x + 8 54 = 2x x = 27

"M & "Rm"M = m"R 58 = 3y ! 2 60 = 3y y = 20

33. Dm"Y = 180 ! (m"X + m"Z) = 180 ! (m"A + m"C) = 180 ! 60.9 = 119.1°

34. JP = MN + NR + RM = SP + QP + SR + RQ = 33 + 30 + 10 + 24 = 97



35. P = TU + UV + VW + TW149 = 6x + 7x + 3 + 9x ! 8 + 8x ! 11149 = 30x ! 16165 = 30x x = 5.5Yes; UV = WV = 41.5, and UT = WT = 33. TV = TV by the Reflex. Prop. of =. It is given that "VWT & "VUT and "WTV & "UTV. "WVT = "UVT by the Third # Thm. Thus $TUV & $TWV by the def. of & ,.

36. "E & "A m"E = m!A y 2 ! 10 = 90 y 2 = 100m"D = m"H = 2 y 2 ! 132 = 2(100) ! 132 = 68°


1. %%

RS & %%

RT ; "S & "T 1. Given

2. %%

ST & %%

TS 2. Reflex. Prop. of &

3. "T & "S 3. Sym. Prop. of &

4. "R & "R 4. Reflex. Prop. of &

5. $RST & $RTS 5. Def. of & ,


38. P(both even) = P(cube 1 even) · P(cube 1 even) = 1 _

2 · 1 _

2 = 1 _

4

39. P(sum is 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1)

= 4 _ 36

= 1 _ 9

40. acute 41. rt.

42. obtuse

43. Step 1 Find x.3x + 20 + 4x + x + 16 = 180 8x + 36 = 180 x = 18Step 2 Find m"Q.m"Q = 4x = 72°

44. m"P = 3x + 20 = 74°

45. m"QRS = m"P + m"Q = 72 + 74 = 146°


1. AB = AD $ABD is isosc. $;

m"A = 90°$ABD is rt. $

2. m"EBD = 1 _ 2 m"EBC

= 1 _ 2 (90) = 45°

( %%

DB bisects rt. "ABC.)

m"BDE = 1 _ 2 m"ADB

= 1 __ 2

( 1 __ 2 m"ADC)

= 1 _ 2 ( 1 __

2 (90)) = 22.5°

( %%

DE bisects "ADB, and %%

DB bisects rt. "ABC.)m"BED = 180 ! (m"EBD + m"BDE) =180 ! (45 + 22.5) = 112.5°($ Sum Thm.)


1. %%

DB bisects "ABC and "EDF 1. Given

2. "EBD & "FBD; "EDB & "FDB

2. Def. of " bisector

3. "DEB & "DFB 3. Third # Thm.

4. %%

BE & %%

BF ; %%

DE & %%

DF 4. Given

5. %%

DB & %%

DB 5. Reflex. Prop. of &

6. $EBD & $FDB 6. Def. of & ,


1. rt. $, since "ACB is rt. "

2. equiangular, since m"BAD = 30 + 30 = 60° = m"B = m"ADB

3. obtuse, since m"ADE = m"B + m"BAD = 120°

4. isosc., since PQ = QR = 5, PR = 8.7

5. equilateral, since PR = RS = PS = 5

6. scalene, since PQ = 8.7, QS = 5 + 5 = 10, PS = 5

7. m"M + m"N = m"NLK 6y + 3 + 84 = 151 ! 2y 8y = 64 y = 8m"M = 6y + 3 = 51°

8. m"C + m"D = m"ABC 90 + 5x = 20x ! 15 105 = 15x x = 7m"ABC = 20x – 15 = 125°

9. m"RTP = m"R + m"T = 55 + 37 = 92°

10. %%

EF 11. %%

JL

12. "E 13. "L

14. %%

PR & %%

SU PR = SU 14 = 3m + 2 12 = 3m m = 4PQ = 2m + 1 = 9

15. "S & "Pm"S = m"P 2y = 46 y = 23



1. - ./ AB * - ./ CD 1. Given

2. "BAD & "CDA 2. Alt. Int. # Thm.

3. %%

AC + %%

CD , %%

DB + %%

AB 3. Given.

4. "ACD and "DBA are rt. # 4. Def. of +

5. "ACD & "DBA 5. Rt. " & Thm.

6. "CAD & "BDA 6. Third # Thm.

7. %%

AB & %%

CD , %%

AC & %%

DB 7. Given

8. %%

AD & %%

DA 8. Reflex. Prop. of &

9. $ACD & $DBA 9. Def. of & ,

GEOMETRY LAB: EXPLORE SSS AND SAS TRIANGLE CONGRUENCE, PAGES 240–241

TRY THIS, PAGE 240–241 1. yes

2. It is not possible. Once the lengths of the 3 straws are determined, only 1 $ can be formed.

3. To prove that 2 , are &, check to see if the 3 pairs of corr. sides are &.

4. Three sides of 1 $ are & to 3 sides of the other $.

5. yes

6. No; once 2 side lengths and the included " measure are determined, only 1 length is possible for the third remaining side.

7. To prove that 2 , are &, check to see if there are 2 pairs of & corr. sides and that their included # are &.


9. Two sides and the included " of 1 $ are & to 2 sides and the included " of the other $.

4-4 TRIANGLE CONGRUENCE: SSS AND SAS, PAGES 242–249


1. It is given that %%

AB & %%

CD and %%

BC & %%

DA . By the Reflex. Prop. of &,

%% AC &

%% CA . So $ABC & $CDA

by SSS.


AB & %%

BD and "ABC & "DBC. By Reflex. Prop. of &,

%% BC &

%% BC . So $ABC & $DBC

by SAS.

3. DA = 3t + 1 = 3(4) + 1 = 13DC = 4t ! 3 = 4(4) ! 3 = 13m"ADB = 32°m"CDB = 2 t 2 = 2(4 ) 2 = 32° %%

DA & %%

DC , %%

DB & %%

DB , and "ADB & "CDBSo $ADB & $CDB by SAS.


1. %%

QR & %%

QS 1. Given

2. ../ QP bisects "RQS 2. Given

3. "RQP & "SQP 3. Def. of " bisector

4. %%

QP & %%%

QP 4. Reflex. Prop. of &

5. $RQP & $SQP 5. SAS Steps 1, 3, 4

THINK AND DISCUSS, PAGE 245 1. Show that all six pairs of corr. parts are &; SSS;

SAS

2. The SSS and SAS Post. are methods for proving , & without having to prove & of all 6 corr. parts.

3.


1. "T


DA & %%

BC and %%

AB & %%

CD . %%

BD & %%

DB by the Reflex. Prop. of &. Thus $ABD & $CBD by SSS.

3. It is given that %%%

MN & %%%

MQ and %%

NP & %%

QP . %%

MP & %%

MP by the Relex. Prop. of &. Thus $MNP & $MQP by SSS.


JG & %%

LG , and %%

GK & %%

GH . "JGK & "LGH by the Vert. # Thm. So $JGK & $LGH by SAS.

5. When x = 4, HI = GH = 3, and IJ = GJ = 5. %%

HJ & %%

HJ by the Reflex. Prop. of &. Therefore, $GHJ & $IHJ by SSS.

6. When x = 18, RS = UT = 61, and m"SRT = m"UTR = 36°.

%% RT &

%% TR by the Reflex.

Prop. of &. So $RST & $TUR by SAS.


1. %%

JK & %%

ML 1. Given

2. "JKL & "MLK 2. Given

3. %%

KL & %%

LK 3. Reflex. Prop. of &

4. $JKL & $MLK 4. SAS Steps 1, 2, 3



8. It is given that BC = ED = 4 in. and BD = EC = 3 in. So by the def. of &,

%% BC &

%% ED ,

and %%

BD & %%

EC . %%

DC & %%

CD by the Reflex. Prop. of &. Thus $BCD & $EDC by SSS.


KJ & %%

LJ and %%

GK & %%

GL . %%

GJ & %%

GJ by the Reflex. Prop. of &. So $GJK & $GJL by SSS.

10. It is given that "C and "B are rt. # and %%

EC & %%

DB . "C & "B by the Rt. " & Thm. %%

CB & %%

BC by the Reflex. Prop. of &. So $ECB & $DBC by SAS.

11. When y = 3, NQ = NM = 3, and QP = MP = 4.So by the def. of &,

%% NQ &

%%% NM , and

%% QP &

%% MP .

m"M = m"Q = 90˚, so "M & "Q by the def. of &. Thus $MNP & $QNP by SAS.

12. When t = 5, YZ = 24, ST = 20, and SU = 22. So by the def. of &,

%% XY &

%% ST , %%

YZ & %%

TU , and %%

XZ & %%

SU . This $XYZ & $STU by SSS.


1. B is mdpt. of %%

DC 1. Given

2. %%

DB & %%

CB 2. Def. of mdpt.

3. %%

AB + %%

DC 3. Given

4. "ABD and "ABC are rt. #

4. Def. of +

5. "ABD & "ABC 5. Rt. " & Thm.

6. %%

AB & %%

AB 6. Reflex. Prop. of &

7. $ABD & $ABC 7. SAS Steps 2, 5, 6

14. SAS (with Reflex. Prop of &)

15. SAS (with Vert. # Thm.)

16. neither 17. neither

18a. To use SSS, you need to know that %%

AB & %%

DE and

%% CB &

%% CE .

b. To use SAS, you need to know that %%

CB & %%

CE .

19. QS = ' (((( 1 2 + 2 2 = ' ( 5

SR = ' (((( 4 2 + 0 2 = 4

QR = ' (((( 3 2 + 2 2 = ' (( 13

TV = ' (((( 1 2 + 2 2 = ' ( 5

VU = ' (((( 4 2 + 0 2 = 4

TU = ' (((( 3 2 + 2 2 = ' (( 13 The , are & by SSS.

20. AB = ' (((( 1 2 + 4 2 = ' (( 17

BC = ' (((( 4 2 + 3 2 = 5

AC = ' (((( 5 2 + 1 2 = ' (( 26

DE = ' (((( 1 2 + 4 2 = ' (( 17

EF = ' (((( 4 2 + 3 2 = 5

DF = ' (((( 4 2 + 0 2 = 4The , are not &.


1. "ZVY & "WYV, "ZVW & "WYZ

1. Given

2. m"ZVY = m"WYV, m"ZVW = m"WYZ

2. Def. of &

3. m"ZVY + m"ZVW = m"WYV + m"WYZ

3. Add. Prop. of =

4. m"WVY = m"ZYV 4. " Add. Post.

5. "WVY & "ZYV 5. Def. of &

6. %%%

WV & %%

YZ 6. Given

7. %%

VY & %%

YV 7. Reflex. Prop. of &

8. $ZVY & $WYV 8. SAS Steps 6, 5, 7

22a. Measure %%

AB and %%

AC on 1 truss and measure %%

DE and

%% DF on the other. If

%% AB &

%% DE and

%% AC &

%% DF ,

then the trusses are & by SAS.

b. 3.5 ft; by the Pyth. Thm., BC 0 3.5 ft. Since the , are congruent,

%% EF &

%% BC .

23.

24. AB = AC 4x = 6x ! 11 11 = 2x x = 5.5

BC = DCx + 4 = 5x ! 7 11 = 2x x = 5.5 1

By the def. of &, AB & BD, and BC & DC. AC & AC by the Reflex. Prop. of &. Thus $ABC & $ADC by SSS.

25. Measure the lengths of the logs. If the lengths of the logs in 1 wing deflector match the lengths of the logs in the other wing deflector, the , will be & by SAS or SSS.

26. Yes; if the , have the same 2 side lengths and the same included " measure, the , are & by SAS.

27. Check students’ constructions; yes; if each side is & to the corr. side of the second $, they can be in any order.

TEST PREP, PAGE 248

28. CIn I and III, two sides are congruent with an congruent angle in between so I and III are similar by SAS.

29. GSAS proves $ABC & $ADC, soAB + BC + CD + DA = AB + CD + CD + AB = 12.1 + 7.8 + 7.8 + 12.1 = 39.8 cm

30. A"F and "J are the included #, so "F & "J proves SAS.


31. J

%% EF &

%% EH

EF = EH4x + 7 = 6x ! 4 11 = 2x x = 5.5



1. Draw %%

DB . 1. Through any 2 pts. there is exactly one line.

2. "ADC and "BCD are supp.

2. Given

3. %%

AD * %%

CB 3. Conv. of Same-Side Int. # Thm.

4. "ADB & "CBD 4. Alt. Int. # Thm.

5. %%

AD & %%

CB 5. Given

6. %%

DB & %%

BD 6. Reflex Prop. of &

7. $ADB & $CBD 7. SAS Steps 5, 4, 6


1. "QPS & "TPR 1. Given

2. "RPS & "RPS 2. Reflex. Prop. of &

3. "QPR & "TPS 3. Subst. Prop. of &

4. %%

PQ & %%

PT , %%

PR & %%

PS 4. Given

5. $PQR & $PTS 5. SAS Steps 3, 4

34. m"FKJ + m"KFJ + m"FJK = 180 2x + 3x + 10 + 90 = 180 5x = 80 x = 16KJ = HJ = 72, so

%% KJ &

%% HJ by the def. of &.

"FJK & "FJH by the Rt. " & Thm. %%

FJ & %%

FJ by the Reflex. Prop. of &. So $FJK & $FJH by SAS.

35. m"KFJ = m"HFJ 2x + 6 = 3x ! 21 27 = xFK = FH = 171, so

%% FK &

%% FH by the def. of &.

"KFJ & "HFJ by the def. of " bisector. %%

FJ & %%

FJ by the Reflex. Prop. of &. So $FJK & $FJH by SAS.


36. x _ 2 ! 8 2 5

x ! 16 2 10 x 2 26

37. 2a + 4 > 3a 4 > a a < 4

38. !6m ! 1 2 ! 13 12 2 6m m 3 2

39. 4x ! 7 = 21 Given 4x ! 7 + 7 = 21 + 7 Add. Prop. of = 4x = 28 Simplify.

4x _ 4 = 28 _

4 Div. Prop. of =

x = 7 Simplify.

40. a _ 4 + 5 = !8 Given

a _ 4 + 5 ! 5 = !8 ! 5 Subtr. Prop. of =

a _ 4 = !13 Simplify.

4 ( a _ 4 ) = 4(!13) Multi. Prop. of =

a = !52 Simplify.

41. 6r = 4r + 10 Given 6r ! 4r = 4r ! 4r + 10 Subtr. Prop. of = 2r = 10 Simplify.

2r _ 2 = 10 _

2 Div. Prop. of =

r = 5 Simplify.

42. "H & "F m"H = m"Fx + 24 = 110 x = 86

43. m"FGE = m"GEH = 36m"FEG + m"F + m"FGE = 180 m"FEG + 110 + 36 = 180 m"FEG = 180 ! 146 = 34°

44. m"FGH = m"FGE + m"EGH = m"GEH + m"FEG = 36 + 34 = 70°

USING TECHNOLOGY, PAGE 249

1. Check students’ drawings.

2. They stay the same size and shape.

3. $ABC & $DEF

4. Check students’ measurements.

TECHNOLOGY LAB: PREDICT OTHER TRIANGLE CONGRUNECE RELATIONSHIPS, PAGES 250–251

TRY THIS, PAGE 250–251 1. Yes; the $ stays the same shape and size if you do

not change AD, m"A, or m"D.

2. no 3. Third # Thm.

4. No; the " measures must stay the same but the side lengths can change.

5. Check students’ constructions; yes; yes, AAS.

6. You need 1 side length of the $. If 2 " pairs and 1 (non-included) side pair are & (AAS), the , are &.

7. many; no 8. 1

9. rt. 10. rt. #


4-5 TRIANGLE CONGRUENCE: ASA, AAS, AND HL, PAGES 252–259

CHECK IT OUT! PAGES 253–255 1. Yes; the $ is uniquely determined by AAS.

2. By the Alt. Int. # Thm., "KLN & "MNL. %%

LN & %%

NL by the Reflex. Prop. of &. No other congruence relationships can be determined, so ASA cannot be applied.

3.

4. Yes; it is given that %%

AC & %%

DB . %%

CB & %%

BC by the Reflex. Prop. of &. Since "ABC and "DCB are rt. #, $ABC & $DCB by HL.

THINK AND DISCUSS, PAGE 255 1. No; the & sides are not corr. sides.

2. Possible answer: corr. # and sides

3.


1. The included side %%

BC is enclosed between "ABC and "ACB.

2.

3. Yes; the $ is determined by AAS.

4. Yes; by the Def. of " bisector, "TSV & "RSV and "TVS & "RVS.

%% SV &

%% SV by the Reflex. Prop. of &.

So $VRS & $VTS by ASA.

5. No; you need to know that a pair of corr. sides are &.

6a. %%

QS & %%

SQ b. "RQS & "PSQ

c. Rt. " & Thm. d. AAS

7. Yes; it is given that "D and "B are rt. # and %%

AD & %%

BC . $ABC and $CDA are rt. , by def. %%

AC & %%

CA by the Reflex. Prop. of &. So $ABC & $CDA by HL.

8. No; you need to know that %%

VX & %%

VZ .

PRACTICE AND PROBLEM SOLVING, PAGE 257–258

9.

10. Yes; the $ is uniquely determined by ASA.

11. No; you need to know that "MKJ & "MKL.

12. Yes; by the Alt. Int. # Thm., "SRT & "UTR, and "STR & "URT.

%% RT &

%% TR by the Reflex. Prop. of &.

So $RST & $TUR by ASA.

13a. "A & "D b. Given

c. "C & "F d. AAS

14. No; you need to know that "K and "H are rt. #.

15. Yes; E is a mdpt. So by def., %%

BE & %%

CE , and %%

AE & %%

DE . "A and "D are & by the Rt. " & Thm. By def., $ ABE and $DCE are rt. ,. So $ABE & $DCE by HL.

16. AAS proves $ADB & $CDB; reflection

17. $FEG & $QSR; rotation

18.

19a. No; there is not enough information given to use any of the congruence theorems.

b. HL can be used, since also %%

JL & %%

JL .


20. Proof B is incorrect. The corr. sides are not in the correct order.

21.

It is given that $ABC and $DEF are rt. ,. %%

AC & %%

DF , %%

BC & %%

EF , and "C and "F are rt. #. "C & "F by the Rt. " & Thm. Thus $ABC & $DEF by SAS.


1. %%

AD * %%

BC 1. Given

2. "DAE & "BCE 2. Alt. Int. # Thm.

3. "AED & "CEB 3. Vert. # Thm.

4. %%

AD & %%

CB 4. Given

5. $AED & $CEB 5. AAS Steps 2, 3, 4


1. %%

KM + %%

JL 1. Given

2. "JKM and "LKM are rt. # 2. Def. of +

3. "JKM & "LKM 3. Rt. " & Thm.

4. %%

JM & %%

LM , "JMK & "LMK 4. Given

5. $JKM & $LKM 5. AAS Steps 3, 4

24. Since 2 sides and the included " are equal in measure and therefore &, you could prove the , & using SAS. You could also use HL since the , are rt. ,.



26. ANeed "XVZ & "XWY for ASA.

27. JFrom figure, 2 corr. side pairs and included " pair are &, i.e., SAS.

28. C Alt. Int. # Thm. gives two & " pairs, and one non-included & side pair is given. AAS proves $AED & $CEB.

29. GFor AAS, need

%% RT &

%%% UW . So:

RT = UW6y ! 2 = 2y + 7 4y = 9 y = 2.25

30. No; check students’ drawings and constructions; since the lengths of the corr. sides of the 2 , are not equal, the 2 , are not & even if the corr. # have the same measure.


31. Yes; the sum of the " measures in each $ must be 180°, which makes it possible to solve for x and y. The value of x is 15, and the value of y is 12. Each $ has # measuring 82°, 68°, and 30°.

%% VU &

%% VU by

the Reflex. Prop. of &. So $VSU & $VTU by ASA or AAS.


1. $ABC is equil. 1. Given

2. %%

AC & %%

BC 2. Def. of equil. $

3. C is mdpt. of DE. 3. Given

4. %%

DC & %%

EC 4. Def. of mdpt.

5. "DAC and "EBC are &. and supp.

5. Given

6. "DAC and "EBC are rt. #.

6. # that are & and supp. are rt. #.

7. $DAC and $EBC are rt. ,.

7. Def. of rt. $

8. $DAC & $EBC 8. HL Steps 4, 2

33.

Case 1: Given rt. $ABC and rt. $DEF with "A & "D and

%% AB &

%% DE

Statements Reasons

1. "A & "D 1. Given

2. %%

AB & %%

DE 2. Given

3. "B & "E 3. Rt. " & Thm.

4. $ABC & $DEF 4. ASA Steps 1, 2, 3

Case 2; given rt. $ABC and rt. $DEF with "A & "D and

%% BC &

%% EF

Statements Reasons

1. "A & "D 1. Given

2. %%

BC & %%

EF 2. Given

3. "B & "E 3. Rt. " & Thm.

4. $ABC & $DEF 4. ASA Steps 1, 3, 2

34. Third # Thm.; if the third " pair is &, then the , are also & by AAS.


35. x-intercept:0 = 3x ! 66 = 3xx = 2

y-intercept:y = 3(0) ! 6y = !6


36. x-intercept: 0 = ! 1 _

2 x + 4

1 _ 2 x = 4

x = 8

y-intercept:y = ! 1 _

2 (0) + 4

y = 4

37. x-intercept: 0 = !5x + 55x = 5 x = 1

y-intercept:y = !5(0) + 5y = 5

38. AC = 10 x 2 ! 6 = 10 x 2 = 16 x = 4(Discard x = !4 since AB > 0.)AB = x + 2 = 4 + 2 = 6BC = x 2 ! 2x = 4 2 ! 2(4) = 8

39. m"A + m"B + m"C = 180 53.1 + 90 + m"C = 180 143.1 + m"C = 180 m"C = 36.9°

4-6 TRIANGLE CONGRUENCE: CPCTC, PAGES 260–265

CHECK IT OUT! PAGES 260–261 1. JL = NL and KL = ML, so

%% JL &

%% NL and

%% KL &

%% ML .

By Vert. # Thm., "MLN & "KLJ. By SAS, $MLN & $KLJ. By CPCTC, JK = NM = 41 ft

2.


1. J is mdpt. of %%

KM and %%

NL . 1. Given

2. %%

KJ & %%

MJ and %%

LJ & %%

NJ 2. Def. of mdpt.

3. "KJL & "MJN 3. Vert. # Thm.

4. $KJL & $MJN 4. SAS Steps 2, 3

5. "LKJ & "NMJ or "JLK & "JNM

5. CPCTC

6. %%

KL * %%%

MN 6. Conv. of Alt. Int. # Thm.

4. Use Distance Formula to find side lengths.

JK = 4 (((((((((((( (2 ! (!1) ) 2 + ((!1) ! (!2) ) 2 = ' ((( 9 + 1 = ' (( 10

KL = 4 ((((((((((( ((!2) ! 2 ) 2 + (0 ! (!1) ) 2 = ' ((( 16 + 1 = ' (( 17

JL = 4 (((((((((((( ((!2) ! (!1) ) 2 + (0 ! (!2) ) 2 = ' ((( 1 + 4 = ' ( 5

RS = 4 (((((((( (5 ! 2 ) 2 + (2 ! 3 ) 2 = ' ((( 9 + 1 = ' (( 10

ST = 4 (((((((( (1 ! 5 ) 2 + (1 ! 2 ) 2 = ' ((( 16 + 1 = ' (( 17

RT = 4 (((((((( (1 ! 2 ) 2 + (1 ! 3 ) 2 = ' ((( 1 + 4 = ' ( 5

So %%

JK & %%

RS , %%

KL & %%

ST , and %%

JL & %%

RT . Therefore, $JKL & $RST by SSS, and "JKL & "RST by CPCTC.

THINK AND DISCUSS PAGE 262 1. SAS;

%%% UW &

%% XZ ; "U & "X; "W & "Z

2.


EXERCISES, PAGES 262–265

GUIDED PRACTICE, PAGE 262–263

1. Corr. # and corr. sides

2. "BCA & "DCE by Vert. # Thm, "CBA &"CDE by Rt. " & Thm., and

%% BC &

%% DC (given). Therefore

$ABC & $EDC by ASA. By CPCTC, %%

AB & %%

DE , so AB = DE = 6.3 m.

3a. Def. of + b. Rt. " & Thm.

c. Reflex. Prop. of & d. Def. of mdpt.

e. $RXS & $RXT f. CPCTC


1. %%

AC & %%

AD , %%

CB & %%

DB 1. Given

2. %%

AB & %%

AB 2. Reflex. Prop. of &

3. $ACB & $ADB 3. SSS Steps 1, 2

4. "CAB & "DAB 4. CPCTC

5. %%

AB bisects "CAD. 5. Def. of " bisector


EF = 4 ((((((((((( ((!1) ! (!3) ) 2 + (3 ! 3 ) 2

= ' ((( 4 + 0 = 2

FG = 4 ((((((((((( ((!2) ! (!3) ) 2 + (0 ! 3 ) 2

= ' ((( 1 + 9 = ' (( 10

EG = 4 ((((((((((( ((!2) ! (!1) ) 2 + (0 ! 3 ) 2

= ' ((( 1 + 9 = ' (( 10

JK = 4 ((((((((((( (0 ! 2 ) 2 + ((!1) ! (!1) ) 2

= ' ((( 4 + 0 = 2

KL = 4 ((((((((( (1 ! 2 ) 2 + (2 ! (!1) ) 2

= ' ((( 1 + 9 = ' (( 10

JL = 4 ((((((((( (1 ! 0 ) 2 + (2 ! (!1) ) 2

= ' ((( 1 + 9 = ' (( 10 So

%% EF &

%% JK , %%

FG & %%

KL , and %%

EG & %%

JL . Therefore $EFG & $JKL by SSS, and "EFG & "JKL by CPCTC.


AB = 4 (((((((( (4 ! 2 ) 2 + (1 ! 3) 2 = ' ((( 4 + 4 = 2 ' ( 2

BC = 4 ((((((((( (1 ! 4 ) 2 + ((!1) ! 1) 2 = ' ((( 9 + 4 = ' (( 13

AC = 4 ((((((((( (1 ! 2 ) 2 + ((!1) ! 3) 2 = ' ((( 1 + 16 = ' (( 17

RS = 4 (((((((((((( ((!3) ! (!1) ) 2 + ((!2) ! 0 ) 2 = ' ((( 4 + 4 = 2 ' ( 2

ST = 4 (((((((((((( (0 ! (!3) ) 2 + ((!4) ! (!2) ) 2 = ' ((( 9 + 4 = ' (( 13

RT = 4 ((((((((((( (0 ! (!1) ) 2 + ((!4) ! 0 ) 2 = ' ((( 1 + 16 = ' (( 17 So

%% AB &

%% RS , %%

BC & %%

ST , and %%

AC & %%

RT . Therefore $ABC & $RST by SSS, and "ACB & "RTS by CPCTC.

PRACTICE AND PROBLEM SOLVING,PAGES 263–264

7. "ABC & "EDC by Rt. " & Thm., "ACB & "ECD by Vert. # Thm., and

%% BC &

%% DC . So $ABC & $EDC

by ASA. By CPCTC, AB = DE = 420 ft.


1. M is mdpt. of %%

PQ and %%

RS . 1. Given

2. %%

PM & %%%

QM , %%%

RM & %%

SM 2. Def. of mdpt.

3. "PMS & "QMR 3. Vert. # Thm.

4. $PMS & $QMR 4. SAS Steps 2, 3

5. %%

QR & %%

PS 5. CPCTC


1. %%%

WX & %%

XY & %%

YZ & %%%

ZW 1. Given

2. %%

ZX & %%

XZ 2. Reflex. Prop. of &

3. $WXZ & $YZX 3. SSS, steps 1, 2

4. "W & "Y 4. CPCTC


1. G is mdpt. of %%

FH . 1. Given

2. FG = HG 2. Def. of mdpt.

3. %%

FG & %%

HG 3. Def. of &.

4. Draw %%

EG . 4. Exactly 1 line through any 2 pts.

5. %%

EG & %%

EG 5. Reflex. Prop. of &

6. %%

EF & %%

EH 6. Given

7. $EGF & $EGH 7. SSS Steps 3, 5, 6

8. "EFG & "EHG 8. CPCTC

9. "1 & " 2 9. & Supp. Thm.



1. %%

LM bisects "JLK. 1. Given

2. "JLM & "KLM 2. Def. of " bisector

3. %%

JL & %%

KL 3. Given

4. %%

LM & %%

LM 4. Reflex. Prop. of &

5. $JLM & $KLM 5. SAS Steps 3, 2, 4

6. %%

JM & %%

KM 6. CPCTC

7. M is mdpt. of %%

JK . 7. Def. of mdpt.

12. RS = 4 (((((((( (2 ! 0 ) 2 + (4 ! 0 ) 2

= 4 ((( 4 + 16 = 2 4 ( 5

ST = 4 ((((((((( ((!1) ! 2 ) 2 + (4 ! 3 ) 2

= 4 ((( 9 + 1 = 4 (( 10

RT = 4 ((((((((( ((!1) ! 0 ) 2 + (3 ! 0 ) 2

= 4 ((( 1 + 9 = 4 (( 10

UV = 4 (((((((((((( ((!3) ! (!1) ) 2 + (( ! 4) ! 0 ) 2

= 4 ((( 4 + 16 = 2 4 ( 5

VW = 4 ((((((((((((( ((!4) ! (!3) ) 2 + ((!1) ! (!4) ) 2

= 4 ((( 1 + 9 = 4 (( 10

UW = 4 (((((((((((( ((!4) ! (!1) ) 2 + ((!1) ! 0 ) 2

= 4 ((( 9 + 1 = 4 (( 10 So

%% RS &

%% UV , %%

ST & %%%

VW , and %%

RT & %%%

UW . Therefore, $RST & $UVW by SSS, and "RST & "UVW by CPCTC.

13. AB = 4 ((((((((( (2 ! (!1) ) 2 + (3 ! 1 ) 2

= ' ((( 9 + 4 = ' (( 13

BC = 4 ((((((((( (2 ! 2 ) 2 + ((!2) ! 3 ) 2

= ' ((( 0 + 25 = 5

AC = 4 ((((((((((( (2 ! (!1) ) 2 + ((!2) ! 1 ) 2

= ' ((( 9 + 9 = 3 ' ( 2

DE = 4 (((((((((((( ((!1) ! 2 ) 2 + ((!5) ! (!3) ) 2

= ' ((( 9 + 4 = ' (( 13

EF = 4 (((((((((((( ((!1) ! (!1) ) 2 + (0 ! (!5) ) 2

= ' ((( 0 + 25 = 5

DF = 4 ((((((((((( ((!1) ! 2 ) 2 + (0 ! (!3) ) 2

= ' ((( 9 + 9 = 3 ' ( 2 So

%% AB &

%% DE , %%

BC & %%

EF , and %%

CA & %%

DF . Therefore, $ABC & $DEF by SSS, and "BAC & "EDF by CPCTC.


1. $QRS is adj. to $QTS. %%

QS bisects "RQT. "R & "T.

1. Given

2. "RQS & "TQS 2. Def. of " bisector

3. %%

QS & %%

QS 3. Reflex. Prop. of &

4. $RSQ & $TSQ 4. AAS Steps 1, 2, 3

5. %%

RS & %%

TS 5. CPCTC

6. %%

QS bisects %%

RT . 6. Def. of bisector


1. E is the mdpt. of %%

AC and

%% BD .

1. Given

2. %%

AE & %%

CE , %%

BE & %%

DE 2. Def. of mdpt.

3. "AEB & "CED 3. Vert # Thm.

4. $AEB & $CED 4. SAS Steps 2, 3

5. "A & "C 5. CPCTC

6. %%

AB * %%

CD 6. Conv. of Alt. Int. # Thm.

16a. "ADB, "ADC are rt. #, hyp. lengths are =, corr. leg lengths are =. So HL proves $ADB & $ADC.


1. %%

AD + %%

BC 1. Given

2. " ADB and "ADC are rt. #.

2. Def. of +

3. $ADB and $ADC are rt. ,

3. Def. of rt. $

4. AB = AC = 20 in. 4. Given

5. %%

AB & %%

AC 5. Def. of &

6. %%

AD & %%

AD 6. Reflex. Prop. of &

7. $ADB & $ADC 7. HL Steps 5, 6

8. %%

BD & %%

CD 8. CPCTC

c. BD 2 + AD 2 = AB 2 BD 2 + 10 2 = 20 2

BD = ' ((((( 400 ! 100 0 17.3 in.BC = 2BD 0 34.6 in.

17. , are & by SAS.x + 11 = 2x ! 3 14 = x

18. , are & by ASA.4x + 1 = 6x ! 41 42 = 2x x = 21


1. PS = RQ 1. Given

2. %%

PS & %%

RQ 2. Def. of &

3. m"1 = m"4 3. Given

4. "1 & "4 4. Def. of &

5. %%

SQ & %%


6. $PSQ & $RQS 6. SAS Steps 2, 4, 5

7. "3 & "2 7. CPCTC

8. m"3 = m"2 8. Def. of &



1. m"1 = m"2, m"3 = m"4

1. Given

2. "1 & "2, "3 & "4 2. Def. of &

3. %%

SQ & %%

SQ 3. Reflex. Prop. of &

4. $PSQ & $RSQ 4. ASA Steps 2, 3

5. %%

PS & %%

RS 5. CPCTC

6. PS = RS 6. Def. of &


1. PS = RQ, PQ = RS 1. Given

2. %%

PS & %%

RQ , %%

PQ & %%

RS 2. Def. of &

3. %%

SQ & %%


4. $PSQ & $RQS 4. SSS Steps 2, 3

5. "3 & "2 5. CPCTC

6. %%

PQ * %%

RS 6. Conv. of Alt. Int. # Thm.

22. Yes; $JKM & $LMK by SSS, so "JKM & "LMK by CPCTC. Therefore,

%% JK * %%

ML by Conv. of Alt. Int. # Thm.

23.

The segs. %%

CA , %%

CD , and %%

CB must be &. "ACB & "DCB. If $ACB & $DCB by SAS, then AB = DB.


24. COnly way to get a second " pair & is first to prove , are & and then to use CPCTC. But you would use CPCTC to prove

%% AC &

%% AD directly.

25. GLNK &NLM, so by CPCTC "LNK & "NLM.

26. C6x = x + 5 _

2

5x = 5 _ 2

x = 1 _ 2

10x + y = 40 y = 40 ! 10x = 40 ! 10 · 1 _

2

= 35

27. GOnly corr. parts are ever used. & ,, * lines, + lines all are used.

28. BRS = 4 (((((((( (3 ! 2 ) 2 + (3 ! 6 ) 2 = ' (( 10

ST = 4 (((((((( (2 ! 6 ) 2 + (6 ! 6 ) 2 = 4

RT = 4 (((((((( (6 ! 3 ) 2 + (6 ! 3 ) 2 = 3 ' ( 2 These lengths only match the $ coordinates in B.


29. Any diagonal on any face of the cube is the hyp. of a rt. $ whose legs are edges of the cube. Any 2 of these , are & by SAS (or LL). Therefore, any 2 diagonals are & by CPCTC.


1. Draw %%

MK . 1. Through any 2 pts. there is exactly 1 line.

2. %%

MK & %%

KM 2. Reflex. Prop. of &

3. %%

JK & %%

LM , %%

JM & %%

LK 3. Given

4. $JKM & $LMK 4. SSS Steps 2, 3

5. "J & "L 6. CPCTC


1. R is the mdpt. of %%

AB . 1. Given

2. %%

AR & %%

BR 2. Def. of mdpt.

3. %%

RS + %%

AB 3. Given

4. "ARS and "BRS are rt. ,

4. Def. of +

5. "ARS & "BRS 5. Rt. " & Thm.

6. %%

RS & %%

RS 6. Reflex. Prop. of &

7. $ARS & $BRS 7. SAS Steps 2, 5, 6

8. %%

AS & %%

BS 8. CPCTC

9. "ASD & "BSC 9. Given

10. S is the mdpt. of %%

DC . 10. Given

11. %%

DS = %%

CS 11. Def. of mdpt.

12. $ASD & $BSC 12. SAS Steps 8, 9, 11

32. "A & "E (given), "B and "D are rt. # (from figure), and BC & CD (from figure). Therefore, $ ABC & $EDC by HL. By CPCTC, AB = DE. By Pythag. Thm., CD 2 + DE 2 = CE 2 DE 2 = 21 2 ! 10 2 AB = DE = ' ((((( 441 ! 100 0 18 ft


33. x = 5x _ n

90 = 90 + 84 + 93 + 88 + 91 + x/6 x = 6(90) ! (90 + 84 + 93 + 88 + 91) = 94

34. P 1 = 3.95 + 0.08m P 1 (75) = 3.95 + 0.08(75) = 9.95 P 2 = 0.10 · min(m, 50) + 0.15 · max(m ! 50, 0) P 2 (75) = 0.10(50) + 0.15(75 ! 50) = 8.75The second plan is cheaper.

35. reflection across the x-axis

36. translation (x, y) 6 (x ! 3, y ! 4)

37. Yes; it is given that "B & "D and %%

BC & %%

DC . By Vert. " Thm., "BCA & "DCE. Therefore, $ABC & $EDC by ASA.


CONNECTING GEOMETRY TO ALGEBRA: QUADRATIC EQUATIONS, PAGE 266

TRY THIS, PAGE 266 1. Method 1: Factoring

FG = FE x 2 ! 3x = 18 x 2 ! 3x ! 18 = 0 (x ! 6)(x + 3) = 0 x = 6 or !3

2. Method 2: Quadratic Formula

x = !4 ± ' (((((( 16 ! 4(1)(!12)

__ 2(1)

= !4 ± 8 _ 2

= !6 or 2

3. Method 1: Factoring YX = YZ x 2 ! 4x = 12 x 2 ! 4x ! 12 = 0 (x ! 6)(x + 2) = 0x = 6 or !2

4. Method 2: Quadratic Formula

x = !2 ± ' (((((( 4 ! 4(1)(!3)

__ 2(1)

= !2 ± 4 _ 2

= !3 or 1

4-7 INTRODUCTION TO COORDINATE PROOF, PAGES 267–272

CHECK IT OUT! PAGES 267–269 1. You can place the longer leg along the

y-axis and the other leg along the x-axis.

2. Proof: $ABC is a rt. $ with height AB and base BC.The area of $ABC = 1 _

2 bh

= 1 _ 2 (4)(6) = 12 square units

By Mdpt. Formula, coordinates of

D = ( 0 + 4 _ 2 , 6 + 0 _

2 ) = (2, 3). The x-coord. of D is

height of $ADB, and base is 6 units.The area of $ADB = 1 _

2 bh

= 1 _ 2 (2)(6) = 6 square units

Since 6 = 1 _ 2 (12), area of $ADB is 1 __

2 area of $ABC.

3. Possible answer:

4. $ABC is a rt. $ with height 2j and base 2n.

The area of $ABC = 1 _ 2 bh

= 1 _ 2 (2n)(2j ) = 2nj square units

By the Mdpt. Forumla, the coords. of D are (n, j). The base of $ABC is 2j units and the heightis n units.

So the area of $ADB = 1 _ 2

bh

= 1 _ 2

(2j )(n) = nj square units

Since nj = 1 _ 2 (2nj ), the area of $ADB is 1 __

2 the area of

$ABC.

THINK AND DISCUSS, PAGE 269 1. Possible answer: By using variables, your results

are not limited to specific numerical values.

2. Possible answer: The way you position the figure will affect the coords. assigned to the vertices and therefore, your calculations.

3. Possible answer: If you need to calculate the coords. of a mdpt., assigning 2p allows you to avoid using fractions.

4.


1. Possible answer: In coordinate geometry, a coord. proof is one in which you position figures in the coord. plane to prove a result.

2.

3.


4. By the Mdpt. Forumla, the coords of A are (0, 3) and the coords. of B are (4, 0).By the Dist. Formula,

PQ = ' (((((((( (0 ! 8) 2 + (6 ! 0) 2

= ' ((((( (!8) 2 + 6 2 = ' (((( 64 + 36

= 4 (( 100 = 10 units.

AB = ' (((((((( (0 ! 4) 2 + (3 ! 0) 2

= ' ((((( (!4) 2 + 3 2 = ' ((( 16 + 9

= ' (( 25 = 5 units.

So AB = 1 _ 2 PQ.

5. Possible answer:

6. Possible answer:

7.

By the Mdpt. Formula, the coords. of A are (0, a) and the coords of B are (b, 0).By the Dist. Formula,

PQ = ' ((((((( (0 ! 2b) 2 + (2a) 2

= ' (((((( (!2b) 2 + (2a) 2

= ' (((( 4 b 2 + 4 a 2

= 2 ' (((( b 2 + a 2 units

AB = ' (((((((( (0 ! b ) 2 + (a ! 0 ) 2

= ' ((((( (!b ) 2 + a 2

= ' (((( b 2 + a 2 units

So AB = 1 _ 2 PQ.


8. Possible answer:

9. Possible answer:

10.

E = ( 0 + 0 _ 2 , 0 + 10 _

2 ) = (0, 5)

F = ( 6 + 6 _ 2 , 0 + 10 _

2 ) = (6, 5)

BC = 4 ((((((((( (6 ! 0 ) 2 + (10 ! 10 ) 2 = 6 units.

EF = 4 (((((((( (6 ! 0 ) 2 + (5 ! 5 ) 2 = 6 units.So EF = BC.



13.

By the Mdpt. Formula, the coords. of E are (0, a) and the coords of F are (2c, a).By the Dist. Formula,

AD = ' ((((((((( (2c ! 0) 2 + (2a ! 2a) 2

= ' (( (2c) 2 = 2c units.Simlarly,

EF = ' (((((((( (2c ! 0) 2 + (a ! a) 2

= ' (( (2c) 2 = 2c units.So EF = AD.

14. Let endpts. be (x, y) and (z, w). By Mdpt. Formula,

(0, 0) = ( x + z _ 2 ,

y + w _

2 )

x + z _ 2 = 0

x + z = 0 z = !x

y + w

_ 2 = 0

y + w = 0 w = !yEndpts are (x, y) and (!x, !y).


15a.

b. Total distance = EW + WC

= 4 (((((((( (3 ! 0 ) 2 + (3 ! 0 ) 2 + 4 (((((((( (6 ! 3 ) 2 + (0 ! 3 ) 2 = 3 ' ( 2 + 3 ' ( 2 = 6 ' ( 2 0 8.5

16. Let A = (0, 0), B = (a, 0), and C = (0, 2a).Perimeter = AB + BC + CA

= a + 4 (((((((( (0 ! a ) 2 + (2a ! 0 ) 2 + 2a = a (3 + ' ( 5 ) units$ABC has base AB, height AC.Area = 1 _

2 bh

= 1 _ 2 (a)(2a) = a 2 square units

17. Let A = (0, 0), B = (s, 0), C = (s, t), and D = (0, t).Perimeter = AB + BC + CD + DA

= s + t + s + t = 2s + 2t unitsArea = "w = st square units

18. (n, n) 19. (p, 0)

20. 4 (((((((((((((( (!23.2 ! (!25) ) 2 + (31.4 ! 31.5 ) 2 0 1.8 units

4 (((((((((((((( (!24 ! (!23.2) ) 2 + (31.1 ! 31.4 ) 2 0 0.9 units

4 ((((((((((((( (!24 ! (!25) ) 2 + (31.1 ! 31.5 ) 2 0 1.1 units 1.8 is twice 0.9. The dist. between 2 of the locations is approx. twice the dist. between another 2 locations.

21. AB = 4 (((((((((((( (70 ! (!30) ) 2 + ((!30) ! 50 ) 2 0 128 nautical miles

Mdpt. of AB = ( !30 + 70 _ 2 ,

50 + (!30) _

2 )

= (20,10)

So, P is the mdpt of %%

AB .

22.

The area of the rect. is A = "w = 3(2) = 6 square units. For $RST, the base is 3 units, and the height is 1 unit. So the area of

$RST = 1 _ 2

bh = 1 _ 2 (3)(1) = 1.5 square units.

Since 1 _ 4

(6) = 1.5, the area of $RST is 1 _ 4 of the area

of the rect.

23. By Dist. Formula,

AB = 4 ((((((((( ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 and

AM = 4 ((((((((((((((

( x 1 + x 2 _

2 ! x 1 ) 2 + ( y 1 + y 2

_ 2 ! y 1 ) 2

= 4 (((((((((((((((

( x 1 + x 2 _

2 !

2 x 1 _

2 ) 2 + ( y 1 + y 2

_ 2 !

2 y 1 _

2 ) 2

= 4 ((((((((((( 1 _ 4 ( x 2 ! x 1 ) 2 + 1 _

4 ( y 2 ! y 1 ) 2

= 1 _ 2 4 ((((((((( ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2

So AM = 1 _ 2 AB.

24.

Proof: By Dist. Formula,

KL = 4 ((((((((( (!2 + 2 ) 2 + (1 ! 3 ) 2 = ' ((( 0 + 4 = 2

MP = 4 (((((((( (1 ! 1 ) 2 + (3 ! 1 ) 2 = ' ((( 0 + 4 = 2

LM = 4 ((((((((( (!2 ! 1 ) 2 + (3 ! 3 ) 2 = ' ((( 9 + 0 = 3

PK = 4 (((((((( (1 + 2 ) 2 + (1 ! 1 ) 2 = ' ((( 9 + 0 = 3Thus KL = MP and LM = PK by Trans. Prop. of &. %%

KL & %%

MP and %%

LM & %%

PK by def. of &, and %%

KM & %%

MK by Reflex. Prop. of &. Thus $KLM & $MPK by SSS.

25. You are assuming the figure has a rt. ".

26a. BD = BC + CD= AE + CD= 28 + 10 = 38 in.

By Dist. Formula,

DE = ' ((((( CD 2 + CE 2 CE 2 = DE 2 ! CD 2

CE = ' (((( 26 2 ! 10 2 = 24 in.

b. B = (24, 0); C = (24, 28); D = (24, 38); E = (0, 28)

TEST PREP, PAGE 272

27. B; Mdpt. Formula shows B is true.

28. F; G, H, and J are all possible vertices.

29. D; Perimeter = a + b + a + b = 2a + 2b

30. H; ( !1 + 7 _ 2 , 2 + 8 _

2 ) = (3, 5) = C


31. (a + c, b)

32. (n + p – n, h – h) = (p, 0)

33. Possible answer: Rotate $ 180° about (0, 0) and translate by (0, 2s). The new coords. would be (0, 0), (2s, 0), (0, 2s).


34. E is intersection of 2 given lines. At E, y = g _

f x and

y = ! g __

f x + 2g.

g _

f x = !

g __

f x + 2g

2 g _

f x = 2g

x = f

y = g _

f x

y = g _

f f

y = g E = (f, g)

Set eqns. = to each other.

Combine like terms.

Simplify.

Given

Subst.

Simplify.


35. x = !18 ± ' (((((( 1 8 2 ! 4(8)(!5)

__ 2(8)

= !18 ± 22 _ 16

= 1 _ 4

or !2 1 _ 4

36. x = !3 ± 4 ((((((( (!3 ) 2 ! 4(1)(!5)

___ 2(1)

= !3 ± ' (( 29 _ 2

0 1.19 or !4.19

37. x = 1 ± ' (((((((( (!1) 2 ! 4(3)(!10)

____________________ 2(3)

= 1 ± 11 _ 6 = 2 or ! 1 2 _

3

38. Think: Use Supp. Int. # Thm.x + 68 = 180 x = 112

39. Think: Use Alt. Int. # Thm.2y + 24 = 68 2y = 44 y = 22

40. AB = 3 BC = 4 ((((((((( (!3 + 1 ) 2 + (1 ! 3 ) 2 = 2 ' ( 2

AC = 4 ((((((((( (!3 + 4 ) 2 + (1 ! 3 ) 2 = ' ( 5 ED = 3DF = 4 ((((((((( (2 ! 0 ) 2 + (!4 + 2 ) 2 = 2 ' ( 2

EF = 4 ((((((((( (2 ! 3 ) 2 + (!4 + 2 ) 2 = ' ( 5 Therefore, $ABC & $EDF by SSS, and "ABC & "EDF by CPCTC.

4-8 ISOSCELES AND EQUILATRAL TRIANGLES, PAGES 273–279

CHECK IT OUT! PAGES 274 –275 1. 4.2 7 10 13 ; since there are 6 months between

September and March, the " measures will be approx. the same between Earth and the star. By the Conv. of the Isosc. $ Thm., the , created are isosc., and the dist. is the same.

2a. m"G = m"H = x m"F + m"G + m"H = 180 48 + x + x = 180 2x = 132 x = 66 Thus m"H = x = 66°.

b. m"N = m"P 6y = 8y ! 16 16 = 2y y = 8 Thus m"N = 6y = 6(8) = 48°.

3. $JKL is equilateral.4t ! 8 = 2t + 12t = 9t = 4.5JL = 2t + 1 = 2(4.5) + 1 = 10

4. Proof: By Mdpt. Formula, coords. of X are

( !2a + 0 ________ 2 , 0 + 2b ______

2 ) = (!a, b), coords. of Y are

( 2a + 0 ______ 2 , 0 + 2b ______

2 ) = (a, b), and coords of Z are

( !2a + 2a _________ 2 , 0 + 0 _____

2 ) = (0, 0).

By Dist. Formula,

XZ = 4 (((((((( (0 + a ) 2 + (0 ! b ) 2 = 4 (((( a 2 + b 2 , and

YZ = 4 (((((((( (0 ! a ) 2 + (0 ! b ) 2 = 4 (((( a 2 + b 2 Since XZ = YZ,

%% XZ &

%% YZ by definition. So $XYZ is

isosc.

THINK AND DISCUSS, PAGE 276 1. An equil. $ is also an equiangular $, so the 3 #

have the same measure. They must add up to 180° by the $ Sum Thm. So each " must measure 60°.

2.


1. legs: %%

KJ and %%

KL base:

%% JL

base #: "J and "L

2.

By the Ext. " Thm., m"R = 35°. Since m"R = m"S by the Conv. of the Isosc. $ Thm., QR = QS = 41 m.


3. Think: Use Isosc. $ Thm., $ " Sum Thm., and Vert. " Thm.m"B = m"A = 31°m"A + m"B + m"ABC = 180 31 + 31 + m"ABC = 180 m"ABC = 118°m"ECD = m"ABC =118°

4. Think: Use Isosc. $ Thm. and $ " Sum Thm.m"J = m"K m"J + m"K + m"L = 180 2m"K + 82 = 180 2m"K = 98 m"K = 49°

5. Think: Use Isosc. $ Thm. m"X = m"Y 5t ! 13 = 3t + 3 2t = 16 t = 8 m"X = 5t ! 13 = 27°

6. Think: Use Isosc. $ Thm. and $ " Sum Thm.m"B = m"C = 2xm"A + m"B + m"C = 180 4x + 2x + 2x = 180 8x = 180 x = 22.5 m"A = 4x = 90°

7. Think: Use Equilat. $ Thm. and $ " Sum Thm."R & "S & "T m"R + m"S + m"T = 180 12y + 12y + 12y = 180 36y = 180 y = 5

8. Think: Use Equilat. $ Thm. and $ " Sum Thm."L & "M & "Nm"L + m"M + m"N = 180 3(10x + 20) = 180 30x = 120 x = 4

9. Think: Use Equiang. $ Thm. %%

AB & %%

BC & %%

AC BC = AC6y + 2 =! y + 23 7y = 21 y = 3 BC = 6y + 2

= 6(3) + 2 = 20

10. Think: Use Equiang. $ Thm. %%

HJ & %%

JK & %%

HK HJ = JK 7t + 15 = 10t 15 = 3t t = 5 JK = 10t

= 10(5) = 50

11. Proof: It is given that $ABC is rt. isosc.,

%% AB &

%% BC , and X

is the mdpt. of %%

AC . By Mdpt. Formula, coords. of X

are ( 0 + 2a ______ 2 , 2a + 0 ______

2 ) = (a, a). By Dist. Formula,

AX = 4 (((((((( (a ! 0 ) 2 + (a ! 2a ) 2 = a ' ( 2 and

BX = 4 (((((((( (a ! 0 ) 2 + (a ! a ) 2 = a ' ( 2 = AX. So $AXB is isosc. by def. of an isosc. $.

PRACTICE AND PROBLEM SOLVING, PAGE 277–278

12. By " Add. Post., m"ATB = 80 ! 40 = 40°. m"BAT = 40° by Alt. Int. # Thm. "ATB & "BAT by def. of &. Since $ABT is isosc. by Conv. of Isosc. $ Thm., BT = BA = 2.4 mi.

13. Think: use Isosc. $ Thm., $ " Sum Thm., and Vert. # Thm.m"B = m"ACBm"A + m"B + m"ACB = 180 96 + 2m"ACB = 180 m"ACB = 42°m"DCE = m"ACB = 42°m"D = m"E m"D + m"E + m"DCE = 180 2m"E + 42 = 180 m"E = 69°

14. Think: Use Isosc. $ Thm. and $ " Sum Thm.m"U = m"S = 57° m"SRU + m"S + m"U = 180 m"SRT + m"TRU + 57 + 57 = 180 2m"TRU = 66 m"TRU = 33°

15. m"D = m"E x 2 = 3x + 10 x 2 ! 3x ! 10 = 0 (x ! 5)(x + 2) = 0 x = 5 or !2 m"D + m"E + m"F = 180 x 2 + 3x + 10 + m"F = 180 m"F = 180 ! 50 or 180 ! 8 = 130° or 172°

16. Think: Use Isosc. $ Thm. and $ " Sum Thm.m"A = m"B = (6y + 1)°m"A + m"B + m"C = 180 2(6y + 1) + 21y + 3 = 180 33y = 165 y = 5°m"A = 6y + 1 = 31°

17. Think: Use Equilat. $ Thm. and $ " Sum Thm."F & "G & "Hm"F + m"G + m"H = 180

3 ( z _ 2 + 14) = 180

z + 28 = 120 z = 92

18. Think: Use Equilat. $ Thm. and $ " Sum Thm."L & "M & "Nm"L + m"M + m"N = 180 3(1.5y ! 12) = 180 y ! 8 = 40 y = 48


19. Think: use Equiang. $ Thm.

%%

BC & %%

CD & %%

BD BC = CD

3 _ 2 x + 2 = 5 _

4 x + 6

6x + 8 = 5x + 24 x = 16 BC = 3 _

2 x + 2

= 3 _ 2 (16) + 2 = 26

20. Think: use Equiang. $ Thm. %%

XY & %%

YZ & %%

XZ XY = XZ

2x = 5 _ 2 x ! 5

5 = 1 _ 2 x

x = 10 XZ = XY

= 2x= 2(10) = 20

21. Proof: It is given that $ABC is isosc., %%

AB & %%%

AC, P is mdpt. of

%% AB , and Q is mdpt. of

%% AC . By Mdpt.

Formula, coords. of P are (a, b), and coords. of Q are (3a, b). By Dist. Formula,

PC = QB = 4 (((( 9 a 2 + b 2 , so %%

PC & %%

QB by def. of &.

22. always

23. sometimes

24. sometimes

25. never

26. No; if a base " is obtuse, the other base " must also be obtuse since they are &. But the sum of the " measures of the $ cannot be > 180°.

27a. %%

PS & %%

PT , so by Isosc. $ Thm., m"PTS = m"PST = 71°. By $ " Sum Thm, m"SPT + m"PTS + m"PST = 180 m"SPT + 71 + 71 = 180 m"SPT = 38°

b. %%

PQ & %%

PR , so by Isosc. $ Thm.,m"PQR = m"PRQ. By $ " Sum Thm, m"PQR + m"PRQ + m"QPR = 180 2m"PQR + (m"QPS + m"SPT + m"TPR) = 180 2m"PQR + 18 + 38 + 18 = 180 2m"PQR = 106 m"PQR = 53° m"PRQ = 53°

28. Let 3rd " of $ be "4.m"1 = m"4 = 58° (Alt. Int. # Thm., Isosc. $ Thm.)m"2 + 58 + 58 = 180 m"2 = 64°m"2 + m"3 = 180 (supp. #) 58 + m"3 = 180 m"3 = 122°

29. Let 3rd " of left $ be "4.m"3 = m"4 (Isosc. $ Thm.)m"3 + m"4 + 74 = 180 2m"3 = 106 m"3 = 53°m"1 + m"4 = 180 (supp. #) m"1 + 53 = 180 m"1 = 127°Let 3rd " of right $ be "5.m"2 = m"5 (Isosc. $ Thm.)m"1 + m"2 + m"5 = 180 127 + 2m"2 = 180 m"2 = 26.5°

30. Proof: It is given that $ABC is isosc., %%

BA & %%

BC , and X is the mdpt. of

%% AC . Assign the coords.

A(0, 2a), B(0, 0), and C(2a, 0). By the Mdpt. Formula, coords. of X are (a, a). By Dist. Formula, AX = XB = XC = a ' ( 2 . So $AXB & $CXB by SSS.

31. Check students’ drawings. The # are approx. 34°, 34°, and 112°. Conjecture should be that $ is isosc. Conjecture is correct since two short sides have equal measure ( ' (( 65 units).

32. List all (unordered) triples of natural numbers such that:• at least two are equal• sum of leg lengths > base length • perimeter is 184 ,: (5, 5, 8), (6, 6, 6), (7, 7, 4), (8, 8, 2).

33. In left $: 40 + x + x = 180 2x = 140 x = 70In right $: x + 2(3y ! 5) = 180 60 + 6y = 180 y = 20

34. In left $: all "s measure 60°.In right $: obtuse " measures180° – 60° = 120°.2(5x + 15) + 120 = 180 10x + 150 = 180 x = 3

35.

Statements Reasons

1. $DEF 1. Given

2. Draw the bisector of "EDF so that it intersects

%% EF at X.

2. Every " has a unique bisector.

3. "EDX & "FDX 3. Def. of " bisector

4. %%

DX & %%

DX 4. Refl. Prop. of &

5. "E & "F 5. Given

6. $EDX & $FDX 6. AAS Steps 3, 5, 4

7. %%

DE & %%

DF 7. CPCTC

36a. "B & "C

b. Isosc. $ Thm

c. Trans. Prop. of &

37. $DEF with "D & "E & "F is given. Since "E & "F,

%% DE &

%% DF by Conv. of Isosc. $ Thm.

Similarly, since "D & "F, %%

EF & %%

DE . By the Trans. Prop. of &,

%% EF &

%% DF . Combining the & statements,

%%

DE & %%

DF & %%

EF , and $DEF is equil. by def.

38. By the Ext. " Thm., m"C = 45°, so "A & "C. BC = AB by the Conv. of the isosc. $ Thm. So the distance to island C is the same as the distance traveled from A to B.


39. 1. !ABC " #CBA (Given)2.

$$ AB "

$$ CB (CPCTC)

3. !ABC (Def. of Isosc. %)

40. Two sides of a ! are " if and only if the & opp. those sides are ".


1. !ABC and !DEF 1. Given

2. Draw ''( EF so that

FG = CB. 2. Through any 2 pts.

there is exactly 1 line.

3. $$

FG " $$

CB 3. Def. of " segs.

4. $$

AC " $$

DF 4. Given

5. #C, #F are rt. &. 5. Given

6. $$

DF ) $$

EG 6. Def. of ) lines

7. #DFG is rt. # 7. Def. of rt. #

8. #DFG " #C 8. Rt. # " Thm.

9. !ABC " !DGF 9. SAS Steps 3, 8, 4

10. $$

DG " $$

AB 10. CPCTC

11. $$

AB " $$

DE 11. Given

12. $$

DG " $$

DE 12. Trans. Prop. of "

13. #G " #E 13. Isosc. ! Thm.

14. #DFG " #DFE 14. Rt. # " Thm.

15. !DGF " !DEF 15. AAS Steps 13, 14, 12

16. !ABC " !DEF 16. Trans. Prop. of "

42. Am#VUT = m#VTU 2m#VUT + m#VTU + m#TUV = 180 2m#VUT + 20 = 180 m#VUT = 80°m#VUR + m#VUT = 90 m#VUR + 80 = 90 m#VUR = 10°

43. Hy + 10 = 3y * 5 15 = 2y

y = 7 1 _ 2

44. 13.56t * 9 + 4t + 4t = 180 14t = 189 t = 13.5


45. It is given that $$

JK " $$

JL , $$

KM " $$

KL , and m#J = x°. By the ! Sum Thm., m#JKL + m#JLK + x ° = 180°. By the Isosc. ! Thm., m#JKL = m#JLK. So 2(m#JLK) + x ° = 180°.

or m#JLK = ( 180 * x _ 2 ) °. Since m#KML = m#JLK,

m#KML = ( 180 * x _ 2 ) ° by the Isosc. ! Thm. By the

! Sum Thm., m#MKL + m#JLK + m#KML = 180°

or m#MKL = 180° * ( 180 * x _ 2 ) ° * ( 180 * x _

2 ) °.

Simplifying gives m#MKL = x °.

46. Let A = (x, y).4 a 2 = A B 2

= x 2 + y 2 = AC 2 = (x – 2a ) 2 + y 2 = x 2 – 4ax + 4 a 2 + y 2 = 4 a 2 * 4ax + 4 a 2

4ax = 4 a 2 x = ay = ±

+ ,,,, 4 a 2 * x 2

= ± a - , 3

(x, y) = (a, a - , 3 ) 47. (2a, 0), (0, 2b), or any pt. on the ) bisector of

$$ AB .


48. x 2 + 5x + 4 = 0

(x + 4)(x + 1) = 0 x = *4

or *1

49. x 2 * 4x + 3 = 0 (x * 3)(x * 1) = 0 x = 3 or 1

50. x 2 * 2x + 1 = 0 (x * 1)(x * 1) = 0 x = 1

51. m = y 2 * y 1

_ x 2 * x 1

= 5 * (*1)

_ 0 * 2

= 6 _ *2

= *3

52. m = y 2 * y 1

_ x 2 * x 1

= *10 * (*10)

__ 20 * (*5)

= 0

53. m = y 2 * y 1

_ x 2 * x 1

= 11 * 7 _ 10 * 4

= 4 _ 6 = 2 _

3



1. Measure $$

AB , $$

BC , and $$

CA . If these three lengths are the same for each truss, then the trusses all have the same size and shape by SSS.


1. $$

CD ) $$

AB 1. Given

2. #CDA and #CDB are rt. &.

2. Def. of )

3. !CDA and !CDB are rt. ..

3. Def. of rt. .

4. $$

AC " $$

BC 4. Given

5. $$

CD " $$

CD 5. Reflex. Prop. of "

6. !CDA " !CDB 6. HL Steps 4, 5

3. $$

AD " $$

DB by CPCTC. AD = BD = 12 in. and

AC = BC = + ,,,, 9 2 + 12 2 = 15 in.

4. Possible answer: A(0, 0), B(24, 0), C(12, 9)



5. m"A = m"B = 37°; base # of an isosc. $ are &, so 2m"A + 106 = 180

6. Length of wood = 4(AB + BC + AC)= 4(24 + 15 + 15)= 216 in. = 18 ft = 3(6 ft)

Cost = (18 ft)($0.80/ft) = $14.40



AC & %%

BC , and %%

DC & %%

DC by Reflex. Prop. of &. By the Rt. " & Thm., "ACD & "BCD. Therefore, $ACD & $BCD by SAS.


1. %%

JK bisects "MJN. 1. Given

2. "MJK & "NJK 2. Def. of " bisector

3. %%

MJ & %%

NJ 3. Given

4. %%

JK & %%

JK 4. Reflex. Prop of &

5. $MJK & $NJK 5. SAS Steps 3, 2, 4

3. Yes, since %%

SU & %%

US . 4. No; need %%

AC & %%

DB .

5. 6. Yes; the $ is uniquely determined by ASA.


1. %%

CD * %%

BE and %%

DE * %%

CB 1. Given

2. "DEC & "BCE and "DCE & "BEC

2. Alt. Int. # Thm.

3. %%

CE & %%

EC 3. Reflex. Prop of &

4. $DEC & $BCE 4. ASA Steps 2, 3

5. "D & "B 5. CPCTC

8. Check students’ drawings; possible answer: vertices at (0, 0), (9, 0), (9, 9), and (0, 9).

9. It is given that ABCD is a rect. M is the mdpt. of %%%

AB, and N is the mdpt. of

%% AD . Use coords. A(0, 0),

B(2a, 0), C(2a, 2b), and D(0, 2b). By Mdpt. Formula,

coords. of M are ( 0 + 2a ______ 2 , 0 + 0 _____

2 ) = (a, 0), and

coords. of N are ( 0 + 0 _____ 2

, 0 + 2b ______ 2 ) = (0, b).

Area of rect. ABCD = "w = (2a)(2b) = 4ab.

Area of $AMN = 1 _ 2 bh = 1 _

2 ab, which is 1 _

8 the area

of rect. ABCD.

10. m"E = m"D = 2x °m"C + m"D + m"E = 180 5x + 2x + 2x = 180 9x = 180 x = 20 m"C = 5x = 100°

11. By Equiang. $ Thm.,

%% RS &

%% RT &

%% ST

RS = RT 2w + 5 = 8 ! 4w 6w = 3 w = 0.5 ST = RS = 2(0.5) + 5 = 6

12. It is given that isosc. $JKL has coords. J(0, 0), K(2a, 2b), and L(4a, 0). M is mdpt. of

%% JK , and N is

mdpt. of %%

KL . By Mdpt. Formula, coords. of M are

( 0 + 2a ______ 2 , 0 + 2b ______

2 ) = (a, b), and coords. of N are

( 2a + 4a _______ 2 , 2b + 0 ______

2 ) = (3a, b). By Dist. Formula,

MK = 4 ((((((((( (2a ! a ) 2 + (2b ! b ) 2 = 4 (((( a 2 + b 2 , and

NK = 4 ((((((((( (2a ! 3a ) 2 + (2b ! b ) 2 = 4 (((( a 2 + b 2 .

Thus %%

MK & %%

NK . So $KMN is isosc. by def. of isosc. $.


1. isosceles 2. corresponding angles

3. included side

LESSON 4-1, PAGE 284 4. equiangular; equilat. 5. obtuse; scalene

LESSON 4-2, PAGE 284 6. Think: Use Ext. " Thm.

m"N + m"P = m(ext. "Q) y + y = 120 y = 60m"N = y = 60°

7. Think: Use $ " Sum Thm. m"L + m"M + m"N = 180 8x + 2x + 1 + 6x ! 1 = 180 16x = 180 x = 11.25m"N = 6x – 1 = 66.5°

LESSON 4-3, PAGE 285 8.

%% PR &

%% XZ 9. "Y & "Q

10. m"CAD = m"ACB 2x ! 3 = 47 2x = 50 x = 25

11. CD = AB 3y + 1 = 15 ! 4y 7y = 14 y = 2 CD = 3y + 1 = 7



1. %%

AB & %%

DE , %%

DB & %%

AE 1. Given

2. %%

DA & %%

AD 2. Reflex. Prop. of &

3. $ADB & $DAE 3. SSS Steps 1, 2



1. %%

GJ bisects %%

FH , and

%% FH bisects

%% GJ .

1. Given

2. %%

GK & %%

JK , %%

FK & %%

HK 2. Def. of seg. bisector

3. "GKF & "JKH 3. Vert. # Thm.

4. $FGK & $HJK 4. SAS Steps 2, 3

14. BC = x 2 + 36 = (!6 ) 2 + 36 = 72YZ = 2 x 2 = 2(!6 ) 2 = 72 = BC

%%

BC & %%

YZ ; "C & "Z; %%

AC & %%

XZ . So $ABC & $XYZ by SAS.

15. PQ = y ! 1 = 25 ! 1 = 24QR = y = 25PR = y 2 ! (y – 1 ) 2 ! 42 = (25 ) 2 ! (24 ) 2 ! 42 = 7

%%

LM & %%

PQ ; %%%

MN & %%

QR ; %%

LN & %%

PR . So $LMN & $PQR by SSS.



1. C is mdpt. of %%

AG . 1. Given

2. %%

GC & %%

AC 2. Def. of mdpt

3. %%

HA * %%

GB 3. Given

4. "HAC & "BGC 4. Alt. Int. # Thm.

5. "HCA & "BCG 5. Vert. # Thm.

6. $HAC & $BGC 6. ASA Steps 4, 2, 5


1. %%%

WX + %%

XZ , %%

YZ + %%

ZX 1. Given

2. "WXZ, "YZX are rt. #. 2. Def. of +

3. $WXZ, $YZX are rt. ,. 3. Def. of rt. $

4. %%

XZ & %%

ZX 4. Reflex. Prop. of &

5. %%%

WZ & %%

YX 5. Given

6. $WZX & $YXZ 6. HL Steps 5, 4


1. "S, "V are rt. #. 1. Given

2. "S & "V 2. Rt. " & Thm.

3. RT = UW 3. Given

4. %%

RT & %%%

UW 4. Def. of &

5. m"T = m"W 5. Given

6. "T & "W 6. Def. of &

7. $RST & $UVW 7. AAS Steps 2, 6, 4



1. M is mdpt. of %%

BD . 1. Given

2. %%

MB & %%%

DM 2. Def. of mdpt.

3. %%

BC & %%

DC 3. Given

4. %%%

CM & %%%

CM 4. Reflex. Prop. of &

5. $CBM & $CDM 5. SSS Steps 2, 3, 4

6. "1 & "2 6. CPCTC


1. %%

PQ & %%

RQ 1. Given

2. %%

PS & %%

RS 2. Given

3. %%

QS & %%


4. $PQS & $RQS 4. SSS Steps 1, 2, 3

5. "PQS & "RQS 5. CPCTC

6. %%

QS bisects "PQR. 6. Def. of " bisector


1. H is mdpt. of %%

GJ , L is mdpt. of

%% MK .

1. Given

2. GH = JH, ML = KL 2. Def. of mdpt.

3. %%

GH & %%

JH , %%

ML & %%

KL 3. Def. of &

4. %%

GJ & %%

KM 4. Given

5. %%

GH & %%

KL 5. Div. Prop. of &

6. %%%

GM & %%

KJ , "G & "K 6. Given

7. $GMH & $KJL 7. ASA Steps 5, 6

8. "GMH & "KJL 8. CPCTC

22. Check students’ drawings; e.g., (0, 0), (r, 0), (0, s)

23. Check students’ drawings; e.g., (0, 0), (2p, 0), (2p, p), (0, p)

24. Check students’ drawings; e.g., (0, 0), (8m, 0), (8m, 8m), (0, 8m)


LESSON 4-7, PAGE 287 25. Use coords. A(0, 0), B(2a, 0), C(2a, 2b), and

D(0, 2b). Then by Mdpt. Formula, the mdpt. coords are E(a, 0), F(2a, b), G(a, 2b), and H(0, b). By Dist.

Formula, EF = 4 (((((((( (2a ! a ) 2 + (b ! 0 ) 2 = 4 (((( a 2 + b 2 ,

and GH = 4 (((((((( (0 ! a ) 2 + (b ! 2b ) 2 = 4 (((( a 2 + b 2 .

So %%

EF & %%

GH by def. of &.

26. Use coords. P(0, 2b), Q(0, 0), and R(2a, 0). By Mdpt. Formula, mdpt. coords are M(a, b). By Dist.

Formula, QM = 4 (((((((( (a ! 0 ) 2 + (b ! 0 ) 2 = 4 (((( a 2 + b 2 ,

PM = 4 (((((((( (a ! 0 ) 2 + (b ! 2b ) 2 = 4 (((( a 2 + b 2 , and

RM = 4 (((((((( (2a ! a ) 2 + (0 ! b ) 2 = 4 (((( a 2 + b 2 . So QM = PM = RM. By def., M is equidistant from vertices of $PQR.

27. In a rt. $, a 2 + b 2 = c 2 .

' (((((((( (3 ! 3) 2 + (5 ! 2) 2 = 3,

' (((((((( (3 ! 2) 2 + (2 ! 5) 2 = ' (( 10 ,

' (((((((( (2 ! 3) 2 + (5 ! 5) 2 = 1, and 3 2 + 1 2 = ( ' (( 10 ) 2 .Since 9 + 1 = 10, it is a rt. $.

LESSON 4-8, PAGE 287 28. Think: Use Equilat. $ Thm. and $ " Sum Thm.

m"K = m"L = m"Mm"K + m"L + m"M = 180 3m"M = 180 3(45 ! 3x) = 180 !45 = 9x x = !5

29. Think: Use Conv. of Isosc. $ Thm. %%

RS & %%

RT RS = RT 1.5y = 2y ! 4.5 4.5 = 0.5y y = 9 RS = 1.5y = 13.5

30. %%

AB & %%

BC AB = BCx + 5 = 2x ! 3 8 = x Perimeter = AC + CD + AD

= 2AB + CD + CD= 2(x + 5) + 2(2x + 6)= 6x + 22= 6(8) + 22 = 70 units


1. Rt. $

2. scalene $ (AC = 4 by Pythag. Thm)

3. isosc. $ (AC = BC = 4)

4. scalene $ (BD = 4 + 3 = 7)

5. m"RTP = 2m"RTSm"RTP + m"RTS = 180 3m"RTS = 180 m"RTS = 60°m"RTS + m"R + m"S = 180 60 + m"R + 43 = 180 m"R = 77°

6. %%

JL & %%

XZ 7. "Y & "K

8. "L & "Z 9. %%

YZ & %%

KL


1. T is mdpt. of %%

PR and %%

SQ . 1. Given

2. %%

PT & %%

RT , %%

ST & %%

QT 2. Def. of mdpt.

3. "PTS & "RTQ 3. Vert. # Thm.

4. $PTS & $RTQ 4. SAS Steps 2, 3


1. "H & "K 1. Given

2. %%

GJ bisects "HGK. 2. Given

3. "HGJ & "KGJ 3. Def. of " bisector

4. %%

JG & %%

JG 4. Reflex. Prop. of &

5. $HGJ & $KGJ 5. AAS Steps 1, 3, 4


1. %%

AB + %%

AC , %%

DC + %%

DB 1. Given

2. "BAC, "CDB are rt. #. 2. Def. of +

3. $ABC and $DCB are rt. ,.

3. Def. of rt. $

4. %%

AB & %%

DC 4. Given

5. %%

BC & %%

CB 5. Reflex. Prop. of &

6. $ABC & $DCB 6. HL Steps 5, 4


1. %%

PQ * %%

SR 1. Given

2. "QPR & "SRP 2. Alt. Int. # Thm.

3. "S & "Q 3. Given

4. %%

PR & %%

RP 4. Reflex. Prop. of &

5. $QPR & $SRP 5. AAS Steps 2, 3, 4

6. "SPR & "QRP 6. CPCTC

7. %%

PS * %%

QR 7. Conv. of Alt. Int. # Thm.

14.


15. Use coords. A(0, 0), B(a, 0), C(a, a), and D(0, a). By Dist. Formula,

AC = ' (((((((( (a ! 0) 2 + (a ! 0) 2 = a ' ( 2 , and

BD = ' (((((((( (0 ! a) 2 + (a ! 0) 2 = a ' ( 2 . Since AC = BD,

%% AC &

%% BD by def. of &.

16. Think: By Equilat. $ Thm., m"F = m"G = m"H. 3m"G = 180 3(5 ! 11y) = 180 5 ! 11y = 60 ! 11y = 55 y = !5

17. Think: Use $ " Sum and Isosc. $ Thms.m"P + m"Q + m"PRQ = 180 2(56) + m"PRQ = 180 m"PRQ = 68°By Vert. " and Isosc. $ Thms., m"T = m"SRT = m"PRQ = 68°.Using $ " Sum and Isosc. Thms.m"S + m"T + m"SRT = 180 m"S + 2(68) = 180 m"S = 44°

18. It is given that $ABC is isosc. with coords. A(2a, 0), B(0, 2b), and C(!2a, 0). D is mdpt. of

%% AC , and E is

mdpt. pf %%

AB . By Mdpt. Formula, coords. of

D are ( !2a + 2a ________ 2

, 0) = (0, 0), and coords. of E are

( 2a + 0 ______ 2 , 0 + 2b ______

2 ) = (a, b). By Dist. Formula,

AE = ' (((((((( (a ! 2a) 2 + (b ! 0) 2 = ' ((( a 2 + b 2 , and

DE = ' (((((((( (a ! 0) 2 + (b ! 0) 2 = ' ((( a 2 + b 2 .

Therefore, %%

AE & %%

DE and $AED is isosc.


1. Cm"EFG = m"DEF + m"EDFTherefore III is false. Also, since m"EDF > 0, I is true. II is true as marked in diagram.

2. HBy CPCTC, m"A = m"C 2x + 14 = 3x ! 15 29 = x m"A + m"DBA + m"BDA = 180 (2x + 14) + 49 + m"BDA = 180 121 + m"BDA = 180 m"BDA = 59°

3. DSide lengths are ' (( 10 , 5 ' ( 2 , and 2 ' ( 5 .

4. H131° + 49° = 180° (supp. "s)136° + 44° = 180° (supp. "s)y = 49 + 44 = 93 (Ext. " Thm.)

5. DBy Equi-" $ Thm., RS & ST RS = ST 2x + 10 = 3x ! 2 12 = x







= 2.25(0.8) = 1.8 mi





c. T









n _ 28,800

= 7 _ 18,327

n = 7 ______ 18,327

(28,800) ! 11 mo





2 (4) = 2 in.; base is b = 6 in.;


6 (40 % 2 & ' 3 )





4 & '' 9.8 ! 12.52 cm.


438.4 ! 20t t ! 22 s



Solutions KeyProperties and Attributes of Triangles5

CHAPTER


1. E 2. C

3. A 4. D

5. B

6. acute 7. right

8. acute 9. obtuse

10. 8 2 = 64 11. (!12 ) 2 = 144

12. " ## 49 = 7 13. ! " ## 36 = !6

14. " ### 9 + 16 = " ## 25 = 5 15. " #### 100 ! 36 = " ## 64 = 8

16. $ ## 81 _ 25

= " ## 81 _ " ## 25

= 9 _ 5

17. $

# 2 2 = 2

18. d + 5 < 1

d < !4

19. !4 % w ! 7 3 % w w & 3

20. !3s & 6 s % !2

21. !2 > m _ 10

!20 > m m < !20

22. Let p and q represent the following:p: Lines ! and m intersect.q: Lines ! and m are not parallel.Given: p ' q, and p. So by the Law of Detachment, q is true: Lines ! and m are not parallel.

23. Let p, q, and r represent the following:p : M is the midpoint of

(( AB

q : AM = MBr : AM = 1 _

2 AB and MB = 1 _

2 AB

Given: p ' q and q ' r. So by the Law of Syllogism,

p ' r : If M is the midpoint of ((

AB , then AM = 1 _ 2 AB and

MB = 1 _ 2 AB.

5-1 PERPENDICULAR AND ANGLE BISECTORS, PAGES 300–306

CHECK IT OUT! PAGES 301–303 1a. DG = EG

DG = 14.6

b. Since DG = GE and ! ) ((

DE , ! is the ) bisector of ((

DE by the Conv. of the ) Bisector Thm.EF = 1 _

2 DE

EF = 1 _ 2 (20.8) = 10.4

2a. WX = WZ WX = 3.05

b. Since XW = ZW, (((

XW ) ((

XY , and (((

ZW ) ((

ZY , **+ YW

bisects ,XYZ by the Conv. of the , Bisector Thm.m,XYZ = 2m,WYZ m,XYZ = 2(63°) = 126°

3. By the Conv. of the , Bisector Thm., **+ QS bisects

,PQR.

4.

Step 1 Graph ((

PQ .The ) bisector of

(( PQ is ) to

(( PQ at its midpoint

Step 2 Find the midpoint of ((

PQ .

midpoint of ((

PQ = ( 5 + 1 _ 2 ,

2 + (!4) _

2 ) = (3, !1)

Step 3 Find the slope of the perpendicular bisector.

slope of ((

PQ = !4 ! 2 _ 1 ! 5

= !6 _ !4

= 3 _ 2

Since the slopes of ) lines are opposite reciprocals,

the slope of the ) 1bisector is ! 2 _ 3 .

Step 4 Use point-slope form to write an equation.

The ) bisector of ((

PQ has slope ! 2 _ 3 and passes

through (3, !1). y ! y 1 = m(x ! x 1 )

y ! (!1) = ! 2 _ 3 (x ! 3)

y + 1 = ! 2 _ 3 (x ! 3)

THINK AND DISCUSS, PAGE 303 1. Yes; no; since PY = QY = 3, Y is the midpoint

of ((

PQ , and thus by the def. of bisector, ! is a bisector of

(( PQ . If ! is the ) bisector of

(( PQ , then PX

would equal QX by the ) Bisector Thm. However, PX = 8.5 and QX = 8.4, so ! is not the ) bisector of ((

PQ .

2. No; although MJ = ML, to apply the Conv. of the , Bisector Thm., you must know that

(( MJ ) ((

KJ and ((

ML ) ((

KL .


3.


1. perpendicular bisector

2. Since PS = QS and m ) ((

PQ , m is the ) bisector of ((

PQ by the Conv. of the ) Bisector Thm.PQ = 2QT

PQ = 2(47.7) = 95.4

3. SP = SQ SP = 25.9

4. PS = QS 4a = 2a + 26 2a = 26 a = 13So QS = 2(13) + 26 = 52.

5. AD = CDAD = 21.9

6. Since AD = CD, ((

AC ) ((

AB , and ((

CD ) ((

BC , **+ BD

bisects ,ABC by the Conv. of the , Bisector Thm.m,CBD = 1 _

2 m,ABC

m,CBD = 1 _ 2 (48°) = 24°

7. Since DA = DC, ((

AD ) ((

AB , and ((

CD ) ((

BC , **+ BD

bisects ,ABC by the Conv. of the , Bisector Thm.m,DBC = m,DBA 10y + 3 = 8y + 10 2y + 3 = 10 2y = 7 y = 7 _

2

So m,DBC = [10 ( 7 _ 2 ) + 3]° = 38°

8. The braces can be installed so that ((

PK ) ((

JL ,

((

PM ) ((

NL , and PK = PM. Then by the Conv. of the , Bisector Thm., P will be on the bisector of ,JLN.

9.

Step 1 Graph (((

MN .The ) bisector of

((( MN is ) to

((( MN at its midpoint.

Step 2 Find the midpoint of (((

MN .

midpoint of (((

MN = ( !5 + 1 _ 2 ,

4 + (!2) _

2 ) = (!2, 1)

Step 3 Find the slope of the perpendicular bisector.slope of

((( MN = !2 ! 4 _

1 ! (!5) = !6 _

6 = !1

Since the slopes of ) lines are opposite reciprocals, the slope of the ) bisector is 1.Step 4 Use point-slope form to write an equation. The ) bisector of

((( MN has slope 1 and passes

through (!2, 1).y ! y 1 = m(x ! x 1 ) y ! 1 = 1[x ! (!2)] y ! 1 = x + 2

10.

Step 1 Graph ((

UV .The ) bisector of

(( UV is ) to

(( UV at its midpoint


UV .

midpoint of ((

UV = ( 2 + 4 _ 2 , !6 + 0 _

2 ) = (3, !3)


slope of ((

UV = 0 ! (!6)

_ 4 ! 2

= 6 _ 2 = 3

Since the slopes of ) lines are opposite reciprocals, the slope of the ) bisector is ! 1 _

3 .

Step 4 Use point-slope form to write an equation. The ) bisector of

(( UV has slope ! 1 _

3 and passes

through (3, !3). y ! y 1 = m(x ! x 1 )

y ! (!3) = ! 1 _ 3 (x ! 3)

y + 3 = ! 1 _ 3 (x ! 3)

11.

Step 1 Graph ((

JK .The ) bisector of

(( JK is ) to

(( JK at its midpoint


JK .

midpoint of ((

JK = ( !7 + 1 _ 2 ,

5 + (!1) _

2 ) = (!3, 2)


slope of ((

JK = !1 ! 5 _ 1 ! (!7)

= !6 _ 8 = ! 3 _

4

Since the slopes of ) lines are opposite reciprocals, the slope of the ) bisector is 4 _

3 .



JK has slope 4 _ 3 and passes

through (!3, 2).

y ! y 1 = m(x ! x 1 )

y ! 2 = 4 _ 3

[x ! (!3)]

y ! 2 = 4 _ 3

(x + 3)



12. GJ = GKGJ = 8.25

13. JG = KG x + 12 = 3x ! 17 12 = 2x ! 17 29 = 2x 14.5 = xSo KG = 3(14.5) ! 17 = 26.5.

14. Since GJ = GK and t ) ((

JK , t is the ) bisector of ((

JK by the Conv. of the ) Bisector Thm.JK = 2JH JK = 2(26.5) = 53

15. RQ = TQRQ = 1.3

16. Since RQ = TQ, ((

RQ ) ((

RS , and ((

TQ ) ((

TS , **+ SQ

bisects ,RST by the Conv. of the Bisector Thm.m,RST = 2m,RSQm,RST = 2(58°) = 116°

17. m,QSR = m,QST 9a + 48 = 6a + 50 3a + 48 = 50 3a = 2 m,QST = 6 ( 2 _

3 ) + 50 = 54°

18. They can position Main St. so that the , formed by Elm St. and Main St. is - to the , formed by Grove St. and Main St. Then by the , Bisector Thm., every point on Main St. will be equidistant from Elm St. and Grove St.

19. Step 1 Graph ((

EF .The ) bisector of

(( EF is ) to

(( EF at its midpoint


EF .

( x 1 + x 2 _

2 ,

y 1 + y 2 _

2 )

midpoint of ((

EF = ( !4 + 0 _ 2 , !7 + 1 _

2 ) = (!2, !3)


slope = y 2 ! y 1

_ x 2 ! x 1

slope of ((

EF = 1 ! (!7)

_ 0 ! (!4)

= 8 _ 4 = 2


the slope of the ) bisector is ! 1 _ 2 .



EF has slope ! 1 _ 2 and passes

through (–2, –3). y ! y 1 = m(x ! x 1 )

y ! (!3) = ! 1 _ 2 [x ! (!2)]

y + 3 = ! 1 _ 2 (x + 2)

20. Step 1 Graph ((

XY .The ) bisector of

(( XY is ) to

(( XY at its midpoint


XY .

( x 1 + x 2 _

2 ,

y 1 + y 2 _

2 )

midpoint of ((

XY = ( !7 + (!1) _

2 ,

5 + (!1) _

2 ) = (–4, 2)


slope = y 2 ! y 1

_ x 2 ! x 1

slope of ((

XY = !1 ! 5 _ !1 ! (!7)

= !6 _ 6 = –1

Since the slopes of ) lines are opposite reciprocals, the slope of the ) bisector is 1.Step 4 Use point-slope form to write an equation.The ) bisector of

(( XY has slope ! 1 _

2 and passes

through (!2, !3). y ! y 1 = m(x ! x 1 ) y ! 2 = 1[x ! (!4)] y ! 2 = x + 4

21. Step 1 Graph (((

MN .The ) bisector of

((( MN is ) to

((( MN at its midpoint

Step 2 Find the midpoint of (((

MN .

( x 1 + x 2 _

2 ,

y 1 + y 2 _

2 )

midpoint of (((

MN = ( !3 + 7 _ 2 ,

1 + (!5) _

2 ) = (2, !3)

Step 3 Find the slope of the ) bisector.

slope = y 2 ! y 1

_ x 2 ! x 1

slope of (((

MN = !5 ! (!1)

_ 7 ! (!3)

= !4 _ 10

= ! 2 _ 5


the slope of the bisector is 5 _ 2 .

Step 4 Use point-slope form to write an equation.The bisector of ) has slope ! 1 _

2 and passes through

(!2, !3). y ! y 1 = m(x ! x 1 )

y ! (!3) = 5 _ 2 (x ! 2)

y + 3 = 5 _ 2 (x ! 2)

22. PS = PT 3m + 9 = 5m ! 13 9 = 2m ! 13 22 = 2m 11 = m

QS = QT 6n ! 3 = 4n + 14 2n ! 3 = 14 2n = 17 n = 8.5

23. JK = LK JK = 38

24. GN = 2GZGN = 2(36) = 72

25. MK = HK ML + LK = HJ + JK ML = HJ ML = 38

26. HY = MY HY = 24

27. JL = 2LXJL = 2(12) = 24

28. NK = GK NM + ML + LK = 114 NM + 38 + 38 = 114 NM = 38

29. Possible answer: C(3, 2); AC = " ## 26 ; BC = $ ## 26 ; so AC = BC, and by the Conv. of the ) Bisector Thm., C is on the ) bisector of

(( AB .

30. Draw line ! ) to ((

AB through X. So m,AYX = 90° and m,BYX = 90° by the def. of ). It is given that AX = BX. So

(( AX - ((

BX by def. of - segs. By the Reflex. Prop. of -,

(( XY - ((

XY . So .AYX - .BYX by HL. Then

(( AY - ((

BY by CPCTC. By the def. of midpoint, Y is the midpoint of

(( AB . Since ! is ) to

(( AB

at its midpoint, ! is the ) bisector of ((

AB . Therefore X is on the ) bisector of

(( AB .



1. **+ PS bisects ,QPR, ((

SQ ) **+ PQ ,

(( SR )

**+ PR

1. Given

2. ,QPS - ,RPS 2. Def. of , bisector

3. ,SQP and ,SRP are rt. /. 3. Def. of )4. ,SQP - ,SRP 4. Rt. , - Thm.5. ((

PS - ((

PS 5. Reflex. Prop. of -6. .PQS - .PRS 6. AAS7. ((

SQ - ((

SR 7. CPCTC8. SQ = SR 8. Def. of - segs.

32. Possible answer: By stating that the point must be in the int. of the ,, the thm. implies that it must be in the same plane as the ,. It is possible for a point to be equidistant from the sides of an , but to lie in a different plane. In the diagram, ,ABC is in plane Z, and P is equidistant from the sides of ,ABC, but P does not lie in plane Z. Thus P cannot be on the bisector of the ,, because the bisector must lie in the same plane as the ,.

33a. Step 1 Graph ((

AC .The ) bisector of

(( AC is ) to

(( AC at its midpoint


AC .

( x 1 + x 2 _

2 ,

y 1 + y 2 _

2 )

midpoint of ((

AC = ( !3 + 3 _ 2 , !2 + 6 _

2 ) = (0, 2)


slope = y 2 ! y 1

_ x 2 ! x 1

slope of ((

AC = 6 ! (!2)

_ 3 ! (!3)

= 8 _ 6 = 4 _

3

Since the slopes of ) lines are opposite reciprocals, the slope of the ) bisector is ! 3 _

4 .



AC has slope ! 3 _ 4 and passes

through (0, 2). y ! y 1 = m(x ! x 1 )

y ! 2 = ! 3 _ 4 (x – 0)

y = ! 3 _ 4 x + 2

b. There are 2 points on the ) bisector that are 4 mi dist. from the midpoint of

(( AC .

c. Distance of warehouse from midpoint of ((

AC = 4 mi.

Distance of midpoint of ((

AC from A

= $ ### 3 2 + 4 2 = 5 mi. Distance of warehouse from A = $ ### 4 2 + 5 2 0 6.4 mi.

34. In the construction of the perpendicular bisector

of ((

AB , the same compass setting is used to draw an arc from each end point of the segment. So in the diagram, AX = BX and AY = BY. By the Converse of the Perpendicular Bisector Theorem, both X and Y lie on the perpendicular bisector of

(( AB . So ! is the

perpendicular bisector of ((

AB .

TEST PREP, PAGE 306

35. D;J is on the perpendicular bisector of

(( XY , so by the

Perpendicular Bisector Theorem, JX = JY.

36. G;

37. Possible answer: All locations that are equidistant from Park St. and Washington Ave. lie on the bisector of the , formed by the 2 streets. All locations that are equidistant from the museum and the library lie on the perpendicular bisector of a segment formed by the museum and the library. So the visitor center should be built at point V, where the angle bisector and the perpendicular bisector intersect.


38a. The dist. from P to **+ BA and from P to

**+ BC are both

2 " # 5. So P is equidistant from **+ BA and

**+ BC , and

therefore by the Converse of the Angle Bisector Theorem, P is on the bisector of ,ABC.

b. Possible answer: y = 3x ! 6.

39. The distance of a point (x, y) from x-axis is 1y2 , and its distance from y-axis is 1x2 . So locus is 1y2 = 1x2 , or the lines y = x and y = !x.


1. ((

VX ) **+ YX ,

(( VZ )

**+ YZ ,

VX = VZ1. Given

2. ,VXY and ,VZY are rt. /.

2. Def. of )

3. ((

YV - ((

YV 3. Reflex. Prop. of -4. .YXV - .YZV 4. HL5. ,XYV - ,ZYV 5. CPCTC

6. 3 *+ YV bisects ,XYZ. 6. Def. of , bisector


41. It is given that ((

KN is the perpendicular bisector of ((

JL and

(( LN is the perpendicular bisector of

((

KM . By the Perpendicular Bisector Theorem, JK = KL and KL = ML. Thus JK = ML by the Trans. Prop. of =. By the definition of - segs.,

(( JK - ((

ML . By the Seg. Add. Post., JR + RL = JL and .MT + TK = MK. By the definition of the perpendicular bisector, R is the midpoint of

(( JL and T is the midpoint of

((

MK . Thus ((

JR - ((

RL and ((

MT - ((

TK . By the definition of cong segs., JR = RL and MT = TK. By Subst., JR + JR = JL and MT + MT = MK. It is given that

(( JR - ((

MT . So JR = MT by definition of - segs. By Subst., JR + JR = MK. By the Trans. Prop. of =, JL = MK, so

(( JL - ((

MK by the definition of - segs. By the Reflex. Prop. of -,

(( JM - ((

JM . Therefore .JKM - .MLJ by SSS, and ,JKM - ,MLJ by CPCTC.


42. C

43. slope of 3 *+ RS = 4 ! 2 _ 1 + 4

= 2 _ 5

slope of 3 *+ VT = !5 + 1 _ !7 ! 3

= 2 _ 5


44. slope of 3 *+ RV = !5 ! 2 _ !7 + 4

= 7 _ 3

slope of 3 *+ ST = !1 ! 4 _ 3 ! 1

= ! 5 _ 2

The slopes are not the same, so the lines are not parallel. The product of the slopes is not !1, so the lines are not perpendicular.

45. slope of 3 *+ RT = !1 ! 2 _ 3 + 4

= ! 3 _ 7

slope of 3 *+ VR = !5 ! 2 _ !7 + 4

= 7 _ 3

The product of the slopes is !1, so the lines are perpendicular.

46. m = !9 ! (!1)

_ 2 ! 1

= !8

y ! y 1 = m(x ! x 1 ) y + 1 = !8(x ! 1) y = !8x + 7

47. y ! y 1 = m(x ! x 1 )y + 15 = !0.5(x ! 10)y + 15 = !0.5x + 5

y = ! 1 _ 2 x ! 10

48. m = 5 ! 0 _ 0 ! (!4)

= 5 _ 4

y = mx + b

y = 5 _ 4

x + 5

5-2 BISECTORS OF TRIANGLES, PAGES 307–313

CHECK IT OUT! PAGES 308–310 1a. GM = MJ = 14.5

b. GK = KH = 18.6

c. Z is circumcenter of .GHJ, By the Circumcenter Theorem, Z is equidistant from the vertices of .GHJ.JZ = GZ = 19.9

2.

Step 1 Graph the ..Step 2 Find equations for two perpendicular bisectors. Since two sides of . lie along the axes, use the graph to find the perpendicular bisectors of these two sides. the perpendicular bisector of GO is y = !4.5, and the perpendicular bisector of OH is x = 4.Step 3 Find the intersection of the two equations. The lines y = !4.5 and x = 4 intersect at (4, !4.5), the circumcenter of .GOH.

3a. X is the incenter of .PQR. By the Incenter Theorem, X is euqidistant from the sides of .PQR. The distance from X to

(( PR is 19.2, so

the distance from X to ((

PQ is also 19.2.

b. m,PRQ = 2m,PRXm,PRQ = 2(12°) = 24°m,RQP + m,PRQ + m,QPR = 180° 52 + 24 + m,RQP = 180° m,RQP = 104°

m,PQX = 1 _ 2 m,RQP

m,PQX = 1 _ 2 (104°) = 52°

4. By the Incenter Theorem, the incenter of a . is equidistant from the sides of the .. Draw the . formed by the streets and draw the , bisectors to find the incenter, point M. The city should place the monument at point M.

THINK AND DISCUSS, PAGE 310 1. Possible answer:

2. Q; P. Possible answer: the incenter is always inside the ., so Q cannot be the incenter. Therefore P must be the incenter, and Q must be the circumcenter.


3.


1. They do not intersect at a single point.

2. circumscribed about

3. N is the circumcenter of .PQR. By the Circumcenter Theorem, N is equidistant from vertices of .PQR. NR = NP = 5.64

4. RV = PV = 5.47

5. TR = QT = 3.95

6. N is the circumcenter of .PQR. By the Circumcenter Theorem, N is equidistant from vertices of .PQR.QN = NP = 5.64

7.

Step 1 Graph ..Step 2 Find equations for the two perpendicular bisectors. Since the two sides of the . lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of KO is y = 6, and the perpendicular bisector of OL is x = 2.Step 3 Find the intersection of the two equations. The lines y = 6 and x = 2 intersect at (2, 6), the circumcenter of .KOL.

8.

Step 1 Graph the ..Step 2 Find equations for the two perpendicular bisectors. Since the two sides of the . lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of AO is x = !3.5, and the perpendicular bisector of OL is y = !5.Step 3 Find the intersection of the two equations. The lines x = !3.5 and y = !5 intersect at (!3.5, !5), the circumcenter of .AOB.

9. F is the incenter of .CDE. By the Incenter Theorem, F is equaldistant from the sides of .CDE. The distance from F to

(( DE is 42.1, so the

distance from F to ((

CD is also 42.1.

10. m,DCE = 2m,FCDm,DCE = 2(17°) = 34°m,DCE + m,CDE + m,CED = 180° 34 + 54 + m,CED = 180° m,CED = 92°

m,FED = 1 _ 2 m,CED

m,FED = 1 _ 2 (92°) = 46°

11. The largest possible 4 in the int. of the . is its inscribed 4, and the center of the inscribed 4 is the incenter. Draw the . and its , bisectors. Center the 4 at E, the point of concurrency of the , bisectors.


12. CF = FA = 59.7 13. YC = YB = 63.9

14. DB = AD = 62.8 15. AY = YB = 63.9

16. Step 1 Write equations of the perpendicular bisectors of (((

MO and ((

NO .The perpendicular bisector of

((( MO is x = !2.5; the


NO is y = 7. Step 2 Find the circumcenter of the ..The circumcenter is at the intersection of the perpendicular bisectors, (!2.5, 7).

17. Step 1 Write equations of the perpendicular bisectors of ((

OV and (((

OW .The perpendicular bisector of

(( OV is y = 9.5; the


WO is x = !1.5. Step 2 Find the circumcenter of the ..The circumcenter is at the intersection of the perpendicular bisectors, (!1.5, 9.5).

18. J is the incenter of .RST. By the Incenter Theorem, J is equaldistant from the sides of .RST. The distance from J to

(( ST is 8.37, so the distance

from J to ((

RS is also 8.37.

19. m,TSR = 2m,JSRm,TSR = 2(14°) = 28°m,TSR + m,SRT + m,RTS = 180° 28 + 42 + m,TSR = 180° m,TSR = 110°

m,RTJ = 1 _ 2 m,TSR

m,RTJ = 1 _ 2 (110°) = 55°

20. By the Circumcenter Theorem, the circumcenter of the . is equidistant from the vertices. Draw the . formed by the cities, and draw the perpendicular bisectors of the sides. The main office should be located at M, the circumcenter.


21. Possible answer: if ,JML is a rt. ,, then m,MJL + m,MLJ = 90° because the acute / of a rt. . are comp. Since M is the incenter of .JKL,

(( JM

and ((

LM are , bisectors of .JKL. So by the def. of , bisector, m,KJL = 2m,MJL and m,KLJ = 2m,MLJ. By subst., m,KJL + m,KLJ = 2(m,MJL + m,MLJ) = 2(90°) = 180°. But by the . Sum Theorem, m,K = 180° ! (m,KJL + m,KLJ) = 180° ! 180° = 0°. This would mean that .JKL is not a .. Therefore ,JML cannot be a rt. ,.

22. The angle bisector; m,BAE = m,EAC

23. The perpendicular bisector; AD = BD, ((

AD ) ((

DG and

(( BD ) ((

DG

24. The angle bisector; m,ABG = m,GBC

25. The angle bisector; since ((

AE and ((

BG are , bisectors, R is the incenter.

(( CR intersects the

incenter, so it is an , the bisector.

26. neither 27. neither

28. never


31. never

32. sometimes

33. The slope of ((

OA is 2; the midpoint of ((

OA is (2, 4).

The perpendicular bisector of ((

OA is y ! 4 = ! 1 _

2 (x ! 2).


OB is x = 4.

At the intersection, x = 4 and y ! 4 = ! 1 _ 2 (4 ! 2) =

!1, so y = 3. The circumcenter is at (4, 3).

34. The perpendicular bisector of ((

OY is y = 6.

The slope of ((

OZ is = 1; the midpoint of ((

OZ is (3, 3).


OZ is y ! 3 = !(x ! 3).At the intersection, y = 6 and 6 ! 3 = 3 = !x + 3, so x = 0. The circumcenter is at (0, 6).

35a. , Bisector Theorem b. the bisector of ,B

c. PX = PZ


1. ((

QS bisects ,PQR, ((

PQ - ((

RQ . 1. Given 2. ,PQS - ,RQS 2. Def. of ,

bisector 3. ((

QS - ((

QS 3. Reflex. Prop. of -

4. .PQS - .RQS 4. SAS 5. ,PSQ - ,RSQ 5. CPCTC 6. ,PSQ and ,RSQ are supp. 6. Lin. Pair Thm. 7. ,PSQ and ,RSQ are rt. /. 7. - / supp.

' rt. / 8. ,PSQ = ,RSQ = 90° 8. Def. of rt. ,

9. 3 *+ QS ) ((

PR 9. Def. of )10. ((

PS - ((

RS 10. CPCTC11. S is midpoint of

(( PR . 11. Def. of

midpoint

12. 3 *+ QS is the perpendicular bisector of

(( PR .

12. Def. of the perpendicular bisector

37a. The new store is at the circumcenter of .ABC.The perpendicular bisector of

(( AB is x = 4.

The slope of ((

AC is 3 _ 4 ; the midpoint of

(( AC is (2, 3 _

2 ) .


AC is

y ! 3 _ 2 = ! 4 _

3 (x ! 2).

At the intersection, x = 4 and y ! 3 _ 2 = ! 4 _

3 (4 ! 2)

= ! 8 _ 3 , so y = 9 ! 16 _

6 = ! 7 _

6 .

The new store is located at (4, ! 7 _ 6

) . b. outside, since y > 0 for all int. poins. of the ., but

! 7 _ 6 < 0

c. distance from each store = distance from store C

= 3 ! (! 7 _ 6 ) = 4 1 _

6 0 4.2 mi

38. Possible answers: Similarities: Both are circles. Both intersect the triangle in exactly 3 points.Differences: The inscribed circle is smaller than the circumscribed circle. Except for the points of intersection, the inscribed circle lies inside the triangle, while the circumscribed circle lies outside. The center of the inscribed circle always lies inside the triangle, while the center of the circumscribed circle may be inside, outside, or on the triangle. The center of the inscribed circle is the point of concurrency of the angle bisectors, while the center of the circumscribed circle is the point of concurrency of the perpendicular bisectors.

39a. Check students’ constructions. b. Check students’ constructions.

TEST PREP, PAGE 313

40. B;PX = PY by the Incenter Theorem.

41. F;m = 1, y + 2 = x ! 5, or y = x ! 7.


42. 14.75 KN = MN5z ! 4 = z + 11 4z = 15 z = 3.75LN = MN = 3.75 + 11 = 14.75


43a. Possible answer: Given: M is the midpoint of

(( QR .

Prove: PM = QM = RMProof: The coordinates of M are

( 0 + 2a _ 2

, 2b + 0 _ 2 ) = (a, b).

By the Distance Formula,

PM = " ######## (a ! 0 ) 2 + (b ! 0 ) 2

= $ #### a 2 + b 2 ,

QM = " ######## (a ! 0 ) 2 + (b ! 2b ) 2

= " ##### a 2 + (!b ) 2 = $ #### a 2 + b 2 , and

RM = " ######## (a ! 2a ) 2 + (b ! 0 ) 2

= " ##### (!a ) 2 + b 2 = $ #### a 2 + b 2 . Therefore, PM = QM = RM.

b. Possible answer: The circumcenter of a rt. . is the midpoint of the hyp.

44. Let C be the circumcenter of the .. Given: AC = 28 cm; so by the properties of 30-60-90 5, BC = 1 _

2 AC.

So AB = AC + BC

= 3 _ 2

AC

= 3 _ 2

(28) = 42 cm.


45. t _ 26

= 10 _ 65

65t = 260 t = 4

46. 2.5 _ 1.75

= 6 _ x

2.5x = 10.5 x = 4.2

47. 420 _ y = 7 _ 2

840 = 7y y = 120

48. m,AFB + m,BFE = 180° 55 + m,BFE = 180° m,BFE = 125°

49. m,AFB + m,BFD + m,DFE = 180° 55 + 90 + m,DFE = 180° m,DFE = 35°So m,BFC = m,DFE = 35°

50. m,BFC + m,CFE = m,BFD + m,DFE 35 + m,CFE = 90 + 35 m,CFE = 90°

51. slope of ((

ST = 8 _ !4

= !2; midpoint of ((

ST = M(2, 4)

slope of ((

MX = 4 ! 3 _ 2 ! 0

= ! 1 _ 2 = opposite reciprocal

of 2; so X is on the perpendicular bisector

52. slope of ((

MY = 1 ! 4 _ !4 ! 2

= 1 _ 2 = opposite reciprocal of

2; so Y is on the perpendicular bisector

53. slope of ((

MZ = !8 ! 4 _ !2 ! 2

= !3 6 opposite reciprocal of 2;

so Z is not on the ) bisector

5-3 MEDIANS AND ALTITUDES OF TRIANGLES, PAGES 314–320

CHECK IT OUT! PAGES 315–316 1a. KZ + ZW = KW

2 _ 3 KW + ZW = KW

ZW = 1 _ 3 KW

7 = 1 _ 3 KW

21 = KW

b. LZ = 2 _ 3 LX

= 2 _ 3 (8.1)

= 5.4

2. 3; 4; possible answer: the x-coordinate of the centroid is the average of the x-coordinates of the vertices of the ., and the y-coordinate of the centroid is the average of the y-coordinates of the vertices of the ..

3. Possible answer: An equation of altitude to ((

JK is y = ! 1 _

2 x + 3. It is true that 4 = ! 1 _

2 (!2) + 3, so

(!2, 4) is a solution of this equation. Therefore this altitude passes through the orthocenter.

THINK AND DISCUSS, PAGE 317 1. Possible answer: . is isosc.

2. Possible answer: . is a rt. ..

3. The ratio of the length of the longer segment to the length of the shorter segment is 2 : 1.

4.


1. centroid 2. altitude


3. VW = 2 _ 3 VX

= 2 _ 3 (204) = 136

4. WX = 1 _ 3 VX

= 1 _ 3 (204) = 68

5. RW = 2 _ 3 RY

104 = 2 _ 3 RY

3 _ 2 (104) = RY

RY = 156

6. WY = 1 _ 2 RW

= 1 _ 2 (104) = 52

7. 1 Understand the ProblemAnswer will be the coordinates of the centroid of the .. Important information is the location of vertices, A(0, 2), B(7, 4), and C(5. 0).2 Make a PlanThe centroid of the . is the point of intersection of the three medians. So write the equations for two medians and find their point of intersection.3 SolveLet M be the midpoint of

(( AB and N be the midpoint

of ((

BC .

M = ( 0 + 7 _ 2 , 2 + 4 _

2 ) = (3.5, 3)

N = ( 5 + 7 _ 2 , 0 + 4 _

2 ) = (6, 2)

((

AN is horizontal. Its equation is y = 2. Slope of

((( CM = 3 ! 0 _

3.5 ! 5 = !2. Its equation is

y = !2(x ! 5). At the centroid, y = 2 = !2(x ! 5), so x = 5 + (!1) = 4. The coordinates of the centroid are D(4, 2).4 Look BackLet L be the midpoint of

(( AC . Equation for

(( BL is

y ! 4 = 2 _ 3 (x ! 7), which intersects y = 2 at (4, 2).

8.

Step 1 Graph the ..Step 2 Find an equation of the line containing the

altitude from L to ((

KM . Since ((

KM is horizontal, the altitude is vertical, so the equation is x = 4.Step 3 Find an equation of the line containing the

altitude from K to ((

LM .

Slope of LM = !2 ! 6 _ 8 ! 4

= !2.

Equation is y + 2 = 1 _ 2 (x ! 2).

Step 4 Solve the system to find the coordinates of the orthocenter.x = 4 and y + 2 = 1 _

2 (4 ! 2) = 1, so y = !1.

The coordinates of the orthocenter are (4, !1).

9.


altitude from W to ((

UV . Since ((

UV is vertical, the altitude is horizontal, so the equation is y = !3.Step 3 Find an equation of the line containing the

altitude from U to (((

VW .

Slope of (((

VW = !3 ! 6 _ 5 + 4

= !1.

Equation is y + 9 = x + 4, or y = x ! 5.Step 4 Solve the system to find the coordinates of the orthocenter.y = !3 and !3 = x ! 5, so x = 2. The coordinates of the orthocenter are (2, !3).

10.


altitude from P to ((

QR . ((

QR is horizontal, the altitude is vertical, so the equation is x = !5.Step 3 Find an equation of the line containing the

altitude from Q to ((

PR .

Slope of ((

PR = 5 ! 8 _ !2 + 5

= !1.

Equation is y ! 5 = x ! 4, or y = x + 1.Step 4 Solve the system to find the coordinates of the orthocenter.x = !5 and y = !5 + 1 = !4. The coordinates of the orthocenter are (!5, !4).

11.


altitude from E to ((

CD . ((

CD is vertical, the altitude is horizontal, so the equation is y = 2.Step 3 Find an equation of the line containing the

altitude from C to ((

DE . ((

DE is horizontal, altitude is vertical, so the equation is x = !1.Step 4 y = 2 and x = !1. The coordinates of the orthocenter are (!1, 2).



12. PC = 1 _ 3

HC

= 1 _ 3 (10.8) = 3.6

13. HC = 2 _ 3 Hp

= 2 _ 3 (10.8) = 7.2

14. JA = 3PA= 3(2.9) = 8.7

15. JP = 2PA= 2(2.9) = 5.8

16. Support should be attached at the centroid. Equation of the median through (4, 0) is x = 4. The median through (0, 10) also passes through

( 4 + 8 _ 2 , 0 + 14 _

2 ) = (6, 7), and has slope 10 ! 7 _

0 ! 6 = ! 1 _

2 .

Equation of the second median is y = ! 1 _ 2 x +

10. At intersection, x = 4, so y = ! 1 _ 2 (4) + 10 = 8.

The coordinates of the centroid are (4, 8).

17. Step 1 Find an equation of the line containing the altitude through X.

(( YZ is vertical, the altitude is

horizontal, so the equation is y = !2.Step 2 Find an equation of the line containing the altitude through Z.

Slope of ((

XY is 10 + 2 _ 6 + 2

= 3 _ 2 .

Equation is y + 6 = ! 2 _ 3 (x ! 6).

Step 3 Find the coordinates of the orthocenter. y = !2, so !2 + 6 = 4 = ! 2 _

3 (x ! 6),

or x = !6 + 6 = 0. The coordinates are (0, !2).

18. Step 1 Find an equation of the line containing the

altitude through J. ((

GH is horizontal, the altitude is vertical, so the equation is x = 4.Step 2 Find an equation of the line containing the altitude through H. Slope of

(( GJ is !1 ! 5 _

4 + 2 = !1.

Equation is y ! 5 = x ! 6.Step 3 Find the coordinates of the orthocenter.x = 4, so y ! 5 = 4 ! 6, or y = 5 ! 2 = 3. The coordinates are (4, 3).

19. Step 1 Find an equation of the line containing the altitude through T.

(( RS is horizontal, the altitude is

vertical, so the equation is x = !2.Step 2 Find an equation of the line containing the altitude through R.

(( ST is vertical, the altitude is

horizontal, so the equation is y = 9.Step 3 Find the coordinates of the orthocenter.x = !2 and y = 9.The coordinates are (!2, 9).

20. Step 1 Find an equation of the line containing the

altitude through A. ((

BC is vertical, the altitude is horizontal, so the equation is y = !3.Step 2 Find an equation of the line containing the altitude through C.

Slope of ((

AB is 5 + 3 _ 8 ! 4

= 2.

Equation is y + 8 = ! 1 _ 2 (x ! 8).

Step 3 Find the coordinates of the orthocenter. y = !3, so !3 + 8 = 5 = ! 1 _

2 (x ! 8),

or x = !10 + 8 = !2. The coordinates are (!2, !3).

21. GL = 3 _ 2

GP

= 3 _ 2 (8) = 12

22. PL = 1 _ 2 GP

= 1 _ 2 (8) = 4

23. HL = LJ = 5

24. ((

GL is the perpendicular bisector of ((

HJ , so ((

GJ - ((

GH .GJ = GH

= 2GK= 2(6.5) = 13

25. P = GJ + GH + HJ= 2GH + 2LJ= 2(13) + 2(5) = 36 units

26. A = 1 _ 2 (HJ)(GL)

= 1 _ 2 (10)(12) = 60 square units

27. G = ( 1 _ 3

(0 + 14 + 16), 1 _ 3 (!4 + 6 ! 8)) = (10, !2)

28. G = ( 1 _ 3

(8 + 2 + 5), 1 _ 3 (!1 + 7 ! 3)) = (5, 1)

29. PZ = 2ZX= 2(27) = 54

30. PX = 3ZX= 3(27) = 81

31. Step 1 Find n.2n + 17 = 54 2n = 54 ! 17 n = 18.5Step 2 Find QZ.QZ = 4n ! 26

= 4(18.5) ! 26 = 48

32. YZ = 1 _ 2 (QZ)

= 1 _ 2 (48) = 24

33. Possible answer: the perpendicular bisector of base; the bisector of vertex ,; the median to the base; the altitude to the base


34. sometimes 35. always

36. never 37. always


1. ((

PS and ((

RT are medians of .PQR.

(( PS - ((

RT 1. Given

2. PS = RT 2. Def. of - segs.

3. 2 _ 3

PS = 2 _ 3 RT 3. Mult. Prop. of =

4. PZ = 2 _ 3 PS, RZ = 2 _

3 RT 4. Centroid Thm.

5. PZ = RZ 5. Subst.

6. ((

PZ - ((

RZ 6. Def. of - segs.

7. ,SPR - ,TRP 7. Isosc. . Thm.

8. ((

PR - ((

PR 8. Reflex. Prop. of -

9. .PTR - .RSP 9. SAS

10. ,QPR - ,QRP 10. CPCTC

11. ((

PQ - ((

RQ 11. Con. of Isosc. . Theorem

12. .PQR is an isosc .. 12. Def. of isosc. .

39. Possible answer: The centroid of a . is also called its center of gravity because the weight of the . shape is evenly distributed in every direction from this point. This means the . shape will rest in a horizontal position when supported at this point.

40a. G = ( 1 _ 3 (0 + 0 + 8), 1 _

3 (0 + 8 + 0)) = (2 2 _

3 , 2 2 _

3 )

b. DG = $ #####

( 8 _ 3 ) 2 + ( 8 _

3 ) 2

= 8 _ 3 " # 2 0 3.8 mi

c. Perpendicular from G crosses ((

EF at H(4, 4),

distance = $ #####

( 4 _ 3 ) 2 + ( 4 _

3 ) 2

= 4 _ 3 " # 2 0 1.9 mi

TEST PREP, PAGE 319

41. D 42. G;I, III true since incenter, centroid always inside .II false since . obtuse

43. D


44a. Possible answer: .ABC is equil., and ! is the perpendicular bisector of

((( BC. Since .ABC is

equil., ((

AB - ((

AC by definition. So AB = AC by the definition of - segs. Therefore by the Converse of the Perpendicular Bisector Theorem, A is on line !. Similarly, B is on the perpendicular bisector of

(( AC , and C is on the perpendicular

bisector of ((

AB .

b. Possible answer: By the definition of the perpendicular bisector,

(( BD - ((

CD . So D is the midpoint of

(( BC by definition, and

(( AD is a median

of .ABC by the definition of median. Therefore ! contains the median of .ABC through A. Also by the definition of the perpendicular bisector,

(( AD

) ((

BC . So ((

AD is the altitude of .ABC by the definition. Therefore ! contains the altitude of .ABC through A. Again by the definition of the perpendicular bisector,

(( BD - ((

CD . ((

AB - ((

AC by the definition of equil. ., and

(( AD - ((

AD by the Reflex. Prop. of -. So .ABD - . ACD by SSS. Then ,DAB - ,DAC by CPCTC, and

(( AD is the bisector

of ,BAC by the definition of , bisector. Therefore ! contains the , bisector of .ABC through A. The same reason can be applied to the other two ) bisectors.

c. Possible answer: The perpendicular bisectors of a . are concurrent at the circumcenter, and the , bisectors are concurrent at the incenter. The medians of a . are concurrent at the centroid, and the altitudes of a . are concurrent at the orthocenter. But in an equil. ., the perpendicular bisector through a given vertex also contains the , bisector, the median, and the altitude through that vertex. So the points of concurrency must all be the same point That is, the circumcenter, the incenter, the centroid, and the orthocenter in an equil. . are the same point.

45a. slope of RS = c _ b ; slope of ST = c _

b ! a ;

slope of RT = 0.

b. Since ! ) ((

RS , slope of ! = ! b _ c . Since m ) ((

ST ,

slope of m = ! b ! a _ c = a ! b _ c . Since n ) ((

RT , n is

a vertical line, and its slope is undefined.


c. An equation of ! is

y ! 0 = ! b _ c (x ! a)

y = ! b _ c x + ab _ c

An equation of m is

y ! 0 = a ! b _ c (x ! 0)

y = a ! b _ c x

An equation of n is x = b.

d. (b, ab ! b 2 _ c ) e. Since the equation of line n is x = b and the x-coordinate of P is b, P lies on n.

f. Lines !, m, and n are concurrent at P.


46. Let x, y be prices of peanuts and popcorn.x = y + 0.75 or y = x ! 0.75 5x + 3y = 21.755x + 3(x ! 0.75) = 21.75 5x + 3x ! 2.25 = 21.75 8x = 24 x = 3Price of peanuts is $3.00.

47. F; Possible answer: a rectangle with width 5 and length 8.

48. T 49. KL = 2KP= 2(7.0) = 14.0

50. QJ = QL = 9.1

51. m, JLQ = m,LJQ = 36°m,JQL + m,LJQ + m, JLQ = 180° m,JQL + 36 + 36 = 180° m,JQL = 108°

CONSTRUCTION, PAGE 320


2. Possible answer: The orthocenter of an acute . is inside the .. The orthocenter of an obtuse . is outside the .. The orthocenter of a rt. . is the vertex of the rt. ,.

SPECIAL POINTS IN TRIANGLES, PAGE 321

TRY THIS, PAGE 321 1. the circumcenter, the orthocenter, and the centroid

2. The centroid; possible answer: distance from the orthocenter to the centroid is twice the distance from the centroid to the circumcenter; that is, CO = 2CU.

3. isosc. . 4. equil. .

5-4 THE TRIANGLE MIDSEGMENT THEOREM, PAGES 322–327

CHECK IT OUT! PAGES 322–323 1. The midpoints are M(1, 1), N(3, 4);

slope of (((

MN = 3 _ 2 ; slope of

(( RS = 6 _

4 = 3 _

2 ;

since the slopes are =, (((

MN 7 ((

RS .

MN = $ #### 2 2 + 3 2 = " ## 13 ;

RS = $ #### 4 2 + 6 2 = " ## 52 = 2 " ## 13 ;

and MN = 1 _ 2 RS.

2a. JL = 2PN= 2(36) = 72

b. PM = 1 _ 2 KL

= 1 _ 2 (97) = 48.5

c. m,MLK = m,JMP = 102°

3. HF = 1 _ 2 AE

= 1 _ 2 (1550) = 775

The distance she measures between H and F is 775 m.

THINK AND DISCUSS, PAGE 324 1. The endpoints of

(( XY are not the midpoints of

the sides of the ..

2.


1. midpoints

2. The midpoints are S(!1, 4), T(4, 6);slope of

(( ST = 2 _

5 ; slope of

(( PR = 4 _

10 = 2 _

5 ;

since the slopes are =, ((

ST 7 ((

PR .

ST = $ #### 2 2 + 5 2 = " ## 29 ;

PR = $ #### 4 2 + 10 2 = " ## 116 = 2 " ## 29 ;and ST = 1 _

2 PR.

3. NM = 1 _ 2 XY

= 1 _ 2 (10.2) = 5.1

4. XZ = 2LM= 2(5.6) = 11.2

5. NZ = 1 _ 2 XZ

= 1 _ 2 (11.2) = 5.6

6. m,LMN = m,MNZ = 29°


7. m,YXZ = m,MNZ = 29°

8. m,XLM + m,LMN = 180 m,XLM + 29 = 180 m,XLM = 151°

9. CD < 1 _ 2 XZ = 15 ft = 5 yd

The width of the 2nd floor is less than 5 yd.


10. The midpoints are D(!4, 3), E(0, 4);slope of

(( DE = 1 _

4 ; slope of

(( CB = 2 _

8 = 1 _

4 ;

since the slopes are =, ((

DE 7 ((

CB .

DE = $ #### 1 2 + 4 2 = " ## 17 ;

CB = $ #### 2 2 + 8 2 = " ## 68 = 2 " ## 17 ;and DE = 1 _

2 CB.

11. GJ = 2PQ= 2(19) = 38

12. RQ = 1 _ 2 GH

= 1 _ 2 (27) = 13.5

13. RJ = 1 _ 2 GJ

= 1 _ 2 (38) = 19

14. m,PQR = m,QRJ= 55°

15. m,HGJ = m,QRJ= 55°

16. m,GPQ + m,HGJ = 180° m,GPQ + 55 = 180° m,GPQ = 125°

17. Yes; ((

DE is a midsegment of .ABC, so its length is half of 4 1 _

2 ft, or 2 1 _

4 ft, which is 27 in. This is less than

30 in. So the carpenter can use the 30 in. timber to make the crossbar.

18. P = GH + HJ + GJ= 12 + 2LJ + 2KL= 12 + 2(4) + 2(7) = 34

19. P = KL + LM + KM= 7 + 1 _

2 GH + 1 _

2 HJ

= 7 + 1 _ 2 (12) + 1 _

2 (8)

= 17

20. The perimeter of . GHJ is twice the perimeter of .KLM.

21. 3n = 2(54)3n = 108 n = 36

22. 2(n ! 9) = 352n ! 18 = 35 2n = 53 n = 26.5

23. 2(4n + 5) = 74 8n + 10 = 74 8n = 64 n = 8

24. 2n ! 23 = 2(9.5)2n ! 23 = 19 2n = 42 n = 21

25. 6n = 2(n + 8)6n = 2n + 164n = 16 n = 4

26. 2(5n) = 8n + 10 10n = 8n + 10 2n = 10 n = 5

27. B; possible answer: in .ABC, ((

DE is a midsegment and

(( BC is the side 7 to it. By the . Midsegment

Theorem, the length of a midsegment is half the length of the 7 side, so DE = 1 _

2 BC.

28. ,D - ,FZY - ,YXE - ,ZYX;,E - ,ZYF - ,DXZ - ,XZY;,F - ,XYE - ,DZX - ,ZXY

29. Possible answer: about 18 parking spaces; the new street is along the midsegment of the triangle plot of land. The length of the street is half of 440 ft, or 220 ft. Estimate the quotient 220 ÷ 23 by rounding 220 to 225 and 23 to 25. Since 225 ÷ 25 = 9, city can put about 9 parking spaces on one side of the street. So the total number of parking spaces is about 2(9), or 18.

30. CG = 1 _ 2 AB

= 1 _ 2 (33) = 16.5

31. EH = 1 _ 2 DC

= 1 _ 2 CB

= 1 _ 2 (22) = 11

32. FJ = 1 _ 2 GH

= 1 _ 4 CG

= 1 _ 4 (16.5) = 4.125

33. m,DCG = m,CBA = 57°

34. m,GHE = m,HCD = m,ABC = 57°

35. m,FJG + m,GHE = 180 m,FJG + 57 = 180 m,FJG = 123°

36. Yes; possible answer: let x be the length of each - side of an isosc. .. By the . Midsegment Theorem, the length of the midsegment 7 to each

of those sides is 1 _ 2 x. Since these two midsegments

are equal in length, they are -.

37a. WX = 1 _ 2 XY

= 1 _ 4 AB

= 1 _ 4 (9) = 2.25 mi

b. XA = CX = 3.5 mi, BC = 2BY = 8 mitrip length = WX + XA + AB + BC + CX + XW

= 2.25 + 3.5 + 9 + 8 + 3.5 + 2.25 = 28.5 mi

38a. M = ( 0 + 2a _ 2 , 0 + 2b _

2 ) = (a, b)

b. N = ( 2a + 2c _ 2 , 2b + 0 _

2 ) = (a + c, b)

c. slope of ((

PR = 0 ! 0 _ 2c ! 0

= 0

slope of (((

MN = b ! b _ a + c ! a = 0

Slopes of (((

MN and ((

PR are =, so MN 7 PR.


d. PR = 2c; MN = a + c ! a = c; the length of ((

PR is

twice length of (((

MN , so MN = 1 _ 2 PR.

TEST PREP, PAGE 326

39. D; RT = 2PQ4x ! 27 = 2(x + 9)4x ! 27 = 2x + 18 2x = 45 x = 22.5RT = 4(22.5) ! 27

= 63 m

40. H

41. D

CHALLENGE AND EXTEND, PAGES 326–327

42. Let the coordinates of the vertices be (u, v), (w, x), (y, z). By the Midpoint Formula,u + w = 2(!6) = !12w + y = 2(2) = 4 u + y = 2(0) = 0 2w = !12 + 4 ! 0 2w = !8 w = !4

v + x = 2(3) = 6x + z = 2(1) = 2v + z = 2(!3) = !6 2x = 6 + 2 ! (!6) 2x = 14 x = 7

So (w, x) = (!4, 7).u + (!4) = !12 u = !8

v + 7 = 6 v = !1

So (u, v) = (!8, !1).!4 + y = 4

y = 87 + z = 2

z = !5So (y, z) = (8, !5).

43. The midsegment . is equilateral and equiangular

44. n 2 ! 3 = 2(39) n 2 = 81 n = ±9

45. n 2 ! 6n + 3 = 2(3n ! 16) n 2 ! 6n + 3 = 6n ! 32 n 2 ! 12n + 35 = 0 (n ! 7)(n ! 5) = 0 n = 7 or 53(7) ! 16 = 5 > 03(5) ! 16 = !1 < 0So n = 7 is the only possible solution.

46. .QXY - .XPZ - .YZR - .ZYX; area of .XYZ = 1 _

4 (area of .PQR)

47a. Number of Midsegment

1 2 3 4

Length ofMidsegment

32 16 8 4

b. length of midsegment 8 = (length of midsegment 4) ( 1 _

2 ) 4

= 4 ( 1 _ 16

) = 1 _ 4

c. length of midsegment

n = AB ( 1 _ 2

) n = 64 ( 1 _

2 )

n = 2 6 ! n


48. concentration = 2% + 3% _ 2

= 2.5%

49. concentration = 3(2%) + 1(3%)

__ 3 + 1

= 2.25%

50. G(!3, !2) ' G8(!3, 2) = G8(!3 + 0, !2 + 4) H(0, 0) ' H8(0 + 0, 0 + 4) = H8(0, 4) J(4, 1) ' J8(4 + 0, 1 + 4) = J8(4, 5) K(1, !2) ' K8(1 + 0, !2 + 4) = K8(1, 2)

51. G(!3, !2) ' G8(1, !4) = G8[!3 + 4, !2 + (!2)] H(0, 0) ' H8[0 + 4, 0 + (!2)] = H8(4, !2) J(4, 1) ' J8[4 + 4, 1 + (!2)] = J8(8, !1) K(1, !2) ' K8[1 + 4, !2 + (!2)] = K8(5, !4)

52. G(!3, !2) ' G8(3, 0) = G8(!3 + 6, !2 + 2) H(0, 0) ' H8(0 + 6, 0 + 2) = H8(6, 2) J(4, 1) ' J8(4 + 6, 1 + 2) = J8(10, 3) K(1, !2) ' K8(1 + 6, !2 + 2) = K8(7, 0)

53. NX = 2XS= 2(3) = 6

54. MR = 3 _ 2 MX

= 3 _ 2 (5.5) = 8.25

55. NP = 2NR= 2(4.5) = 9


1. XY = 1 _ 2

AC

2. Possible answer: Find m,BXY and m,BAC and confirm that they are =. This means the two segments are 7 by the Converse of the Corr. / Post.


1. G = ( 1 _ 3 (0 + 30 + 15), 1 _

3 (0 + 0 + 32)) = (15, 10 2 _

3 )

2. GA = GB = $ ######

15 2 + (10 2 _ 3 ) 2 0 18.4

GC = 32 ! 10 2 _ 3 0 21.3

about 21.3 mi from Carson; about 18.4 mi from Benton and Ashville

3. The equation of the altitude to ((

AB is x = 15;slope of

(( AC = 32 _

15 ' slope of the altitude to

(( AC =

! 15 _ 32

; 2 points on the altitude are the orthocenter

O (15, y) and B(30, 0); therefore

y ! 0

_ 15 ! 30

= ! y _

15 = ! 15 _

32 ,

so y = 225 _ 32

= 7 1 _ 32

, and O = (15, 7 1 _ 32

) . 4. OA = OB = $

###### 15 2 + (7 1 _

32 ) 2 0 16.6

OC = 32 ! 7 1 _ 32

0 25.0

about 25.0 mi from Carson; about 16.6 mi from Benton and Ashville


5. The equation of the perpendicular bisector of ((

AB is x = 15, so let the circumcenter be D(15, y).The slope of the perpendicular bisector of

(( AC =

! 15 _ 32

, and the midpoint of ((

AC is M (7 1 _ 2 , 16) , so

the slope of (((

MC = y ! 16

_ 15 ! 15 _

2

= 2y ! 32

_ 15

=

! 15 _ 32

; therefore y = 16 ! 225 _ 64

= 799 _ 64

= 12 31 _ 64

, and

D = (15, 12 31 _ 64

) . 6. DA = DB = DC = 32 ! 12 31 _

64 0 19.5;

about 19.5 mi from all 3 cities

7. They should choose the circumcenter.


1. PQ = 2PR= 2(4.8) = 9.6

2. JM = ML = 58 3. AB = AC5z + 16 = 8z ! 5 21 = 3z 7 = zAC = 8(7) ! 5 = 51

4. Slope of (((

MN = 4 _ 8 = 1 _

2 , so the slope of the

perpendicular bisector = !2;

equation is y + 1 = !2(x ! 3).

5. PS = PT = 83.9XT = RX = 46.7

6. m,GJK + m,KJH + m,JHK + m,KHL + m,LGJ = 180 2m,GJK + 2(16) + 50 = 180 2m,GJK = 98 m,GJK = 49°The distance from K to

(( HJ = KL = 21.

7. The equations of the two perpendicular bisectors are x = 4.5 and y = !2. So C = (4.5, !2).

8. BW = 1 _ 3 BD

= 1 _ 3 (87) = 29

CE = 3CW = 3(19) = 57

CW = 1 _ 2 WE

= 1 _ 2 (38) = 19

9. G = ( 1 _ 3 (0 + 8 + 10), 1 _

3 (4 + 0 + 8)) = (6, 4)

10. ((

PS is horizontal, the altitude is vertical, so the equation is x = 4; the slope of

(( SV = 4 _

4 = 1, so the

slope of the altitude to it is !1; the equation of this altitude is y ! 4 = !(x ! 2); at he orthocenter O, x = 4, so y = 4 ! (4 ! 2) = 2, and O = (4, 2).

11. ZV = 1 _ 2 JM

= RM = 45

PM = 2ZR= 2(53) = 106

m,RZV = m,PVZ = 36°

12. XY = 2MN= 2(39) = 78

The distance across the pond is 78 m.

CONNECTING GEOMETRY TO ALGEBRASOLVING COMPOUND INEQUALITIES, PAGE 330

TRY THIS, PAGE 330 1. !4 + x > 1 OR !8 + 2x < !6

x > 5 OR x < 12 rays

2. 2x ! 3 & !5 OR x ! 4 > !1 x & !1 OR x > 3 x & !1ray

3. !6 < 7 ! x % 12!13 < !x % 5 !5 % x < 13segment

4. 22 < !2 ! 2x % 54 24 < !2x % 56!28 % x < !12segment

5. 3x & 0 OR x + 5 < 7 x & 0 OR x < 2all real numbersline

6. 2x ! 3 % 5 OR !2x + 3 % !9 x % 4 OR x & 62 rays

GEOMETRY LABEXPLORE TRIANGLE INEQUALITIES, PAGE 331

ACTIVITY 1, TRY THIS, PAGE 331 1. Check students’ work.

2. Possible answer: The longest side is opposite the largest ,, and the shortest side is opposite the smallest ,.


ACTIVITY 2, TRY THIS, PAGE 331 4. only set with lengths 3 in., 4 in., and 6 in.

5. Possible answer: If the sum of any 2 lengths is > the 3rd length, then the set of stems can form a .. If the sum of any 2 lengths is % the 3rd length, then the set of stems cannot form a ..



5-5 INDIRECT PROOF AND INEQUALITIES IN ONE TRIANGLE, PAGES 332–339

CHECK IT OUT! PAGES 332–335 1. Possible answer:

Given: .RSTProve: .RST cannot have 2 rt /.Proof: Assume that .RST has 2 rt /. Let ,R and ,S be the rt. /. By the def. of rt. ,, m,R = 90° and m,S = 90°. By the . Sum Thm., m,R + m,S + m,T = 180°. But then 90 + 90 + m,T = 180° by subst., so m,T = 0°. However, a . cannot have an , with a measure of 0°. So there is no .RST, which contradicts the given information. This means the assumption is false, and .RST cannot have 2 rt. /.

2a. The shortest side is ((

AC , so the smallest , is ,B.The longest side is

(( AB , so the greatest , is ,C.

/ from smallest to greatest are ,B, ,A, ,C.

b. m,F = 90°, m,E = 180 ! (22 + 90) = 68°The smallest , is ,D, so the shortest side is

(( EF .

The greatest , is ,F, so the longest side is ((

DE .Sides from shortest to longest are

(( EF , ((

DF , ((

DE .

3a. 8 + 13 9 21 21 : 21No; 8 + 3 = 21, which is not greater than the third side length.

b. 6.2 + 7 9 9 13.2 > 9 ;

6.2 + 9 9 7 15.2 > 7 ;

7 + 9 9 6.2 16 > 6.2 ;

Yes; the sum of each pair of the lengths is greater than the third length.

c. When t = 4, t ! 2 = 2, 4t = 16, t 2 + 1 = 17.2 + 16 9 17 18 > 17 ;

2 + 17 9 16 19 > 16 ;

16 + 17 9 2 33 > 2 ;

Yes; the sum of each pair of the lengths is greater than the third length.

4. Let s be the length of the 3rd side. Apply the . Inequal. Theorem.s + 17 > 22 s > 5

s + 22 > 17 s > !5

17 + 22 > s 39 > s

Combine the inequals. So 5 < s < 39. the length of the 3rd side is > 5 in. and < 39 in.

5. Let d be the distance from Seguin to Johnson City.d + 22 > 50 d > 28

d + 50 > 22 d > !28

22 + 50 > d 72 > d

28 < d < 72The distance from Seguin to Johnson City is > 28 mi and < 72 mi.


1. No; possible answer: the student must consider 2 cases and assume that either the , is acute or the , is rt.

2. Possible answers: 2 cm, 4 cm, 5 cm; 2 cm, 4 cm, 8 cm

3.


1. Possible answer: To prove something indirectly, you assume the opposite of what you are trying to prove. Then you use logic to lead to a contradiction of given information, a definition, a postulate, or a previously proven theorem. You can then conclude that the assumption was false and the original statement is true.

2. Possible answer:Given: .ABC is a scalene triangle.Prove: .ABC cannot have 2 - /.Proof: Assume that .ABC does have 2 - /. Let ,A and ,C be the - /. Then

(( AB - ((

CB by the Converse of the Isosc. . Theorem. However, a scalene . by definition has no - sides. So .ABC is not scalene, which contradicts the given information. This means the assumption is false, and therefore .ABC can not have 2 - /.

3. Possible answer:Given: .PQR is an isosc. . with base

(( PR .

Prove: .PQR cannot have a base , that is a rt. ,.Proof: Assume that .PQR has a base , that is a rt. ,. Let ,P be the rt. ,. By the Isosc. . Theorem, ,P - ,R, so ,R is also a rt. ,. By the definition of rt. /, m,P = m,R = 90°. By the . Sum Theorem, m,P + m,Q + m,R = 180°. By Subst. m,Q = 0°. However, a . cannot have an , with a measure of 0°. So there is no .PQR, which contradicts the given information. This means the assumption is false, and therefore .PQR can not have a base , that is rt.

4. The shortest side is ((

PQ , so the smallest , is ,R.The longest side is

(( PR , so the greatest , is ,Q.

/ from smallest to greatest are ,R, ,P, ,Q.

5. m,Z = 180 ! (39 + 46) = 95°The smallest , is ,X, so the shortest side is

(( YZ .

The greatest , is ,Z, so the longest side is ((

XY .Sides from shortest to longest are

(( YZ , ((

XZ , ((

XY .


6. 4 + 7 9 10 11 > 10 ;

4 + 10 9 7 14 > 7 ;

7 + 10 9 4 17 > 4 ;

Yes; the sum of each pair of 2 lengths is greater than the third length.

7. 2 + 9 9 12 11 : 12No; 2 + 9 = 11, which is not greater than the third side length.

8. 3.5 + 3.5 9 6 7 > 6 ;

3.5 + 6 9 3.5 9.5 > 3.5 ;

Yes; the sum of each pair of 2 lengths is greater than the third length.

9. 1.1 + 1.7 9 3 2.8 : 3No; 1.1 + 1.7 = 2.8, which is not greater than the third side length.

10. When x = 5, 3x = 15, 2x ! 1 = 9, x 2 = 25.9 + 15 9 25 24 : 25No; when x = 5, the value of 3x is 15, the value of 2x ! 1 is 9 and the value of x 2 is 25. 15 + 9 = 24, which is not greater than the third side length.

11. When c = 2, 7c + 6 = 20, 10c ! 7 = 13, 3 c 2 = 12.12 + 13 9 20 25 > 20 ;

12 + 20 9 13 32 > 13 ;

13 + 20 9 12 33 > 12 ;

Yes; when c = 2, the value of 7c + 6 is 20, the value of 10c ! 7 is 13, and the value of 3 c 2 is 12. The sum of each pair of 2 lengths is greater than the third length.


s + 12 > 8 s > !4

8 + 12 > s 20 > s

Combine the inequals. So 4 < s < 20. The length of the 3rd side is > 4 mm and < 20 mm.


16 + 16 > s 32 > s

Combine the inequals. So 0 < s < 32. The length of the 3rd side is > 0 ft and < 32 ft.

14. Let s be the length of the 3rd side. Apply the . Inequal. Theorem.s + 11.4 > 12 s > 0.6

s + 12 > 11.4 s > !0.6

11.4 + 12 > s 23.4 > s

Combine the inequals. So 0.6 < s < 23.4. the length of the 3rd side is > 0.6 cm and < 23.4 cm.

15a. The longest side is opposite the greatest ,. So the longest side is between the refrigerator and the stove.

b. No; 4 + 5 = 9 : 9. By the . Inequal. Theorem, a . cannot have these side lengths.


16. Possible answer:Given: .ABC is scalene.

(( XZ and

(( YZ are

midsegments of .ABC.Prove: .ABC cannot have 2 - midsegments.Proof: Assume that .ABC does have 2- midsegments. Let

(( XZ and

(( YZ be the -

midsegments. By the def. of - segs., XZ = YZ. By

the . Mid segment Thm., XZ = 1 __ 2 BC and

YZ = 1 __ 2 BA. So 1 __

2 BC = 1 __

2 BA by subst. But then

BC = BA, and by the def. of - segs., ((

BC - ((

BA .However, a scalene . by def. has no - sides. So.ABC is not scalene, which contradicts the given information. This means the assumption is false, and therefore a scalene . cannot have 2 - midsegments.

17. Possible answer:Given: ,J and ,K are supp.Prove: ,J and ,K cannot both be obtuse.Proof: Assume that ,J and ,K are both obtuse. Then m,J > 90° and m,K > 90° by the definition of obtuse. Add the 2 inequals., m,J + m,K > 180°. However, by the definition of supp. /, m,J + m,K = 180°. So m,J + m,K > 180° contradicts the given information. This means the assumption is false, and therefore a pair of supp. / cannot both be obtuse.

18. The shortest side is ((

KL , so the smallest , is ,J.The longest side is

(( JL , so the greatest , is ,K.

/ from smallest to greatest are ,J, ,L, ,K.

19. m,S = 90, m,T = 90 ! m,R = 24°The smallest , is ,T, so the shortest side is

(( RS .

The greatest , is ,S, so the longest side is ((

RT .Sides from shortest to longest are

(( RS , ((

ST , ((

RT .

20. 6 + 10 9 15 16 > 15 ;Yes; the sum of each pair of 2 lengths is greater than the third length.

21. 14 + 18 9 32 32 : 32No; 14 + 18 = 32, which is not greater than the third side length.

22. 5.8 + 5.8 9 11.9 11.6 : 11.9No; 5.8 + 5.8 = 11.6, which is not greater than the third side length.

23. 41.9 + 62.5 9 103 104.4 > 103 ;Yes; the sum of each pair of 2 lengths is greater than the third.


24. When z = 6, z + 8 = 14, 3z + 5 = 23, 4z ! 11 = 13.13 + 14 9 23 27 > 23 ;Yes; when z = 6, the value of z + 8 is 14, the value of 3z + 5 is 23, and the value of 4z ! 11 is 13. The sum of each pair of 2 lengths is greater than the third side length.

25. When m = 3, m + 11 = 14, 8m = 24, m 2 + 1 = 10.10 + 14 9 24 24 : 24No; when m = 3, the value of m + 11 = 14, the value of 8m is 24, and the valu e of m 2 + 1 is 10. the sum of 14 and 10 is 24, which is not greater than the third side length.

26. b + c > ss + 4 > 19 s > 15

4 + 19 > s 23 > s

15 yd < s < 23 yd

27. s + 23 > 28 s > 55 km < s < 51 km

23 + 28 > s51 > s

28. s + 3.8 > 9.2 s > 5.45.4 cm < s < 13.0 cm

3.8 + 9.2 > s 13.0 > s

29. s + 1.89 > 3.07 s > 1.181.18 m < s < 4.96 m

1.89 + 3.07 > s4.96 > s

30. s + 2 1 _ 8 > 3 5 _

8

s > 1 1 _ 2

1 1 _ 2

in. < s < 5 3 _ 4 in.

2 1 _ 8 + 3 5 _

8 > s

5 3 _ 4 > s

31. s + 3 5 _ 6 > 6 1 _

2

s > 2 2 _ 3

2 2 _ 3

ft < s < 10 1 _ 3

ft

3 5 _ 6 + 6 1 _

2 > s

10 1 _ 3 > s

32. ((

AD , ((

BD , ((

AB , ((

BC , ((

CD ; possible answer: in .ABD, m,ABD = 50°. In .BCD, m,DBC = 74°. In .ABD, the order of the tubes from shortest to longest is

(( AD , ((

BD , ((

AB . In .BCD, the order of the tubes from shortest to longest is

(( BD , ((

BC , ((

CD . So AD < BD < AB, and BD < BC < CD. Since AB = 50.8 and BC = 54.1, it is also true that AB < BC. So ((

AD < ((

BD < ((

AB < ((

BC < ((

CD .

33. a > 7.5, where a is the length of a leg. Possible answer: By the . Inequal. Thm., a + a > 15 and a + 15 > a. The solution of the first inequality is

a > 7.5. The second inequality simplifies to 15 > 0, which is always a true statement.

34. Step 1 Find x.2x + 5x ! 1 = 90 7x = 91 x = 13Step 2 Find , measures and order sides.m,A = 90°, m,B = 5(13) ! 1 = 64°, m,C = 2(13) = 26°m,C < m,B < m,A, so order is

(( AB , ((

AC , ((

BC .

35. Step 1 Find x.4.5x ! 5 + 10x ! 2 + 5x ! 8 = 180

19.5x = 195 x = 10

Step 2 Find , measures and order sides.m,D = 4.5(10) ! 5 = 40°, m,E = 10(10) ! 2 = 98°, m,F = 5(10) ! 8 = 42°m,D < m,F < m,E, so order is

(( EF , ((

DE , ((

DF .

36. A rt. , cannot be an acute ,. So the 1st and the 3rd statements contradict each other.

37. An obtuse , measures > 90°. So the 2nd and the 3rd statements contradict each other.

38. If 1st statement is true, JK = LK. So the 1st and the 3rd statements contradict each other.

39. 2 line segs. cannot be both ) and 7. So the 1st and the 3rd statements contradict each other.

40. A figure cannot be both a . and a quad. So the 2nd and the 3rd statements contradict each other.

41. 4 is not a prime number, so no multiple of 4 is prime. So the 2nd and the 3rd statements contradict each other.

42. m,P > m,PQS, so QS > PS.

43. m,PSQ = 180 ! (54 + 75) = 51°m,PSQ < m,P, so PQ < QS.

44. m,R < m,RSQ, so QS < QR.

45. m,RQS = 180 ! (51 + 78) = 51° = m,RBy Converse of Isosc. . Theorem, QS = RS.

46. PQ < QS and QS = RS, so PQ < RS.

47. RS = QS and QS > PS, so RS > PS.

48. AE > BA, so m,ABE > m,BEA.

49. CE > BC, so m,CBE > m,CEB.

50. CD = DE, so by Isosc. . Theorem, m,DCE = m,DEC.

51. DE < CE, so m,DCE < m,CDE.

52. AE < BE, so m,ABE < m,EAB.

53. BE = CE, so by Isosc. . Theorem, m,EBC = m,ECB.

54. JK = $ #### 6 2 + 8 2 = 10; KL = $ #### 5 2 + 8 2 0 9.4; JL = 1!3 ! 82 = 11KL < JK < JL, so order is ,J, ,L, ,K.

55. JK = 1!10 ! 22 = 12; KL = $ #### 12 2 + 7 2 0 13.9;

JL = $ #### 12 2 + 5 2 = 13JK < JL < KL, so order is ,L, ,K, ,J.

56. JK = $ #### 1 2 + 7 2 0 7.1; KL = $ #### 6 2 + 4 2 0 7.2;

JL = $ #### 7 2 + 3 2 0 7.6JK < KL < JL, so order is ,L, ,J, ,K.

57. JK = $ #### 10 2 + 7 2 0 12.2; KL = $ #### 2 2 + 11 2 0 11.2;

JL = $ #### 12 2 + 4 2 0 12.6KL < JK < JL, so order is ,J, ,L, ,K.


58. Possible answer: Assume that the client committed the burglary. A person who commits a burglary must be present at the scene when the crime is committed. However, a witness saw the client in a different city at the time that the crime was committed. This means that the assumption that the client committed the burglary is false. Therefore the client did not commit the burglary.

59a. AR + 400 > 600 AR > 200 mi

400 + 600 > AR 1000 mi > AR

Time to travel 200 mi is 200 _ 500

= 0.4 h. Time to travel

1000 mi is 1000 _ 500

= 2 h. So the range of time is

0.4 h < t < 2 h.

b. No; AR < 1000, so by the . Inequal. Theorem, AM must be less than 1800.

60. Step 1 Write and solve 2 inequals. for n.n + 6 > 8 n > 2

6 + 8 > n 14 > n

Step 2 Combine the inequals.2 < n < 14

61. Step 1 Write and solve 2 inequals. for n.2n + 5 > 7 2n > 2 n > 1

5 + 7 > 2n 12 > 2n 6 > n


62. Step 1 Write and solve 2 inequals. for n.n + 1 + 3 > 6 n > 2

3 + 6 > n + 1 8 > n

Step 2 Combine the inequal.s2 < n < 8

63. Step 1 n + 1 < n + 2 < n + 3, so only need to check 1 inequal. for n.n + 1 + n + 2 > n + 3 2n + 3 > n + 3 n > 0

64. Step 1 Write and solve 2 inequals. for n. Use the fact that n + 2 < n + 3.3n ! 2 + n + 2 > n + 3 4n > n + 3 3n > 3 n > 1

n + 2 + n + 3 > 3n ! 2 2n + 5 > 3n ! 2 7 > n


65. Step 1 Write and solve 2 inequals. for n. Use fact that n < n + 2.2n + 1 + n > n + 2 3n + 1 > n + 2 2n > 1 n > 0.5

n + n + 2 > 2n + 1 2 > 1 always true

66. Possible answer:Given: P is in the int. of .XYZ.Prove: XY + XP + PZ > YZ.Proof: By the . Inequal. Theorem, PY + PZ > YZ and XY + XP > YP. Since PZ > 0, the second inequal. is equivalent to XY + XP + PZ > YP + PZ. But then YP + PZ > YZ, so XY + XP + PZ > YZ by the Trans. Prop of Inequal.

67a. definition of - segs.

c. definition of - /

e. subst.

g. trans. Prop. of Inequal.

b. Isosc. . Theorem

d. m,1 + m,3

f. m,S

68a. .ABC

c. Isosc. . Theorem

e. m,3

g. in ., longer side is opp. larger ,

i. AC + BC > AB

b. AD

d. definition of - /

f. subst.

h. subst.

69. Possible answer: A rt. . has a rt. , and 2 acute /. By definition, the rt. , has the greatest measure. Since the hyp. is the side opposite the rt. ,, the hyp. is the longest side by Thm 5-5-2. Similarly, the diagonal of a square forms 2 rt. 5, with the diagonal being the hyp. of each. Since the diagonal is longer than the leg lengths in both 5, the diagonal is longer than the side length of the square.

TEST PREP, PAGE 339

70. A;3 + 3 = 6 > 5 ;

71. H;GH + HJ < GJ contradicts the . Inequal. Theorem

72. C;,S must be the largest ,, so

(( RT is the longest side.


73. The total number of choices is ( 5 3 ) = 10. The

choices that form a .:1 + 3 = 4 : 5, 7, or 9 <1 + 5 = 6 : 7 or 9 <1 + 7 = 9 : 9 <

3 + 5 = 8 > 7 ;3 + 5 = 8 : 9 <3 + 7 = 10 > 9 ;5 + 7 = 12 > 9 ;

. is possible for 3 choices. So prob. = 3 _ 10

or 30%.

74a. " # 2 is rational

c. 2 q 2

b. p 2

_ q 2

d. (2 x) 2 = 4 x 2

e. q 2 = 1 _ 2 p 2 and p 2 is divisible by 4



1. ((

PX ) !; Y is any point on ! other than X.

1. Given

2. m,1 = 90° 2. Def. of )3. ,1 is a rt. ,. 3. Def. of rt. ,4. .XPY is a rt. .. 4. Def. of rt. .5. ,2 and ,P are comp. 5. Acute / of rt. .

are comp.6. 90° = m,2 + m,P 6. Def. of comp. /7. 90° > m,2 7. Comparison

Prop. of Inequal.8. m,1 > m,2 8. Subst.9. PY > PX 9. In ., longer side

is opp. larger ,


76. slope = !2 ! 2 _ !1 + 3

= !2

y ! y 1 = m(x ! x 1 ) y ! 2 = !2[x ! (!3)] y ! 2 = !2(x + 3) y ! 2 = !2x ! 62x + y = !4

77. y ! y 1 = m(x ! x 1 ) y ! 0 = 2[x ! (!3)] y ! 0 = 2(x + 3) y = 2x + 6!2x + y = 6

78. QP = 5(!1 ) 2 ! 2 = 3, ST = !1 + 7 = 6, SU = 3(!1 ) 2 + 1 = 4, so .PQR - .TUS by SSS.

79. BC = 6 2 ! 5(6) + 4 = 10, EF = 2(6) ! 1 = 11, m,ABC = 14(6) + 18 = 102°, so .ABC - .EFD by SAS.

80. Equation of altitude from S is y = 3. Slope of RS is 3 ! 5 _ 4 ! 0

= ! 1 _ 2 , so slope of altitude from T is 2;

equation is y ! 1 = 2x, or y = 2x + 1. At O, y = 3 and therefore 3 = 2x + 1, so x = 1; thus O = (1, 3).

81. The altitudes from N and P lie along the x- and y-axes, respectively. Therefore O = (0, 0).

5-6 INEQUALITIES IN TWO TRIANGLES, PAGES 340–345

CHECK IT OUT! PAGES 341–342 1a. Compare the side lengths in .EFG and .EHG.

EF > EH EG = EG FG = HGBy the Converse of the Hinge Theorem, m,EGF > m,EGH.

b. Compare the sides and the / in .ABD and .CBD.AD = CD BD = BD m,CDB > m,ADBBy the Hinge Theorem, BC > AB.

2. The , of swing at full speed is greater than the , of swing at low speed.

3a. Statements Reasons

1. C is the midpoint of ((

BD ; m,1 = m,2 m,3 > m,4

1. Given

2. ((

BC - ((

DC 2. Def. of midpoint3. ,1 - ,2 3. Def. of - /4. ((

AC - ((

EC 4. Con. of Isosc. . Thm.

5. AB > ED 5. Hinge Thm.


1. ,SRT - ,STR, TU > RU

1. Given

2. ((

ST - ((

SR 2. Con. of Isosc. . Thm.

3. ((

SU - ((

SU 3. Reflex. Prop. of -4. m,TSU > m,RSU 4. Con. of the Hinge

Thm.


1. Possible answer: kitchen tongs

2. No; in this case, 2 sides of the 1st . are - to 2 sides of the 2nd ., but the given , measures are not the measures of / included between the - sides. Thus you cannot apply the Hinge Theorem.

3.


1. Compare the sides and the / in .ABC and .XYZ.AB = YZ BC = XY m,B < m,YBy the Hinge Theorem, AC < XZ.

2. Compare the side lengths in .SRT and .QRT.RT = RT RS = RQ ST > QTBy the Converse of Hinge Theorem, m,SRT > m,QRT.

3. Compare the sides and / in .KLM and .KNM.KM = KM LM = NM m,KML > m,KMNBy the Hinge Theorem, KL > KN.


4. Step 1 Compare the side lengths in 5. By the Converse of the Hinge Theorem, 2x + 8 < 25 2x < 17 x < 8.5.Step 2 Since (2x + 8)° is an angle in a .,2x + 8 > 0 2x > !8 x > !4.Step 3 Combine the inequals.The range of values is !4 < x < 8.5.

5. Step 1 Compare the sides and the / in 5. By the Hinge Theorem, 5x ! 6 < 9 5x < 15 x < 3.Step 2 Since 5x ! 6 is a length, 5x ! 6 > 0 5x > 6 x > 1.2.Step 3 Combine the inequals.The range of values is 1.2 < x < 3.

6. Step 1 Compare the sides and the / in 5. By the Hinge Theorem, 2x ! 5 < x + 7 x < 12.Step 2 Since 2x ! 5 is a length, 2x ! 5 > 0 2x > 5 x > 2.5.Step 3 Combine the inequals.The range of values is 2.5 < x < 12.

7. The 2nd position; the lengths of the upper and lower arm are the same in both positions, but the distance from the shoulder to the wrist is greater in the 2nd position. So the included , measure is greater by the Converse of the Hinge Theorem.


1. ((

FH is a median of .DFG; m,DHF > m,GHF

1. Given

2. H is midpoint of ((

DG . 2. Def. of median3. ((

DH - ((

GH 3. Def. of midpoint4. ((

FH - ((

FH 4. Reflex. Prop. of -5. DF > GF 5. Hinge Thm.


9. BC = CD, CA = CA, AD > AB; by Converse of Hinge Theorem, m,DCA > m,BCA.

10. GH = KL, HJ = LM, GJ < KM; by Converse of Hinge Theorem, m,GHJ < m,KLM.

11. ST = UV, SU = SU, m,UST > m,SUV; by Hinge Theorem, TU > SV.

12. 4z ! 12 < 16 4z < 28 z < 7

4z ! 12 > 0 4z > 12 z > 3

Combining, 3 < z < 7.

13. 2z + 7 < 72 2z < 65 z < 32.5

2z + 7 > 0 2z > !7 z > !3.5

Combining, !3.5 < z < 32.5.

14. 4z ! 6 < z + 11 3z < 17

z < 17 _ 3

4z ! 6 > 0 4z > 6

z > 3 _ 2

Combining, 3 _ 2

< z < 17 _ 3 .

15. The lengths of the arms are the same in both positions, but the included , measure is greater in the 2nd position. Therefore, by the Hinge Theorem, the distance from the cab to the bucket is greater in the 2nd position.


1. ((

JK - (((

NM , ((

KP - (((

MQ , JQ > NP

1. Given

2. ((

QP - ((

QP 2. Reflex. Prop. of -3. QP = QP 3. Def. of - segs.4. JQ + QP > NP + QP 4. Add. Prop. of

Inequal.5. JQ + QP = JP,

NP + QP = NQ5. Segment Add.

Post.6. JP > NQ 6. Subst.7. m,K > m,M 7. Con. of the Hinge

Thm.

17. BC = YZ 18. m,QRP < m,SRP 19. m,QPR > m,QRP 20.m,PRS < m,RSP 21. m,RSP = m,RPS 22. m,QPR > m,RPS

23. m,PSR < m,PQR

24. Corr. sides are -, and the included / are ,B and ,E. By the Hinge Theorem, m,B > m,E ' AC > DF.

25. ((

SR - ((

ST by definition, and ((

SV - ((

SV . So by the Converse of the Hinge Theorem, RV < TV ' m,RSV < m,TSV.

26. Corr. sides are -, and the included / are ,G and ,K. m,G = 90° > m,K, so by the Hinge Theorem, HJ > LM.

27. ((

YM - ((

MZ by definition, and ((

XM - ((

XM . So by the Converse of the Hinge Theorem, YX > ZX ' m,YMX > m,ZMX.

28. Possible answer: As the angle made by a door hinge gets larger, the width of the door opening increases. As the angle made by the hinge gets smaller, the width of the door opening decreases. This is like the side opposite an angle in a triangle getting larger as the measure of the angle increases or getting smaller as the angle decreases.


29. Possible answer: Similarities: Both the SAS - Post. and the Hinge Theorem concern the relationship between 2 5. Both involve 2 sides and the included , of each ..Differences: To apply the SAS - Post., you must know that 2 sides and the included , of one . are - to 2 sides and the included , of the 2nd .. To apply the Hinge Theorem, you must know that 2 sides of one . are - to 2 sides of the 2nd ., but the included / are 6 in measure. The SAS - Post. allows you to conclude that the 2 5 are -; then by CPCTC, you can show that the sides opposite the - / are -. The Hinge Theorem involves 2 5 that are =; in this case, the sides opposite the included / are 6 in length, and the exact relationship between the lengths is determined by the sizes of the included /.

30a. Newton Springs; ((

NS - ((

HS , ((

SJ - ((

SJ , and m,NSJ < m,HSJ, so NJ < JH by the Hinge Theorem.

b. By . Inequal. Theorem, NJ + SJ > SNNJ + 182 > 300 NJ > 118 miMin. distance = SN + NJ

> 300 + 118 = 418 mi

TEST PREP, PAGE 345

31. D;0 < 3x ! 9 < 2x + 19 < 3x or 3 < x, and x < 103 < x < 10

32. H;D lies on

(( AB ; AD = DB by the definition of median.

33. Group A is closer to the camp.Possible answer: The 6.5-mi and 4-mi paths together with the distance lines back to the camp form 2 5. 2 sides of 1 . are - to 2 sides of the other .. In the . for Group A, the measure of the included , is 90° + 35° = 125°. In the . for Group B, the measure of the included , is 90° + 45° = 135°. By the Hinge Theorem, the side opposite the 125° , is shorter than the side opposite the 135° ,. So Group A is closer to the camp.


34. Step 1 Apply Hinge Theorem.By Converse of Isosc. . Theorem,

(( VZ - ((

VY ; ((

VX - ((

VX ; m,XVZ > m,XVY. So XZ > XY.Step 2 Write and solve 2 inequals.5x + 15 > 8x ! 6 21 > 3x 7 > x

8x ! 6 > 0 8x > 6 x > 0.75

Step 3 Combine the inequals.0.75 < x < 7

35a. Locate point P outside .ABC so that ,ABP - ,DEF and

(( BP - ((

EF . It is given that ((

AB -

(( DE , so .ABP - .DEF by SAS. Thus

(( AP - ((

DF by CPCTC.

b. Locate point Q on ((

AC so that ((

BQ bisects ,PBC. By the definition of , bisector, ,QBC - ,QBP. It is given that

(( BC - ((

EF. Since ((

BP - ((

EF from part a, ((

BC - ((

BP by the Trans. Prop. of -. By the Reflex. Prop. of -,

(( BQ - (((

BQ. So .BQP - .BQC by SAS, and ((

QP - ((

QC by CPCTC.

c. AQ + QP > AP by the . Inequal. Theorem in .AQP. AQ + QC = AC by the Segment Add. Post. From part b,

(( QP - ((

QC , so QP = QC by the definition of - segs. Thus AQ + QC > AP by subst., and so AC > AP by subst. From part a,

(( AP

- ((

DF . So by the definition of - segs., AP = DF. Therefore AC > DF by subst.


36. range: 5 ! 0.5 = 4.5 mode: 2

37. range: 99 ! 85 = 14mode: none

38. range: 9 ! 4 = 5modes: 4, 5, 7

39. m,2 = 3(5) + 21 = 36°, m,6 = 7(5) + 1 = 36° = m,2; m 7 n by the Converse of the Corr. / Post.

40. m,4 = 2(7) + 34 = 48°, m,7 = 15(7) + 27 = 132°; so m,4 + m,7 = 180°; m 7 n by the Converse of the Same-Side Int. / Theorem.

41. By Similar Triangles Theorem:DF = 1 _

2 AB

= AE = 2.5

42. BC = 2DE= 2(2.3) = 4.6

43. m,BFD = 180 ! m,CFD= 180 ! m,CBA= 180 ! 95 = 85°

CONNECTING GEOMETRY TO ALGEBRASIMPLEST RADICAL FORM, PAGE 346

TRY THIS, PAGE 346 1. " ## 720

$ #### (144)(5)

" ## 144 · " # 5 12 " # 5

2. $ ## 3 _ 16

" # 3 _

" ## 16

" # 3 _ 4

3. 10 _ " # 2

10 _ " # 2

( " # 2 _ " # 2

) 10 " # 2 _

2

5 " # 2

4. $ # 1 _ 3

" # 1 _ " # 3

1 _ " # 3

( " # 3 _ " # 3

) " # 3 _ 3


5. " ## 45

$ ### (9)(5)

" # 9 · " # 5

3 " # 5

GEOMETRY LAB

HANDS-ON PROOF OF THE PYTHAGOREAN THEOREM, PAGE 347

TRY THIS, PAGE 347 1. a 2 + b 2 = c 2


5-7 THE PYTHAGOREAN THEOREM, PAGES 348–355

CHECK IT OUT! PAGES 349–351 1a. a 2 + b 2 = c 2

4 2 + 8 2 = x 2 80 = x 2 " ## 80 = x x = " ### (16)(5) = 4 " # 5

b. a 2 + b 2 = c 2 x 2 + 12 2 = (x + 4 ) 2 x 2 + 144 = x 2 + 8x + 16 128 = 8x x = 16

2. Let y be the distance in ft from the foot of the ladder to the base of the wall. Then 4y is the distance in ft from the top of the ladder to the base of the wall.

a 2 + b 2 = c 2 (4y ) 2 + y 2 = 30 2 17 y 2 = 900

y 2 = 900 _ 17

y = " ## 900 _ 17

4y = 4 " ## 900 _ 17

0 29 ft 1 in.

3a. a 2 + b 2 = c 2 8 2 + 10 2 = c 2 164 = c 2 c = " ## 164 = 2 " ## 41 The side lengths do not form a Pythagorean triplebecause 2 " ## 41 is not a whole number.

b. a 2 + b 2 = c 2 24 2 + b 2 = 26 2 b 2 = 100 b = 10The side lengths are nonzero whole numbers that satisfy the equation a 2 + b 2 = c 2 , so they form a Pythagorean triple.

c. a 2 + b 2 = c 2 1 2 + 2.4 2 = c 2 6.76 = c 2 c = 2.6The side lengths do not form a Pythagorean triple because 2.4 and 2.6 are not whole numbers.

d. a 2 + b 2 = c 2 16 2 + 30 2 = c 2 1156 = c 2 c = 34The side lengths are nonzero whole numbers that satisfy the equation a 2 + b 2 = c 2 , so they form a Pythagorean triple.

4a. Step 1 Determine if the measures form a ..By the . Inequal. Theorem, 7, 12, and 16 can be the side lengths of a ..Step 2 Classify the .. c 2 > a 2 + b 2 16 2 > 7 2 + 12 2 256 > 49 + 144256 > 193Since c 2 > a 2 + b 2 , . is obtuse.

b. Step 1 Determine if the measures form a ..Since 11 + 18 = 29 : 34, these cannot be the side lengths of a ..

c. Step 1 Determine if the measures form a ..By the . Inequal. Theorem, 3.8, 4.1, and 5.2 can be the side lengths of a ..Step 2 Classify the .. c 2 > a 2 + b 2 5.2 2 > 3.8 2 + 4.1 2 27.04 > 14.44 + 16.8127.04 < 31.25Since c 2 < a 2 + b 2 , . is acute.

THINK AND DISCUSS, PAGE 352 1. The greatest number is substituted for c. The other

2 numbers are substtituted for a and b in any order.

2. Possible answer: The sum of the areas of the 2 smaller squares equals the area of the largest square. So

3 2 + 4 2 = 5 2 , or 9 + 16 = 25.

3. Must be nonzero and whole numbers, and must satisfy the equation a 2 + b 2 = c 2

4.



1. No; although it is true that (2.7) 2 + ( 3.6) 2 = ( 4.5) 2 , the numbers 2.7, 3,6, and 4.5 are not whole numbers.

2. a 2 + b 2 = c 2 3 2 + 9 2 = x 2 90 = x 2 " ## 90 = x x = " ### (9)(10)

= 3 " ## 10

3. a 2 + b 2 = c 2 x 2 + 7 2 = 11 2 x 2 = 72 x = " ## 72 x = " ### (36)(2)

= 6 " # 2

4. a 2 + b 2 = c 2 (x ! 2 ) 2 + 8 2 = x 2 x 2 ! 4x + 4 + 64 = x 2 !4x + 68 = 0 68 = 4x x = 17

5. Let the width and the height of the monitor be w = 5x and h = 4x, respectively.

a 2 + b 2 = c 2

(4x ) 2 + (5x ) 2 = 19 2

41 x 2 = 361

x 2 = 361 _ 41

x = " ## 361 _ 41

w = 5x = 5 " ## 3 61 _ 41

0 14.8 in.

h = 4x = 4 " ## 361 _ 41

0 11.9 in.

6. a 2 + b 2 = c 2 4 2 + 5 2 = c 2 41 = c 2 c = " ## 41 The side lengths do not form a Pythagorean triple

because " ## 41 is not a whole number.

7. a 2 + b 2 = c 2 12 2 + b 2 = 20 2 b 2 = 256 b = 16The side lengths are nonzero whole numbers that satisfy the equation a 2 + b 2 = c 2 , so they form a Pythagorean triple.

8. a 2 + b 2 = c 2 1.5 2 + b 2 = 1.7 2 b 2 = 0.64 b = 0.8The side lengths do not form a Pythagorean triple because they are not whole numbers.

9. Step 1 Determine if the measures form a ..By the . Inequal. Theorem, 7, 10, and 12 can be the side lengths of a ..Step 2 Classify the .. c 2 > a 2 + b 2 12 2 > 7 2 + 10 2 144 > 49 + 100144 < 149Since c 2 < a 2 + b 2 , . is acute.

10. Step 1 Determine if the measures form a ..By the . Inequal. Theorem, 9, 11, and 15 can be the side lengths of a ..Step 2 Classify the .. c 2 > a 2 + b 2 15 2 > 9 2 + 11 2 225 > 81 + 121225 > 202Since c 2 > a 2 + b 2 , . is obtuse.

11. Step 1 Determine if the measures form a ..By the . Inequal. Theorem, 9, 40, and 41 can be the side lengths of a ..Step 2 Classify the .. c 2 > a 2 + b 2 41 2 > 9 2 + 40 2 1681 > 81 + 16001681 = 1681Since c 2 = a 2 + b 2 , . is a rt. ..

12. Step 1 Determine if measures form a ..Since 1 1 _

2 + 1 3 _

4 = 3 1 _

4 : 3 1 _

4 , these cannot be

the side lengths of a ..

13. Step 1 Determine if the measures form a ..By the . Inequal. Theorem, 5.9, 6, and 8.4 can be the side lengths of a ..Step 2 Classify the .. c 2 > a 2 + b 2 8.4 2 > 5.9 2 + 6 2 70.56 > 34.81 + 3670.56 < 70.81Since c 2 < a 2 + b 2 , . is acute.

14. Step 1 Determine if the measures form a ..By the . Inequal. Theorem, 11, 13, and 7 " # 6 can be the side lengths of a ..Step 2 Classify the .. c 2 > a 2 + b 2 (7 " # 6 ) 2 > 11 2 + 13 2 294 > 121 + 169 294 > 290Since c 2 > a 2 + b 2 , . is obtuse.


15. 6 2 + 8 2 = x 2 100 = x 2 x = 10

16. 9 2 + x 2 = 13 2 81 + x 2 = 169 x 2 = 88 x = " ## 88 = 2 " ## 22

17. x 2 + 7 2 = (x + 1 ) 2 x 2 + 49 = x 2 + 2x + 1 48 = 2x x = 24


18. (3x ) 2 + (5x ) 2 = 8 2 34 x 2 = 64 x 2 = 32 _

17

3x = 3 $ ## 32 _ 17

0 4 ft 1 in.

5x = 5 $ ## 32 _ 17

0 6 ft 10 in.

19. 2.5 2 + b 2 = 6.5 2 6.25 + b 2 = 42.25 b 2 = 36 b = 6The side lengths cannot form a . because 2.5 and 6.5 are not whole numbers.

20. 15 2 + 20 2 = c 2 625 = c 2 c = 25Yes; the three side lengths are nonzero whole

numbers that satisfy a 2 + b 2 = c 2 .

21. 2 2 + b 2 = 7 2 4 + b 2 = 49 b 2 = 45 b = " ## 45 = 3 " # 5 The side lengths cannot form a . because 3 " # 5 is not a whole number.

22. 10 + 12 = 22 > 15 ; 15 2 > 10 2 + 12 2 225 < 244The side lengths form an acute ..

23. 8 + 13 = 21 : 23 <The side lengths can not form a ..

24. 9 + 14 = 23 > 17 ; 17 2 > 9 2 + 14 2 289 > 277The side lengths form an obtuse ..

25. 1 1 _ 2 + 2 = 3 1 _

2 > 2 1 _

2 ;

(2 1 _ 2 ) 2 > (1 1 _

2 ) 2 + 2 2

6 1 _ 4 = 6 1 _

4

The side lengths form a rt. ..

26. 0.7 + 1.1 = 1.8 > 1.7 ; 1.7 2 > 0.7 2 + 1.1 2 2.89 > 1.7The side lengths form an obtuse ..

27. 7 + 12 = 19 > 6 " # 5 ;

(6 " # 5 ) 2 > 7 2 + 12 2 180 < 193The side lengths form an acute ..

28. Possible answer: Shape the rope into a . with side lengths of 3, 4, and 5. Because 3 2 + 4 2 = 5 2 , the . is a rt. . with the rt. , opposite the side 5.

29. B; (x + 3 ) 2 + 4

2 = ( x

2 + 6x + 9) + 16. In the

solution shown, the 6x term was omitted. 30. Let a and b be the horizontal leg lengths of the left-

and right-hand 5.

9 2 + b 2 = 15 2 b 2 = 144

b = 12a + b = 25 a = 25 ! 12 = 13

13 2 + 9 2 = x 2 250 = x 2 x = " ## 250 = 5 " # 10

31. Let a and b be the horizontal leg lengths of the left- and right-hand 5.

a 2 + 6 2 = 10 2 a 2 = 64

a = 8

6 2 + b 2 = 7 2 b 2 = 13 b = " ## 13

x = a + b = 8 + " ## 13

32. Let d be the length of the shared side. 2 2 + d 2 = 7 2 d 2 = 45 x 2 + 5 2 = d 2 x 2 = 20 x = " ## 20 = 2 " # 5

33. Let h be the height of the 5. 3 2 + h 2 = ( " ## 34 ) 2 h 2 = 25 x 2 + h 2 = 11 2 x 2 = 96 x = " ## 96 = 4 " # 6

34. Let h be the height of the 5. 5 2 + h 2 = 13 2 h 2 = 144 (x + 5 ) 2 + h 2 = 20 2 x 2 + 10x + 25 + 144 = 400 x 2 + 10x ! 231 = 0 (x ! 11)(x + 21) = 0Since x > 0, the only possible solution is x = 11.

35. Let b be the base length of each .. 18 2 + (2b ) 2 = 30 2 4 b 2 = 576 b 2 = 144 18 2 + b 2 = x 2 468 = x 2 x = " ## 468 = 6 " ## 13

36. 3963 2 + x 2 = (3963 + 250 ) 2 x 2 = 2,044,000 x = " #### 2,044,000 0 1430 mi

37. Possible answer: Outer figure: The length of each side is a + b, so the outer figure has 4 - sides. Each , is a rt. , from one of the rt. 5, so the outer figure has 4 rt. /. By definition, it is a square.Inner figure: The length of each side is c, so the inner figure has 4 - sides. The 2 acute / of a rt. . are comp., so the measure of each , in the inner figure is 90°. Therefore the inner figure has 4 rt. /. By definition, it is a square.

38. Let b be the base of the .. 8 2 + b 2 = 17 2 b 2 = 225 b = 15P = 8 + 15 + 17

= 40 unitsA = 1 _

2 (15)(8)

= 60 square units

39. Let 2b be the base of the .. 6 2 + b 2 = 8 2 b 2 = 28 b = 2 " # 7 2b = 4 " # 7 P = 8 + 8 + 4 " # 7 = 16 + 4 " # 7 units

A = 1 _ 2

(4 " # 7 ) (6)

= 12 " # 7 square units


40. Let h be the height of the .. 4 2 + h 2 = 12 2 h 2 = 128 h = " ## 128 = 8 " # 2 P = 12 + 12 + 8

= 32 unitsA = 1 _

2 (8)8 " # 2

= 32 " # 2 square units

41. Let h be the height and c be the 3rd side length of the .. 3 2 + h 2 = 5 2 h 2 = 16 h = 4 6 2 + 4 2 = c 2 52 = c 2 c = " ## 52 = 2 " ## 13 P = 5 + (3 + 6) + 2 " ## 13

= 14 + 2 " ## 13 unitsA = 1 _

2 (3 + 6)(4)

= 18 square units

42. Let a + b = 15 be the 2nd side length and c be the 3rd side length of the .. a 2 + 12 2 = 15 2 a 2 = 81 a = 9b = 15 ! 9 = 6 6 2 + 12 2 = c 2 180 = c 2 c = " ## 180 = 6 " # 5 P = 15 + 15 + 6 " # 5

= 30 + 6 " # 5 unitsA = 1 _

2 (15)(12) = 90 square units

43. P = 4 + 5 + 5 + 8 = 22 units

A = 1 _ 2

(a + b)h

= 1 _ 2

(5 + 8)(4)

= 26 square units

44. Possible answer: When you use Pythagorean Theorem, you know that the . is a rt. .. You substitute the known values into a 2 + b 2 = c 2 and solve for the unknown side length. When you use the Converse of Pythagorean Theorem, you are trying to find out whether a given . is a rt. .. Usually all side lengths are known. You substitute all the values into a 2 + b 2 = c 2 to determine whether the resulting equation is true. If it is true, then you know that the . is a rt. ..

45. Draw .PQR with ,R as the rt. ,, leg lengths of a and b, and hyp. length of x. In .ABC, it is given

that a 2 + b 2 = c 2 . In .PQR, a 2 + b 2 = x 2 by the Pythagorean Theorem. Since a 2 + b 2 = c 2 , and

a 2 + b 2 = x 2 , it follows by subst. that x 2 = c 2 . Take the positive square root of both sides, and

x = c. So AB = PQ, BC = QR, and AC = PR. By the definition of - segs.,

(( AB - ((

PQ , ((

BC - ((

QR , and

(( AC - ((

PR . Then .ABC - .PQR by SSS, and ,C - ,R by CPCTC. By the definition of rt. ,, m,R = 90°. So by the definition of - /, m,C = 90°. Therefore ,C is a rt. , by definition, and .ABC is a rt. . by definition.

46a. ( x 2 , y 1 ) b. JL = x 2 ! x 1 ,LK = y 2 ! y 1

c. JK 2 = JL

2 + LK

2

= ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2

JK = $ ######### ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2

47a. KR 2 + 500 2 = 1300 2 KR 2 = 1,440,000 KR = 1200 mi KM 2 + 390 2 = 1200 2 KM 2 = 1,287,900 KM 0 1135 mi SK + KM > SR + RM 500 + 1135 > 1300 + 390 1635 < 1690 She should fly first to King City.

b. S M 2 > SR 2 + RM 2 1360 2 > 1300 2 + 390 2 1,849,600 > 1,842,100 So by the Pythagorean Inequals. Theorem,

m,SRM > 90°.

TEST PREP, PAGE 355

48. GX = HX = 6 GM 2 + MX 2 = GX 2 GM 2 + 4 2 = 6 2 GM 2 = 20 GJ = 2GM

= 2 " ## 20 0 8.9

49. B; 7 2 + 24 2 > 25 2 625 = 625 ;

50. H; 11 2 9 7 2 + 9 2 121 : 130

51a. PA 2 = 1 2 + 1 2 = 2 PA = " # 2 PC 2 = 1 2 + PB 2 = 4 PC = 2 PE 2 = 1 2 + PD 2 = 6 PE = " # 6

PB 2 = 1 2 + PA 2 = 3 PB = " # 3 PD 2 = 1 2 + PC 2 = 5 PD = " # 5 PF 2 = 1 2 + PE 2 = 7 PF = " # 7

b. " ## 10 ; possible answer: for each . added to the pattern, the number under the radical symbol increases by 1. So the length of the hyp. of the 7th . would be " # 8 , the length of the hyp. of the 8th . would be " # 9 , and the length of the hyp. of the 9th . would be " ## 10 .

c. " ### n + 1 ; possible answer: the length of the hyp. is the square root of the whole number 1 greater than the number of the ., or " ### n + 1 .



52. Let 3 points be A(!1, 2), B(!10, 5), and C(!4, k).

AB 2 = 9 2 + 3 2 = 90 BC 2 = 6 2 + (k ! 5 ) 2

= 36 + k 2 ! 10k + 25= k 2 ! 10k + 61

AC 2 = 3 2 + (k ! 2 ) 2 = 9 + k 2 ! 4k + 4= k 2 ! 4k + 13

If AB 2 + BC 2 = AC 2 ,90 + k 2 ! 10k + 61 = k 2 ! 4k + 13 138 = 6k k = 23If AB 2 + AC 2 = BC 2 ,90 + k 2 ! 4k + 13 = k 2 ! 10k + 61 6k = !42 k = !7If AC 2 + BC 2 = AB 2 , k 2 ! 4k + 13 + k 2 ! 10k + 61 = 90 2 k 2 ! 14k ! 16 = 0 k 2 ! 7k ! 8 = 0 (k ! 8)(k + 1) = 0 k = 8 or !1So k = !7, !1, 8, or 23

53. By the . Inequal. Theorem, a + b > c.

By the Pythagorean Theorem, c = $ #### a 2 + b 2 .

By subst., a + b > $ #### a 2 + b 2 .

54. c 2 = a 2 + b 2

c = $ #### a 2 + b 2

A = 1 _ 2 ab = 1 _

2 hc

ab = hc

ab = h $ #### a 2 + b 2 ab _ $ #### a 2 + b 2

= h

55a. No; possible answer: let a = 3, b = 4, and c = 5. So a + 1 = 4, b + 1 = 5, and c + 1 = 6. 3, 4, and 5 form a Pythagorean triple, but 4, 5 and 6 do not

because 4 2 + 5 2 6 6 2 .

b. Yes; possible answer: if a, b and c form a Pythagorean triple, a 2 + b 2 = c 2 is true. Multiply both sides by 4 to get the equation 4 a 2 + 4 b 2 =

4 c 2 . This is equivalent to (2a ) 2 + (2b ) 2 = (2 c) 2 . So by def., 2a, 2b, and 2c also form a Pythagorean triple.

c. No; possible answer: let a = 3, b = 4, and c = 5. So a 2 = 9, b 2 = 16, and c 2 = 25. 3, 4, and 5 form a Pythagorean triple, but 9, 16, and 25 do not because 9 2 + 1 6 2 6 2 5 2 .

d. No; possible answer: let a = 3, b = 4, and c = 5. So " # a = " # 3 , " # b = 2, and " # c = " # 5 . 3, 4, and 5 form a Pythagorean triple, but " # 3 , 2 and " # 5 do not because ( " # 3 ) 2 + 2 2 6 ( " # 5 ) 2 .


56. (4 + x)12 ! (4x + 1)6 = 0 48 + 12x ! 24x ! 6 = 0 42 ! 12x = 0 42 = 12x x = 3.5

57. 2x ! 5 _ 3 = x

2x ! 5 = 3x !5 = x

58. 4x + 3(x + 2) = !3(x + 3) 4x + 3x + 6 = !3x ! 9 10x = !15 x = !1.5 = ! 3 _

2

59. By the Midpoint Formula, the coordinates of M are (a, b). By the Distance Formula,

AM = $ ######## (a ! 0 ) 2 + (b ! 0 ) 2 = $ #### a 2 + b 2 and

MB = $ ######## (0 ! a ) 2 + (2b ! b ) 2 = $ #### a 2 + b 2 . So by subst., AM = MB.

60. ((

JK - ((

NP , ((

JL - (((

NM , KL < MP m,J < m,N4x ! 6 < 68 4x < 74 x < 18.5

m,J > 0°4x ! 6 > 0 4x > 6 x > 1.5

So 1.5 < x < 18.5.

61. ((

BA - ((

BC , ((

BD - ((

BD , m,ABD < m,CBD3x + 1 < 7 3x < 6 x < 2

3x + 1 > 0 3x > !1 x < ! 1 _

3

So ! 1 _ 3

< x < 2.

5-8 APPLYING SPECIAL RIGHT TRIANGLES, PAGES 356–362

CHECK IT OUT! PAGES 357–359 1a. The . is an isosc. rt. ., which is a 45°-45°-90° ..x = (10 " # 2 ) " # 2 = 20

b. The . is an isosc. rt. ., which is a 45°-45°-90° .. 16 = x " # 2 16 _

" # 2 = x

16 " # 2 _ 2 = x

x = 8 " # 2

2. Tessa needs a 45°-45°-90° . with hyp. of length [C + 2(8)] cm and leg length of 42 cm.C + 2(8) = 42 " # 2 C = !16 + 42 " # 2 0 43 cm

3a. 18 " # 3 = 2x 9 " # 3 = xy = x " # 3 y = (9 " # 3 ) " # 3 y = 27

b. x = 5 " # 3 y = 2(5) = 10

c. 24 = 2x12 = xy = x " # 3 y = 12 " # 3

d. 9 = y " # 3 9 _

" # 3 = y

9 " # 3 _ 3 = y

3 " # 3 = yx = 2yx = 6 " # 3


4. Step 1 Divide the equil. . into two 30°-60°-90° 5. The height of the frame is the length of the longer leg.Step 2 Find the length x of the shorter leg. 30 = x " # 3

30 _ " # 3

= x

30 " # 3 _ 3

= x

10 " # 3 = xStep 3 Find the length s of each side of the frame.s = 20 " # 3 0 34.6 cm

THINK AND DISCUSS, PAGE 359 1. Possible answer: The . is a rt. ., so the measure

of one , is 90°, and the other 2 acute / are comp. The . is isosc., so its base / are -. So the measure of each of the base / is 45°.

2. In figure I, use the relationship x = 2(8). In figure II, first use the relationship 8 = " # 3 (shorter leg), and then use the relationship x = 2(shorter leg).

3.


1. The . is an isosc. rt. ., which is a

45°-45°-90° ..x = 14 " # 2


45°-45°-90° ..9 " # 2 = x " # 2 x = 9


45°-45°-90° .. 12 = x " # 2

12 _ " # 2

= x

12 " # 2 _ 2 = x

x = 6 " # 2

4. The sign forms a right .. Using the Pyth. Thm., we get

d = " ###### 19.5 2 + 19.5 2 d 0 27.6 in.

5. 6 = 2x3 = xy = x " # 3 y = 3 " # 3

7. x = (7 " # 3 ) " # 3 x = 21y = 2 (7 " # 3 ) y = 14 " # 3

6. 15 = x " # 3

15 _ " # 3

= x

15 " # 3 _ 3 = x

5 " # 3 = xy = 2xy = 10 " # 3

8. Step 1 Divide the equil. . into two 30°-60°-90° 5. The height of the frame is the length of the longer leg.Step 2 Find the length x of the shorter leg.5(2.25) = 2x 5.625 = xStep 3 Find the length h of the longer leg.h = 5.625 " # 3 0 9.75 in.



45°-45°-90° ..

15 = x " # 2

15 _ " # 2

= x

15 " # 2 _ 2 = x


45°-45°-90° ..

x = (4 " # 2 ) " # 2 = 8


45°-45°-90° ..18 " # 2 = x " # 2 18 = x

12. The tabletop is a 45°-45°-90° .. 48 = w " # 2 48 " # 2 = 2w w = 24 " # 2 0 33.9 in.

13. x = 2(24) = 48y = 24 " # 3

14. 10 " # 3 = 2x 5 " # 3 = xy = x " # 3

= (5 " # 3 ) " # 3 = 15

15. 2 = x " # 3 2 " # 3 = 3x

2 " # 3 _ 3 = x

y = 2x = 4 " # 3 _ 3

16a. The ramp forms a 30°-60°-90° .. Let the length of the ramp be x.

x = 2(4.5) = 9 ft

b. Length of the dog walk = x + 12 + x= 9 + 12 + 9 = 30 ft

17. 12 = a " # 2 12 " # 2 = 2a 6 " # 2 = ab = a = 6 " # 2 P = a + b + c

= 6 " # 2 + 6 " # 2 + 12= (12 + 12 " # 2 ) in.

A = 1 _ 2 ab

= 1 _ 2 (6 " # 2 ) (6 " # 2 )

= 36 in. 2

18. 28 = 2a14 = ab = a " # 3 = 14 " # 3 P = a + b + c

= 14 + 14 " # 3 + 28= (42 + 14 " # 3 ) cm

A = 1 _ 2 ab

= 1 _ 2 (14) (14 " # 3 )

= 98 " # 3 cm 2


19. 18 = s " # 2 18 " # 2 = 2s 9 " # 2 = sP = 4s

= 4 (9 " # 2 ) = 36 " # 2 m

A = s 2 = (9 " # 2 ) 2 = 81(2) = 162 m 2

20. h = ( s _ 2 ) " # 3

= 2 " # 3 P = 3s

= 3(4) = 12 ft

A = 1 _ 2 sh

= 1 _ 2 (4) (2 " # 3 )

= 4 " # 3 ft 2

21. h = ( s _ 2

) " # 3

2h = s " # 3 60 = s " # 3 60 " # 3 = 3s20 " # 3 = s

P = 3s= 3 (20 " # 3 ) = 60 " # 3 yd

A = 1 _ 2 sh

= 1 _ 2 (20 " # 3 ) (30)

= 300 " # 3 yd 2

22. Let s be the leg length. 18 = s " # 2 9 " # 2 = sHyp.: 18 ÷ 1 _

2 = 36 nails

Legs: 2(9 " # 2 ) ÷ 1 _ 4 = 72 " # 2 0 102 nails

The total is approximately 138 nails.

23. No; possible answer: if the , measures are in ratio 1 : 2 : 3 , then the measures of the angles are

30°-60°-90°, and the . is a 30°-60°-90° .. Assume the length of the shortest leg is 1. Then the length of the hyp. is 2, and the length of the longer leg is " # 3 . So the side lengths would be in the ratio 1 : " # 3 : 2.

24.

Let P = (x, y). ((

QR is the hyp. From the diagram,

(( QR is a 45° , to the axes. P is in

quad II ' ((

PQ is horizontal ' y = y-coordinate of Q = 6;

(( PR is vertical ' x = x-coordinate of R = !6.

So P = (!6, 6).

25.

Let P = (x, y). ((

PT is the hyp. From the diagram,

(( ST is a 45° , to the axes. P is in

quad I ' ((

PT is horizontal ' y = y-coordinate of

T = 3; PT = ST " # 2 = ( $ #### 6 2 + 6 2 ) " # 2 = " ## 144 = 12;

x = (x-coordinate of T ) + 12 = !2 + 12 = 10. So P = (10, 3).

26.

Let P = (x, y). ((

PX is the hyp. From the diagram, P is in quad. II '

((( PW is vertical

' x = x-coordinate of W = !1; PX = WX " # 3 = 5 " # 3 ; y = (y-coordinate of W ) + 5 " # 3 = !4 + 5 " # 3 . So P = (!1, !4 + 5 " # 3 ) .

27.

Let P = (x, y). ((

PY is the hyp. From the diagram, P is in quad. IV '

(( PZ is vertical

' x = x-coordinate of Z = 5; PZ = YZ " # 3 = 12 " # 3 ; y = (y-coordinate of Z ) ! 12 " # 3 = 10 ! 12 " # 3 .So P = (5, 10 ! 12 " # 3 ) .

28. Possible answer: Both types of . are rt. 5. In each one, there is a unique relationship among the side lengths. For each type of ., if you know 1 side length, you can find the other 2.

29a. NB = 2NL= 2(320) = 640 mi

b. IN = NL " # 2 0 453 mi

c. BI = BL ! IL= NL " # 3 ! NL= 320 " # 3 ! 320 0 234 mi

TEST PREP, PAGE 362

30. C 31. F;(5, 12, 13) is a Pythagorean triple, and 5 + 13 = 18.

32. B;24 = a " # 2 a = 12 " # 2 0 17.0 in.

33. 32 = 2w w = 16 ! = w " # 3 = 16 " # 3 A = !w

= 16 " # 3 (16)= 443.4 in. 2


34. Step 1 Identify the pattern.The length of each hyp. is " # 2 times the length of the previous hyp.Step 2 Write and solve an equation for x.4 = ( " # 2 ) 4 x 4 = 4xx = 1


35. Step 1 Identify the pattern.

The length of each hyp. is 2 _ " # 3

times the length of

the previous hyp. The length of the first hyp. is 2.Step 2 Find x.

x = ( 2 _ " # 3

) 4 (2) = 32 _ 9

36a. Let f be the length of the face diagonal. Then f = e " # 2 .e = 1: f = " # 2 , so

d = $ #### e 2 + f 2 = " ### 1 + 2 = " # 3

e = 2: f = 2 " # 2 , so

d = $ #### e 2 + f 2 = " ### 4 + 8 = " ## 12 = 2 " # 3

e = 3: f = 3 " # 2 , so

d = $ #### e 2 + f 2 = " ### 9 + 18 = " ## 27 = 3 " # 3

b. d = $ ### e 2 + f 2

= $ #### e 2 + 2 e 2

= $

## 3 e 2 = e " # 3

37. Possible answer:Given: .ABC is a 30°-60°-90° . with m,A = 30° and m,B = 60°.

(( CD is the altitude to the hyp.

Prove: AD = 3DBProof: It is given that

(( CD is the altitude to the hyp.

Thus ((

CD ) ((

AB by the definition of altitude. So ,ADC and ,BDC are rt. / by the definition of ), and .ADC and .BDC are rt. 5 by definition. It is given that m,A = 30° and m,B = 60°. Since the acute / of a rt. . are comp., m,DCA = 60° and m,DCB = 30° by Subtr. Prop. of =. So .ADC and .BDC are both 30°-60°-90° 5. By the 30°-60°-90° . Theorem, AD = " # 3 (DC) and DC = " # 3 (DB). By subst., AD = " # 3 ( " # 3 (DB)) . This simplifies to AD = 3DB.


38. y = x 2 + 4x + 0= (x + 2 ) 2 + 0 ! 2 2 = (x + 2 ) 2 ! 4

Axis of symmetry: x = !2

39. y = x 2 ! 10x ! 2= (x ! 5 ) 2 ! 2 ! 5 2 = (x ! 5 ) 2 ! 27

Axis of symmetry: x = 5

40. y = x 2 + 7x + 15= (x + 3.5 ) 2 + 15 ! 3.5 2 = (x + 3.5 ) 2 + 2.75

Axis of symmetry: x = !3.5

41. m,ADB ! 180 ! 70 = 110° is obtuse. So .ADB is obtuse.

42. m,DBC = 180 ! (60 + 70) = 50°. All 3 / are acute, so .BDC is acute.

43. m,ABC = m,ABD + m,DBC= 180 ! (30 + 110) + 50 = 90°

,ABC is a rt. ,, so .ABC is a rt. ..

44. ,PSQ and ,PQS are comp. By the Converse of the

, Bisector Theorem, **+ QS is the bisector of ,PQR.

So m,PQR = 2m,PQS= 2(90 ! m,PSQ)= 2(90 ! 65) = 50°.

45. ,QTV and ,VTS are supp., and ,TQV and ,QTV are comp. By the Converse of the , Bisector

Theorem, **+ QS is the bisector of ,PQR. So

m,VTS = 180 ! m,QTV= 180 ! (90 ! m,TQV)= 180 ! (90 ! m,PQS)= 180 ! (90 ! 42) = 132°.

46. By the , Bisector Theorem, PS = SR and TU = TV. Substitute in the given equation. SR = 3TUPS = 3TV 7.5 = 3TV TV = 2.5

GEOMETRY LABGRAPH IRRATIONAL NUMBERS, PAGE 363

TRY THIS, PAGE 363 1. 1 2 + 1 2 = 1 + 1 = 2 = ( " # 2 ) 2 ;

1 2 + ( " # 2 ) 2 = 1 + 2 = 3 = ( " # 3 ) 2 ;

2. Check students’ constructions to confirm that " # 4 lies at 2 on the number line.

3. Check students’ constructions to confirm that " # 9 lies at 3 on the number line.

4. Check students’ constructions to confirm that 2 " # 2 lies at " # 8 on the number line.


1. H is the intersection of ((

AC and ((

BD . Assume that a different point X, not on

(( AC and not on

(( BD , results

in the least combined distance to the cities. Then AX + CX > AC by the . Inequal. Theorem, so AX + CX > AH + CH. Similarly, BX + DX > BH + DH. If the inequalities are added, AX + CX + BX + DX > AH + CH + BH + DH. This contradicts the assumption that X results in the least combined distance. Therefore, H must result in the least combined distance.

2. AB + BC = AB + AB = 400 miAC = AB " # 2

= 200 " # 2 0 282.8 midistance saved 0 400 ! 283 0 117.2 mi

3. BC + CD = BC + BC " # 3 = 200 + 200 " # 3 0 546.4 mi

BD = 2BC = 400 midistance saved 0 546.4 ! 400 0 146.4 mi


4. The longest route is the one between Dumas and Colfax. DC = 200 " # 3 0 346.4 mi., DB = 400 mi, AB = 200 mi, BC = 200 mi, and AC 0 282.8 mi. So

DC > AB, DC > BC, and DC > AC. Since AC > AH and AC > CH, it follows that DC > AH and

DC > CH.In .BCH, m,BHC = 30 + 45 = 75° and m,BCH = 45°, so BC > BH. Since DC > BC, it follows that DC > BH. Similarly in .DCH, m,DHC = 60 + 45 = 105° and m,DCH = 45°, soDC > DH. Also

(( AB - ((

BC , ((

DB - (((

DB, and m,ABD = 30° < m,DBC = 60°. So by the HingeTheorem, DC > AD. Thus

(( DC is the longest route.


1. Possible answer: Given: ,A and ,B are supplementary. ,A is an acute angle.Prove: ,B cannot be an acute angle. Proof: Assume that ,B is an acute angle. By the def. of acute, m,A < 90° and m,B < 90°. When the 2 inequalities are added. m,A + m,B < 180°. However, by the def. of supp., m,A + m,B = 180°. So m,A + m,B < 180° contradicts the given information, and the assumption that ,B is an acute , is false. Therefore ,B cannot be acute.

2. ((

KM is the shortest side, so ,L is the least ,. ((

KL is the longest side, so ,M is the greatest ,. From smallest to greatest, the order is ,L, ,K, ,M.

3. m,D = 90 ! 48 = 42°, m,E = 90°,D is the least ,, so

(( EF is the shortest side.

,E is the greatest ,, so ((

DF is the longest side.From shortest to longest, the order is

(( EF , ((

DE , ((

DF .

4. No; possible answer: the sum of 8.3 and 10.5 is 18.8, which is not greater than 18.8. By the . Inequality Thm., a . cannot have these side lengths.

5. Yes; possible answer: when s = 4, the value of 4s is 16, the value of s + 10 is 14, and the value of s 2 is 16. The sum of each pair of 2 lengths is greater than the third length. So a . can have sides with these lengths.

6. Let d be the distance from the theater to the zoo.d + 9 > 16 d > 16 ! 9 = 7

9 + 16 > d 25 > d

Range of the distances: greater than 7 km and less than 25 km.

7. ((

PQ - ((

ST , ((

QR - ((

TV , and m,Q > m,T. By the Hinge Theorem, PR > SV.

8. ((

JK - ((

JM , ((

JL - ((

JL , and KL < ML. By the Converse of the Hinge Theorem, m,KJL < m,MJL.

9. ((

AD - ((

BC , ((

BD - ((

BD , and m,ADB < m,DBC. By the Hinge Theorem, AB < CD4x ! 13 < 15 4x < 28 x < 7

AB > 04x ! 13 > 0 4x > 13 x > 3.25

3.25 < x < 7

10. x 2 = 5 2 + 9 2 x 2 = 106 x = " ## 106

11. a 2 + 9 2 = 11 2 a 2 + 81 = 121 a 2 = 40 a = " ## 40 = 2 " ## 10 The side lengths do not form a Pythagorean triple, because2 " ## 10 is not a whole number.

12. 10 + 12 = 22 > 16 ;The side lengths can form a .. 16 2 > 10 2 + 12 2 256 > 100 + 144256 > 244The . is obtuse.

13. Length of the walkway = " #### 50 2 + 80 2 = " ## 8900 0 94 ft 4 in.

14. Length of the shorter leg of a 30°-60°-90° . is 36 ÷ 2 = 18 in. So h = 18 " # 3 0 31 in.

15. x = 8 " # 2 16. 22 = x " # 2 22 " # 2 = 2x 11 " # 2 = x

17. 5 " # 3 = x " # 3 5 = xy = 2x

= 2(5) = 10


1. equidistant 2. midsegment

3. incenter 4. locus

LESSON 5-1, PAGE 366 5. BD = 2CD = 2(3.7) = 7.4

6. XY = YZ3n + 5 = 8n ! 9 14 = 5n n = 2.8YZ = 8(2.8) ! 9 = 13.4

7. HT = FT = 5.8

8. m,MNV = m,PNV 2z + 10 = 4z ! 6 16 = 2z z = 8m,MNP = 2m,MNV

= 2[2(8°) + 10°] = 52°

9. The midpoint of ((

AB is (1, 0);slope of

(( AB = !10 _

10 = !1, so the slope of the

perpendicular bisector is 1;

the equation of the perpendicular bisector is y = x !1.


10. The midpoint of ((

XY is (4, 6);

slope of ((

XY = 8 _ 2

= 4, so the slope of the

perpendicular bisector is !0.25;

the equation of the perpendicular bisector is y ! 6 = !0.25(x ! 4).

11. No; to apply the Converse of the Angle Bisector Theorem, you need to know that

(( AP ) ((

AB and ((

CP ) ((

CB .

12. Yes; since ((

AP ) ((

AB , ((

CP ) ((

CB , and ((

AP - ((

CP , P is on the bisector of ,ABC by the Converse of the Angle Bisector Theorem.

LESSON 5-2, PAGE 367 13. GY = HY = 42.2 14. GP = JP = 46

15. GJ = 2GX= 2(28.8) = 57.6

16. PH = JP = 46

17. distance from A to ((

UV = distance from A to (((

UW = 18

18. m,WVU + m,VUW + m,UWV = 180 2m,WVA + 2(20)+ 66 = 180 2m,WVA = 74 m,WVA = 37°

19. (((

MO is vertical, so the equation of the horizontal perpendicular bisector is y = 3;

((

NO is horizontal, so the equation of the vertical perpendicular bisector is x = 4.The circumcenter is at (4, 3).

20. ((

OR is vertical, so the equation of the horizontal perpendicular bisector is y = !3.5;

((

OS is horizontal, so the equation of the vertical perpendicular bisector is x = !6.The circumcenter is at (!6, !3.5).

LESSON 5-3, PAGE 367 21. DZ = 2 _

3 DB

= 2 _ 3 (24.6) = 16.4

22. DB = 3ZB24.6 = 3ZB ZB = 8.2

23. EZ = 2ZC11.6 = 2ZC ZC = 5.8

24. EC = 3ZC = 3(5.8) = 17.4

25. ((

JK is vertical, so the equation of the altitude from L is y = 0; ((

KL is horizontal, so the equation of the altitude from J is x = !6.The orthocenter is at (!6, 0).

26. ((

AB is horizontal, so the equation of the altitude from C is x = 1; ((

AC is vertical, so the equation of the altitude from B is y = 2.The orthocenter is at (1, 2).

27. ((

RT is horizontal, so the equation of the altitude from S is x = 7; ((

RS has slope 5 _ 5 = 1, so the equation of the altitude

from T is y ! 3 = !(x ! 8).At the orthocenter, x = 7 and y ! 3 = !(7 ! 8) = 1 ' y = 4, so the orthocenter is at (7, 4).

28. ((

XY is horizontal, so the equation of the altitude from Z is x = 3; ((

XZ has slope 6 _ !6

= !1, so the equation of the altitude

from Y is y ! 2 = x ! 5 or y = x ! 3.

At the orthocenter, x = 3 and y = x ! 3 = 0, so the orthocenter is at (3, 0).

29. G = ( 1 _ 3 (0 + 3 + 6), 1 _

3 (4 + 8 + 0)) = (3, 4)

LESSON 5-4, PAGE 368 30. BC = 1 _

2 XY

= 1 _ 2 (70.2) = 35.1

31. XZ = 2AB= 2(32.4) = 64.8

32. XC = 1 _ 2 XZ

= AB = 32.4

33. m,BCZ = m,ABC = 42°

34. m,BAX = 180° ! m,ABC = 180° ! 42° = 138°

35. m,YXZ = m,BCZ = 42°

36. V = (!1, !1); W = (6, 1); slope of (((

VW = 2 _ 7 ;

slope of ((

GJ = 4 _ 14

= 2 _ 7 ; since the slopes are the

same, (((

VW 7 ((

GJ .

VW = $ #### 2 2 + 7 2 = " ## 53 ;

GJ = $ #### 4 2 + 14 2 = 2 " ## 53 , so VW = 1 _ 2 GJ.

LESSON 5-5, PAGE 368 37. ,A is the smallest ,, so

(( BC is the shortest side;

,C is the largest ,, so ((

AB is the longest side;From shortest to longest, the order is

(( BC , ((

AC , ((

AB .

38. ((

GH is the shortest side, so ,F is the smallest ,; ((

FH is the longest side, so ,G is the largest ,;From smallest to largest, the order is ,F, ,H, ,G.

39. x + 4.5 > 13.5 x > 9

4.5 + 13.5 > x 18 > x

Range of the values: > 9 cm and < 18 cm

40. 6.2 + 8.1 > 14.2 14.3 > 14.2 Yes; possible answer: the sum of each pair of 2 lengths is greater than the third length.

41. z + z 9 3z 2z : 3zNo; possible answer: when z = 5, the value of 3z is 15. So the 3 lengths are 5, 5, and 15. the sum of 5 and 5 is 10, which is not greater than 15. By the . Inequality Thm., a . cannot have these side lengths.


42. Possible answer: Given: .ABCProve: .ABC cannot have 2 obtuse /.Proof: Assume that .ABC has 2 obtuse /. Let ,A and ,B be the obtuse /. By the definition of obtuse, m,A > 90° and m,B > 90°. If the 2 inequalities are added, m,A + m,B > 180°. However, by the . Sum Theorem, m,A + m,B + m,C = 180°. So m,A + m,B = 180° ! m,C. But then 180° ! m,C > 180° by subst., and thus m,C < 0°. A . cannot have an , with a measure less than 0°. So the assumption that .ABC has 2 obtuse / is false. Therefore a . cannot have 2 obtuse /.

LESSON 5-6, PAGE 368 43. ((

PQ - ((

QR , ((

QS - ((

QS , and m,PQS < m,RQS. By the Hinge Theorem, PS < RS.

44. ((

BC - ((

DC , ((

AC - ((

AC , and AB < AD. By the Converse of the Hinge Theorem, m,BCA < m,DCA.

45. m,GFH < m,EFH 5n + 7 < 22 5n < 15 n < 3!1.4 < n < 3

m,GFH > 0 5n + 7 > 0 5n > !7 n > !1.4

46. XZ < JK4n ! 11 < 39 4n < 50 n < 12.5

XZ > 04n ! 11 > 0 4n > 11 n > 2.75

2.75 < n < 12.5

LESSON 5-7, PAGE 369 47. a 2 + b 2 = c 2

2 2 + 6 2 = x 2 40 = x 2 x = 2 " ## 10

48. a 2 + b 2 = c 2 x 2 + 8 2 = 14 2 x 2 = 132 x = 2 " ## 33

49. a 2 + b 2 = c 2 x 2 + (4.5 ) 2 = (7.5 ) 2 x 2 = 36 x = 6The side lengths do not form a Pythagorean triple because 4.5 and 7.5 are not whole numbers.

50. a 2 + b 2 = c 2 24 2 + 32 2 = x 2 1600 = x 2 x = 40The side lengths form a Pythagorean triple because they are nonzero whole numbers that satisfy

a 2 + b 2 = c 2 .

51. 9 + 12 = 21 > 16 ; 16 2 > 9 2 + 12 2 256 > 225The side lengths can form an obtuse ..

52. 11 + 14 = 25 : 27The side lengths cannot form a ..

53. 1.5 + 3.6 = 5.1 > 3.9 ; 3.9 2 > 1.5 2 + 3.6 2 15.21 = 15.21The side lengths can form a rt. ..

54. 2 + 3.7 = 5.7 > 4.1 ; 4.1 2 > 2 2 + 3.7 2 16.81 < 17.69The side lengths can form an acute ..

LESSON 5-8, PAGE 369 55. 45°-45°-90° .

x = 26 " # 2 56. 45°-45°-90° .

12 = x " # 2 12 " # 2 = 2x x = 6 " # 2

57. 45°-45°-90° .x = (16 " # 2 ) " # 2 = 32

58. 30°-60°-90° .48 = 2x x = 24y = x " # 3 = 24 " # 3

59. 30°-60°-90° .x = 6 " # 3 y = 2(6) = 12

60. 30°-60°-90° . 14 = x " # 3 14 " # 3 = 3x

x = 14 " # 3 _ 3

y = 2x

= 2 ( 14 " # 3 _ 3 ) = 28 " # 3 _

3

61. The diagonal forms two 45°-45°-90° 5. 30 = s " # 2 30 " # 2 = 2s s = 15 " # 2 0 21 ft 3 in.

62. The altitude forms two 30°-60°-90° 5. The shorter legs measure 9 ft.h = 9 " # 3 0 15 ft 7 in.


1. KL = JK = 9.8

2. m,WXY = 2m,WXZ = 2(17) = 34°

3. AC = BC2n + 9 = 5n ! 9 18 = 3n n = 6BC = 5(6) ! 9 = 21

4. RS = 2MS= 2(3.4) = 6.8

RQ = SQ = 4.9

5. m,DEF + m,EFD + m,FDE = 180 2m,GEF + 2(25) + 42 = 180 2m,GEF = 88 m,GEF = 44°distance from G to

(( DF = distance from G to

(( DE

= 3.7

6. XW = 2 _ 3 XC

= 2 _ 3 (261) = 174

BW = 1 _ 2 ZW

= 1 _ 2 (118) = 59

BZ = 3BW = 3(59) = 177


7. ((

JK is vertical, so the equation of the horizontal altitude is y = 4;

slope of ((

KL is !6 _ 6 = !1, so the slope of the altitude

is 1, and its equation is y ! 2 = x + 5, or y ! 7 = x.At the orthocenter, y = 4 and x = 4 ! 7 = !3. The orthocenter is at (!3, 4).

8. PR = 1 _ 2

HJ

= QJ = 51GJ = 2PQ

= 2(74) = 148m,GRP = m,GJH = 71°

9. Possible answer:Given: ,1 and ,2 form a lin. pair.Prove: ,1 and ,2 cannot both be obtuse /.Proof: Assume ,1 and ,2 are both obtuse /. By the definition of obtuse, m,1 > 90° and m,2 > 90°. If the 2 inequalities are added, m,1 + m,2 > 180°. However, by the Lin. Pair Theorem, ,1 and ,2 are supp. /. By the definition of supp. /, m,1 + m,2 = 180°. So m,1 + m,2 > 180° contradicts the given information. The assumption that ,1 and ,2 are both obtuse / is false. Therefore ,1 and ,2 cannot both be obtuse.

10. ((

BH is the shortest side, so ,E is the smallest ,. ((

BE is the longest side, so ,H is the largest ,.From smallest to largest, the order is ,E, ,B, ,H.

11. ,R is the smallest ,, so ((

TY is the shortest side.,Y is the largest ,, so

(( RT is the longest side.

From shortest to longest, the order is ((

TY , ((

RY , ((

RT .

12. AC + 114 > 247 AC > 133

114 + 247 > AC 361 > AC

Range of the distance: > 133 mi and < 361 mi.

13. ((

PS - ((

PZ , ((

PV - ((

PV , and SV < ZV. By the Converse of the Hinge Theorem, m,SPV < m,ZPV.

14. ((

DH - ((

KH , ((

HN - ((

HN , and DN < KN. m,DHN < m,KHN 4x ! 10 < 24 4x < 34 x < 8.5

m,DHN > 0° 4x ! 10 > 0 4x > 10 x > 2.5

2.5 < x < 8.5

15. x 2 + 21 2 = 24 2 x 2 = 135 x = 3 " ## 15 The side lengths do not form a Pythagorean triple

because 3 " ## 15 is not a whole number.

16. 18 + 20 = 38 > 27 ; 27 2 > 18 2 + 20 2 729 > 724The side lengths can form an obtuse ..

17. c 2 = 62 2 + 82 2 = 10,568 c = " ### 10,568 0 102 ft 10 in.

18. 20 = x " # 2 20 " # 2 = 2x x = 10 " # 2

19. 32 = 2x x = 16y = x " # 3 = 16 " # 3

20. 8 = x " # 3 8 " # 3 = 3x

x = 8 " # 3 _ 3

y = 2x

= 2 ( 8 " # 3 _ 3 ) = 16 " # 3 _

3


1. C;.ABC is a 30°-60°-90° ..AB = CB " # 3

= 3 " # 3 0 5.20 units

2. D;AB = 2DB = 12 cmAC = 2DE = 8 cmBC = 2BE = 8 cmP = 12 + 8 + 8 = 28 cm

3. B; c 2 = 2 2 + 5 2 c 2 = 29 c = " ## 29

4. E;8 + 10 = 18 : 18.1

5. D;The segment is horizontal, so the perpendicular bisector is vertical, with equation x = 6,


Solutions KeyPolygons and Quadrilaterals6

CHAPTER


1. F 2. B

3. A 4. D

5. E

6. Use ! Sum Thm.x ° + 42° + 32° = 180° x ° = 180° " 42° " 32° x ° = 106°

7. Use ! Sum Thm.x ° + 53° + 90° = 180° x ° = 180° " 53° " 90° x ° = 37°

8. Use ! Sum Thm.x ° + x ° + 32° = 180° 2 x ° = 180° " 34° 2 x ° = 146° x ° = 73°

9. Use ! Sum Thm.2 x ° + x ° + 57° = 180° 3 x ° = 180° " 57° 3 x ° = 123° x ° = 41°

10. By Lin. Pair Thm.,m#1 + 56 = 180 m#1 = 124°By Vert. $ Thm.,m#2 = 56°By Corr. $ Post.,m#3 = m#1 = 124°By Alt. Int. $ Thm., m#4 = 56°

11. By Alt. Ext. $ Thm.,m#2 = 101°By Lin. Pair Thm.,m#1 + m#2 = 180 m#1 + 101 = 180 m#1 = 79°Since ! % m, m & n ' ! % n, m#3 = m#4 = 90°.

12. By Same-Side Int. $ Thm., 3x + 2x = 180 5x = 180 x = 36By Lin. Pair Thm., m#1 + 3(36) = 180 m#1 + 108 = 180 m#1 = 72°By Corr. $ Post., m#2 = 3(36) = 108°

13. 45°"45°"90° !x = (11 ( ) 2 ) ( ) 2 = 11(2) = 22

14. 30°-60°-90° !14 = 2x x = 7

15. 45°-45°-90° !x = 3 ( ) 2

16. 30°-60°-90° !x = 2(8) = 16

17. T (Lin. Pair Thm.); if 2 $ are supp., then they form a lin. pair; F (counterexample: any supp. but non-adj. pair of $).

18. F (counterexample: a pair of $ with measure 30°); if 2 $ are rt $, then they are *; T (Rt. # * Thm.).

19. F (counterexample: ! with side lengths 5, 6, 10);if a triangle is an acute triangle, then it is a scalene triangle; F (counterexample: any equilateral triangle).

GEOMETRY LAB: CONSTRUCT REGULAR POLYGONS, PAGES 380–381

ACTIVITY 1: TRY THIS, PAGE 380 1. Possible answer: Draw a line !. Draw A and D on

!. Construct m % to ! through A. Construct n % to ! through D. With compass pt. at A, draw an arc that intersects m above !; label pt. of intersection B. With compass pt. at D and same compass setting, draw an arc that intersects n above !; label pt. of intersection C. Draw

++ BC . Polygon ABCD is a reg. quad.

2. The circle is circumscribed about the polygon.

3. Check students’ work. Possible answer:

ACTIVITY 2: TRY THIS, PAGE 380 4. Possible answer: 6 sides of ABCDEF were marked

off with same compass setting, so they are *. When

++ AD , ++

BE , and ++

CF are drawn, the 6 , formed are * and equil. Each # measure in an equil. triangle is 60°. Each # of the hexagon is formed by 2 of these 60° $, so the 6 $ of ABCDEF are *. Since it has 6 * sides and 6 * $, ABCDEF is a reg. hexagon.


5. Check students’ work. Possible answer: Draw

++ AD, ++

BE, and ++

CF . Construct the bisectors of #APB, #BPC, and #CPD, which are also bisectors of #DPE, #EPF, and #FPA. Connect the 6 pts. where the # bisectors intersect the circle to pts. A, B, C, D, E, and F in order around circle.

ACTIVITY 3: TRY THIS, PAGE 381 6. Check students’ work. Possible answer: Bisect

#EPF, #FPG, #GPH, #HPJ, and #JPE. Connect the 5 pts. where the # bisectors intersect circle to pts. E, F, G, H, and J in order around the circle.

7. 1st row: 90°, 108°, 120°2nd row: 360°, 540°, 720°

8. (n " 2)180° 9. (n " 2)180°

_ n

6-1 PROPERTIES AND ATTRIBUTES OF POLYGONS, PAGES 382–388

CHECK IT OUT! PAGES 382–385 1a. not a polygon b. polygon, nonagon

c. not a polygon

2a. regular, convex b. irregular, concave

3a. Think: Use Polygon # Sum Thm.(n " 2)180°(15 " 2)180°2340°

b. (10 " 2)180° = 1440°m#1 + m#2 + … + m#10 = 1440° 10m#1 = 1440° m#1 = 144°

4a. Think: Use Polygon Ext. # Sum Thm.m#1 + m#2 + … + m#12 = 360° 12m#1 = 360° m#1 = 30°

b. 4r + 7r + 5r + 8r = 360 24r = 360 r = 15

5. By Polygon Ext. # Sum Thm., sum of ext. # measures is 360°. Think: There are 8 * ext. $, so divide sum by 8.

m(ext. #) = 360° _ 8 = 45°

THINK AND DISCUSS, PAGE 385 1. Possible answers:

Concave pentagon Convex pentagon

A concave polygon seems to “cave in” or have a dent. A convex polygon does not have a dent.

2. Since polygon is not regular, you cannot assume that each of the ext. $ has the same measure.

3.


1. Possible answer: If a polygon is equil., all its sides are *, but all its $ are not necessarily *. For a polygon to be regular, all its sides must be *, and all its $ must be *.

2. polygon, decagon 3. not a polygon

4. polygon, quadrilateral 5. not a polygon

6. regular, convex 7. irregular, concave

8. irregular, convex

9. Think: Use Polygon # Sum Thm. (5 " 2)180° = 540°3z + 4z + 5z + 3z + 5z = 540 20z = 540 z = 27m#A = m#D = 3(27) = 81°m#B = 4(27) = 108°m#C = m#E = 5(27) = 135°

10. Think: Use Polygon # Sum Thm. (12 " 2)180° = 1800°m#1 + m#2 + … + m#12 = 1800° 12m#1 = 1800° m#1 = 150°

11. Think: Use Polygon # Sum Thm.(n " 2)180°(20 " 2)180°3240°

12. Think: Use Polygon Ext. # Sum Thm.4y + 2y + 6y + 4y = 360 16y = 360 y = 22.5


13. Think: Use Polygon Ext. # Sum Thm.m#1 + m#2 + … + m#5 = 360° 5m#1 = 360° m#1 = 72°

14. pentagon

15. By Polygon # Sum Thm., sum of # measures is (5 " 2)180 = 540°. Think: m#Q = m#S by def. of * $.m#P + m#Q + m#R + m#S + m#T = 540 90 + m#S + 90 + m#S + 90 = 540 2m#S = 270 m#Q = m#S = 135°


16. polygon, hexagon 17. not a polygon

18. polygon, quadrilateral 19. irregular, concave

20. regular convex 21. irregular, convex

22. 2n + 6n + 2n + 5n = (4 " 2)180 15n = 360 n = 24

m#R = m#T = 2(24) = 48° m#S = 6(24) = 144° m#V = 5(24) = 120°

23. 18m# = (18 " 2)180 18m# = 2880 m# = 160°

24. (7 " 2)180 = 900°

25. 9m(ext. #) = 360 m(ext. #) = 40°

26. 5a + 4a + 10a + 3a + 8a = 360 30a = 360 a = 12

27. 6m#JKM = (6 " 2)180 6m#JKM = 720 m#JKM = 120°

28. 6m#MKL = 360 m#MKL = 60°

29. x + x " 3 + 110 + 130 = (4 " 2)180 2x + 237 = 360 2x = 123 x = 61.5

30. 2(90) + 2x + 2(x + 22) = (6 " 2)180 4x + 224 = 720 4x = 496 x = 124

31. 5x = 360° x = 72°

32. m# = m(ext. #) n(m#) = n(m(ext. #))(n " 2)180 = 360 180n = 720 n = 4

33. m# = 4m(ext. #) n(m#) = 4n(m(ext. #))(n " 2)180 = 4(360) 180n = 1800 n = 10

34. m(ext. #) = 1 _ 8 m#

8n(m(ext. #)) = n(m#) 8(360) = (n " 2)180 3240 = 180n n = 18

35. (n " 2)180 = 540 n " 2 = 3 n = 5pentagon

36. (n " 2)180 = 900 n " 2 = 5 n = 7heptagon

37. (n " 2)180 = 1800 n " 2 = 10 n = 12dodecagon

38. (n " 2)180 = 2520 n " 2 = 14 n = 1616-gon

39. 360 = n(120) n = 3 m# = 180 " 120 = 60°

40. 360 = n(72) n = 5 m# = 180 " 72 = 108°

41. 360 = n(36) n = 10 m# = 180 " 36 = 144°

42. 360 = n(24) n = 15 m# = 180 " 24 = 156°

43. A; possible answer: this is not a plane figure, so it cannot be a polygon.

44.

Check students’ estimates; possible answer: pentagon is not equiangular; m#A = 100°; m#B = 113°; m#C = 113°; m#D = 101°; m#E = 113°; yes, pentagon is not equiangular.

45a. heptagon b. (7 " 2)180 = 900°

c. m#A + m#B + m#C + m#D + m#E + m#F + m#G = 90095 + 125 + m#F + 130 + 130 + m#F + m#F = 900 3m#F + 480 = 900 3m#F = 420 m#F = 140°

46. Let n be number of sides and s (= 7.5) be side length. P = ns45 = n(7.5) n = 6Polygon is a (regular) hexagon.

47. 48. Possible answer:

49. Possible answer: 50. Possible answer:

51. The figure has 6 sides, so it is a hexagon. The 6 sides are *, so the hexagon is equilateral. The 6 $ are *, so the hexagon is equiangular. Since the hexagon is equilateral and equiangular, it is regular. No diagonal contains pts. in the interior, so it is convex.


52. As number of sides increases, isosc. , formed by each side become thinner, and dists. from any pt. on base of each triangle to its apex approach same value. For a circle, each pt. is the same dist. from center. So polygon begins to resemble a circle.

TEST PREP, PAGE 388

53. A 54. H(16 "2)180 = 2520°

- 2880°

55. D49 + 107 + 2m#D + m#D = (4 " 2)180 3m#D = 204 m#D = 68° m#C = 2(68) = 136°


56. # measures are a, a + 4, …, a + 16, where a is a multiple of 4.a + a + 4 + … + a + 16 = (5 " 2)180 5a + 40 = 540 5a = 500 a = 100# measures are 100°, 104°, 108°, 112°, and 116°.

57. ++

PQ * ++

ST , ++

QR * ++

RS , and #Q * #S. So by SAS, !PRQ * !SRT. By CPCTC,

++ PR * ++

RT , so !PRT is isosc. By Isosc. ! Thm., #RTP * #RPT, so m#RTP = m#RPT = z°.By ! Sum Thm., 2z + y = 180 (1)By CPCTC and Isosc. ! Thm.,#PRQ * #SRT * # QPR * #RTSm#PRQ = m#SRT = m#QPR = m#RTS = x°Since PQRST is reg.,5m#QRS = (5 " 2)180 5(2x + y) = 540 2x + y = 108 (2) 5m#PTS = (5 " 2)180 5(y + z) = 540 y + z = 108 (3)

Subtr. (3) from (1):z = 180 " 108 = 72°Subst: in (3): y + 72 = 108 y = 36°Subst. in (2):2x + 36 = 108 2x = 72 x = 36°

58. ++

KA & ++

EF & ++

LC . By Alt. Int. $ Thm., #BLC * ext. #A and #CLD * ext. #Em#ALC = m#CLE = 360 _

10 = 36°

m#BLD = m#BLC + m#CLD = 72°

59. Yes, if you allow for # measures greater than 180°.

m#A + m#B + m#C + m#D + m#E + m#F = 720°


60. x 2 + 3x " 10 = 0(x + 5)(x " 2) = 0x = "5 or x = 2

61. x 2 " x " 12 = 0(x " 4)(x + 3) = 0x = 4 or x = "3

62. x 2 " 12x = "35 x 2 " 12x + 35 = 0 (x " 7)(x " 5) = 0 x = 7 or x = 5

63. x + 4 > 4 x > 04 + 4 > x 8 > x 0 < x < 8

64. Check x + 6 > 12 and 6 + 12 > x, since 6 < 12.x + 6 > 12 x > 6

65. Check x + 3 > 7 and 3 + 7 > x, since 3 < 7. x + 3 > 7 x > 43 + 7 > x 10 > x4 < x < 10

6 + 12 > x 18 > x6 < x < 18

66. c = 2a = 2(6) = 12

67. c = 2a10 = 2a a = 5 b = a ( ) 3 = 5 ( ) 3

CONNECTING GEOMETRY TO ALGEBRA RELATIONS AND FUNCTIONS, PAGE 389

TRY THIS, PAGE 389 1. D: all real numbers; R: all real numbers; function

2. D: all real numbers; R: 360; function

3. y = (x " 2)180

_ x = 180x " 360 _ x D: all real numbers except 0; R: all real numbers except 180; function

4. D: all real numbers except 0; R: all real numbers except 0; function

5. D: all real numbers; R: all real numbers; function

6. x 2 = 9 " y 2 . 9, and similarly y 2 . 9D: "3 . x . 3; R: "3 . y . 3; not a function

7. D: "2; R: all real numbers; not a function

8. y = x 2 + 4 / 4D: all real numbers; R: y / 4; function

9. D: all real numbers; R: all real numbers; function

GEOMETRY LAB: EXPLORE PROPERTIES OF PARALLELOGRAMS, PAGE 390

TRY THIS, PAGE 390 1. Check students’ work. They should obtain the same

results.

2. Possible answers: If a quad. is a 0, then its opp. sides are *. If a quad. is a 0, then its opp. $ are *. If a quad. is a 0, then its consecutive $ are supp. If a quad. is a 0, then its diags. intersect at their mdpts.

6-2 PROPERTIES OF PARALLELOGRAMS, PAGES 391–397

CHECK IT OUT! PAGES 392–394 1a.

++ KN * ++

LM KN = LM = 28 in.

b. #NML * #LKNm#NML = m#LKN

= 74°


c. O is mdpt. of ++

LN LO = 1 _

2 LN

= 1 _ 2 (26) = 13 in.

2a. ++

EJ * ++

JG EJ = JG3w = w + 8 2w = 8 w = 4 JG = (4) + 8 = 12

b. ++

FJ * ++

JH FJ = JH

4z " 9 = 2z 2z = 9

FH = 2JH = 2(2z) = 2(9) = 18

3. Step 1 Graph given pts.Step 2 Find slope of ++

PQ by counting units from P to Q.Rise from "2 to 4 is 6.Run from "3 to "1 is 2.Step 3 Start at S and count same # of pts.Rise of 6 from 0 is 6.Run of 2 from 5 is 7.Step 4 Use slope formula to verify that

++ QR & ++

PS .

slope of ++

QR = 6 " 4 _ 7 + 1

= 1 _ 4

slope of ++

PS = 0 + 2 _ 5 + 3

= 1 _ 4

Coords. of vertex R are (7, 6).


1. GHJN and JKLM are 1. 1. Given2. #N and #HJN are supp.;#K and #MJK are supp.

2. 0 ' cons. $ are supp.

3. #HJN * #MJK 3. Vert. $ Thm.4. #N * #K 4. * Supps. Thm.

THINK AND DISCUSS, PAGE 394 1. Measure of opp. # is 71°. Measure of each cons. #

is 180 " 71 = 109°.

2. XY = 21, WZ = 18, and YZ = 18; possible answer:since VWXY is a 0, opp. sides are *, so XY = VW = 21.

+++ WY is a diag., and by Thm. 6-2-4, the other

diag. bisects it, so WZ = YZ = 36 ÷ 2 = 18.

3.


1. Only 1 pair of sides is &. By def., a 0 has 2 pairs of & sides.

2. Possible answer:opp. sides:

++ PQ and

++ RS, ++

QR and

++ SP ; opp. $: #P and

#R, #Q and #S

3. E is mdpt. of ++

BD BD = 2DE = 2(18) = 36

4. ++

CD * ++

AB CD = AB = 17.5

5. E is mdpt. of ++

BD BE = DE = 18

6. #ABC and #BCD are supp.m#ABC = 180 " m#BCD = 180 " 110 = 70°

7. #ADC * #ABCm#ADC = m#ABC = 70°

8. #DAB * BCDm#DAB = m#BCD = 110°

9. ++

JK * ++

LM JK = LM 7x = 3x + 14 4x = 14 x = 3.5 JK = 7(3.5) = 24.5

10. LM = 3(3.5) + 14 = 24.5

11. #L and #M are supp. m#L + m#M = 180 2z " 3 + 5z " 6 = 180 7z = 189 z = 27m#L = 2(27) " 3 = 51°

12. m#M = 5(27) " 6 = 129°

13. Step 1 Graph given pts.Step 2 Find slope of ++

FG by counting units from F to G.Rise from 5 to 0 is "5.Run from "1 to 2 is 3.Step 3 Start at D and count same # of pts.Rise of "5 from 4 is "1.Run of 3 from "9 is "6.Step 4 Use slope formula to verify that

++ DF & ++

GH .

slope of ++

DF = 5 " 4 _ "1 + 9

= 1 _ 8

slope of ++

GH = "1 " 0 _ "6 " 2

= 1 _ 8

Coords. of vertex H are ("6, "1).


1. PSTV is a 0; ++

PQ * ++

RQ . 1. Given2. #STV * #P 2. 0 ' opp. $ *3. #P * #R 3. Isoc. ! Thm.4. #STV * #R 4. Trans. Prop. of *


15. JN = 1 _ 2 JL

= 1 _ 2 (165.8) = 82.9

16. LM = JK = 110

17. LN = JN = 82.9 18. m#JKL = m#JML = 50°

19. m#KLM = 180 " m#JML = 180 " 50 = 130°

20. m#MJK = m#KLM = 130°

21. WV = VYb + 8 = 5b 8 = 4b b = 2WV = (2) + 8 = 10


22. YW = 2WV = 2(10) = 20

23. XV = ZV3a " 7 = 2a a = 7 XZ = 2ZV = 2(2(7)) = 28

24. ZV = 2(7) = 14

25. slope from P to R: rise = 4, run = "6slope from V to T: rise of 4 from "1 is 3, run of "6 from 5 is "1Coords. of T are ("1, 3).


1. ABCD and AFGH are 1. 1. Given2. #C * #A, #A * #G 2. 0 ' opp. $ *3. #C * #G 3. Trans. Prop. of *

27. PQ = RS and SP = QR; given PQ = QR, all 4 side lengths are =. P = PQ + QR + RS + SP 84 = 4PQ PQ = QR = RS = SP = 21

28. PQ = RS, SP = QR = 3RS P = PQ + QR + RS + PS 84 = RS + 3RS + RS + 3RS 84 = 8RS PQ = RS = 10.5 SP = QR = 3(10.5) = 31.5

29. PQ = RS = SP " 7, QR = SP P = PQ + QR + RS + PS 84 = SP " 7 + SP + SP " 7 + SP 84 = 4SP " 14 98 = 4SPQR = SP = 24.5, PQ = RS = 24.5 " 7 = 17.5

30. PQ = RS, QR = SP = RS 2 P = PQ + QR + RS + PS84 = RS + RS 2 + RS + RS 2 84 = 2RS + 2RS 2 0 = RS 2 + RS " 42 0 = (RS + 7)(RS " 6)Since RS > 0, PQ = RS = 6, and QR = SP = 6 2 = 36.

31a. #3 * #1 (Corr. $ Post.)#6 * #1 (0 ' opp. $ *)#8 * #1 (0 ' opp. $ *)

b. #2 is supp. to #1 (0 ' cons. $ supp.), #4 is supp. to #1 (0 ' cons. $ supp.), #5 is supp. to #1 (0 ' cons. $ supp.), and #7 is supp. to #1 (Subst.).

32. #MPR * #RKM (0 ' opp. $ *)

33. #PRK * #KMP (0 ' opp. $ *)

34. ++

MT * ++

RT (0 ' diags. bisect each other)

35. ++

PR * ++

KM (0 ' opp. sides *)

36. ++

MP & ++

RK (Def. of 0)

37. ++

MK & ++

RP (Def. of 0)

38. #MPK * #RKP (Alt. Int. $ Thm.)

39. #MTK * #RTP (Vert. $ Thm.)

40. m#MKR + m#PRK = 180° (0 ' cons $ supp.)

41. By props. of 1, y = 61°x + 61 = 18° x = 119 z = x = 119°

42. By Alt. Int. $ Thm.,x = 90°By props. of 1,z = 53°By def. of comp. $,y = 90 " 53 = 37°

43. By Vert. $ Thm. and ! Sum Thm., 31 + 125 + x = 180 x = 24°y + (75 + 31) + 24 = 180 y = 50°By Alt. Int. $ Thm., z = y = 50°

44a. ++

CD b. #2

c. #4 d. opp. sides of a 0 are *

e. ASA f. CPCTC

g. bisect

45. Given: ABCD is a 0,Prove: #A and #B are supp.#B and #C are supp.#C and #D are supp.#D and #A are supp.

Statements Reasons

1. ABCD is a 0. 1. Given2. ++

AB & ++

CD , ++

BC & ++

DA 2. Def. of 0.3. #A and #B are supp., #B and #C are supp., #C and #D are supp., #D and #A are supp.

3. Same-side Int. $ Thm.

46. 2x = y 4x = 2y x = 2y " 93x = 9 x = 3 y = 2(3) = 6

47. y = x + 3y + 7 = 3x 7 = 2x " 3 10 = 2x x = 5y = 5 + 3 = 8

48a. #B * #D m#B = m#D6x + 12 = 9x " 33 45 = 3x x = 15m#B = 6(15) + 12 = 102°

b. m#A = m#C = 180 " m#B = 180 " 102 = 78°(0 ' cons. $ supp.)


a. No; possible answer: Drawings show acounterexample, since all side pairs are * but 1 are 2.

b. No; possible answer: For any given set of sidelengths, a 0 could have many different shapes.


50. Possible answer: a quad. is a 4-sided polygon. Since every 0 is a polygon with 4 sides, every 0 is a quad. A 0 has 2 pairs of & sides. Since sides of a quad. are not necessarily &, a quad. is not necessarily a 0.

TEST PREP, PAGE 397

51. A m#Q = m#S3x + 25 = 5x " 5 30 = 2x x = 15

52. J

53. 26.4P = AB + BC + CD + DA = CD + BC + CD + BC = 2(5 + 8.2) = 26.4


54. Let given pts. be A(0, 5), B(4, 0), C(8, 5), and possible 4th pts. be X, Y, Z. ++

AC is horiz., and AC = 38 " 04 = 8. So ++

XB is horiz., and X = (4 " 8, 0) = ("4, 0). Similarly,

++ BY is horiz., and Y = (4 + 8, 0) = (12, 0).

From C to B is rise of 5 and run of 4; rise of 5 from A is 5 + 10 = 10, run of 4 from A is 0 + 4 = 4; so Z = (4, 10).

55. Let given pts. be A("2, 1), B(3, "1), C("1, "4), and possible 4th pts. be X, Y, Z. From C to B is rise of 3 and run of 4; rise of 3 from A is 1 + 3 = 4, run of 4 from A is "2 + 4 = 2, so X = (2, 4).From B to C is rise of "3 and run of "4; rise of "3 from A is 1 " 3 = "2, run of "4 from A is "2 " 4 = "6, so Y = ("6, "2).From A to B is rise of "2 and run of 5; rise of "2 from C is "4 " 2 = "6, run of 5 from C is "1 + 5 = 4; so Z = (4, "6).

56.

Draw AD. ABCD and AFED are 1, so BC & AD and FE & AD by def. So #1 * #2 and #3 * #4 by the Alt. int. $ Thm. Thus m#1 = m#2 and m#3 = m#4. Then m#1 + m#3 = m#2 + m#4 by the Add. Prop. of =. By the # Add. Post., m#2 + m#4 = m#CDE. So m#1 + m#3 = #CDE. Since ABCD and AFED are *, with #1 corr. to #3. So m#1 + m#1 = m#CDE by subst. So 2m#1 = m#CDE, or y = 2x.

57.

Given: ABCD is a 0. 556 AE bisects #DAB.

556 BE bisects #CBA.

Prove: 556 AE %

556 BE

Statements Reasons

1. ABCD is a 0. 556 AE bisects #DAB. 556 BE bisects #CBA.

1. Given

2. #1 * #2, #3 * #4 2. Def. of # bisector 3.

++ BC & ++

AD 3. Def. of 0 4. #4 * #7 4. Alt. Int. $ Thm. 5. #3 * #7 5. Trans. Prop. of * 6.

++ AE *

++ AE 6. Reflex. Prop. of *

7. !ABE * !AFE 7. AAS 8. #5 * #6 8. CPCTC 9. #5 and #6 are supp. 9. Lin. Pair Thm.10. #5 and #6 are rt. $. 10. * $ supp. ' rt. $11.

556 AE %

556 BE 11. Def. of %


58. negative correlation 59. no correlation

60. alt. int. $ 61. alt. ext. $

62. same-side int. $ 63. corr. $

64. (n)120 = (n " 2)180 360 = 60n n = 6 sides6m(ext. #) = 360m(ext. #) = 60°

65. (n)135 = (n " 2)180 360 = 45n n = 8 sides

8m(ext. #) = 360 m(ext. #) = 45°

66. (n)156 = (n " 2)180 360 = 24n n = 15 sides

15m(ext. #) = 360 m(ext. #) = 24°

6-3 CONDITIONS FOR PARALLELOGRAMS, PAGES 398–405

CHECK IT OUT! PAGES 399–401 1. Think: Show that

++ PQ and

++ RS . are & and *.

a = 2.4 ' PQ = 7(2.4) = 16.8, RS = 2(2.4) + 12 = 16.8, so PQ * RSb = 9 ' m#Q = 10(9) " 16 = 74°, m#R = 9(9) + 25= 106°m#Q + m#R = 180°, so #Q and #R are supp.By Conv. of Alt. Int. $ Thm.,

++ PQ & ++

RS . So PQRS is a 0 by Thm. 6-3-1 since 1 pair of opposite sides is & and *..


2a. Yes; possible answer: the diag. of the quad. forms 2 , with 2 * pairs of $. By 3rd $ Thm., 3rd pair of $ in , are *. So both pairs of opp. $ of the quad. are *. By Thm. 6-3-3, quad. is a 0 .

b. No; 2 pairs of cons. sides are *, but none of the sets of conditions for a 0 are met.

3. Possible answers: Find slopes of both pairs of opp. sides.slope of

++ KL = 7 " 0 _ "5 + 3

= " 7 _ 2

slope of +++

MN = "2 " 5 _ 5 " 3

= " 7 _ 2

slope of ++

LM = 5 " 7 _ 3 + 5

= " 1 _ 4

slope of ++

KN = "2 " 0 _ 5 + 3

= " 1 _ 4

Since both pairs of opp. sides are &, KLMN is a 0 by definition.

4. Possible answer: by Thm. 6-3-2, ABRS is a 0.Since

++ AB is vert. and

++ RS & ++

AB , ++

RS is vert., so # of binoculars stays the same.

THINK AND DISCUSS, PAGE 401 1. Possible answer: Conclusion of each thm. is “The

quad. is a 0.”

2. Possible answer: In Lesson 6-2, “A quad. is a 0,” is the hypothesis of each thm., rather than the conclusion.

3.


1. Step 1 Find ++

EJ and ++

JG .EJ = t + 12 JG = 3tEJ = 6 + 12 = 18 JG = 3(6) = 18Step 2 Find

++ FJ and

++ JH .

FJ = 2s JG = s + 5FJ = 2(5) = 10 JG = 5 + 5 = 10Since EJ = JG and FJ = JH, EFGH is a 0 by Thm. 6-3-5 since its diagonals bisect each other.

2. #L = 5m + 36#L = 5(14) + 36 = 106°#P = 6n " 1#P = 6(12.5) " 1 = 74°#Q = 4m + 50#Q = 4(14) + 50 = 106°Since 106° + 74° = 180°, #P is supp. to both #L and #Q. KLPQ is a 0 by Thm. 6-3-4 since an angle is supp. to both its cons $.

3. Yes; both pairs of opp. $ of the quad. are *. By Thm. 6-3-3, the quad. is a 0.

4. No; 1 pair of opp. sides of quad. are *. 1 diag. is bisected by other diag. None of the conditions for a 0 are met.

5. Yes; possible answer: a pair of alt. int. $ are *, so 1 pair of opp. sides are &. The same pair of opp. sides are *. By Thm. 6-3-1, quad. is a 0.

6. Possible answer: Find slopes of both pairs of opp. sides.slope of

+++ WX = 3 + 2 _ "3 + 5

= 5 _ 2

slope of ++

YZ = 0 " 5 _ 1 " 3

= 5 _ 2

slope of ++

XY = 5 " 3 _ 3 + 3

= 1 _ 3

slope +++

WZ = 0 + 2 _ 1 + 5

= 1 _ 3

Since both pairs of opp. sides are &, WXYZ is a 0 by definition.

7. Possible answer: Find slopes of both pairs of opp. sides.slope of

++ RS = "1 + 5 _ "2 + 1

= "4

slope of ++

TU = "5 + 1 _ 5 " 4

= "4

slope of ++

ST = "1 + 1 _ 4 + 2

= 0

slope of ++

RU = "5 + 5 _ 5 + 1

= 0

Since both pairs of opp. sides are &, RSTU is a 0 by def.

8. Since ++

AD & ++

BC and ++

AD * ++

BC , ABCD is a 0 by Thm. 6-3-1, so

++ AB & ++

CD by def. of 0.


9. BC = 3(3.2) + 7 = 16.6, GH = 8(3.2) " 9 = 16.6BH = 3(7) + 7 = 28, CG = 6(7) " 14 = 28BCGH is a 0 by Thm. 6-3-2.


10. UV = 10(19.5) " 6 = 189, TW = 8(19.5) + 33 = 189m#V = 2(22) + 41 = 85°, m#W = 7(22) " 59 = 95°

++

UV * +++

TW ; #V and #W are supp., so by Conv. of Same-Side Int. $ Thm.,

++ UV & +++

TW . TUVW is a 0 by Thm. 6-3-1.

11. Yes; both pairs of opp. sides are *, since all sides are *, so quad is a 0 by Thm. 6-3-2.

12. Yes; by # Add. Post., 1 pair of opp. $ are *, and by 3rd $ Thm., 2nd pair of opp. $ are *. So quad is a 0 by Thm. 6-3-3.

13. No; by looking at the angles, we can be sure that one pair of sides are &. This is not enough.

14. slope of ++

JK = 7 _ "2

= " 7 _ 2 ; slope of

++ LM = "7 _

2 = " 7 _

2

slope of ++

KL = "1 _ 5 = " 1 _

5 ; slope of

++ JM = "1 _

5 = " 1 _

5

both pairs of opp. sides have the same slope, so ++

JK & ++

LM and ++

KL & ++

MJ ; JKLM is a 0 by definition.

15. slope of ++

PQ = 5 _ 3 ; slope of

++ RS = "5 _ "3

= 5 _ 3

slope of ++

QR = "6 _ 6 = "1; slope of

++ PS = "6 _

6 = "1.

So, ++

PQ & ++

RS and ++

PS & ++

QR .PQRS is a 0 by def.

16. Possible answer: The brackets are always the same length, so it is always true that AB = CD. The bolts are always the same dist. apart, so it is always true that BC = DA. By Thm. 6-3-2, ABCD is always a 0. The side

++ AD stays horiz. no matter how you move

the tray. Since ++

BC & ++

AD , ++

BC stays horiz. Since ++

BC holds the tray in position, the tray will stay horiz. no matter how it is moved.

17. No; the given # measures only indicate that 1 # of the quad. is supp. to 1 of its cons. $. By Thm. 6-3-4, you must know that 1 # is supp. to both of its cons. $ in order to conclude that quad. is a 0.

18. No; you are only given the measures of the 4 $ formed by the diags. None of the sets of conditions for a 0 are met.

19. Yes; diags. of the quad. bisect each other. By Thm. 6-3-5, the quad. is a 0.

20. Think: Opp. sides must be *.2a + 6 = 3a " 10 16 = a

6b " 3 = 5(16) + 1 6b = 84 b = 14

21. Think: Middle # must be supp. to cons. $. 4a " 8 + 8a " 10 = 180 12a = 198 a = 16.54(16.5) " 8 + 5b + 6 = 180 5b = 116 b = 23.2

22. Think: Diags. must bisect each other.5b " 7 = 3b + 6 2b = 13 b = 6.5

2a = 3(6.5) " 5 2a = 14.5 a = 7.25

23. Think: 1 pair of opp. sides must be * and &. For conditions of Conv. of Alt. Int. $ Thm., given $ must be *.3a + 1.8 = 4a " 6.6 8.4 = a

1.4b = b + 80.4 b = 8

b = 20

24. Possible answer:If the diags. of a quad. are *, you cannot necessarily conclude that the quad. is a 0.

25. Possible answer: The red and green , are isosc. rt. ,, so the measure of each acute # of the , is 45°. Each of the smaller $ of the yellow stripe is comp. to 1 of the acute $ of the rt. ,, so the measure of each of the smaller $ of the yellow stripe is 90° " 45° = 45°. Each of the larger $ of the yellow stripe is supp. to 1 of the acute $ of the rt. ,, so the measure of each of the larger $ of the yellow stripe is 180° " 45° = 135°. So the yellow stripe is quad. in which both pairs of opp. $ are *. By Thm. 6-3-3, the shape of the yellow stripe is a 0.

26a. Reflex. Prop. of * b. !BCD

c. SSS d. #3

e. #2 f. Conv. of Alt. Int. $ Thm.

g. def. of 0

27a. #Q b. #S

c. ++

SP d. ++

RS

e. 0

28. Given: ABCD is a 0, E is the mdpt. of ++

AB , and F is the mdpt. of

++ CD .

Prove: AEFD and EBCF are 1.Proof: Since ABCD is a 0,

++ AB & ++

CD , so ++

AE & ++

DF and

++ EB & ++

FC . Since opp. sides of a 0 are *, ++

AB * ++

CD . It is given that E is the mdpt. of ++

AB , and F is the mdpt. of

++ CD . Because these two segs. are *,

it follows that ++

AE * ++

EB * ++

DF * ++

FC . Since ++

AE & ++

DF and

++ AE * ++

DF , AEFD is a 0. Similarly, EBCF is a 0.


1. #E * #G, #F * #H 1. Given2. m#E = m#G, m#F = m#H 2. Def. of * $3. m#E + m#F + m#G

+ m#H = 360°3. Polygon Sum

Thm.4. m#E + m#F + m#E

+ m#F = 360°, m#E + m#H + m#E + m#H = 360°

4. Subst.

5. 2m#E + 2m#F = 360°, 2m#E + 2m#H = 360°

5. Distrib. Prop.of =

6. m#E + m#F = 180°, m#E + m#H = 180°

6. Div. Prop. of =

7. #E is supp. to #F and #H. 7. Def. of supp. $8. ++

EF & ++

GH , ++

FG & ++

HE 8. Conv. of Same-Side Int. $ Thm.

9. EFGH is a 0. 9. Def. of 0



1. ++

JL and ++

KM bisect each other. 1. Given2. ++

JN * ++

LN , ++

KN * ++

MN 2. Def. of bisect3. #JNK * #LNM, #KNL * #MNJ

3. Vert. $ Thm.

4. !JNK * !LNM, !KNL * !MNJ

4. SAS

5. #JKN * #LMN, #KLN * #MJN

5. CPCTC

6. ++

JK & ++

LM , ++

KL & ++

MJ 6. Conv. of Alt Int. $ Thm.

7. JKLM is a 0. 7. Def. of 0


Given: ++

DE and ++

EF are midsegments of !ABC.Prove: ADEF is a 0.

Statements Reasons

1. ++

DE and ++

EF are midsegs. of !ABC.

1. Given

2. ++

DE & ++

FA , ++

AD & ++

EF 2. ! Midseg. Thm.3. ADEF is a 0. 3. Def. of 0

32. Possible answer: A quad. is a 0 if and only if both pairs of opp. sides are *. A quad. is a 0 if and only if both pairs of opp. $ are *. A quad. is a 0 if and only if its diags. bisect each other.


Draw line !. Draw P, not on !. Draw a line through P that intersects ! at Q. Construct m & to ! through P. Place the compass point at Q and mark off a seg. on !. Label the second endpoint of this seg as R. Using the same compass setting, place the compass point at P and mark off a * seg. on m. Label the second endpoint of this seg. S. Draw

++ RS .

Since ++

PS & ++

QR and ++

PS * ++

QR , PSRQ is a 0 by Thm. 6-3-1.

34a. No; none of sets of conditions for a 0 are met.

b. Yes; since #S and #R are supp., ++

PS & ++

QR . Thus PQRS is a 0 by Thm. 6-3-1.

c. Yes; draw ++

PR . #QPR * #SRP (Alt. Int. $ Thm.) and ++

PR * ++

PR (Reflex Prop. of *). So !QPR * !SRP (AAS), and

++ PQ * ++

SR (CPCTC). Since ++

PQ & ++

SR and

++ PQ * ++

SR , PQRS is a 0 by Thm. 6-3-1.

TEST PREP, PAGE 405

35. BBy Conv. of Alt. Int. $ Thm.,

+++ WX & ++

YZ ; need +++

WX * ++

YZ to meet conditions of Thm. 6-3-1.

36. GSlope of

++ AB : rise of 4 and run of 2, or rise of "4 and

run of "2; rise of ±4 from C is 1 ± 4 = 5 or "3, run of ±2 from C is 6 ± 2 = 8 or 4. So D could be at (8, 5) or (4, "3).

37. No; possible answer: slope of ++

RS = 3 _ 4 , slope of

++

TV = 1; ++

RS and ++

TV do not have same slope, so ++

RS 7 +++

TV; ++

RS and ++

TV are opp. sides of RSTV; by def., both pairs of opp. sides of a 0 are &, so RSTV is not a 0.


38. The top and bottom of each step form a small 0 with the back of the stairs and the base of the railing. The vertices of each 0 have joints that allow the pieces to move. But the lengths of the sides of 1 stay the same. Since they start out as 1 with opp. sides that are *, and the lengths do not change, they remain 1. Therefore the top and bottom of each step, and thus also the upper platform, remain & to ground regardless of the position of the staircase.

39. Let intersection and vertices be P("2, 1.5), A("7, 2), B(2, 6.5), C(x, y), and D(u, v). P is mdpt. of

++ AC

and ++

BD . ("2, 1.5) = ( "7 + x _

2 ,

2 + y _

2 )

x = 3, y = 1; C = (3, 1)

("2, 1.5) = ( 2 + u _ 2 , 6.5 + v _

2 )

u = "6, v = "3.5; D = ("6, "3.5)


Draw F collinear with D and E such that ++

DE * ++

EF . Since E is the mdpt. of

++ BC , ++

CE * ++

EB . By the Vert. $ Thm., #CED * #BEF. Thus !CED * !BEF by SAS. By CPCTC,

++ CD *

++ FB . Since D is the mdpt.

of ++

AC , ++

CD * ++

AD . So by the Trans. Prop. of *, ++

AD *

+++ FB. Also by CPCTC, #CDE * #BFE. By Conv.

of Alt. Int. $ Thm., ++

AC & ++

FB . Thus DFBA is a 0 since 1 pair of opp. sides are & and *. Since DFBA is a 0,

++ DE & ++

AB by definition. Since opp. sides of a 0 are *,

++ AB * ++

DF and AB = DF by the def. of * segs. Since

++ DE *

++ EF , E is the mdpt. of

++ DF , and DE

= 1 _ 2 DF. By subst., DE = 1 _

2 AB.


41. x "5 "2 0 0.5y "38 "17 "3 0.5


42. x "5 "2 0 0.5y "1.5 0 1 1.25

43. x "5 "2 0 0.5y 77 14 2 2.75

44. It is given that ++

BC * ++

DA and that #DBC * #BDA. By Reflex. Prop. of *,

++ DB * ++

DB . Therefore !ABD * !CDB by SAS.

45. Possible answer: It is given that +++

TW * +++

VW . Because #UWV is a rt. # and is supp. to #UWT, #UWT is also a rt. #. Thus #UWV * #UWT. By the Reflex. Prop. of *,

+++ UW *

+++ UW . Therefore

!TUW * !VUW by SAS.

46. KN = NMx + 6 = 2x " 2 8 = xNM = 2(8) " 2 = 14

47. JK = LM16z " 4 = 8z + 4 8z = 8 z = 1LM = 8(1) + 4 = 12

48. JN = NL 4y = 3y + 2 y = 2 JL = 2JN = 2(4y) = 8y = 8(2) = 16

49. JK = LM = 12

6A MULTI-STEP TEST PREP, PAGE 406

1. pentagon;By Same-Side Int. $ Thm., m#D + m#E = 180 m#D + 65 = 180 m#D = 115° = m#C

m#A + m#B + m#C + m#D + m#E = (5 " 2)180 m#A + 120 + 115 + 115 + 65 = 540 m#A = 125°

2. Since FGHJ is a 0, #F and #J are supp. So9x " 13 + 7x + 1 = 180 16x = 192 x = 12

Opp. $ are *, so m#G = m#J = 7(12) + 1 = 85°

3. Yes; #NQP * #QNM, so +++

MN & ++

QP by Conv. of Alt. Int. $ Thm. So MNPQ is a 0 by Thm. 6-3-1.

6A READY TO GO ON? PAGE 407

1. polygon; octagon 2. not a polygon

3. not a polygon 4. polygon; pentagon

5. (n " 2)180°(16 " 2)180°2520°

6. (n)m# = (n " 2)180 6m# = 720 m# = 120°

7. 14z + 8z + 7z + 11z = 360 40z = 360 z = 9Ext. # measures are 14(9) = 126°, 8(9) = 72°, 7(9) = 63°, and 11(9) = 99°.

8. 10 m(ext. #) = 360m(ext. #) = 36°

9. N is mdpt. of ++

KM .KM = 2KN = 2(13.5) = 27 cm

10. ++

KJ * ++

LM KJ = LM = 17 cm

11. MN = KN = 13.5 cm

12. #JKL and #KJM are supp.m#JKL + m#KJM = 180 m#JKL + 102 = 180 m#JKL = 78°

13. #JML * #JKLm#JML = m#JKL

= 78°

14. #KLM * #KJMm#KLM = m#KJM

= 102°

15. slope from B to C: rise of "5 and run of 1rise of "5 from A is 1 " 5 = "4; run of 1 from A is "3 + 1 = "2; D = ("2, "4)

16. WX = YZ6b " 7 = 10b " 19 12 = 4b b = 3WX = 6(3) " 7 = 11

17. YZ = WX = 11

18. #X and #W are supp. m#X + m#Y = 1805a " 39 + 3a + 27 = 180 8a = 192 a = 24 m#X = 5(24) " 39 = 81°

19. m#W = 3(24) + 27 = 99°

20. x = 6 ' RS = 7(6) + 6 = 48, TV = 9(6) " 6 = 48y = 4.5 ' RV = 8(4.5) " 8 = 28, ST = 6(4.5) + 1 = 28RS * TV, ST * RV ' RSTV is a 0 (Thm. 6-3-2)

21. m = 12 ' m#G = 2(12) + 31 = 55°, m#J = 7(12) " 29 = 55°n = 9.5 ' m#K = 12(9.5) + 11 = 125°#K supp. to #G, and #J ' GHJK is a 0 (Thm. 6-3-4).

22. Yes; both pairs of opp. sides are &, so quad. is a 0 by definition.

23. No; one pair of opposites $ of the quad. are *. None of the sets of conditions for a 0 are met.

24. No; the diagonals are divided into two segments at their point of intersection, and each segment of one diagonal is * to a segment of the other diagonal. None of the sets of conditions for a 0 are met.

25. Slope of ++

CD = 4 _ 5 ; slope of

++ EF = "4 _ "5

= 4 _ 5

slope of ++

DE = "2 _ 6 = " 1 _

3 ; slope of

++ FC = "2 _

6 = " 1 _

3

Both pairs of opp. sides are &, so quad. is a 0 by definition.


6-4 PROPERTIES OF SPECIAL PARALLELOGRAMS, PAGES 408–415

CHECK IT OUT! PAGES 408–411 1a. Think: rect. ' 0

' opp. sides * ++

HJ * ++

GK HJ = GK = 48 in.

b. Think: rect. ' diags. * ++

HK * ++

GJ HK = GJ = 2JL = 2(30.8) = 61.6 in.

2a. CG = GF 5a = 3a + 17 2a = 17 a = 8.5 CD = CG = 5(8.5) = 42.5

b. Think: rhombus ' 0 ' cons. $ supp.m#GCD + m#CDF = 180 b + 3 + 6b " 40 = 180 7b = 217 b = 31

Think: Use Thm. 6-4-5.m#GCH = 1 _

2 m#GCD

= 1 _ 2

(31 + 3) = 17°

3. Step 1 Show that ++

SV and +++

TW are *.

SV = 8 )))) 11 2 + 1 2 = ( )) 122

TW = 8 )))) 1 2 + 11 2 = ( )) 122 Since SV = TW,

++ SV *

+++ TW .

Step 2 Show that ++

SV and +++

TW are %.slope of

++ SV = 1 _

11 ; slope of

+++ TW = "11 _

1 = "11

Since ( 1 _ 11

) ("11) = "1, ++

SV and +++

TW are %.

Step 3 Show that ++

SV and +++

TW bisect each other.

mdpt. of ++

SV = ( "5 + 6 _ 2 , "4 " 3 _

2 ) = ( 1 _

2 , " 7 _

2 )

mdpt. of +++

TW = ( 0 + 1 _ 2 , 2 " 9 _

2 ) = ( 1 _

2 , " 7 _

2 )

Since ++

SV and +++

TW have same mdpt., they bisect each other.Diags. are * % bisectors of each other.

4. Possible answer:

Statements Reasons

1. PQTS is a rhombus. 1. Given2. ++

PT bisects #QPS. 2. Thm. 6-4-53. #QPR * #SPR 3. Def. of # bisector4. ++

PQ * ++

PS 4. Def. of rhombus5. ++

PR * ++

PR 5. Reflex. Prop. of *6. !QPR * !SPR 6. SAS7. ++

RQ * ++

RS 7. CPCTC

THINK AND DISCUSS, PAGE 411 1. Thm. 6-4-2; possible answer: when the thm. is

written as a conditional statement, it is easier to identify the hypothesis and the conclusion.

2. Same properties: 2 pairs of & sides, opp. sides *, opp. $ *, cons. $ supp., diags. bisect each other;Special properties: 4 * sides, % diags., each diag. bisects a pair of opp. $

3.


1. rhombus; rectangle; square

2. rect. 0 ' diags. bisect each otherTQ = 1 _

2 QS

= 1 _ 2 (380) = 190 ft

3. ++

PQ * ++

RS PQ = RS = 160 ft

4. T is mdpt. of ++

QS .ST = TQ = 190 ft

5. rect. ' diags. are * ++

PR * ++

QS PR = QS = 380 ft

6. BC = CD4x + 15 = 7x + 2 13 = 3x x = 4 1 _

3

AB = BC

= 4 (4 1 _ 3 ) + 15 = 32 1 _

3

7. ++

AC % ++

BD m#AFB = 90 12y = 90 y = 7.5

m#ABC + m#BCD = 180 m#ABC + 2m#FCD = 180m#ABC + 2(4(7.5) " 1) = 180 m#ABC + 58 = 180 m#ABC = 122°


JL and ++

KM are *.

JL = 8 )))) 5 2 + 7 2 = ( )) 74

KM = 8 )))) 7 2 + 5 2 = ( )) 74 Since JL = KM,

++ JL * ++

KM .Step 2 Show that

++ JL and

++ KM are %.

slope of ++

JL = 7 _ 5 ; slope of

++ KM = "5 _

7 = " 5 _

7

Since ( 7 _ 5 ) (" 5 _

7 ) = "1,

++ JL and

++ KM are %.

Step 3 Show that ++

JL and ++

KM bisect each other.

mdpt. of ++

JL = ( "3 + 2 _ 2 , "5 + 2 _

2 ) = (" 1 _

2 , " 3 _

2 )

mdpt. of ++

KM = ( "4 + 3 _ 2 , 1 " 4 _

2 ) = (" 1 _

2 , " 3 _

2 )

Since ++

JL and ++

KM have same mdpt., they bisect each other.Diags. are * % bisectors of each other.


9. Possible answer:

Statements Reasons

1. RECT is a rect.; ++

RX * ++

TY 1. Given

2. ++

XY * ++

XY 2. Reflex. Prop. of * 3. RX = TY, XY = XY 3. Def. of * segs. 4. RX + XY = TY + XY 4. Add. Prop. of = 5. RX + XY = RY,

TY + XY = TX 5. Seg. Add. Post.

6. RY = TX 6. Subst. 7.

++ RY *

++ TX 7. Def. of * segs.

8. #R and #T are rt. $. 8. Def. of rect. 9. #R * #T 9. Rt. # * Thm.10. RECT is a 0. 10. Rect. ' 011.

++ RE *

++ CT 11. 0 ' opp. sides *

12. !REY * !TCX 12. SAS


10. JL = 2JP = 2(14.5) = 29 in.

11. KL = JM = 25 in.

12. KM = JL = 29 in. 13. MP = 1 _ 2

KM

= 1 _ 2

(29) = 14 1 _ 2 in.

14. WX = XY9a " 18 = 3a + 15 6a = 33 a = 5.5

VW = WX = 9(5.5) " 18 = 31.5

15. m#XZW = 90 10b " 5 = 90 10b = 95 b = 9.5

m#VWX + m#WVY = 180m#VWX + 4(9.5) + 10 = 180 m#VWX = 132°

m#WYX = 1 _ 2 m#VYX

= 1 _ 2 m#VWX

= 1 _ 2 (132) = 66°


PR and ++

QS are *.

PR = 8 )))) 11 2 + 5 2 = ( )) 146

QS = 8 )))) 5 2 + 11 2 = ( )) 146 Since PR = QS,

++ PR * ++

QS .Step 2 Show that

++ PR and

++ QS are %.

slope of ++

PR = "5 _ 11

= " 5 _ 11

; slope of ++

QS = "11 _ "5

= 11 _ 5

Since (" 5 _ 11

) ( 11 _ 5 ) = "1,

++ PR and

++ QS are %.

Step 3 Show that ++

PR and ++

QS bisect each other.

mdpt. of ++

PR = ( "4 + 7 _ 2 , 0 " 5 _

2 ) = ( 3 _

2 , " 5 _

2 )

mdpt. of ++

QS = ( 4 " 1 _ 2 , 3 " 8 _

2 ) = ( 3 _

2 , " 5 _

2 )

Since ++

PR and ++

QS have same mdpt., they bisect each other.Diags. are * % bisectors of each other.


Statements Reasons

1. RHMB is a rhombus. ++

HB is a diag. of RHMB.1. Given

2. +++

MH * ++

RH 2. Def. of rhombus3. ++

HB bisects #RHM. 3. Rhombus ' each diag. bisects opp. $

4. #MHX * #RHX 4. Def. of # bisector5. ++

HX * ++

HX 5. Reflex. Prop. of *6. !MHX * !RHX 6. SAS7. #HMX * #HRX 7. CPCTC

18. m#1 = 90 " 61 = 29° (comp. $)m#2 = 61° (Alt. Int. $ Thm.)m#3 = 90° (def. of rect.)m#4 = m#1 = 29° (Alt. Int. $ Thm.)m#5 = 90° (def. of rect.)

19. m#1 = 90 " 36 = 54° (comp. $)m#2 = 36° (diags. * ' , * by SSS, $ * by CPCTC)m#3 = 90 " m#2 = 54° (comp. $)m#4 = 180 " (m#2 + 36) = 108° (! Sum Thm., Alt. Int. $ Thm.)m#5 = 180 " m#4 = 72° (supp. $)

20. m#1 = 90° (rect. is a rhombus, Thm. 6-4-4)m#2 = m#3, m#2 + m#3 = 90°' m#2 = m#3 = 45° (Thm 6-4-5, comp. $)m#4 = 45° (same reasoning as #2, #3)m#5 = m#3 = 45° (rect. ' 0, Alt. Int. $ Thm.)

21. m#2 = 27° (Isosc. ! Thm.)m#1 = 180 " (27 + 27) = 126° (! Sum Thm.)m#3 = m#2 = 27° (Thm. 6-4-5)m#4 = m#1 = 126° (rhombus ' 0 ' opp. $ *)m#5 = 27° (Thm. 6-4-5)

22. m#1 = m#2, m#1 + m#2 + 70 = 180 ' m#1 = m#2 = 55° (Isosc. ! Thm.)m#3 = m#2 = 55° (Thm. 6-4-5)m#4 = 70° (rhombus ' 0 ' opp. $ *)m#5 = m#1 = 55° (Thm. 6-4-5)

23. m#1 = 90 " 26 = 64° (Thm. 6-4-4, comp. $)m#2 = m#1 = 64° (Thm. 6-4-5)m#3 = 26° (rhombus ' 0, Alt. Int. $ Thm.)m#4 = 90° (Thm. 6-4-4)m#5 = m#2 = 64° (rhombus ' 0, Alt. Int. $ Thm.)

24. always (Thm. 6-4-1) 25. sometimes


28. always (all 4 sides *) 29. always (has 4 sides)

30. always (4 rt. $) 31. sometimes

32. No; possible answer: a rhombus with int $ that measure 70°, 110°, 70°, and 110° is equliateral, but it is not equiangular. A rect. with side lengths 5, 7, 5, and 7 is equiangular, but it is not equilateral.


33a. 1. polygon

2. polygon

3. polygon

4. polygon

5. Not a polygon

b. 1. triangle; reg. 2. quad.; reg. 3. hexagon; reg. 4. quad.; irreg.

c. Shape 2 appears to be a square.Shape 4 appears to be a rhombus.

d. Assume polygon is reg.6m# = (6 " 2)180 = 720 m# = 120°

34. You cannot use Thm. 6-2-1 to justify the final statement because you do not know that JKLM is a 0. That is what is being proven. Instead, Thm. 6-3-2 states that if both pairs of opp. sides of a quad. are *, then the quad. is a 0. So JKLM is a 0 by Thm. 6-3-2.

35a. rect. ' 0

c. reflex. Prop. of *

e. #GHE

g. CPCTC

b. ++

HG

d. def. of rect.

f. SAS

36a. slope of ++

AB = "2 _ 2 = "1; slope of

++ CD = 2 _ "2

= "1

slope of ++

BC = "5 _ "5

= 1; slope of ++

AD = "5 _ "5

= 1

b. Rect.; adj. sides are %.

c. By Thm. 6-4-2, the diags. of a rect. are *.


Statements Reasons

1. VWXY is a rhombus. 1. Given 2.

+++ WX * ++

YX 2. Def. of rhombus 3. VWXY is a 0. 3. Rhombus ' 0 4.

++ WZ *

++ YZ 4. 0 ' diags. bisect

each other 5.

++ XZ *

++ XZ 5. Reflex. Prop. of *

6. !WZX * !YZX 6. SSS 7. #WZX * #YZX 7. CPCTC 8. #WZX and #YZX are

supp. 8. Lin. Pair Thm.

9. #WZX and #YZX are rt. $.

9. * $ supp. ' rt. $

10. m#WZX = m#YZX = 90°

10. Def. of rt. #

11. ++

VX % +++

WY 11. Def. of %

38. Possible answer: It is given that ABCD is a rect. By def. of a rect., #A, #B, #C, and #D are rt. $. So #A * #C and #B * #D because all rt. $ are *. By Thm. 6-3-3, ABCD is a 0.


Statements Reasons

1. ABCD is a rhombus. 1. Given2. ABCD is a 0. 2. Rhombus ' 03. #B * #D, #A * #C 3. 0 ' opp. $ *4. ++

AB * ++

BC * ++

CD * ++

DA 4. Def. of rhombus5. E, F, G and H are the

mdpts. of sides.5. Given

6. ++

EB * ++

BF * ++

HD * ++

DG, ++

EA * ++

AH * ++

FC * ++

CG 6. Def. of mdpt.

7. !BEF * !DGH, !AEH * !CGF

7. SAS

8. ++

EF * ++

GH , ++

EH * ++

GF 8. CPCTC9. EFGH is a 0. 9. Quad. with opp.

sides * ' 0

40. 5 = 2ww = 2.5 cm ! = w ( ) 3 = 2.5 ( ) 3 cm P = 2! + 2w = 2(2.5) + 2 (2.5 ( ) 3 ) = 5 + 5 ( ) 3 cm 9 13.66 cm A = !w = (2.5 ( ) 3 ) (2.5) = 6.25 ( ) 3 cm 9 10.83 cm 2

41. s = 7 ( ) 2 in.P = 4s = 28 ( ) 2 in. 9 39.60 in. A = s 2 = (7 ( ) 2 ) 2 = 98 in. 2

42. s = 8 )))) 3 2 + 4 2 = 5 cmP = 4s = 20 cm

A = 4 ( 1 _ 2 (3)(4)) = 24 cm

2

43a. By def., a square is a quad. with 4 * sides. So it is true that both pairs of opp. sides are *. Therefore, a square is a 0 by Thm. 6-3-2.

b. By def., a square is a quad. with 4 rt. $ and 4 * sides. So a square is a rect., because by def., a rect. is a quad. with 4 rt. $.

c. By def., a square is a quad. with 4 rt. $ and 4 * sides. So a square is a rhombus, because by def., a rhombus is a quad. with 4 * sides.

44. (1) Both pairs of opp. sides are &. Both pairs of opp. sides are *. Both pairs of opp. $ are *. All pairs of cons. $ are supp. Its diags. bisect each other.

(2) Its diags. are *.(3) Its diags. are %. Each diag. bisects a pair of

opp. $.


45. DBy Thm. 6-4-5, #LKM * #JKM.

++ JK * ++

JM , so !JKM is isosc.; by Isosc. ! Thm., #JMK * #JKM. So m#J + m#JMK + m#JKM =180m#J + x + x = 180m#J = (180 " 2x)°


46. The perimeter of !RST is 7.2 cm.Possible answer: Opp. sides of a rect. are *, so RS = QT = 2.4 and ST = QR = 1.8. Diags. of a rect. bisect each other, so QS = 2QP = 2(1.5) = 3. The diags. of a rect. are *, so TR = QS = 3. Therefore the perimeter of !RST is 2.4 + 1.8 + 3 = 7.2.

47. HCons. sides need not be *.


48. Think: By Alt. Int. $ Thm. and Thm. 6-4-4, given $. 3 x 2 " 15 + x 2 + x = 90 4 x 2 + x " 105 = 0 (4x + 21)(x " 5) = 0 x = 5 or "5.25


Given: ABCD is a rhombus. X is mdpt. of ++

AB . Y is mdpt. of

++ AD .

Prove: ++

XY & ++

BD ; ++

XY % ++

AC Proof: Since X is the mdpt. of

++ AB and Y is the

mdpt. of ++

AD , ++

XY is a midseg. of !ABD by def. By the ! Midsegment Thm.,

++ XY & ++

BD . By Thm 6-4-4, since ABCD is a rhombus then its diags. are %. So

++ AC % ++

BD . Since also ++

BD & ++

XY , it follows by the % Transv. Thm. that

++ AC % ++

XY .

50. Possible answer: The midseg. of a rect. is a seg. whose endpoints are mdpts. of opp. sides of the rect.

Given: ABCD is a rect. X is mdpt. of

++ AB .

Y is mdpt. of ++

CD .Prove: AXYD * BXYCProof: A rect. is a 0, so ABCD is a 0. Since opp. sides of a 0 are *,

++ AB * ++

CD and ++

AD * ++

BC . Since X is the mdpt. of

++ AB , ++

AX * ++

XB . Since Y is the mdpt. of ++

CD, ++

DY * ++

YC . But because ++

AB * ++

CD , you can conclude that

++ AX * ++

XB * ++

DY * ++

YC . Opp. sides of a 0 are & by def., so

++ AX & ++

DY . Since also ++

AX * ++

DY , AXYD is a 0 by Thm. 6-3-1. But since ABCD is a rect. #A is a rt. #. So 0AXYD contains a rt. # and is therefore a rect. By similar reasoning, you can conclude that BXYC is a rect. Since

++ XY

* ++

XY by the Reflex. Prop. of *, all corr. sides are *. Also, all rt. $ are *, so all corr. $ are *. Therefore AXYD * BXYC by def. of *.

51. 11 1-by-1s, 8 1-by-2s, 5 1-by-3s, 2 1-by-4s, 1 1-by-5, 6 2-by-1s, 4 2-by-2s, 2 2-by-3s, 3 3-by-1s, 2 3-by-2s, 1 3-by-345 rects.


52. change = 20 " 1.1c = 20 "1.1(2 + 1.8(5)) = 7.9change is $7.90

53. T (a = ("3)b ' a = 3("b))

54. F; possible answer: suppose a : has a diam. of 4 cm and an area of 4" cm 2 . If diam. is doubled to 8 cm, area of : changes to 16" cm 2 . New area is 4 times as large as original area.

55. No; none of the conditions for a 0 are met.

56. Yes; 135° # is supp. to both of its cons. $, so by Thm. 6-3-4, quad. is a 0.


Check students’ constructions.

TECHNOLOGY LAB PREDICT CONDITIONS FOR SPECIAL PARALLELOGRAMS, TECHNOLOGY LAB, PAGES 416–417

ACTIVITY 1: TRY THIS, PAGE 416 1. Both pairs of opp. sides are &, so ABCD is a 0 by def.

2. Possible answers: If a 0 has a rt. #, then it is a rect. If a 0 has * diags., then it is a rect.

ACTIVITY 2: TRY THIS, PAGE 417 3. Possible answers: If a pair of cons. sides of a 0

are *, then the 0 is a rhombus. If the diags. of a 0 are %, then the 0 is a rhombus. If a diag. of a 0 bisects opp. $, then the 0 is a rhombus.

4. Possible answer: If a 0 is a rect. and a rhombus, then 0 is a square.

6-5 CONDITIONS FOR SPECIAL PARALLELOGRAMS, PAGES 418–425

CHECK IT OUT! PAGES 419–421 1. Both pairs of opp. sides of WXYZ are *, so WXYZ

is a 0. The contractor can use the carpenter’s square to see if one # of WXYZ is a rt. #. If so, then by Thm. 6-5-1, the frame is a rect.

2. Not valid; by Thm. 6-5-1, if one # of a 0 is a rt. #, then the 0 is a rect. To apply this thm., you need to know that ABCD is a 0.


3a. Step 1 Graph 0KLMN.

Step 2 Determine if KLMN is a rect.

KM = 8 )))) 8 2 + 2 2 = ( )) 68 = 2 ( )) 17

LN = 8 )))) 2 2 + 8 2 = ( )) 68 = 2 ( )) 17 Since KM = LN, diags. are *. KLMN is a rect.Step 3 Determine if KLMN is a rhombus.slope of KM = 2 _

8 = 1 _

4

slope of LN = "8 _ 2

= "4

Since ( 1 _ 4 ) ("4) = "1,

++ KM % ++

LN . KLMN is a rhombus.

Since KLMN is a rect. and a rhombus, KLMN is a square.

b. Step 1 Graph 0PQRS.

Step 2 Determine if PQRS is a rect.

PR = 8 )))) 7 2 + 7 2 = ( )) 98 = 7 ( ) 2

QS = 8 )))) 5 2 + 5 2 = ( )) 50 = 5 ( ) 2 Since PR - QS, PQRS is not a rect. Thus PQRS is not a square.Step 3 Determine if PQRS is a rhombus.slope of

++ PR = "7 _

7 = "1

slope of ++

QS = "5 _ "5

= 1

Since ("1)(1) = "1, ++

PR % ++

QS . PQRS is a rhombus.

THINK AND DISCUSS, PAGE 421 1. rect.; rhombus; square

2. Possible answer:

3. If a quad. is a rect., then it is a 0. If a 0 has one rt. #, then it is a rect. Thus these defs. are equivalent.

4.


1. Possible answer: If WXYZ is both a rhombus and a rect., then it is a square. All 4 sides of WXYZ are *. So WXYZ is a rhombus and therefore a 0. If the diags. of a 0 are *, then by Thm. 6-5-2, 0 is a rect. So the club members can measure the diags., and if they are equal, WXYZ is both a rhombus and a rect., and therefore it is a square.

2. Not valid; by Thm. 6-5-2, if the diags. of a 0 are *, then the 0 is a rect. To apply this thm., you need to know that ABCD is a 0.

3. Valid ( ++

AB & ++

CD , ++

AB * ++

CD ' 0, ++

AB % ++

BC ' #B a rt. # ' rect.)

4. Step 1 Graph 0PQRS.

Step 2 Determine if PQRS is a rect.

PR = 8 )))) 11 2 + 3 2 = ( )) 130

QS = 8 )))) 7 2 + 9 2 = ( )) 130 Since PR = QS, diags. are *. PQRS is a rect. Step 3 Determine if PQRS is a rhombus.slope of

++ PR = "3 _

11 = " 3 _

11

slope of ++

QS = "9 _ "7

= 9 _ 7

Since (" 3 _ 11

) ( 9 _ 7 ) - "1, PQRS is not a rhombus.

Thus PQRS is not a square.


5. Step 1 Graph 0WXYZ.

Step 2 Determine if WXYZ is a rect.

WY = 8

))) 8 2 + 4 2 = ( )) 80 = 4 ( ) 5

XZ = 8

)))) 6 2 + 12 2 = ( )) 180 = 6 ( ) 5 Since WY - XZ, WXYZ is not a rect. Thus WXYZ is not a square.Step 3 Determine if WXYZ is a rhombus.slope of

+++ WY = "4 _

8 = " 1 _

2

slope of ++

XZ = "12 _ "6

= 2

Since (" 1 _ 2 v) (2) = "1,

+++ WY % ++

XZ . WXYZ is a

rhombus.


6. Both pairs of opp. sides of PQRS are *, so PQRS is a 0. Since PZ = QZ and RZ = SZ, it follows that PR = QS by the Seg. Add. Post. Thus

++ PR * ++

QS . So the diags. of 0PQRS are *. The frame is a rect. by Thm. 6-5-2.

7. valid (by Thms. 6-3-5 and 6-5-4)

8. Not valid; by Thm. 6-5-5, if one diag. of a 0 bisects a pair of opp. $, then the 0 is a rhombus. To apply this thm., you need to know that EFGH is a 0.

9. Step 1 Determine if ABCD is a rect.

AC = 8 )))) 14 2 + 2 2 = ( )) 200 = 10 ( ) 2

BD = 8 )))) 2 2 + 14 2 = 10 ( ) 2 AC = BD, so diags. are *. ABCD is a rect.Step 2 Determine if ABCD is a rhombus.

slope of ++

AC = "2 _ 14

= " 1 _ 7

slope of ++

BD = "14 _ "2

= 7

Since (" 1 _ 7 ) (7) = "1, AC % BD. ABCD is a rhombus.

Since ABCD is a rect. and a rhombus, ABCD is a square.

10. Step 1 Determine if JKLM is a rect.

JL = 8 )))) 12 2 + 4 2 = 4 ( )) 10

KM = 8 )))) 2 2 + 6 2 = 2 ( )) 10 JL - KM, so JKLM is not a rect., and therefore not a square.Step 2 Determine if JKLM is a rhombus.

slope of ++

JL = 4 _ 12

= 1 _ 3

slope of ++

KM = "6 _ 2 = "3

Since ( 1 _ 3 ) ("3) = "1, JL % KM. JKLM is a rhombus.

11. diags. bisect each other ' 0; one # a rt. # ' rect.

12. diags. bisect each other ' 0

13. diags. bisect each other ' 0; diags. * ' rect.; diags. % ' rhombus; rect., rhombus ' square

14. both pairs of opp. sides * ' 0

15. both pairs of opp. sides * ' 0; one # a rt. # ' rect.; all 4 sides * ' rhombus; rect., rhombus ' square

16. ASA ' two , are * ' all 4 sides * ' 0, rhombus

17. B; possible answer: it is given that ABCD is a 0.

++ AC and

++ BD are its diags. By Thm. 6-5-2, if diags.

of a 0 are *, you can conclude that the 0 is a rect. There is not enough information to conclude that ABCD is a square.

18. ++

JL and ++

KM bisect each other.

19. ++

PR * ++

QS

20. ++

AB is horiz. and ++

AC is vert. So ++

BD is vert. and ++

CD is horiz. D has same x-coord. as B and same y-coord. as C, so D = ("5, 4).

21. AB = BC = ( )) 50 ; so D is opp. B. A is rise of 5 and run of "5 from B; so D is rise of 5 and run of "5 from C. Therefore D = (7 " 5, 1 + 5) = (2, 6).

22. AB = BC = 4 ( ) 2 ; so D is opp. B.A is rise of 4, run of "4 from B;so D is rise of 4, run of "4 from C.Therefore D = (0 " 4, "6 + 4) = ("4, "2).

23. AB = BC = 5; so D is opp. B.A is rise of "4, run of 3 from B;so D is rise of "4, run of 3 from C.Therefore D = ("5 + 3, 2 " 4) = ("2, "2).

24. Given # must be rt. #.5x " 3 = 90 5x = 93 x = 18.6

25. All 4 sides must be *.14 " x = 2x + 5 9 = 3x x = 3

26. Diags. must be %.13x + 5.5 = 90 13x = 84.5 x = 6.5

27. Rhombus; since diags. bisect each other the quad. is a 0. Since the diags. are %., the quad. is a rhombus.


28a. 0 ' opp. sides * b. ++

EH * ++

EH

c. SSS d. #GHE

e. CPCTC f. 0 ' cons. $ supp.

g. #FEH, #GHE are rt.$ h. 0 with 1 rt. # ' rect.

29a. slope of ++

AB = " 1 _ 3 ; slope of

++ CD = 1 _ "3

= " 1 _ 3

slope of ++

BC = "3 _ 1 = "3; slope of

++ AD = "3 _

1 = "3

b. slope of ++

AC = "4 _ 4 = "1; slope of

++ BD = "2 _ "2

= 1;

the slopes are negative reciprocals of each other, so

++ AC % ++

BD .

c. ABCD is a rhombus, since it is a 0 and its diags. are % (Thm. 6-5-4).

30a. ++

RT b. ++

QT

c. % lines d. Rt. # * Thm.

e. SAS f. ++

QR

g. rhombus


Statements Reasons

1. ABCD is a 0. #A is a rt. #.

1. Given

2. m#A = 90° 2. Def. of rt. # 3. #A and #B are supp. 3. 0 ' cons. $ supp. 4. m#A + m#B = 180° 4. Def. of supp. $ 5. 90° + m#B = 180° 5. Subst. 6. m#B = 90° 6. Subtr. Prop. of = 7. #C * #A, #D * #B 7. 0 ' opp. $ * 8. m#C = m#A,

m#D = m#B 8. Def. of * $

9. m#C = 90°, m#D = 90° 9. Trans. Prop. of =10. #B, #C, and #D are

rt. $.10. Def. of rt. #

11. ABCD is a rect. 11. Def. of rect.

32. Possible answer: It is given that ++

JK * ++

KL . Since opp. sides of a 0 are *,

++ JK *

++ LM and

++ KL * ++

MJ . By Trans. Prop. of *,

++ JK * ++

MJ . So ++

JK is * to each of the other 3 sides of JKLM. Therefore JKLM is a rhombus by definition.

33a. b. Slopes of sides are 2, "1, 2, "1, so quad. is 0 but not. rect. Side lengths are 2 ( ) 5 and 2 ( ) 2 , so quad. is not rhombus.

c. Lines change to n: y = x + 1 and p: y = x + 7; new slopes are 1, "1, 1, "1; new side lengths are all 3 ( ) 2 . So the quad. becomes a square.


Statements Reasons

1. FHJN and GLMF are 1. ++

FG * ++

FN 1. Given

2. ++

FH & ++

NJ , ++

GL & ++

FM 2. Def. of 03. FGKN is a 0. 3. Def. of 04. FGKN is a rhombus. 4. 0 with 1 pair cons.

sides * ' rhombus

35. A 0 is a rect. if and only if its diags. are *;a 0 is a rhombus if and only if its diags. are %;no; possible answer: Thms. 6-4-5 and 6-5-5 are not converses. The conclusion of the conditional in Thm. 6-4-5 refers to both diags. of a 0. The hypothesis of the conditional in Thm 6-5-5 refers to only one diag. of a 0.

36. Possible answer: Draw 2 pairs of arcs from same center on same compass setting. Draw 2 lines, through center and each pair of arcs. 2 lines bisect each other and are *, so they are diags. of a rect. Draw sides to complete rect.

37. Possible answer: Draw seg. for 1st diag. Construct % bisector. Seg. between 2 pairs of arcs in construction is bisected by 1st seg., so segs. are diags. of a rhombus.

38. Possible answer: Draw seg. for 1st diag. Construct % bisector. Set compass to half length of diags., and construct 2nd diag. Diags. are % bisectors of each other and are *, so they are diags. of a square.

TEST PREP, PAGE 425

39. A(condition for Thm. 6-5-2)

40. GSlope of

+++ WX = slope of

++ YZ = 1; slope of

+++ WZ =

slope of ++

XY = "1; WX = 4 ( ) 2 ; XY = 7 ( ) 2 . So WXYZ is a rect. but not a square.


41a. Think: Use Vert. $ Thm.m#KNL = m#JNM 15x = 13x + 12 2x = 12 x = 6

b. Yes; m#JKN = 6(6) = 36° and m#LMN = 5(6) + 6 = 36°, so by Conv. of Alt. Int. $ Thm.,

++ JK & ++

LM . Since

++ KL & ++

JM , JKLM is a 0 by def.

c. No; by subst., Lin. Pair Thm., and Rt. # * Thm., all 4 $ at N are rt. $. Since JKLM is a 0, N is mdpt. of both diags. By SAS, !KNL * !KNJ, so by CPCTC, #LKN * #JKN. Therefore m#JKL = 2m#JKN = 2(36) = 72° - 90°.

d. Yes; from part c., diags of 0JKLM are %. So by Thm. 6-5-4, JKLM is a rhombus.



Statements Reasons

1. ++

AC * ++

DF , ++

AB * ++

DE , ++

AB % ++

BC , ++

DE % ++

EF , ++

BE % ++

EF , ++

BC & ++

EF

1. Given

2. m#ABC = 90°, m#DEF = 90°, m#BEF = 90°

2. Def. of %

3. #ABC, #DEF, and #BEF are rt. $.

3. Def. of rt. #

4. !ABC and !DEF are rt. ,.

4. Def. of rt. !

5. !ABC * !DEF 5. HL6. ++

BC * ++

EF 6. CPCTC7. EBCF is a 0. 7. Thm. 6-3-18. EBCF is a rect. 8. Thm. 6-5-1

43a. Possible answer: If a quad. is a rect., then it has four rt. $. If a quad. is a rhombus, then it has four * sides. By def., a quad. with four rt. $ and four * sides is a square. Therefore statement is true.

b. No; possible answer: if a quad. is a rect., then it is a 0. By Thm. 6-5-3, if 1 pair of cons. sides of a 0 are *, then the 0 is a rhombus. So if 1 pair of cons. sides of a rect. are *, it is a rhombus. If a quad. is a rect. and a rhombus, then it is a square.

c. No; possible answer: if a quad. is a rhombus, then it is a 0. By Thm. 6-5-1, if 1 # of a 0 is a rt. then the 0 is a rect. So if 1 # of a rhombus is a rt. #, it is a rect. If a quad. is a rhombus and a rect., then it is a square.

44. Diags. of the 0 are %, so it is a rhombus.


45.

linear

46.

nonlinear

47.

linear

48. c = 8 )))) 8 2 + 10 2 = ( )) 164 = 2 ( )) 41 P = 20 + 2 ( )) 41 + 2 ( )) 41 = 20 + 4 ( )) 41 9 45.6

49. c = 8 )))))))) (12 " 6 ) 2 + (9 " 4 ) 2 = ( )) 61 P = 6 + 9 + 12 + 4 + ( )) 61 = 31 + ( )) 61 9 38.8

50. Cons. $ are supp. So8x + 10 + 10x + 44 = 180 18x = 126 x = 7

51. Opp. sides are *. So 2y + 7 = 7z + 1 5y = 9z + 2Think: Eliminate z. 18y + 63 = 63z + 9 35y = 63z + 1417y " 63 = 5 17y = 68 y = 4

52. Subst. for y to find z.5(4) = 9z + 2 18 = 9z z = 2

TECHNOLOGY LABEXPLORE ISOSCELES TRAPEZOIDS TECHNOLOGY LAB, PAGE 426

ACTIVITY 1: TRY THIS, PAGE 426 1. Possible answer: If a trap. is isosc., then its base $ are

*.

2. Possible answer: If one pair of base $ of a trap. are *, then the trap. is isosc.

ACTIVITY 2: TRY THIS, PAGE 426 3. Possible answer: If a trap. is isosc., then its diags.

are *.

4. Possible answer: If diags. of a trap. are *, then the trap. is isosc.


6-6 PROPERTIES OF KITES AND TRAPEZOIDS, PAGES 427–435

CHECK IT OUT! PAGES 428–431 1. Perimeter of kite doubles:

P = 20 ( )) 17 + 8 ( )) 185 9 191.3 in.Daryl needs approx. 191.3 in. of binding.

191.3 _ 72

9 2.7 packages of binding

In order to have enough, Daryl must buy 3 packages of binding.

2a. Think: !PQR is isosc., so #QPT * #QRT. m#PQR + m#QPT + m#QRT = 180 78 + 2m#QRT = 180 2m#QRT = 102 m#QRT = 51°

b. Think: Kite ' 1 pair opp. $ *.m#QPS = m#QRS = m#QRT + m#TRS = 51 + 59 = 110°

c. Think: Use Quad. # Sum Thm.m#PSR + m#QRS + m#PQR + m#QPS = 360 m#PSR + 110 + 78 + 110 = 360 m#PSR = 62°

3a. Think: Use Same-Side Int. $ Thm., isosc. trap. ' base $ *.m#G + m#H = 180 m#G + 49 = 180 m#G = 131° #F * #G m#F = m#G = 131°

b. Think: Isosc. trap. ' diags. *. ++

KM * ++

JL KM = JL = JN + NL = 10.6 + 14.8 = 25.4

4. Think: 1 pair base $ * ' trap. isosc. #Q * #S m#Q = m#S2 x 2 + 19 = 4 x 2 " 13 32 = 2 x 2 16 = x 2 x = 4 or "4

5. Think: Use Trap. Midseg. Thm. XY = 1 _

2 (EH + FG)

16.5 = 1 _ 2

(EH + 25)

33 = EH + 25 8 = EH

THINK AND DISCUSS, PAGE 431 1. No; possible answer: if the legs are &, then the trap.

has two pairs of & sides. But by def., a trap. has exactly one pair of & sides, so the figure would be a 0.

2. Possible answer: Similarities: The endpts. of both are the mdpts. of two sides. Both are & to another side. Differences: A ! has three midsegs., while a trap. has just one. To find the length of a midseg. of a !., you find half the measure of just one side; to find the length of a midseg. of a trap., you must average the lengths of two sides.

3.


1. bases: ++

RS and ++

PV ; legs: ++

PR and ++

VS ; midseg.: ++

QT

2. Possible answer: In a 0, two pairs of opp. sides are *. In a kite, exactly two distinct pairs of cons. sides are *.

3. 1 Understand the ProblemAnswer has 2 parts:• Total amount of lead needed• Number of suncatchers that can be sealed2 Make a PlanDiags. of a kite are %, so 4 , are rt. ,. Use Pyth. Thm. and props. of kites to find unknown side lengths. Add these lengths to find perimeter of kite.3 Solve

JK = 8 ))))) JH 2 + KH 2

= 8 )))))) 2.7 5 2 + 2.7 5 2 = ( ))) 15.125 KL = JK = ( ))) 15.125

JM = 8 ))))) JH 2 + MH 2

= 8 ))))) 2.7 5 2 + 5. 5 2 = ( )))) 37.8125 LM = JM = ( )))) 37.8125 perimeter of JKLM = 2 ( ))) 15.125 + 2 ( )))) 37.8125

9 20.1 in.20.1 in. of lead is needed to seal edges.One 3-ft length of lead contains 36 in.

2(36)

_ 20.1

9 3.6

3 sun catchers can be sealed.4 Look BackTo estimate perimeter, change side lengths intodecimals and round. ( ))) 15.125 9 4, and ( )))) 37.8125 9 6. Perimeter of sun catcher is approx. 2(4) + 2(6) = 20. So 20.1 in. is a reasonable answer.

4. Think: Kite ' diags. % ' #VZY and #VYZ are comp.m#VZY + m#VYZ = 90 m#VZY + 49 = 90 m#VZY = 41°


5. Think: !XYZ is isosc., so #VXY * #VZY.m#VXW + m#VXY = m#WXY m#VXW + 41 = 104 #VXW = 63°

6. Think: Use Quad. # Sum Thm., kite ' 1 pair opp. $ *.m#XWZ + m#WXY + m#XYZ + m#YZW = 360 m#XWZ + 104 + 2m#VYZ + 104 = 360 m#XWZ + 104 + 2(49) + 104 = 360 m#XWZ = 54°

7. Think: Use Same-Side Int. $ Thm., isosc. trap. ' base $ *.m#C + m#D = 180 74 + m#D = 180 m#D = 106° #A * #D m#A = m#D = 106°

8. Think: Isosc. trap. ' diags. *.

++ RT * ++

SV RT = SV RW + TW = SV17.7 + TW = 23.3 TW = 5.6

9. Think: 1 pair base $ * ' trap. isosc. #E * #Hm#E = m#H 12 z 2 = 7 z 2 + 20 5 z 2 = 20 z 2 = 4 z = 2 or "2

10. Think: Diags. * ' trap. isosc.

+++ MQ *

++ LP

MQ = LP7y " 6 = 4y + 11 3y = 17 y = 17 _

3 = 5 2 _

3

11. Think: Use Trap. Midseg. Thm.XY = 1 _

2 (PS + QR)

22 = 1 _ 2 (30 + QR)

44 = 30 + QR 14 = QR

12. Think: Use Trap. Midseg. Thm.AZ = 1 _

2 (DF + JK)

= 1 _ 2 (11.9 + 7.1)

= 1 _ 2 (19) = 9.5


13. 1 Understand the ProblemAnswer has 2 parts:• Amount of iron needed to outline 1 kite• Amount of iron needed for 1 complete section2 Make a PlanDiags. of a kite are %., so 4 , are rt. ,. Use Pyth. Thm. and props. of kites to find unknown side lengths. Add these lengths to find perimeter of kite.3 Solveshorter side length = 8 )))) 7 2 + 7 2 = 7 ( ) 2

longer side length = 8 )))) 7 2 + 17 2 = 13 ( ) 2

perimeter = 2 (7 ( ) 2 ) + 2 (13 ( ) 2 ) = 40 ( ) 2 9 56.6 in.

56.6 in. of iron is needed for 1 kite.For 1 complete section, need iron for 4 kites and 1 square of side length 2(7 + 17) = 48 in.Amount of iron = 4 (40 ( ) 2 ) + 4(48) 9 418.3 in.4 Look BackTo estimate perimeter, change side lengths into decimals and round. 7 ( ) 2 9 10, and 13 ( ) 2 9 18. Perimeter of one kite is approx. 2(10) + 2(18) = 56. So 56.6 in. is a reasonable answer.

14. Think: Kite ' diags. %.m#XDA + m#DAX = 90 m#XDA + 32 = 90 m#XDA = 58°

15. Think: Kite ' 1 pair opp. $ *.m#ABC = m#ADC = m#XDA + m#XDC = 58 + 64 = 122°

16. Think: Use Quad. # Sum Thm.m#BCD + m#ADC + m#BAD + m#ABC = 360 m#BCD + 122 + 2m#DAX + 122 = 360 m#BCD + 122 + 2(32) + 122 = 360 m#BCD = 52°

17. Think: Use Same-Side Int. $ Thm., Thm. 6-6-3.m#L + m#M = 180 m#L + 118 = 180 m#L = 62° #Q * #L m#Q = m#L = 62°

18. Think: Use Thm. 6-6-5. ++

RJ * ++

SK RJ = SK = SZ + KZ = 62.6 + 34 = 96.6

19. Think: 1 pair base $ * ' trap. isosc. #X * #W m#X = m#W a 2 + 15 = 2 a 2 " 65 80 = a 2 a = ± ( )) 80 = ±4 ( ) 5

20. Think: Diags. * ' trap. isosc.

++ GJ * ++

FH GJ = FH4x " 1 = 9x " 15 14 = 5x x = 2.8

21. Think: Use Trap. Midseg. Thm.PQ = 1 _

2 (AV + BT)

= 1 _ 2 (4.2 + 3) = 3.6

22. Think: Use Trap. Midseg. Thm. MN = 1 _

2 (JT + KR)

52.5 = 1 _ 2 (32.5 + KR)

105 = 32.5 + KR72.5 = KR

23. Sometimes (opp. $ supp. only when trap. is isosc.)

24. Sometimes (kites with non-* opp. $ of 30° and 150° or 20° and 150°)

25. Never (this 's both opp. pairs of $ are *, so quad. would be a 0, not a kite)

26. ! formed by dashed line is 30°-60°-90°. So shorter leg measures a, where 6 = 2a3 = a4th edge measures 20 " 2a = 20 " 2(3) = 14 ftP = 20 + 6 + 14 + 6 = 46 ftC = 1.29P9 1.3(46) 9 $60

about $60.

27. Think: Use Same-Side Int. $ Thm.m#1 + 98 = 180 m#1 = 82°

m#2 + 52 = 180 m#2 = 128°


28. Think: Use Thm 6-6-2, Quad. # Sum Thm. m#1 = 116°m#1 + m#2 + 116 + 82 = 360 116 + m#2 + 116 + 82 = 360 m#2 + 314 = 360 m#2 = 46°

29. Think: All 4 , are rt. ,; left pair of , is *.m#1 + 39 = 90 m#1 = 51°

m#2 + 74 = 90 m#2 = 16°

30. Think: Top left and bottom right $ of isosc. trap. are supp.; also use Alt. Int. $ Thm.(m#2 + 34) + (72 + 34) = 180 m#2 + 140 = 180 m#2 = 40°Think: By * ,, lower part of top right # measures 40°. By Ext. # Thm. m#1 = 72 + 40 = 112°

31. Think: Use Thm. 6-6-2, Quad. # Sum Thm. 3x = 48 x = 16m#1 + 3(16) + 9(16) + 48 = 360 m#1 + 240 = 360 m#1 = 120°

32. Think: Use Same-Side Int. $ Thm. 40z + 5 + 10z = 180 50z = 175 z = 3.518(3.5) + m#1 = 180 63 + m#1 = 180 m#1 = 117°

33a. EF = 8 )))))))) (3 + 1 ) 2 + (4 " 3 ) 2 = 8 )) 17

FG = 8 )))))))) (2 " 3 ) 2 + (0 " 4 ) 2 = 8 )) 17

GH = ( ))))))))) ("3 " 2 ) 2 + ("2 " 0 ) 2 = 8 )) 29

EH = ( ))))))))) ("3 + 1 ) 2 + ("2 " 3 ) 2 = 8 )) 29

EF * ++

FG 2 ++

GH * ++

EH ' EFGH is a kite.

b. m#E = m#G = 126°

34. 12t = 1 _ 2

(16t + 10)

24t = 16t + 10 8t = 10 t = 1.25length of midseg. = 12(1.25) = 15

35. n + 6 = 1 _ 2 (n + 3 + 3n " 5)

2n + 12 = 4n " 2 14 = 2n 7 = nlength of midseg. = (7) + 6 = 13

36. 4c = 1 _ 2 ( c 2 + 6 + c 2 + 2)

4c = c 2 + 4 c 2 " 4c + 4 = 0 (c " 2 ) 2 = 0 c = 2length of midseg. = 4(2) = 8

37. m#PAQ + m#AQB + m#PBQ + m#APB = 360m#PAQ + 72 + m#PAQ + 72 = 360 m#PAQ + 72 = 180 m#PAQ = 108°m#OAQ + m#AQB + m#OBQ + m#AOB = 360m#OAQ + 72 + m#OAQ + 28 = 360 2m#OAQ + 100 = 360 2m#OAQ = 260 m#OAQ = 130° m#OBQ = m#OBP + m#PBQ m#OAQ = m#OBP + m#PAQ 130 = m#OBP + 108 22° = m#OBP


Given: ABCD is a kite with ++

AB * ++

AD and ++

CB * ++

CD. Prove:

++ AC bisects #DAB and #DCB.

++ AB bisects

++ BD .

Statements Reasons

1. ++

AB * ++

AD and ++

CB * ++

CD. 1. Given2. ++

AC * ++

AC 2. Reflex. Prop. of *3. !ABC * !ADC 3. SSS4. #1 * #2, #3 * #4 4. CPCTC5. ++

AC bisects #DAB and #DCB.

5. Def. of # bisector

6. ++

AE * ++

AE 6. Reflex. Prop. of *7. !ABE * !ADE 7. SAS8. ++

BE * ++

DE 8. CPCTC9. ++

AC bisects ++

BD . 9. Def. of seg. bisector


Given: ABCD is a kite with ++

AB * ++

CB and ++

AD * ++

CD .Prove:

++ BD % ++

AC It is given that

++ AB * ++

CB and ++

AD * ++

CD . This means that B and D are equidistant from A and from C. By the Conv. of the % Bisector Thm., if a pt. is equidist. from the endpts. of a seg., then it is on the % bisector of the seg. Through any two pts. there is exactly one line, so the line containing B and D must be the % bisector of

++ AC . Therefore

++ BD % ++

AC .


40.

++

AB and ++

CD are vert.; ++

BC is horiz.;

slope of ++

DA = 3 _ "6

= " 1 _ 2

Exactly 2 sides are &, so quad. is a trap.

41.

AB = BC = 4; CD = 8 )))) 3 2 + 7 2 = ( )) 58 ;

DA = 8 )))) 7 2 + 3 2 = ( )) 58 Exactly 2 pairs of cons. sides are *, so quad is a kite.

42.

Diags. have equations y = x and x = 1; they intersect at (1, 1), and bisect each other, but are not %. Therefore quad is a 0.

43.

++

AD and ++

BC are horiz.; AB = 8 )))) 4 2 + 6 2 = 2 ( )) 13 ,

CD = 8 )))) 4 2 + 6 2 = 2 ( )) 13 ++

AD & ++

BC and ++

AB * ++

CD ; so quad. is an isosc. trap.

44. Extend ++

BA and ++

CD to meet in center of window, X. #AXC = 360 ÷ 8 = 45°; by Isosc. ! Thm., #XBC * #XCB, so m#XBC = 1 _

2 (180 " m#AXC) = 67.5°

Since ++

BC & ++

AD , by Corr. $ Post., m#B = m#C = 67.5°by Lin. Pair Thm.,m#A = m#D = 180 " 67.5° = 112.5°

45. Possible answer: Common props.: exactly one pair of & sides; two pairs of cons. $ supp.; length of midseg. is the average of the lengths of the bases; special props. of isosc. trap.: * legs; two pairs of * base $; * diags.

46a. ( 0 + 2a _ 2 , 0 + 2b _

2 ) = (a, b)

b. ( c + c + 2d _ 2 , 2b + 0 _

2 ) = (c + d, b)

c. slope of ++

QR = 2b " 2b _ c " 2a

= 0

slope of ++

PS = 0 " 0 _ c + 2d " 0

= 0

slope of +++

MN = b " b _ c + d "a

= 0

All 3 segs. are &.

d. QR = c " 2a; PS = c + 2d " 0 = c + 2d; MN = c + d " a; c + d " a = 1 _

2 (c " 2a + c + 2d ),

so MN = 1 _ 2 (PS + QR)


47. B 1 _ 2 (6 + 26) = 16

48. H

49. 18By ! Midseg. Thm., DE = 1 _

2 (24) = 12 in.

midseg. length = 1 _ 2 (12 + 24) = 18 in.



Statements Reasons

1. WXYZ is a trap. with

++ XZ *

+++ YW .

1. Given

2. Draw ++

XU through X so that

++ XU %

+++ WZ .

Draw ++

YV through Y so that

++ YV %

+++ WZ .

2. There is exactly 1 line through a pt. not on a line that is % to that line.

3. m#XUZ = 90°, m#YVW = 90°

3. Def. of % lines

4. #XUZ and #YVW are rt. $.

4. Def. of rt. #

5. !XUZ and !YVW are rt. ,.

5. Def. of rt. !

6. ++

XU & ++

YV 6. 2 lines % to 3rd line ' 2 lines &

7. ++

XY & +++

WZ 7. Def. of trap. 8. XYVU is a 0. 8. Def. of 0 9.

++ XU *

++ YV 9. 0 ' opp. sides *

10. !XUZ * !YVW 10. HL11. #XZW * #YWZ 11. CPCTC12.

+++ WZ *

+++ WZ 12. Reflex. Prop. of *

13. !XZW * !YWZ 13. SAS14.

+++ XW *

++ YZ 14. CPCTC

15. WXYZ is an isosc. trap. 15. Def. of isosc. trap.

51. BC + AD = 2(8.62) = 17.24 and CD = AB P = AB + BC + CD + AD 27.4 = 2AB + 17.24 10.16 = 2AB 5.08 = ABAB = CD = 5.08 in. BC = 2AB = 2(5.08) = 10.16 in. AD = 17.24 " BC = 17.24 " 10.16 = 7.08 in.



52. x _ 20%

= 25 _ 10

x = 25 _ 10

(20%)

= 50% = 1 _ 2

53. Think: Height of ! is min. dist. from apex to base.2x < x + 6 x < 6

54. 3x " 10 < 30 + x 2x < 40 x < 20

55. slope of ++

AB = 2 _ 2

= 1; slope of ++

CD = "2 _ "2

= 1

slope of ++

BC = "2 _ 2 = "1; slope of

++ AD = "2 _

2 = "1

So 0 is a rect.AB = BC = CD = AD =

8 )))) 2 2 + 2 2 = 2 ( ) 2

So 0 is a rhombus.Rect., rhombus ' square; so 0 is a square.

56. slope of ++

AB = 4 _ 3

; slope of ++

CD = "4 _ "3

= 4 _ 3

++

BC and ++

AD are vert.Cons. edges are not %, so 0 is not a rect. and therefore not a square.

++

AB = 8 )))) 3 2 + 4 2 = 5; ++

CD = 8 )))) 3 2 + 4 2 = 5 ++

BC = 30 " 54 = 5; ++

AD = 3"4 " 14 = 5Since all four sides are *., 0 is a rhombus.


1. Choose B and D both above or both below ++

AC .


1. By the Distance Formula AB = ( )))) 8 2 + 4 2 = 4 ( ) 5 ;

BC = ( )))) 4 2 + 8 2 = 4 ( ) 5

CD = ( )))) 8 2 + 4 2 = 4 ( ) 5 ; AD = ( )))) 4 2 + 8 2 = 4 ( ) 5 Since the four sides of ABCD are *, ABCD is a rhombus.

++ AC and

++ BD are the diags. of the rhombus,

so by Thm. 6-4-4, ++

AC % ++

BD .

2. Vertices are P(3, 6), Q(5, 4), R("1, "2), S("3, 0).

slope of ++

PQ = "2 _ 2

= "1; slope of ++

RS = 2 _ "2

= "1

slope of ++

PS = "6 _ "6

= 1; slope of ++

QR = "6 _ "6

= 1

So opp. sides are &, and PQRS is a 0 by definition.

3. By Dist. Formula, PR = SQ = 4 ( ) 5 , so diags. of PQRS are *. By Thm. 6-5-2, PQRS is a rect.

4. AE = 0 " 4 = 4; CE = "1 " 3 = 4Exactly 2 pairs of cons. sides are *, so ABCE is a kite.

5. ++

AE is vert. and ++

CE is horiz., so #E is a rt. #. m#E = 90°Also, #BAE * #BCE by Thm. 6-6-2. So by Quad. # Sum Thm.,m#B + m#BAE + m#E + m#BCE = 360 37 + m#BAE + 90 + m#BAE = 360 127 + 2m#BAE = 360 2m#BAE = 233 m#BCE = m#BAE = 116.5°


1. Think: ++

QS and ++

RT bisect each other at T.SP = 1 _

2 QS

= 1 _ 2

(80.5) = 40.25

2. ++

QT * ++

RS QT = RS = 36

3. Think: Diags. are *. ++

TR * ++

QS TR = QS = 80.5

4. TP = 1 _ 2 TR

= 1 _ 2 (80.5) = 40.25

5. ++

GH * ++

GK * ++

HJ GH = GK6a " 7 = 3a + 9 3a = 16 HJ = GK = (16) + 9 = 25

6. Think: Rhombus ' diags. are % m#JLH = 4b " 6 90 = 4b " 6 96 = 4b 24 = bThink: All 4 , are *. m#HJG + m#JHK = 90 m#HJG + m#JKH = 90m#HJG + 2(24) + 11 = 90 m#HJG = 31° m#GHJ = m#GHK + m#JHK = 2m#JHK = 2(2(24) + 11) = 118°


1. QSTV is a rhombus. ++

PT * ++

RT 1. Given

2. 556 TQ bisects #PTR. 2. rhombus ' each

diag. bisects opp. $3. #QTP * #QTR 3. Def. of # bisector4. ++

QT * ++

QT 4. Reflex. Prop. of *5. !PQT * ! RQT 5. SAS6. ++

PQ * ++

RQ 6. CPCTC

8. not valid; By Thm. 6-5-4, if the diags. of a 0 are %, then the 0 is a rhombus. But you need to know that ABCD is a 0.

9. valid (By Thm. 6-3-1, ABCD is a 0. By Thm. 6-5-2, ABCD is a rect.)

10. WY = 8 )))) 9 2 + 3 2 = 3 ( )) 10 ; XZ = 8 )))) 3 2 + 9 2 = 3 ( )) 10 The diags. are *, so WXYZ is a rect.

slope of +++

WY = "3 _ 9 = " 1 _

3 ; slope of

++ XZ = "9 _

3 = "3

The diags. are not %, so WXYZ is not a rhombus. Therefore WXYZ is not a square.

11. MP = 8 )))) 7 2 + 3 2 = ( )) 58 ; NQ = ( )))) 3 2 + 7 2 = ( )) 58 The diags. are *, so MNPQ is a rect.slope of

++ MP = " 3 _

7 ; slope of

++ NQ = "7 _ "3

= 7 _ 3

The diags. are %, so MNPQ is a rhombus.Rect., rhombus ' MNPQ is a square.


12. Possible answer: Since ++

VX is a midseg. of !TWY, ++

VX & ++

TZ by ! Midseg. Thm. Similarly, ++

XZ & ++

TV . So TVXZ is a 0 by def. V is mdpt. of

+++ TW ; thus

TV = 1 _ 2 TW by def. of mdpt. Similarly, TZ = 1 _

2 TY.

It is given that +++

TW * ++

TY , so TW = TY by def. of * segs. By subst., TZ = 1 _

2 TW, and thus TZ = TV. By

def. of * segs., ++

TZ * ++

TV . Since TVXZ is a 0 with 1 pair of cons. sides *, TVXZ is a rhombus.

13. Think: 556 EG bisects #FEH.

m#FEJ = 1 _ 2 m#FEH

= 1 _ 2 (62) = 31°

14. Think: #HEJ is * to #FEJ and comp. to #EHJ.m#HEJ + m#EHJ = 90 m#FEJ + m#EHJ = 90 31 + m#EHJ = 90 m#EHJ = 59°

15. Think: #HFG is comp. to #FGJ and * to #FHG.m#HFG + m#FGJ = 90m#FHG + m#FGJ = 90 68 + m#FGJ = 90 m#FGJ = 22°

16. m#EHG = m#EHJ + m#FHG = 59 + 68 = 127°

17. Think: Use Same-Side Int. $ Thm., isosc. trap. ' base $ *. m#U + m#T = 180 m#U + 77 = 180 m#U = 103° #R * #Um#R = m#U = 103°

18. Think: Isosc. trap ' diags. * WY * VX WZ + YZ = VXWZ + 34.2 = 53.4 WZ = 19.2

19. length of midseg. = 1 _ 2 (43 + 23) = 33 in.


1. vertex of a polygon 2. convex

3. rhombus 4. base of a trapezoid

LESSON 6-1, PAGE 438 5. not a polygon 6. polygon; !

7. polygon; dodecagon 8. irregular; concave

9. irregular; convex 10. regular; convex

11. (n " 2)180(12 " 2)1801800°

12. (n)m# = (n " 2)180 20m# = (18)180 20m# = 3240 m# = 162°

13. (n)m(ext. #) = 360 4m(ext. #) = 360 m(ext. #) = 90°

14. m#A + … + m#F = (6 " 2)1808s + 7s + 5s + 8s + 7s + 5s = 720 40s = 720 s = 18m#A = m#D = 8(18) = 144°; m#B = m#E = 7(18) = 126°; m#C = m#F = 5(18) = 90°

LESSON 6-2, PAGE 439 15. Think: Diags. bisect

each other.BE = 1 _

2 BD

= 1 _ 2 (75) = 37.5

16. ++

AD * ++

BC AD = BC = 62.4

17. ED = BE = 37.5 18. #CDA * #ABCm#CDA = m#ABC = 79°

19. Think: Cons. $ supp.m#ABC + m#BCD = 180 79 + m#BCD = 180 m#BCD = 101°

20. #BCD * #DABm#BCD = m#DAB = 101°

21. WX = YZb + 6 = 5b " 8 14 = 4b 3.5 = b WX = 3.5 + 6 = 9.5

22. YZ = 5(3.5) " 8 = 9.5

23. m#W + m#X = 180 6a + 14a = 180 20a = 180 a = 9 m#W = 6(9) = 54°

24. m#X = 14(9) = 126°

25. #Y * #Wm#Y = m#W = 54°

26. #Z * #Xm#Z = m#X = 126°

27. Slope from R to S is rise of 2 and run of 10;rise of 2 from V to T is "7 + 2 = "5;run of 10 from V to T is "4 + 10 = 6; T = (6, "5)


1. GHLM is a 0. #L * #JMG

1. Given

2. #G * #L 2. 0 ' opp. $ *3. #G * #JMG 3. Trans. Prop. of *4. ++

GJ * ++

MJ 4. Conv. Isosc. ! Thm.5. !GJM is isosc. 5. Def. of isosc. !

LESSON 6-3, PAGE 439 29. m = 13 ' m#G = 9(13) = 117°; n = 27 '

m#A = 2(27) + 9 = 63°, m#E = 3(27) " 18 = 63°Since 117° + 63° = 180° #G is supp. to #A and #E, so one # of ACEG is supp. to both of its cons. $. ACEG is a 0 by Thm. 6-3-4.

30. x = 25 ' m#Q = 4(25) + 4 = 104°, m#R = 3(25) + 1 = 76°; so #Q and #R are supp.y = 7 ' QT = 2(7) + 11 = 25, RS = 5(7) " 10 = 25By Conv. of Same-Side Int. $ Thm.,

++ QT & ++

RS ; since

++ QT * ++

RS , QRST is a 0 by Thm. 6-3-1.


31. Yes; The diags. bisect each other. By Thm. 6-3-5 the quad. is a 0.

32. No; By Conv. of Alt. Int. $ Thm., one pair of opp. sides is &, but other pair is *. None of conditions for a 0 are met.

33. slope of ++

BD = 2 _ 10

= 1 _ 5 ; slope of

++ FH = "2 _ "10

= 1 _ 5

slope of ++

BH = "6 _ 1 = "6; slope of

++ DF = "6 _

1 = "6

Both pairs of opp. sides have the same slope, so ++

BD & ++

FH and ++

BH & ++

DF ; by def., BDFH is a 0.


++ AB * ++

CD AB = CD = 18

35. AC = 2CE = 2(19.8) = 39.6

36. ++

BD * ++

AC BD = AC = 39.6

37. ++

BE * ++

CE BE = CE = 19.8

38. WX = WZ7a + 1 = 9a " 6 7 = 2a 3.5 = a WX = 7(3.5) + 1 = 25.5

39. ++

XV * ++

VZ XV = VZ = 3(3.5) = 10.5

40. ++

XY * +++

WX XY = WX = 25.5

41. XZ = 2XV= 2(10.5) = 21

42. m#TZV = 90 8n + 18 = 90 8n = 72 n = 9

Think: 556 RT bisects #SRV.

m#TRS = 1 _ 2 m#SRV

= 1 _ 2

(9(9) + 1) = 41°

43. m#RSV + m#TRS = 90 m#RSV + 41 = 90 m#RSV = 49°

44. #STV * #SRVm#STV = m#SRV = 9(9) + 1 = 82°

45. m#TVR + m#STV = 180 m#TVR + 82 = 180 m#TVR = 98°

46. Think: All 4 , are isosc. ,.2m#1 + m#2 = 1802m#3 + m#4 = 180By Lin. Pair Thm., m#2 + m#4 = 180By Alt. Int $ Thm., m#3 = 33° 2(33) + m#4 = 180 m#4 = 114° m#2 + 114 = 180 m#2 = 66° 2m#1 + 66 = 180 2m#1 = 114 m#1= 57°By Alt. Int $ Thm., #1 * #5 m#5 = m#1 = 57°

47. Think: All 4 , are * rt. ,.m#2 = m#5 = 53° m#3 = 90° m#4 + m#5 = 90 m#4 + 53 = 90 m#4 = 37° #1 * #4 m#1 = m#4 = 37°


RT and ++

SU are congruent.

RT = ( ))))))))))) (("3) " ("5)) 2 + ("6 " 0) 2 = 2 ( )) 10

SU = ( ))))))))))))) (("7) " ("1)) 2 + (("4) " ("2)) 2 = 2 ( )) 10

Since RT = SU, ++

RT * ++

SU

Step 2 Show that ++

RT and ++

SU are perpendicular.

slope of RT: "6 " 0 _ "3 " ("5)

= "3

slope of SU: "4 " ("2)

_ "7 " ("1)

= 1 _ 3

since "3 ( 1 _ 3 ) = "1,

++ RT % ++

SU

Step 3 Show that ++

RT and ++

SU bisect each other.

mdpt. of RT: ( "5 + ("3) _

2 ,

0 + ("6) _

2 ) = ("4, "3)

mdpt. of SU: ( "1 + ("7) _

2 ,

"2 + ("4) _

2 ) = ("4, "3)

Since ++

RT and ++

SU have the same midpoint, they bisect each other. The diagonals are congruent perpendicular biectors of each other.


EG and ++

FH are congruent.

EG = ( ))))))))) (5 " 2) 2 + ("2 " 1) 2 = 3 ( ) 2

FH = ( ))))))))) (2 " 5) 2 + ("2 " 1) 2 = 3 ( ) 2

Since EG = FH, ++

EG * ++

FH

Step 2 Show that ++

EG and ++

FH are perpendicular.

slope of EG: "2 " 1 _ 5 " 2

= "1

slope of FH: "2 " 1 _ 2 " 5

= 1

since "1(1) = "1, ++

EG % ++

FH

Step 3 Show that ++

EG and ++

FH bisect each other.

mdpt. of RT: ( 2 + 5 _ 2 ,

1 + ("2) _

2 ) = ( 7 _

2 , " 1 _

2 )

mdpt. of SU: ( 5 + 2 _ 2 ,

1 + ("2) _

2 ) = ( 7 _

2 , " 1 _

2 )

Since ++

EG and ++

FH have the same midpoint, they bisect each other. The diagonals are congruent perpendicular biectors of each other.


LESSON 6-5, PAGES 440–441 50. Not valid; by Thm. 6-5-2, if the diags. of a 0 are

*, then the 0 is a rect. By Thm. 6-5-4, if the diags. of a 0 are %, then the 0 is a rhombus. If a quad. is a rect. and a rhombus, then it is a square. But to apply this line of reasoning, you must first know that EFRS is a 0.

51. valid (diags. bisect each other ' 0: 0 with diags. * ' rect.)

52. valid (EFRS is a 0 by def.; 0 with 1 pair cons. sides * ' rhombus)

53. BJ = ( )))) 8 2 + 8 2 = 8 ( ) 2 ; FN = ( )))) 6 2 + 6 2 = 6 ( ) 2 Diags. are 2, so 0 is not a rect. Therefore 0 is not a square.slope of

++ BJ = 8 _

8 = 1; slope of

++ FN = "6 _

6 = "1

Diags. are %, so 0 is a rhombus.

54. DL = ( )))) 12 2 + 6 2 = 6 ( ) 5 ; HP = ( )))) 6 2 + 12 2 = 6 ( ) 5 Diags. are *, so 0 is a rect. slope of

++ DL = 6 _

12 = 1 _

2 ; slope of

++ HP = "12 _

"6 = 2

Diags. are not %, so 0 is not a rhombus. Therefore 0 is not a square.

55. QW = ( )))) 12 2 + 8 2 = 4 ( )) 13 ; TZ = ( )))) 8 2 + 12 2 = 4 ( )) 13 Diags. are *, so 0 is a rect. slope of

+++ QW = 8 _

12 = 2 _

3 ; slope of

++ TZ = "12 _

8 = " 3 _

2

Diags. are %, so 0 is a rhombus.Rect., rhombus ' 0 is a square.

LESSON 6-6, PAGE 441 56. Think: All 4 , are rt. ,; left pair of , is *, as is

right pair.m#XYZ = 2m#XYV = 2(90 " m#VXY) = 2(90 " 58) = 64°

57. m#ZWV = 1 _ 2 m#ZWX

= 1 _ 2 (50) = 25°

58. m#VZW = 90 " m#ZWV = 90 " 25 = 65°

59. m#WZY = m#VZW + m#VZY = m#VZW + m#VXY = 65 + 58 = 123°

60. Think: Use Same-Side Int. $ Thm., isosc. trap. ' base $ *. m#V + m#T = 180 m#V + 54 = 180 m#V = 126° #R * #Vm#R = m#V = 126° #S * #T m#S = m#T = 54°

61. Think: Isosc. trap. ' diags. *

++ BH * ++

EK BZ + ZH = EK BZ + 70 = 121.6 BZ = 51.6

62. MN = 1 _ 2 (AD + JG)

= 1 _ 2 (67 + 30)

= 48.5

63. ST = 1 _ 2 (FP + EQ)

2ST = FP + EQ2(3.1) = 2.7 + EQ 3.5 = EQ

64. #P * #Y m#P = m#Y8 n 2 " 11 = 6 n 2 + 7 2 n 2 = 18 n 2 = 9 n = ±3

65.

AB = DA = ( )))) 3 2 + 3 2 = 3 ( ) 2

BC = CD = ( )))) 6 2 + 3 2 = 3 ( ) 5 Exactly 2 pairs of cons. sides * ' kite.

66.

++

AD and ++

BC are vert., so ++

AD & ++

BC

AB = 35 " 14 = 4, CD = 8 )))) 4 2 + 3 2 = 5

++

AB 2 ++

CD and hence ++

AD 7 ++

BC Exactly 1 pair opp. sides &, other pair 2 ' trap. (not isosc.)

67.

++

AD and ++

BC are horiz., but AD = 2 + 6 = 8 - 4 = BC ++

AD & ++

BC , ++

AD 2 ++

BC ' trap. (not 0)

AB = ( )))) 2 2 + 3 2 = ( )) 13 ; CD = ( )))) 2 2 + 3 2 = ( )) 13 ++

AB * ++

CD , not 0 ' isosc. trap


1. not a polygon 2. polygon; decagon

3. Think: Use Quad. # Sum Thm. m#A + m#B + m#C + m#D = 360 12n + 14n + 8n + 11n = 360 45n = 360 n = 8m#A = 12(8) = 96°, m#B = 14(8) = 112°, m#C = 8(8) = 64°, m#D = 11(8) = 88°


4. (9 " 2)180(7)1801260°

5. 15 m(ext. #) = 360m(ext. #) = 24°

6. Think: Z is mdpt. of ++

FH .FH = 2HZ = 2(9) = 18

Think: Cons. $ are supp.m#FEH + m#EHG = 180 m#FEH + 145 = 180 m#FEH = 35°

7. Think: Opp. sides are *. JM = KL4y " 7 = y + 11 3y = 18 y = 6KL = 6 + 11 = 17

Think: Cons. $ are supp. m#M + m#L = 1806x " 1 + 2x + 9 = 180 8x = 172 x = 21.5m#L = 2(21.5) + 9 = 52°

8. Slope from S to R is rise of 4 and run of 1;from P to Q: rise of 4 is "3 to "3 + 4 = 1, run of 1 is "2 to "2 + 1 = "1; so coords. of Q = ("1, 1).

9. a = 4 ' XN = 3(4) = 12, NZ = 4 + 8 = 12b = 3 ' WN = 4(3) + 3 = 15, NY = 5(3) = 15So N is mdpt. of

++ XZ and

+++ WY , and therefore diags.

bisect each other. By Thm. 6-3-5, WXYZ is a 0.

10. No; one pair of opp. sides of the quad. are &. A pair of vert. $ formed by the diags. are *. None of conditions for a 0 are met.

11. Possible answer: slope of ++

KL = 3 _ 9 = 1 _

3 ; slope of

++ ST

= "3 _ "9

= 1 _ 3

slope of ++

KT = " 4 _ 3

; slope of ++

LS = " 4 _ 3

++

KL & ++

ST and ++

KT & ++

LS ' KLST is a 0.

12. PT = 1 _ 2

PC

= 1 _ 2

LM

= 1 _ 2

(23) = 11.5

++

PM * ++

LC PM = LC = 19

13. Think: All 4 , are * rt. ,.m#NQK = 90 7z + 6 = 90 7z = 84 z = 12m#HEQ = 90 " m#EHQ = 90 " m#ENQ = 90 " (5(12) + 1) = 29° m#EHK = 2m#EHQ = 2m#ENQ = 2(5(12) + 1) = 122°

14. Not valid; possible answer: MNPQ is a rhombus by def. However, to show that MNPQ is a square, you need to know that MNPQ is also a rect.

15. valid (MNPQ is a 0 by def.; diags * ' MNPQ is a rect.)

16. AE = ( )))) 12 2 + 8 2 = 4 ( )) 13 ; CG = ( )))) 4 2 + 6 2 = 2 ( )) 13 Diags. are 2, so ACEG is not a rect. Therefore ACEG is not a square.

slope of ++

AE = "8 _ 12

= " 2 _ 3 ; slope of

++ CG = "6 _ "4

= 3 _ 2

Diags. are %, so ACEG is a rhombus.

17. PR = ( )))) 7 2 + 1 2 = ( )) 50 ; QS = ( )))) 5 2 + 5 2 = ( )) 50 Diags. are *, so PQRS is a rect.slope of

++ PR = 1 _ "7

= " 1 _ 7 ; slope of

++ QS = "5 _ "5

= 1

Diags. are not %, so PQRS is not a rhombus. Therefore PQRS is not a square.

18. m#FBN = m#FBR + m#RBN = 90 " m#BFR + 90 " m#RNB = 90 " m#JFR + 90 " 1 _

2 m#JNB

= 180 " 43 " 1 _ 2 (68) = 103°

19. ++

MS * ++

PV MY + YS = PVMY + 24.7 = 61.1 MY = 36.4

20. XY = 1 _ 2

(HR + GS)

2XY = HR + GS2(25.5) = HR + 24 HR = 27 in.


1. E2x + 1 = 3x " 2 3 = x x " 1 ; 2x " 4 3 " 1 ; 2(3) " 4 2 = 2 <

2. DDiags are * and bisect each other, so

++ AE * ++

BE.

3. B2 segs. that are % to a 3rd seg. (and are not collinear) are &. Other pair of opp. sides are not &.

4. CSlope from M to N is rise of 5 and run of "3;slope from Q to P is rise of 5 and run of "3.

5. D5m#C = (5 " 2)180 5m#C = 540 m#C = 108°


Solutions KeySimilarity7

CHAPTER


1. E 2. F

3. B 4. D

5. A

6. 16 _ 20

= 4(4)

_ 4(5)

= 4 _ 5 7. 14 _

21 =

7(2) _

7(3) = 2 _

3

8. 33 _ 121

= 11(3)

_ 11(11)

= 3 _ 11

9. 56 _ 80

= 8(7)

_ 8(10)

= 7 _ 10

10. 18 to 246(3) to 6(4)3 to 4

11. 34 to 182(17) to 2(9)17 to 9

12. Total # of CDs is: 36 + 18 + 34 + 24 = 11236 to 1124(9) to 4(28)9 to 28

13. 112 to 248(14) to 8(3)14 to 3

14. yes; pentagon 15. yes; hexagon

16. no 17. yes; octagon

18. P = 2! + 2w= 2(8.3) + 2(4.2) = 25 ft

19. P = 6s= 6(30) = 180 cm

20. P = 4s= 4(11.4) = 45.6 m

21. P = 5s= 5(3.9) = 19.5 in.

7-1 RATIO AND PROPORTION, PAGES 454–459


1. slope = rise _ run = y 2 ! y 1

_ x 2 ! x 1

= 5 ! 3 _ 6 ! (!2)

= 2 _ 8 = 1 _

4

2. Let " measures be x, 6x, and 13x. Then x + 6x + 13x = 180. After like terms are combined, 20x = 180. So x = 9. The " measures are x = 9°, 6x = 6(9) = 54°, and 13x = 13(9) = 117°.

3a. 3 _ 8 = x _

56

3(56) = x(8) 168 = 8x x = 21

b. 2y

_ 9

= 8 _ 4y

2y(4y) = 9(8) 8 y 2 = 72 y 2 = 9 y = ±3

c. d _ 3 = 6 _

2

d(2) = 3(6) 2d = 18 d = 9

d. x + 3 _ 4 = 9 _

x + 3

(x + 3)(x + 3) = 4(9) x 2 + 6x + 9 = 36 x 2 + 6x ! 27 = 0(x ! 3)(x + 9) = 0 x = 3 or !9

4. 16s = 20t t _ s = 16 _

20

t _ s = 4 _ 5 = 4:5

5. 1 Understand the ProblemAnswer will be height of new tower.2 Make a PlanLet y be height of new tower. Write a proportion that compares the ratios of model height to actual height.

height of 1st tower

__ height of 1st model

= height of new tower

__ height of new model

1328 _ 8 =

y _

9.2

3 Solve

1328 _ 8 =

y _

9.2

1328(9.2) = 8(y) 12,217.6 = 8y y = 1527.2 m4 Look BackCheck answer in original problem. Ratio of actual height to model height is 1328 : 8, or 166 : 1. Ratio of actual height to model height for new tower is 1527.2 : 9.2 In simplest form, this ratio is also 166 : 1. So ratios are equal, and answer is correct.

THINK AND DISCUSS, PAGE 457 1. No; ratio 6 : 7 is < 1, but ratio 7 : 6 is > 1.

2. She can see if cross products are =. Since 3(28) = 7(12), ratios do form a proportion. Therefore ratios are = and fractions are equivalent.

3.


1. means: 3 and 2; extremes: 1 and 6

2. sv; tu


_ x 2 ! x 1

= 4 ! 3 _ 1 ! (!1)

= 1 _ 2


_ x 2 ! x 1

= 2 ! (!2)

_ 2 ! (!2)

= 4 _ 4 = 1 _

1



_ x 2 ! x 1

= !1 ! 1 _ 2 ! (!1)

= !2 _ 3 = ! 2 _

3

6. Let side lengths be 2x, 4x, 5x, and 7x. Then 2x + 4x + 5x + 7x = 36. After like terms are combined, 18x = 36. So x = 2. The shortest side measures 2x = 2(2) = 4 m.

7. Let " measures be 5x, 12x, and 19x. Then 5x + 12x + 19x = 180. After like terms are combined, 36x = 180. So x = 5. The largest " measures 19x = 19(5) = 95°.

8. x _ 2 = 40 _

16

x(16) = 2(40) 16x = 80 x = 5

9. 7 _ y = 21 _ 27

7(27) = y(21) 189 = 21y y = 9

10. 6 _ 58

= t _ 29

6(29) = 58(t) 174 = 58t t = 3

11. y _

3 = 27 _ y

y(y) = 3(27) y 2 = 81 y = ±9

12. 16 _ x ! 1

= x ! 1 _ 4

16(4) = (x ! 1)(x ! 1) 64 = x 2 ! 2x + 1 0 = x 2 ! 2x ! 63 0 = (x ! 9)(x + 7) x = 9 or !7

13. x 2 _ 18

= x _ 6

x 2 (6) = 18(x)6 x 2 ! 18x = 0 6x(x ! 3) = 0 x = 0 or 3

14. 2a = 8b a _

b = 8 _

2

a _ b = 4 _

1 = 4:1

15. 6x = 27y

6 _ 27

= y _ x

y _ x = 2 _

9 = 2:9

16. 1 Understand the ProblemAnswer will be height of Arkansas State Capitol.2 Make a PlanLet x be height of Arkansas State Capitol. Write a proportion that compares the ratios of height to width.

height of U.S. Capitol

__ width of U.S. Capitol

= height of Arkansas Capitol

___ width of Arkansas Capitol

288 _ 752

= x _ 564

3 Solve

288 _ 752

= x _ 564

288(564) = 752(x) 162,432 = 752x x = 216 ft4 Look BackCheck answer in original problem. Ratio of height to width for U.S. Capitol is 288 : 752, or 18 : 47. Ratio of height to width for Arkansas State Capitol is 216 : 564 In simplest form, this ratio is also 18 : 47. So ratios are equal, and answer is correct.


17. slope = 4 ! 1 _ 1 ! 0

= 3 _ 1 18. slope = !4 + 1 _

3 ! 0 = ! 1 _

1

19. slope = 0 + 3 _ 3 ! 1

= 3 _ 2

20. Let side lengths be 4x and 4x, and let base length be 7x.4x + 4x + 7x = 52.5 15x = 52.5 x = 3.5length of base = 7(3.5) = 24.5 cm

21. Let " measures be 2x, 3x, 2x, and 3x. By Quad. " Sum Thm., sum of " measures is 360°.2x + 3x + 2x + 3x = 360 10x = 360 x = 36" measures are 2(36) = 72°, 3(36) = 108°, 72°, and 108°.

22. 6 _ 8 = 9 _ y

6y = 8(9) = 72 y = 12

23. x _ 14

= 50 _ 35

35x = 14(50) = 700 x = 20

24. z _ 12

= 3 _ 8

8z = 12(3) = 36 z = 4.5

25. 2m + 2 _ 3 = 12 _

2m + 2

(2m + 2 ) 2 = 3(12) 4 m 2 + 8m + 4 = 364 m 2 + 8m ! 32 = 0 m 2 + 2m ! 8 = 0 (m ! 2)(m + 4) = 0 m = 2 or !4

26. 5y

_ 16

= 125 _ y

5 y 2 = 16(125)5 y 2 = 2000 y 2 = 400 y = ±20

27. x + 2 _ 12

= 5 _ x ! 2

(x + 2)(x ! 2) = 12(5) x 2 ! 4 = 60 x 2 = 64 x = ±8

28. 5y = 25x

5 _ 25

= x _ y

x _ y = 1 _ 5

29. 35b = 21c b _ c = 21 _

35 = 3 _

5

Ratio is 3 : 5.

30. Let x represent height of actual windmill.

height of windmill

__ width of windmill

= height of model

__ width of model

x _ 20

= 1.2 _ 0.8

0.8x = 20(1.2) = 24 x = 30 m

31. a _ b = 5 _

7

7a = 5b

32. a _ b = 5 _

7

7a = 5b

7 = 5b _ a

7 _ 5 = b _ a


33. a _ b = 5 _

7

7a = 5b

7a _ 5 = b

a _ 5 = b _

7

34. Cowboys lost 16 ! 10 = 6 games.wins : losses = 10 : 6 = 10 _

2 : 6 _

2

= 5 : 3

35. slope = 5 + 4 _ 21 + 6

= 9 _ 27

= 1 _ 3

36. slope = 1 + 5 _ 6 ! 16

= 6 _ !10

= ! 3 _ 5

37. slope = 5.5 + 2 _ 4 ! 6.5

= 7.5 _ !2.5

= !3

38. slope = 0 ! 1 _ !2 + 6

= ! 1 _ 4

39a. 1.25 in. _ 15 in.

= x in. _ 9600 in.

b. 1.25(9600) = 15x 12,000 = 15x x = 800 in. = 66 ft 8 in.

40. Quad. is a rect. because opp. sides are # and diags. are #.

41. Areas are 6 2 = 36 cm 2 and 9 2 = 81 cm 2 . 36 _ 81

= 4 _ 9

42. 5 _ 3.5

= 20 _ w

5w = 3.5(20) = 70 w = 14 in.

43. A ratio is a comparison of 2 numbers by div. A proportion is an eqn. stating that 2 ratios are =.

TEST PREP, PAGE 459

44. Bx + 4x + 5x = 18 10x = 18 x = 1.8 in.4x = 4(1.8) = 7.2 in., 5x = 5(1.8) = 9 in.

45. H

3 _ 5

= x _ y

3y = 5x

y = 5x _ 3

y _

5 = x _

3

46. A

5 _ 2 = 1.25 _ v

5v = 2(1.25) = 2.5v = 1 _

2

47. First, cross multiply:36x = 15(72) = 1080Then divide both sides by 36:

36x _ 36

= 1080 _ 36

Finally, simplify:x = 30You must assum that x $ 0.


48. Perimeters are 2(3) + 2(5) = 16 and 2x + 2(4) = 2x + 8.

4 _ 7 = 16 _

2x + 8

4(2x + 8) = 7(16) 8x + 32 = 112 8x = 80 x = 10

49. Given a _ b = c _

d , add 1 to both sides of eqn:

a _ b + b _

b = c _

d + d _

d

Adding fractions on both sides of eqn. gives

a + b _ b = c + d _

d .

50. Possible proportions are 1 _ 2 = 3 _

6 , 1 _

3 = 2 _

6 , 2 _

1 = 6 _

3 ,

2 _ 6 = 1 _

3 , 3 _

1 = 6 _

2 , 3 _

6 = 1 _

2 , 6 _

2 = 3 _

1 , and 6 _

3 = 2 _

1 .

There are 8 possible proportions. Total number of outcomes = 4! = 24.Probability = 8 _

24 = 1 _

3

51. x 2 + 9x + 18 __ x 2 ! 36

= (x + 6)(x + 3)

__ (x + 6)(x ! 6)

= x + 3 _ x ! 6

, where x $ ±6


52. y ! 6(0) = !3 y = !3

53. (3) ! 6x = !3 !6x = !6 x = 1

54. y ! 6(!4) = !3 y + 24 = !3 y = !27

55. Think: Use Same-Side Ext. % Thm. to find y, then use Vert. % Thm.3y + 2y + 20 = 180 5y = 160 y = 32m"ABD = 3y

= 3(32) = 96°

56. Think: Use Vert. % Thm.m"CDB = 2y + 20

= 2(32) + 20= 84°

57. 9 2 & 5 2 + 8 2 81 & 25 + 6481 < 89' is acute.

58. 20 2 & 8 2 + 15 2 400 & 64 + 225400 > 289' is obtuse.

59. 25 2 & 7 2 + 24 2 625 & 49 + 576625 = 625' is a right triangle.

TECHNOLOGY LAB: EXPLORE THE GOLDEN RATIO, PAGES 460–461

ACTIVITY 1 1. Check students’ work. The equal ratios have the

approximate value of 1.62.

2. The ratios have the same value as the ratios in Step 1.


TRY THIS, PAGE 461 1. If side length of square is 2 units, then MB = 1 unit

and BC = 2 units. (((

MC is hyp. of rt. ' formed by ((

MB and

(( BC . By Pyth. Thm.,

MC = ) * 5 units AE = ) * 5 + 1 units

AE _ EF

= ) * 5 + 1 _

2 + 1.618

2. BE = ) * 5 ! 1 units

BE _ EF

= ) * 5 ! 1 _

2 + 0.618

The sign of the numerator in this fraction is different from that of the fraction in Try This Problem 1.

3. Quotients have values that approach 1.618.

4. There are 1 + 1 = 2 rabbits.

5. There are 8 + 13 = 21 petals on the daisy.

6. No; 5.4 _ 4

+ 1.4 7. Yes; 4.5 _ 2.8

+ 1.6

7-2 RATIOS IN SIMILAR POLYGONS, PAGES 462–467

CHECK IT OUT! PAGES 462–464 1. "C # "H. By Rt. " # Thm., "B # "G.

By 3rd % Thm., "A # "J.

AB _ JG

= 10 _ 5

= 2, BC _ GH

= 6 _ 3 = 2, AC _

JH = 11.6 _

5.8 = 2

2. Step 1 Identify pairs of # %."L # "P (Given)"M # "N (Rt. " # Thm.)"J # "S (3rd % Thm.)Step 2 Compare corr. sides.

JL _ SP

= 75 _ 30

= 5 _ 2

, LM _ PN

= 60 _ 24

= 5 _ 2 , JM _

SN = 45 _

18 = 5 _

2

yes; similarity ratio is 5 _ 2 , and 'LMJ , 'PNS.

3. Let x be length of the model boxcar in inches. Rect. model of boxcar is , to rect. boxcar, so corr. lengths are proportional.

length of boxcar

__ length of model

= width of boxcar __ width of model

36.25 _ x = 9 _ 1.25

36.25(1.25) = 9x 45.3125 = 9x

x = 45.3125 _ 9 + 5 in.

THINK AND DISCUSS, PAGE 464 1. # symbol is formed.

2. Sides of rect. EFGH are 9 times as long as corr. sides of rect. ABCD.

3. Possible answers: reg. polygons of same type; -

4.


1. Possible answer: students’ desks

2. "M # "U and "N # "V. By 3rd % Thm., "P # "W.

MN _ UV

= 4 _ 8

= 1 _ 2 , MP _

UW = 3 _

6 = 1 _

2 , NP _

VW = 2 _

4 = 1 _

2

3. "A # "H and "C # "K. By def. of # %, and taking vertices clockwise in both figures, "B # "J and "D # "L.

AB _ HJ

= 8 _ 12

= 2 _ 3 , BC _

JK = 4 _

6 = 2 _

3 , CD _

KL = 4 _

6 = 2 _

3 ,

DA _ LH

= 8 _ 12

= 2 _ 3

4. Step 1 Identify pairs of # %. Think: All % of a rect. are rt. % and are #."A # "E, "B # "F, "C # "G, and "D # "H.Step 2 Compare corr. sides.

AB _ EF

= 135 _ 90

= 3 _ 2 , AD _

EH = 45 _

30 = 3 _

2

Yes; since opp. sides of a rect. are #, corr. sides are proportional. Similarity ratio is 3 _

2 , and

ABCD , EFGH.

5. Step 1 Identify pairs of # %."M # "W, "P # "U (Given)"R # "X (3rd % Thm.)Step 2 Compare corr. sides.

RM _ XW

= 8 _ 12

= 2 _ 3 , MP _

WU = 10 _

15 = 2 _

3 , RP _

XU = 4 _

6 = 2 _

3

yes; similarity ratio is 2 _ 3 , and 'RMP , 'XWU.

6. Let x be height of reproduction in feet. Reproduction is , to original, so corr. lengths are proportional.

height of reproduction

__ height of original

= width of reproduction

__ width of original

x _ 73

= 24 _ 58

58x = 73(24) = 1752

x = 1752 _ 58

+ 30 ft


7. "K # "T, "L # "U (Given)"J # "S, "M # "V (Rt. " # Thm.)

JK _ ST

= 20 _ 24

= 5 _ 6 , KL _

TU = 14 _

16.8 = 5 _

6 , LM _

UV = 30 _

36 = 5 _

6 ,

JM _ SV

= 10 _ 12

= 5 _ 6

8. "A # "X, "C # "Z (Given)"B # "Y (3rd % Thm.)

AB _ XY

= 8 _ 4 = 2, BC _

YZ = 6 _

3 = 2, CA _

ZX = 12 _

6 = 2


9. Step 1 Identify pairs of # %.m"R = 90 ! 53 = 37°"R # "U (Def. of # %)"S # "Z (Rt. " # Thm.)"Q # "X (3rd % Thm.)Step 2 Compare corr. sides.

QR _ XU

= 35 _ 40

= 7 _ 8 , QS _

XZ = 21 _

24 = 7 _

8 , RS _

UZ = 28 _

32 = 7 _

8

yes; similarity ratio = 7 _ 8 ; 'RSQ , 'UZX

10. Step 1 Identify pairs of # %."A # "M, "B # "J, "C # "K, "D # "L (Rt. " # Thm.)Step 2 Compare corr. sides.

AB _ MJ

= 18 _ 24

= 3 _ 4 , AD _

ML = AD _

JK = 36 _

54 = 2 _

3

no; the rectangles are not similar

11. model length

__ car length

= 1 _ 56

3 _ !

= 1 _ 56

3(56) = ! ! = 168 in. = 14 ft

12. Let x, y be side lengths of squares ABCD and PQRS. Areas are x 2 and y 2 , so

x 2 _ y 2

= 4 _ 36

= 1 _ 9

x _ y = ) * 1 _ 9 = 1 _

3

, ratio of ABCD to PQRS = x _ y = 1 _ 3

, ratio of PQRS to ABCD = y _ x = 3 _

1

13. sometimes (iff acute % are #)

14. always (all (rt.) % are #, all side-length ratios are =)

15. never (in trap., 1 pair sides are ., so opp. pairs of % cannot be #; but in /, they are #)

16. always (by CPCTC, all corr. % are #, and since corr. sides #, , ratio = 1)

17. sometimes (similar polygons are # iff , ratio = 1)

18. By def. of reg. polygons, corr. int. % are #, and side lengths are # and thus proportional. So any 2 reg. polygons with same number of sides are ,.

19. EF _ AB

= FG _ BC

x + 3 _ 4 = 2x ! 4 _

3

3(x + 3) = 4(2x ! 4) 3x + 9 = 8x ! 16 25 = 5x x = 5

20. MP _ XZ

= NP _ YZ

x + 5 _ 30

= 4x ! 10 _ 75

75(x + 5) = 30(4x ! 10) 5(x + 5) = 2(4x ! 10) 5x + 25 = 8x ! 20 45 = 3x x = 15


Statue of Liberty’s nose

__ Statue of Liberty’s hand

+ your nose

_ your hand

x ft _ 16.4 ft

+ 2 in. _ 7 in.

7x + 2(16.4) = 32.8 x + 4.7Estimated length of Statue of Liberty’s nose is 4.7 ft (or between 4.5 ft and 5 ft).

22. If 2 polygons are ,, then their corr. % are # and their corr. sides are proportional. If corr. % of 2 polygons are # and their corr. sides are proportional, then polygons are ,.

23. /JKLM , /NOPQ 0 "O # "K 0 m"O = 75°NOPQ a / 0 "Q # "O 0 m"Q = 75°"O and "Q are 75° %.

24. width on blueprint

__ actual width

= length on blueprint

__ actual length

w _ 14

= 3.5 _ 18

18w = 14(3.5) = 49 w = 49 _

18 + 2.7 in.

25. Polygons must be #. Since polygons are ,, their corr. % must be #. Since , ratio is 1, corr. sides must have same length.

26a. height of tree on backdrop

___ height of tree on flat

= 1 _ 10

0.9 _ h = 1 _

10

0.9(10) = h h = 9 ft

b. height of tree on flat

__ height of actual tree

= 1 _ 2

9 _ H

= 1 _ 2

9(2) = H H = 18 ft

c. , ratio = height of tree on backdrop

___ height of actual tree

= 0.9 _ 18

= 1 _ 20

TEST PREP, PAGE 467

27. C

y _

14.4 = 8.4 _

4.8

4.8y = 14.4(8.4) = 120.96 y = 25.2

28. F 5 _

2 = GL _

PS

5 _ 2 = 20 _

PS

5PS = 20(2) = 40 PS = 8

29. Ratios of sides are not the same: 12 _ 3.5

= 24 _ 7 ,

10 _ 2.5

= 4, 6 _ 1.5

= 4


30. model length

__ building length

= 1 _ 500

! _ 200

= 1 _ 500

500! = 200 ! = 0.4 ft = 4.8 in.

model width __ building width

= 1 _ 500

w _ 140

= 1 _ 500

500w = 140 w = 0.28 ft = 3.36 in.


31. Since ((

QR 1 ((

ST , "PQR # "PST and "PRQ # "PTS by Alt. Int. % Thm. "P # "P by Reflex. Prop. of #. Thus corr. % of 'PQR and 'PST are #. Since PS = 6 and PT = 8, PQ _

PS = PR _

PT = QR _

ST = 1 _

2 .

Therefore 'PQR , 'PST by def. of , polygons.

32a. By HL, 'ABD # 'CBD, so "A # "C, andm"A = m"C = 45°. So 'ABC is a 45°-45°-90° '.AC = AB ) * 2 = 1 ) * 2 = ) * 2 m"CBD = 90 ! "C = 45°, so 'CDB is also a 45°-45°-90° '. SoBC = 1 = DC ) * 2 = DB ) * 2 ) * 2 = 2DC = 2DB

DC = DB = ) * 2 _ 2

b. From part a., corr. % of 'ABC and 'CDB.

AB _ BD

= BC _ DC

= AC _ BC

= ) * 2 . By def. of ,,

'ABC , 'CDB.

33a. rect. ABCD , rect. BCFE

b. ! _ 1 = 1 _

! ! 1

c. !(! ! 1) = 1 ! 2 ! ! = 1 ! 2 ! ! ! 1 = 0

! = 1 ± ) ****** 1 2 ! 4(1)(!1)

__ 2

= 1 ± ) * 5 _ 2

Think: ! > 0, so take positive sq. root.

! = 1 + ) * 5 _ 2

d. ! + 1.6


34. # of orders = # of permutations of 4 things= 4! = 24

35. Think: Kite 0 diags. are 2. So "QTR is a rt. ".m"QTR = 90°

36. Think: 'PST # 'RST. By CPCTC, "PST # "RSTm"PST = m"RST = 20°

37. Think: 'PST is a rt. '. So " PST and "TPS are comp.m"TPS = 90 ! m"PST

= 90 ! 20 = 70°

38. x _ 4 =

y _

10

10x = 4y

39. x _ 4

= y _

10

10x = 4y

10x _ y = 4

10 _ y = 4 _ x

40. x _ 4 =

y _

10

x = 4y

_ 10

x _ y = 4 _ 10

or 2 _ 5

TECHNOLOGY LAB: PREDICT TRIANGLE SIMILARITY RELATIONSHIPS, PAGES 468–469

ACTIVITY 1, PAGE 468 3. The ratios of cor. side lengths are =.

TRY THIS, PAGE 468 1. ' Sum Thm.

2. Yes; in , 3, corr. sides are proportional.

ACTIVITY 2, PAGE 468 3. corr. % are #.

TRY THIS, PAGE 469 3. Yes; if 2 3 have their corr. sides in same ratio, then

they are ,.

4. They are similar in that both allow you to conclude that corr. % are #. They are different in that the conjecture suggests that 3 with corr. sides in same ratio have same shape, but SSS # Thm. allows you to conclude that the 3 have both same shape and same size.

ACTIVITY 3, PAGE 469 3. The ratio of the corr. sides of 'ABC and 'DEF are

proportional.

4. The corr. % of the 3 are #.

TRY THIS, PAGE 469 5. Yes; corr. sides are proportional and corr. % are #.

6. If 3 have 2 pairs of corr. sides in same proportion and included % are #, then 3 are ,. This is related to the SAS # Thm.

7-3 TRIANGLE SIMILARITY: AA, SSS, AND SAS, PAGES 470–477

CHECK IT OUT! PAGES 470–473 1. By ' Sum Thm., m"C = 47°, so "C # "F. "B # "E

by Rt. " # Thm. Therefore 'ABC , 'DEF by AA ,.

2. "TXU # "VXW by Vert. % Thm.

TX _ VX

= 12 _ 16

= 3 _ 4 , XU _

XW = 15 _

20 = 3 _

4

Therefore 'TXU , 'VXW by SAS ,.

3. Step 1 Prove 3 are ,.It is given that "RSV # "T. By the Reflex. Prop. of #, "R # "R. Therefore 'RSV , 'RTU by AA ,. Step 2 Find RT.

RT _ RS

= TU _ SV

RT _ 10

= 12 _ 8

8RT = 10(12) = 120 RT = 15



1. M is mdpt. of ((

JK , N is mdpt. of ((

KL , and P is mdpt. of ((

JL. 1. Given

2. MP = 1 _ 2 KL, MN = 1 _

2 JL,

NP = 1 _ 2 KJ

2. ' Midsegs. Thm.

3. MP _ KL

= MN _ JL

= NP _ KJ

= 1 _ 2 3. Div. Prop. of =

4. 'JKL , 'NPM 4. SSS , Step 3

5. FG _ AC

= BF _ AB

FG _ 5x

= 4 _ 4x

FG(4x) = 4(5x) 4FG = 20 FG = 5


1. "A # "D or "C # "F 2. BA _ ED

= 3 _ 5

3. No; corr. sides need to be proportional but not necessarily # for 3 to be ,.

4.


GUIDED PRACTICE, PAGE 474

1. By def. of " #, "C # "H. By ' Sum Thm., m"A = 47°, so "A # "F. Therefore 'ABC , 'FGH by AA ,.

2. "P # "T (given). "QST is a rt. " by the Lin. Pair Thm., so "QST # "RSP. Therefore 'QST , 'RSP by AA ,.

3. DE _ JK

= 8 _ 16

= 1 _ 2 , DF _

JL = 6 _

12 = 1 _

2 , EF _

KL = 10 _

20 = 1 _

2

Therefore 'DEF , 'JKL by SSS ,.

4. "NMP # "RMQ (given)

MN _ MR

= 4 _ 6 = 2 _

3 , MP _

MQ = 8 _

4 + 8 = 8 _

12 = 2 _

3

Therefore 'MNP , 'MRQ by SAS ,.

5. Step 1 Prove 3 are ,.It is given that "C # "E. "A # "A by Reflex. Prop. of #. Therefore 'AED # 'ACB by AA ,.Step 2 Find AB.

AB _ AD

= BC _ DE

AB _ 6 = 15 _

9

9AB = 15(6) = 90 AB = 10

6. Step 1 Prove 3 are ,.Since

(( UV || ((

XY , by Alt. Int. % Thm., "U # "Y and "V # "X. Therefore 'UVW , 'YXW by AA ,.Step 2 Find WY.

WY _ WU

= WX _ WV

WY _ 9 = 8.75 _

7

7WY = 9(8.75) = 78.75 WY = 11.25


1. ((

MN 1 ((

KL 1. Given2. "JMN # "JKL, "JNM # "JLK 2. Corr. % Post.3. 'JMN , 'JKL 3. AA , Step 2


1. SQ = 2QP, TR = 2RP 1. Given2. SP = SQ + QP,

TP = TR + RP2. Seg. Add. Post.

3. SP = 2QP + QP,TP = 2RP + RP

3. Subst.

4. SP = 3QP, TP = 3RP 4. Seg. Add. Post.

5. SP _ QP

= 3, TP _ RP

= 3 5. Div. Prop. of =

6. "P # "P 6. Reflex. Prop. of #7. 'PQR , 'PST 7. SAS , Steps 5, 6

9. SAS or SSS , Thm.

10. Step 1 Prove 3 are ,."S # "S by Reflex. Prop. of #

SA _ SC

= 733 + 586 _ 586

+ 2.25, SB _ SD

= 800 + 644 _ 644

+ 2.24

Therefore 'SAB , 'SCD by SAS ,.Step 2 Find AB.

AB _ CD

= SA _ SC

AB _ 533

+ 2.25

AB + 2.25(533) + 1200 m or 1.2 km


11. "G # "G by Reflex. Prop. of #. "GLH # "K by Rt. " # Thm. Therefore ' HLG , 'JKG by AA ,.

12. By Isosc. ' Thm., "B # "C and "E # "F. By ' Sum Thm., 32 + 2m"B = 180 2m"B = 148° m"B = 74°By def. of # %, "B # "E and "C # "F. Therefore 'ABC , 'DEF by AA ,.


13. "K # "K by Reflex. Prop. of # KL _ KN

= 6 _ 4 = 3 _

2 , KM _

KL = 5 + 4 _

6 = 3 _

2

Therefore 'KLM , 'KNL by SAS ,.

14. UV _ XY

= VW _ YZ

= WU _ ZX

= 4 _ 5.5

= 8 _ 11

Therefore 'UVW , 'XYZ by SSS ,.

15. Step 1 Prove 3 are ,.It is given that "ABD # "C. "A # "A by Reflex. Prop. of #. Therefore 'ABD # 'ACB by AA ,.Step 2 Find AB.

AB _ AD

= AC _ AB

AB _ 4 = 4 + 12 _

AB

AB 2 = 4(16) = 64 AB = + ) ** 64 = 8

16. Step 1 Prove 3 are ,.Since

(( ST 1 (((

VW , "PST # "V by Corr. % Post. "P # "P by Reflex. Prop. of #. Therefore 'PST , 'PVW by AA ,.Step 2 Find PS.

PS _ PV

= ST _ VW

PS _ PS + 6

= 10 _ 17.5

= 4 _ 7

7PS = 4(PS + 6) 7PS = 4PS + 24 3PS = 24 PS = 8


1. CD = 3AC, CE = 3BC 1. Given

2. CD _ AC

= 3, CE _ BC

= 3 2. Div. Prop. of #

3. "ACB # "DCE 3. Vert. % Thm.4. 'ABC , 'DEC 4. SAS , Steps 2, 3


1. PR _ MR

= QR _ NR

1. Given

2. "R # "R 2. Reflex. Prop. of #3. 'PQR , 'MNR 3. SAS , Steps 1, 24. "1 # "2 4. Def. of , 3

19.

By Vert. % Thm., "1 # "2. Since vert. sides are 1, "3 # "4 by Corr. % Post., so marked 3 are ,. Therefore,

h ÷ 2 _ 1.25

= 32.5 _ 54

54 ( h _ 2 ) = 1.25(32.5)

27h = 40.625 h + 1.5 in.

20.

J a

ka

kb

bK

M

L

N

P

yes; SAS ,

21. J ac

bK

L

ka

kbkc

MN

P

yes; SSS ,

22. J

ab

K

Lkakb

MN

P

no

23. Think: 'PQR # 'PST by AA ,. PS _

PQ = ST _

QR

x + 3 _ 3 = x + 5 _

4

4(x + 3) = 3(x + 5) 4x + 12 = 3x + 15 x = 3

24. Think: 'EFG # 'HJG by AA ,. EG _

GH = FG _

GJ

2x ! 2 _ 15

= x + 9 _ 20

20(2x ! 2) = 15(x + 9) 40x ! 40 = 15x + 135 25x = 175 x = 7

25a. Think: Calculate slant edge lengths

__ base edge length

for each

pyramid.Pyramid A: 12 _

10 = 6 _

5 ; Pyramid B: 9 _

7.2 = 5 _

4 ;

Pyramid C: 9.6 _ 8 = 6 _

5

Since slant edges of each pyramid are #, Pyramids A and C are , by SSS ,.Lengths are =.

b. base of A _ base of C

= 10 _ 8 = 5 _

4


26. Possible answer: Yes; If corr. % are # and corr. sides are prop., 'ABC , 'XYZ.

A

B

C X

Y

ZE

D

F

27. Think: Since all horiz. lines are 1, 3 3 with horiz. bases are , by AA ,. JK _

6 = 3 _

9

9JK = 6(3) = 18 JK = 2 ft

MN _ 6 = 6 _

9

9MN = 6(6) = 36 MN = 4 ft

28. Since 'ABC , 'DEF, by def. of , 3, "A # "D and "B # "E. Similarly, since 'DEF , 'XYZ, "D # "X and "E # "Y. Thus by Trans. Prop. of #, "A # "X and "B # "Y. So 'ABC , 'XYZ by AA ,.


Q

P

R

2

3

T

S

U6

4

30. Since 'KNJ is isosc. with vertex "N, ((

KN # ((

JN by def. of an isosc. '. "NKJ # "NJK by Isosc. ' Thm. It is given that "H # "L, so 'GHJ # 'MLK by AA ,.

31a. The 3 are , by AA , if you assume that camera is 1 to hurricane (that is,

(( YX 1 ((

AB ) .

b. 'YWZ , 'BCZ and 'XWZ , 'ACZ, also by AA ,.

c. XW _ AC

= WZ _ ZC

= 50 _ 150

150XW = 50AC

YW _ BC

= WZ _ ZC

= 50 _ 150

150YW = 50BC150XW + 150YW = 50AC + 50BC 150XY = 50AB 50AB = 150(35) = 5250 AB = 105 mi

32. Solution B is incorrect. The proportion should be

8 _ 10

= 8 + y

_ 14

.

33. Let measure of vertex % be x °. Then by Isosc. '

Thm., base % in each ' must measure ( 180 ! x _ 2 )

4 .

So 3 are , by AA ,.

TEST PREP, PAGE 477

34. C TU _ PQ

= UV _ QR

TU _ 60

= 60 + 20 _ 40 + 60

= 4 _ 5

5TU = 60(4) = 240 TU = 48

35. J FG _ BC

= 10.5 _ 42

= 1 _ 4

GH _ CD

= 14.5 _ 58

= 1 _ 4

36. CRects. , 0

(( BC , ((

FG , "C , "G, and ((

CD , ((

GH , which are conditions for SAS ,.

37. 30 x _ 12

= 20 _ 8

8x = 12(20) = 240 x = 30


38. Assume that AB < DE and choose X on ((

DE so that ((

AB # ((

DX . Then choose Y on ((

DF so that 5 67 XY 1 ((

EF . By Corr. % Post., "DXY # "DEF and "DYX # "DFE. Therefore 'DXY , 'DEF by AA ,. By

def. of , 3, DX _ DE

= XY _ EF

= DY _ DF

. By def. of #, AB = DX.

So AB _ DE

= XY _ EF

. It is given that AB _ DE

= BC _ EF

, so XY = BC.

((

XY # ((

BC by def. of #. Similarly, ((

DY # ((

AC , so 'ABC # 'DXY by SSS # Thm. It follows that 'ABC , 'DXY. Then by Trans. Prop. of ,, 'ABC , 'DEF.

39. Assume that AB < DE and choose X on ((

DE so that ((

XE # ((

AB . Then choose Y on ((

EF so that 5 67 XY 1 ((

DF . "EXY # "EDF by Corr. % Post., "E # "E by Reflex. Prop. of #. Therefore 'XEY , 'DEF by

AA ,. By def. of , 3, XE _ DE

= EY _ EF

. It is given that

AB _ DE

= BC _ EF

. By def. of #, XE = AB, so XE _ DF

= BC _ EF

.

Thus by def. of #, BC = EY and so ((

BC # ((

EY . It is also given that "B # "E, so 'ABC # 'XEY by SAS # Thm. It follows that 'ABC , 'XEY. Then by Trans. Prop. of ,, 'ABC , 'DEF.

40. Think: Use ' Sum Thm. and def. of ,. m"X + m"Y + m"Z = 1802x + 5y + 102 ! x + 5x + y = 180 6x + 6y = 78 x + y = 13 y = 13 ! xThink: Use def. of ,. "A # "Xm"A = m"X 50 = 2x + 5y 50 = 2x + 5(13 ! x) 50 = 65 ! 3x 3x = 15 x = 5y = 13 ! 5 = 8m"Z = 5(5) + 8 = 33°


41. 100 = 96 + 99 + 105 + 105 + 94 + 107 + x ____ 7

700 = 606 + x x = 94

42. Possible answer: (0, 4), (0, 0), (2, 0)

43. Possible answer: (0, k), (2k, k), (2k, 0), (0, 0)


44. 2x _ 10

= 35 _ 25

25(2x) = 10(35) 50x = 350 x = 7

45. 5y _

450 = 25 _

10y

5y(10y) = 450(25) 50 y 2 = 11,250 y 2 = 225 y = ±15

46. b ! 5 _ 28

= 7 _ b ! 5

(b ! 5 ) 2 = 28(7) = 196 b ! 5 = ±14 b = 5 ±14 = 19 or !9


1. height of model

__ height of real engine

= 1 _ 87

2.5 _ x = 1 _

87

2.5(87) = x x = 217.5 in. + 18 ft

2. height of model

__ height of real station

= 1 _ 87

y _

20 = 1 _

87

87y = 20 y + 0.23 ft + 2 3 _

4 in.

3. height of model

___ height of actual restaurant

= 1 _ 87

z _ 24

= 1 _ 87

87z = 24 z + 0.28 ft + 3 in.

4. base of B _ base of G

= 8 _ 14

= 5 _ 7 ; slant of B _

slant of G = 6 _

10 = 3 _

5 ; not ,

base of G _ base of H

= 14 _ 6 = 7 _

3 ; slant of G _

slant of H = 10 _

4.5 = 20 _

9 ; not ,

base of B _ base of H

= 8 _ 6 = 4 _

3 ; slant of B _

slant of H = 6 _

4.5 = 4 _

3 ; ,

Bank’s and hotel’s roofs are ,, by SSS ,.


1. slope = !1 + 2 _ 4 + 1

= 1 _ 5 2. slope = !3 ! 3 _

2 + 1

= !6 _ 3 = !2 _

1

3. slope = 1 ! 3 _ 4 + 4

= !2 _ 8

= !1 _ 4

4. slope = 0

5. y _

6 = 12 _

9

9y = 6(12) = 72 y = 8

6. 16 _ 24

= 20 _ t

16t = 24(20) = 480 t = 30

7. x ! 2 _ 4 = 9 _

x ! 2

(x ! 2 ) 2 = 4(9) = 36 x ! 2 = ±6 x = 2 ± 6

= –4 or 8

8. 2 _ 3y

= y _

24

2(24) = 3y(y) 48 = 3 y 2 16 = y 2 y = ±4

9. length of building

__ length of model

= width of building

__ width of model

! _ 1.4

= 240 _ 0.8

0.8! = 1.4(240) = 336 ! = 420 m

10. AB _ WX

= 64 _ 96

= 2 _ 3 ; AD _

WZ = 30 _

50 = 3 _

5 ; no

11. By def. of comp. %, m"M = 23° and m"K = 67°; so "J # "N, "M # "P, and "R # "K;

JM _ NP

= 24 _ 36

= 2 _ 3 ; MR _

PK = 26 _

39 = 2 _

3 ; JR _

NK = 10 _

15 = 2 _

3

yes; 2 _ 3 ; 'JMR , 'NPK

12. Think: Assume magnet , portrait.

length of magnet

__ length of portrait

= width of magnet

__ width of portrait

! _ 30

= 3.5 _ 21

21! = 30(3.5) = 105 ! = 5 cm


1. ABCD is a /. 1. Given2. ((

AD 1 ((

BC 2. Def. of /3. "EDG # "FBG 3. Alt. Int. % Thm.4. "EGD # "FGB 4. Vert. % Thm.5. 'EDG , 'FBG 5. AA , Steps 3, 4


1. MQ = 1 _ 3 MN, MR = 1 _

3 MP 1. Given

2. MQ _ MN

= 1 _ 3 , MR _

MP = 1 _

3 2. Div. Prop. of =

3. MQ _ MN

= MR _ MP

3. Trans. Prop. of =

4. "M # "M 4. Reflex. Prop. of #5. 'MQR , 'MNP 5. SAS , Steps 3, 4

15. Think: ' XYZ , 'VUZ with ratio of proportion 5 _ 2 ,

by SAS ,.

XY _ UV

= 5 _ 2

2XY = 5UV2XY = 5(16) = 80 XY = 40 ft

TECHNOLOGY LAB: INVESTIGATE ANGLE BISECTORS OF A TRIANGLE, PAGE 480

TRY THIS, PAGE 480

1. BD _ AB

= CD _ AC

or BD _ CD

= AB _ AC

.

2. BD _ CD

= AB _ AC

or BD _ AB

= CD _ AC

.

ACTIVITY 2: 2. Check students’ work.

3. DI _ DG

= DE + DF __ perimeter 'DEF


4. DI _ DG

= DE + DF __ DE + DF + EF

; the length of the seg. from

the vertex of the bisected " to the incenter divided by the length of the seg. from the vertex to the opp. side is = to the sum of the sides of the bisected " divided by the perimeter of the '.

TRY THIS, PAGE 480 3. Check students’ work.


7-4 APPLYING PROPERTIES OF SIMILAR TRIANGLES, PAGES 481–487



PQ 1 ((

LM , so PL _ PN

= QM _ QN

by ' Prop. Thm. 3 _ PN

= 2 _ 5

15 = 2PN PN = 7.5

2. AD = 36 ! 20 = 16 and BE = 27 ! 15 = 12, so

DC _ AD

= 20 _ 16

= 5 _ 4

EC _ BE

= 15 _ 12

= 5 _ 4

Since DC _ AD

= EC _ BE

, ((

DE 1 ((

AB by Conv. of ' Prop. Thm.

3. ((

AK 1 ((

BL 1 (((

CM 1 ((

DN

KL _ LM

= AB _ BC

2.6 _ LM

= 2.4 _ 1.4

2.4(LM) = 2.6(1.4) LM + 1.5 cm KL _

MN = AB _

CD

2.6 _ MN

= 2.4 _ 2.2

2.4(MN) = 2.6(2.2) MN + 2.4 cm

4. BD _ CD

= AB _ BC

by

' " Bis. Thm.

4.5 _ y __ 2 = 9 _ y = 8 _

y ! 2

9(y ! 2) = 8y 9y ! 18 = 8y y = 18

AC = y ! 2= 18 !2 = 16

DC = y _

2 = 18 _

2 = 9


1. Possible answer: AX _ XB

= AY _ YC

; AX _ AB

= XY _ BC

; AY _ AC

= XY _ BC

;

2.

E F

B C

A

A BDF

CE

A

DB C

!Proportionality Thm.:

If ! "# EF " $$ BC , then AE __ EB = AF __ FC .

Conv. of ! Proportionality Thm.:

If AE __ EB = AF __ FC , then ! "# EF " $$ BC .

EF

2-Transv. Proportionality Corollary:

If ! "# AB " ! "# CD " ! "# , then AC ___ CE = BD ___ DF .

!# Bisector Thm.:

If $$ AD bisects #A, then BD ___ DC = AB ___ AC .

E F

B C

A

Proportionality



CD 1 ((

FG , so CE _ CF

= DE _ DG

by ' Prop. Thm.

32 _ 24

= 40 _ DG

32DG = 960DG = 30


QR 1 ((

PN , so QM _ QP

= RM _ RN

by ' Prop. Thm.

8 _ 5 = 10 _

RN

8RN = 50 RN = 6.25

3. EC _ AC

= 1.5 _ 1.5

= 1; ED _ DB

= 1.5 _ 1.5

= 1

Since EC _ AC

= ED _ DB

, ((

AB 1 ((

CD by Conv. of ' Prop. Thm.

4. VU _ US

= 67.5 _ 54

= 5 _ 4 ; VT _

TR = 90 _

72 = 5 _

4

Since VU _ US

= VT _ TR

, ((

TU 1 ((

RS by Conv. of ' Prop. Thm.

5. Let ! represent length of Broadway between 34th and 35th Streets. ! _ 275

= 250 _ 240

240! = 275(250) ! + 286 ft

6. QR _ RS

= PQ _ PS

by ' " Bis. Thm.

x ! 2 _ x + 1

= 12 _ 16

16(x ! 2) = 12(x + 1) 16x ! 32 = 12x + 12 4x = 44 x = 11QR = 11 ! 2 = 9; RS = 11 + 1 = 12


7. BC _ CD

= AB _ AD

by ' " Bis. Thm.

6 _ y ! 1

= 9 _ 2y ! 4

6(2y ! 4) = 9(y ! 1) 12y ! 24 = 9y ! 9 3y = 15 y = 5CD = 5 ! 1 = 4; AD = 2(5) ! 4 = 6


8. GJ _ JL

= HK _ KL

6 _ 4 = 8 _

KL

6KL = 32 KL = 5 1 _

3

9. XY _ YU

= XZ _ ZV

30 ! 18 _ 18

= XZ _ 30

12(30) = 18XZ XZ = 20

10. EC _ CA

= 12 _ 4 = 3, ED _

DB = 14 _

4 2 __ 3 = 42 _

14 = 3

So ((

AB 1 ((


11. PM _ MQ

= 9 ! 2.7 _ 2.7

= 2 1 _ 3 , PN _

NR = 10 ! 3 _

3 = 2 1 _

3

So (((

MN 1 ((

QR by Conv. of ' Prop. Thm.

12. LM _ GL

= HJ _ GH

LM _ 11.3

= 2.6 _ 10.4

LM = 2.6 _ 10.4

(11.3)

+ 2.83 ft

MN _ GL

= JK _ GH

MN _ 11.3

= 2.2 _ 10.4

MN = 2.2 _ 10.4

(11.3) + 2.39 ft

13. BC _ CD

= AB _ AD

z ! 4 _ z __ 2 = 12 _

10

10(z ! 4) = z _ 2 (12)

10z ! 40 = 6z 4z = 40 z = 10BC = 10 ! 4 = 6; CD = 10 _

2 = 5

14. TU _ UV

= ST _ SV

2y _

14.4 =

4y ! 2 _

24

24(2y) = 14.4(4y ! 2) 48y = 57.6y ! 28.8 28.8 = 9.6y y = 3ST = 4(3) ! 2 = 10; TU = 2(3) = 6

15. AB _ BD

= AC _ CE

16. AD _ DF

= AE _ EG

17. DF _ BD

= EG _ CE

18. AF _ AB

= AG _ AC

19. BD _ CE

= DF _ EG

20. AB _ AC

= BF _ CG

21. Let x represent length of 3rd side.either x _ 20

= 12 _ 16

16x = 240 x = 15 in.

or x _ 20

= 16 _ 12

12x = 320

x = 80 _ 3 = 26 2 _

3 in.

22a. AC _ BD

= CE _ DF

b. 81.6 _ 80

= CE _ 70

81.6(70) = 80CE CE = 71.4 cm

c. AJ _ BK

= AC _ BD

AJ __ 80 + 70 + 60 + 40

= 81.6 _ 80

AJ = 81.6 _ 80

(250) = 255 cm


1. AE _ EB

= AF _ FC

1. Given

2. "A # "A 2. Reflex. Prop. of #3. 'AEF , 'ABC 3. SAS , Steps 1, 24. "AEF # "ABC 4. Def. of , '5. 5 67 EF 1

(( BC 5. Conv. of Corr. % Post.


1. 5 67 AB 1 5 67 CD , 5 67 CD 1 5 67 EF 1. Given2. Draw 5 67 EB intersecting

(( CD

at X.2. 2 pts. determine

a line

3. AC _ CE

= BX _ XE

3. ' Prop. Thm.

4. BX _ XE

= BD _ DF

4. ' Prop. Thm.

5. AC _ CE

= BD _ DF


25a. PR _ RT

= QS _ SU

x _ x + 2

= x __ 2 _

x ! 2 = x _

x(x ! 2)

2x(x ! 2) = x(x + 2)

2 x 2 ! 4x = x 2 + 2x x 2 ! 6x = 0 x(x ! 6) = 0 x = 6 (since x > 0)

PR = 6; RT = 6 + 2 = 8; QS = 6 _ 2 = 3;

SU = 6 ! 2 = 4

b. PR _ RT

= QS _ SU

or 6 _ 8

= 3 _ 4

26. Think: Use ' Prop. Thm. and ' " Bis. Thm.

EF _ BE

= CD _ BC

= AD _ AB

EF _ 10

= 24 _ 18

= 4 _ 3

3EF = 40 EF = 13 1 _

3


27. ST _ TQ

= SR _ RQ

= PN _ NM

ST _ 10

= 6 _ 4

4ST = 60 ST = 15

28. Total length along Chavez St. is 150 + 200 + 75 = 425 ft.

x _ 150

= 500 _ 425

= 20 _ 17

17x = 150(20) = 3000 x + 176 ft

y _

200 = 500 _

425 = 20 _

17

17y = 4000 y + 235 ft

z _ 75

= 500 _ 425

= 20 _ 17

17z = 1500 z = 88 ft

29. Draw a seg. on tracing paper whose length is = to the vert. dist. from line 1 to line 6 or no greater than the diag. dist. from line 1 to line 6 of the notebook paper. Place the tracing paper over the notebook paper so that the seg. spans exactly 6 of the lines on the notebook paper. Then mark the spots where the tracing-paper seg. crosses the line on the notebook paper. The method works by the 2-Transv. Proportionality Corollary.

30. Think: Use ' Prop. Thm. First find EX.

EX _ AX

= EY _ DY

EX _ 17

= 16 _ 18

18EX = 272 EX = 15 1 _

9

AE = AX + XE = 17 + 15 1 _

9 = 32 1 _

9

EC _ AE

= DB _ AD

EC _ 3 2 1 / 9

= 7.5 _ 15

= 1 _ 2

2EC = 32 1 _ 9

EC = 16 1 _ 18

31. Possible answer: BD _ CD

= AB _ AC

; ' " Bis. Thm.

TEST PREP, PAGE 487

32. C

US _ SR

= 20 _ 35

= 4 _ 7 , VT _

TR = 16 _

28 = 4 _

7

33. J

AB _ 25

= 16 _ 20

20AB = 400 AB = 20

34. CLet x be dist. to 1st St. x _

2.4 = 2.1 _

2.8 = 3 _

4

4x = 7.2 x = 1.8 mix + 2.4 = 4.2 mi

35. x _ 24

= 20 _ 16

= 5 _ 4

4x = 120 x = 30

y _

15 = 16 _

20 = 4 _

5

5y = 60 y = 12possible answer: 20 _

16 = 15 _

12 ; 20 _

16 = 30 _

24 ; 15 _

12 = 30 _

24 ;

20 + 15 _ 30

= 16 + 12 _ 24

; 20 _ 15 + 30

= 16 _ 12 + 24

;

20 __ 20 + 15 + 30

= 16 __ 16 + 12 + 24


36. P = AB + BC + AC 29 = AB + 9 + AC20 ! AB = AC

AB _ AC

= BD _ CD

AB _ 20 ! AB

= 4 _ 5

5AB = 4(20 ! AB) 9AB = 80 AB = 8 8 _

9

AC = 20 ! 8 8 _ 9 = 11 1 _

9

37. Given: 'ABC , 'XYZ, ((

AD bisects "BAC, and

((( XW bisects "YXZ.

Prove: AD _ XW

= AB _ XY

Statements Reasons

1. 'ABC , 'XYZ 1. Given2. "B # "Y 2. Def. of , polygons3. m"BAC = m"YXZ 3. Def. of , polygons4. ((

AD bisects "BAC and (((

XW bisects "YXZ.4. Given

5. m"BAC = 2m"BAD, m"YXZ = 2m"YXW

5. Def. of " bis.

6. 2m"BAD = 2m"YXW 6. Trans. Prop. of =7. m"BAD = m"YXW 7. Div. Prop. of =8. 'ABD , 'XYW 8. AA , Steps 2, 7

9. AD _ XW

= AB _ XY

9. ' Prop. Thm.


38.

Statements Reasons

1. ((

AD bisects "A. 1. Given 2. Draw

(( BX 1 ((

AD , extending

(( AC to X.

2. 1 Post.

3. BD _ DC

= AX _ AC

3. ' Prop. Thm.

4. "CAD # "AXB 4. Corr. % Post. 5. "CAD # "DAB 5. Def of " bis. 6. "DAB # "ABX 6. Alt. Int. % Thm. 7. "DAB # "AXB 7. Trans. Prop. of # 8. "ABX # "AXB 8. Trans. Prop. of # 9. ((

AX # ((

AB 9. Conv. Isosc. ' Thm.10. AX = AB 10. Def. of # segs.

11. BD _ DC

= AB _ AC

11. Subst.

39. Possible answer: Check students’ work.


40. 5 = 1 + 4, 6 = 2 + 4, … n th term is n + 4

41. 3 = 3(1), 6 = 3(2), … n th term is 3n

42. 1 = 1 2 , 4 = 2 2 , 9 = 3 2 , … n th term is n 2

43. Let C = (x, y).

3 = 1 + x _ 2

6 = 1 + x x = 5C = (5, !18)

!7 = 4 + y

_ 2

!14 = 4 + y y = !18

44. "A # "A (Reflex. Prop. of #)

AB _ AD

= 8 _ 12

= 2 _ 3 , AC _

AE = 6 _

9 = 2 _

3

Therefore 'ABC , 'ADE by SAS ,.

45. "KLJ # "NLM (Vert. % Thm.)"K # "N (' Sum Thm. 0 m"N = 68°)Therefore 'JKL , 'MNL by AA ,.

7-5 USING PROPORTIONAL RELATIONSHIPS, PAGES 488–494

CHECK IT OUT! PAGES 488–490 1. Step 1 Convert measurements to inches.

GH = 5 ft 6 in. = 5(12) in. + 6 in. = 66 in.JH = 5 ft = 5(12) in. = 60 in.NM = 14 ft 2 in. = 14(12) in. + 2 in. = 170 in.Step 2 Find , 3.Because sun’s rays are 1, "J # "N. Therefore 'GHJ # 'LMN by AA ,.Step 3 Find LM.

GH _ LM

= JH _ NM

66 _ LM

= 60 _ 170

60LM = 66(170) LM = 187 in. = 15 ft 7 in.

2. Use a ruler to measure dist. between City Hall and El Centro College. Dist. is 4.5 cm.To find actual dist. y, write a proportion comparing map dist. to actual dist.

4.5 _ y = 1.5 _ 300

1.5y = 4.5(300)1.5y = 1350 y = 900Actual dist. is 900 m, or 0.9 km.

3. Step 1 Set up proportions to find length ! and width w of of scale drawing. ! _ 74

= 1 _ 20

20! = 74 ! = 3.7 in.

w _ 60

= 1 _ 20

20w = 60 w = 3 in.

Step 2 Use a ruler to draw a rect. with new dimensions. (Check students’ work.)

4. Similarity ratio of 'ABC to 'DEF is 4 _ 12

, or 1 _ 3 .

By Proportional Perimeters and Areas Thm., ratio of 3’ perimeters is also 1 _

3 , and ratio of 3’ areas

is ( 1 _ 3 )

2 , or 1 _

9 .

Perimeter P _ 42

= 1 _ 3

3P = 42 P = 14 mm

Area A _ 96

= 1 _ 9

9A = 96 A = 10 2 _

3 mm 2

Perimeter of 'ABC is 14 mm, and area is 10 2 _ 3 mm 2 .

THINK AND DISCUSS, PAGE 490 1. Set up a proportion: 5.5 _ x = 1 _

25 . Then solve for x to

find actual dist.: x = 5.5(25) = 137.5 mi.

2.



1. indirect measurement

2. Step 1 Convert measurements to inches.5 ft 6 in. = 5(12) in. + 6 in. = 66 in.4 ft = 4(12) in. = 48 in.40 ft = 40(12) in. = 480 in.Step 2 Find , 3.Since marked % are #, 3 are , by AA ,.Step 3 Find height of dinosaur, h.

h _ 66

= 480 _ 48

h _ 66

= 10

h = 10(66) = 660 in.Height of dinosaur is 660 in., or 55 ft.

3. Use a ruler to measure to-scale length of ((

AB . Length is 0.25 in.To find actual length AB, write a proportion comparing to-scale length to actual length.

0.25 _ AB

= 1 _ 48

AB = 0.25(48) = 12 ft


CD . Length is 0.75 in.To find actual length CD, write a proportion comparing to-scale length to actual length.

0.75 _ CD

= 1 _ 48

CD = 0.75(48) = 36 ft


EF . Length is 1.25 in.To find actual length EF, write a proportion comparing to-scale length to actual length.

1.25 _ EF

= 1 _ 48

EF = 1.25(48) = 60 ft


FG . Length is 0.5 in.To find actual length FG, write a proportion comparing to-scale length to actual length.

0.5 _ FG

= 1 _ 48

FG = 0.5(48) = 24 ft

7. Step 1 Set up proportions to find length ! and width w of scale drawing. ! _ 10

= 1 _ 1

! = 10 cm

w _ 4.6

= 1 _ 1

w = 4.6 cmStep 2 Use a ruler to draw a rect. with new dimensions. (Check students’ drawings.)

8. Step 1 Set up proportions to find length ! and width w of scale drawing. ! _ 10

= 1 _ 2

2! = 10 ! = 5 cm

w _ 4.6

= 1 _ 2

2w = 4.6 cm w = 2.3 cm

Step 2 Use a ruler to draw a rect. with new dimensions. (Check students’ drawings.)

9. Step 1 Set up proportions to find length ! and width w of scale drawing. b _

10 = 1 _

2.3

2.3b = 10 b = 4.3 cm

w _ 4.6

= 1 _ 2.3

2.3w = 4.6 cm w = 2 cm

Step 2 Use a ruler to draw a rect. with new dimensions. (Check students’ drawings.)

10. Similarity ratio of MNPQ to RSTU is 4 _ 6 , or 2 _

3 .

By Proportional Perimeters and Areas Thm., ratio of

perimeters is also 2 _ 3

.

14 _ P

= 2 _ 3

2P = 14(3) = 42 P = 21Perimeter of RSTU is 21 cm.

11. Ratio of areas is ( 2 _ 3

) 2 , or 4 _

9 .

12 _ A

= 4 _ 9

4A = 12(9) = 108 A = 27Area of RSTU is 27 cm 2 .


12. 5 ft 2 in. = 62 in.; 7 ft 9 in. = 93 in.; 15.5 ft = 186 in. h _ 62

= 186 _ 93

= 2

h = 62(2) = 124 in. = 10 1 _ 3 ft or 10 ft 4 in.

13. map dist. for ((

JK = 6 cm

6 _ JK

= 1 _ 9.4

JK = 6(9.4) + 57 km

14. map dist. for ((

NP = 0.45 cm

0.45 _ NP

= 1 _ 9.4

NP = 0.45(9.4) + 4 km

15. Step 1 Set up proportions to find base b and height h of scale drawing. b _ 150

= 1.5 _ 100

100b = 225 b = 2.25 in.

h _ 200

= 1.5 _ 100

100h = 300 h = 3 in.

Step 2 Use a ruler to draw a rt. ' with new dimensions. (Check students’ drawings.)


= 1 _ 300

300b = 150 b = 0.5 in.

h _ 200

= 1 _ 300

300h = 200 h + 0.67 in.



= 1 _ 150

150b = 150 b = 1 in.

h _ 200

= 1 _ 150

150h = 200 h + 1.3 in.



18. scale factor = 60 _ 90

= 2 _ 3

P _ 381

= 2 _ 3

3P = 762 P = 254 m

19. A _ 1944

= ( 2 _ 3 )

2 = 4 _

9

9A = 7776 A = 864 m 2

20. scale factor = 10 ft _ 0.5 in.

= 20map dist. = 30 _

16 in.

x _ 30 __ 16

= 20

x = 30 _ 16

(20)

+ 38 ft

21. map dist. = 10 _ 8 in.

x _ 10 __ 8 = 20

x = 10 _ 8 (20)

= 25 ft

22. map dist. = 25 _ 16

in.

x _ 25 __ 16

= 20

x = 25 _ 16

(20)

+ 32 ft

23. map dist. = 32 _ 16

in.

x _ 32 __ 16

= 20

x = 32 _ 16

(20)

+ 39 ft

24. By Proportional Perimeters and Areas Thm., , ratio = ratio of perimeters = 8 _

9 .

25. By Proportional Perimeters and Areas Thm., ratio of areas = (, ratio ) 2 .

16 _ 25

= (, ratio ) 2

, ratio = ) ** 16 _ 25

= 4 _ 5

26. ratio of areas = (, ratio ) 2 ratio of areas = (ratio of perims.) 2

4 _ 81

= (ratio of perims. ) 2

ratio of perims. = ) ** 4 _ 81

= 2 _ 9

27. scale width _ model width

= 1 _ 50

w _ 15

= 1 _ 50

w = 15 _ 50

= 0.3 ft

scale length

__ model length

= 1 _ 50

! _ 60

= 1 _ 50

! = 60 _ 50

= 1.2 ft

28a. hyp. of 'PQR = ) **** 3 2 + 4 2 = 5 in.

hyp. of 'WXY = ) **** 6 2 + 8 2 = 10 in.

perimeter of 'PQR

__ perimeter of 'WXY

= 3 + 4 + 5 _ 6 + 8 + 10

= 12 _ 24

= 1 _ 2

b. area of 'PQR __ area of 'WXZ

= 1 __ 2 (4)(4)

_ 1 __ 2 (8)(6)

= 6 _ 24

= 1 _ 4

c. The ratio of areas is square of ratio of perimeters.

29. Let ! 1 and w 1 be dimensions of rect. ABCD; let ! 2 and w 2 be dimensions of rect. EFGH. A 1 = ! 1 w 1 135 = ! 1 (9) ! 1 = 15 in.Think: Rects. are ,; let scale factor be s.

! 2

_ ! 1

= w 2

_ w 1 = s

! 2 = s ! 1 , w 2 = s w 1 A 2 = ! 2 w 2 = (s ! 1 )(s w 1 )

= s 2 A 1 240 = 135 s 2 16 _

9 = s 2

s = 4 _ 3

! 2 = s ! 1

= 4 _ 3 (15) = 20 in.

w 2 = s w 1

= 4 _ 3 (9) = 12 in.


scale length

__ actual length

= ! _ 94

= 0.25 _ 10

10! = 23.5 ! = 2.35 in.

scale width _ actual width

= w _ 50

= 0.25 _ 10

10w = 12.5 w = 1.25 in.

31a., ratio = 1 in. _ 2 ft

= 1 in. _ 24 in.

= 1 _ 24

b. actual dimensions are 24(2) = 48 in. and 24(3) = 72 in.actual area = (48)(72) = 3456 in. 2 model area = (2)(3) = 6 in. 2

model area _ actual area

= 3456 _ 6 = 1 _

576

c. actual area = (4 ft)(6 ft) = 24 ft 2

32. In photo, height of person + 1 _ 2 in. and height of

statue + 1 5 _ 8 in.

actual height of statue

__ height of statue in photo

= actual height of person

___ height of statue in person

h _ 1.625

+ 5 _ 0.5

0.5h + 8 h + 16 ft

33. map length

__ actual length

= scale factor

! _ 1 km

= 1 cm _ 900,000 cm

= 1 cm _ 9 km

! = 1 _ 9 cm


34. By ' Midseg. Thm., def. of mdpt., and SSS #, 'XYZ # ' ZJX; so 3 have same height h. Therefore height of 'JKL = h + h = 2h. Since KL = 2ZX, area of 'JKL = 1 _

2 (2ZX)(2h)

= 2(ZX)h

= 4 ( 1 _ 2 (ZX)(h))

= 4(area of 'XYZ)

area of 'JKL __ area of 'XYZ

= 4 _ 1

35. 1 cm : 5 m; Since each cm will represent 5 m, this drawing will be 1 _

5 size of the 1 cm : 1 m drawing.

36. 4(x ! 2)

_ 4(2x)

= x ! 2 _ 2x

= 4 _ 9

9(x ! 2) = 8x 9x ! 18 = 8x x = 18AB = 18 ! 2 = 16 unitsHE = 2(18) = 36 units

37. With a scale of 1 : 1, drawing is same size as actual object.

38. Suppose x and y are whole-number side lengths of smaller square and larger square. Then 2 x 2 = y 2 . Thus x ) * 2 = y. A whole number that is multiplied by ) * 2 cannot equal a whole number, since ) * 2 is irrational.

TEST PREP, PAGE 493

39. Darea of 'RST = (scale factor ) 2 (area of 'ABC)

= ( 15 _ 10

) 2 (24) = 9 _

4 (24) = 54 m 2

40. G

3.75 _ !

= 0.25 _ 1

3.75 = 0.25! ! = 15 ft

41. C. Ratio of perimeters = , ratio = 4 _ 9

42. Farea of '2 = (, ratio ) 2 (area of '1)

= ( 1 _ 2 )

2 (16) = 4 ft 2


43a. x __ 1.5 8 10 8 km

= 1 km _ 10 9 km

= 1 0 3 m _ 1 0 9 km

x = 1 0 3 m _ 1 0 9 km

(1.5 8 1 0 8 km)

= 1.5 8 1 0 2 m or 150 m

43b. d __ 1.28 8 1 0 4 km

= 1 0 3 m _ 1 0 9 km

d = 1 0 3 m _ 1 0 9 km

(1.28 8 1 0 4 km)

= 1.28 8 1 0 !2 m or 1.28 cm

44. It is given that 'ABC , 'DEF. Let AB _ DE

= x. Then

AB = DEx by Mult. Prop. of =. Similarly, BC = EFx

and AC = DFx. By Add. Prop. of =, AB + BC + AC =DEx + EFx + DFx. Thus AB + BC + AC = x (DE + EF + DF). By Div. Prop. of =,

AB + BC + AC __ DE + EF + DF

= x. By subst., AB + BC + AC __ DE + EF + DF

= AB _ DE

.

45. It is given that 'PQR , 'WXY. Draw 2s from

Q and X to meet ((

PR at S and (((

WY at Z. By def. of

, polygons, PQ _ WX

= QR _ XY

= PR _ WY

, and "P # "W.

In 'PQS and 'WXZ, "PSQ # "WZX. Thus

'PQS , 'WXZ by AA ,. PQ _ WZ

= QS _ XZ

= PS _ WZ

by

def. of , polygons. QR _ XY

= SP _ ZW

by subst.

Area of 'PQR __ area of 'WXY

= PR _ WY

· QS _ XZ

= P R 2 _ W Y 2

.

46a. 6 _ WX

= 1 _ 2

WX = 12

10 _ YZ

= 1 _ 2

YZ = 20

7 _ XY

= 1 _ 2

XY = 14

12 _ WZ

= 1 _ 2

WZ = 24

b. Quad. PQRS Quad. WXYZ

Side Length (m) Side Length (m)

PQ 6 WX 12

QR 7 XY 14

RS 10 YZ 20

PS 12 WZ 24

c.

d. WX = 12 = 2PQ; similarly XY = 2QR, YZ = 2RS, and WZ = 2PS. So eqn. is y = 2x.


47. (x ! 3 ) 2 = 49 x ! 3 = ±7 x = 3 ± 7

= 10 or !4

48. (x + 1 ) 2 ! 4 = 0 (x + 1 ) 2 = 4 x + 1 = ±2 x = !1 ± 2

= !3 or 1

49. 4(x + 2 ) 2 ! 28 = 0 4(x + 2 ) 2 = 28 (x + 2 ) 2 = 7 x + 2 = ± ) * 7 x = !2 ± ) * 7

+ 0.65 or !4.65

50. slope of ((

AB = 2 _ 3 ; slope of

(( CD = !2 _ !3

= 2 _ 3

slope of ((

BC = slope of ((

AD = 0 ((

AB 1 ((

CD and ((

BC 1 ((

AD , so ABCD is a /.


51. slope of ((

JK = 2 _ 2 = 1; slope of

(( LM = !2 _ !2

= 1

slope of ((

KL = !3 _ 3 = !1; slope of

(( JM = !3 _

3 = !1

((

JK 1 ((

LM and ((

KL 1 ((

JM , so JKLM is a /.

52. 58x = 26yy : x = 58 : 26 = 29 : 13

7-6 DILATIONS AND SIMILARITY IN THE COORDINATE PLANE, PAGES 495–500

CHECK IT OUT! PAGES 495–497 1. Step 1 Multiply vertices of photo A(0, 0), B(0, 4),

C(3, 4), D(3, 0) by 1 _ 2

.

Rect. ABCD Rect. A9B9C9D9

A(0, 0) 0 A9 ( 1 _ 2 (0), 1 _

2 (0)) 0 A9(0, 0)

B(0, 0) 0 B9 ( 1 _ 2 (0), 1 _

2 (4)) 0 B9(0, 2)

C(0, 0) 0 C9 ( 1 _ 2 (3), 1 _

2 (4)) 0 C9(1.5, 2)

D(0, 0) 0 D9 ( 1 _ 2 (3), 1 _

2 (0)) 0 D9(1.5, 0)

Step 2 Plot pts. A9(0, 0), B9(0, 2), C9(1.5, 2), and D9(1.5, 0). Draw the rectangle.

Check student’s work

2. Since 'MON , 'POQ,

PO _ MO

= OQ _ ON

!15 _ !10

= 3 _ 2 = !30 _

ON

3ON = !60 ON = !20N lies on y-axis, so its x-coord. is 0. Since ON = !20, its y-coord. must be !20. Coords. of N are (0, !20).

(0, !30) 0 ( 2 _ 3 (0), 2 _

3 (!30)) 0 (0, !20), so scale

factor is 2 _ 3

.

3. Step 1 Plot pts. and draw 3.

Step 2 Use Dist. Formula to find side lengths.

RS = ) ********* (!3 + 2 ) 2 + (1 ! 0 ) 2 = ) * 2

RT = ) ******** (0 + 2 ) 2 + (1 ! 0 ) 2 = ) * 5

RU = ) ********* (!5 + 2 ) 2 + (3 ! 0 ) 2 = ) ** 18 = 3 ) * 2

RV = ) ******** (4 + 2 ) 2 + (3 ! 0 ) 2 = ) ** 45 = 3 ) * 5 Step 3 Find similarity ratio.

RS _ RU

= ) * 2 _

3 ) * 2 = 1 _

3 RT _

RV = ) * 5 _

3 ) * 5 = 1 _

3

Since RS _ RU

= RT _ RV

and "R # "R by Reflex. Prop.

of #, 'RST , 'RUV by SAS ,.

4. Step 1 Multiply each coord. by 3 to find coords of vertices of 'M9N9P9.M(!2, 1) 0 M9(3(!2), 3(1)) = M9(!6, 3)N(2, 2) 0 N9(3(2), 3(2)) = N9(6, 6)P(!1, !1) 0 P9(3(!1), 3(!1)) = P9(!3, !3)Step 2 Graph 'M9N9P9.


MN = ) ******** (2 + 2 ) 2 + (2 ! 1 ) 2 = ) ** 17

M9N9 = ) ******** (6 + 6 ) 2 + (6 ! 3 ) 2 = ) ** 153 = 3 ) ** 17

NP = ) ********* (!1 ! 2 ) 2 + (!1 ! 2 ) 2 = ) ** 18 = 3 ) * 2

N9P9 = ) ********* (!3 ! 6 ) 2 + (!3 ! 6 ) 2 = ) ** 162 = 9 ) * 2

MP = ) ********* (!1 + 2 ) 2 + (!1 ! 1 ) 2 = ) * 5

M9P9 = ) ********* (!3 + 6 ) 2 + (!3 ! 3 ) 2 = ) ** 45 = 3 ) * 5

Step 4 Find similarity ratio.

M9N9 _ MN

= 3 ) ** 17 _ ) ** 17

= 3, N9P9 _ NP

= 9 ) * 2 _ 3 ) * 2

= 3, M9P9 _ MP

= 3 ) * 5 _ ) * 5

= 3

Since M9N9 _ MN

= N9P9 _ NP

= M9P9 _ MP

, 'M9N9P9 , 'MNP by

SSS ,.

THINK AND DISCUSS, PAGE 497 1. The scale factor is 4, since each coord. of preimage

is multiplied by 4 in order to get coords. of image.


2.


1. dilation 2. scale factor

3. Step 1 Multiply vertices of figure A(0, 0), B(0, 2), C(2, 4), D(3, 3), E(3, 4), F(4, 4), G(4, 2), H(4, 0) by 2.Fig. ABCDEFGH Fig. A9B9C9D9E9F9G9H9A(0, 0) 0 A9(2(0), 2(0)) 0 A9(0, 0)B(0, 2) 0 B9(2(0), 2(2)) 0 B9(0, 4)C(2, 4) 0 C9(2(2), 2(4)) 0 C9(4, 8)D(3, 3) 0 D9(2(3), 2(3)) 0 D9(6, 6)E(3, 4) 0 E9(2(3), 2(4)) 0 E9(6, 8)F(4, 4) 0 F9(2(4), 2(4)) 0 F9(8, 8)G(4, 2) 0 G9(2(4), 2(2)) 0 G9(8, 4)H(4, 0) 0 H9(2(4), 2(0)) 0 H9(8, 0)

Step 2 Plot pts. A9, B9, C9, D9, E9, F9, G9, and H9. Draw the figure.

4. Since 'AOB , 'COD,

AO _ CO

= OB _ OD

10 _ CO

= 6 _ 15

150 = 6CO CO = 25C lies on x-axis, so its y-coord. is 0. Since CO = 25, its x-coord. must be 25. Coords. of C are (25, 0).

(10, 0) 0 ( 5 _ 2 (10), 5 _

2 (0)) 0 (25, 0), so scale factor

is 5 _ 2 .

5. Since 'ROS , 'POQ,

RO _ PO

= OS _ OQ

4 _ 10

= OS _ !20

!80 = 10OS OS = !8S lies on y-axis, so its x-coord. is 0. Since OS = !8, its y-coord. must be !8. Coords. of S are (0, !8).

(0, !20) 0 ( 2 _ 5

(0), 2 _ 5 (!20)) 0 (0, !8), so scale

factor is 5 _ 2 .



AB = ) ********* (!1 ! 0 ) 2 + (1 ! 0 ) 2 = ) * 2

AC = ) ******** (3 ! 0 ) 2 + (2 ! 0 ) 2 = ) ** 13

AD = ) ********* (!2 ! 0 ) 2 + (2 ! 0 ) 2 = ) * 8 = 2 ) * 2

AE = ) ******** (6 ! 0 ) 2 + (4 ! 0 ) 2 = ) ** 52 = 2 ) ** 13 Step 3 Find similarity ratio.

AB _ AD

= ) * 2 _

2 ) * 2 = 1 _

2 AC _

AE = ) ** 13 _

2 ) ** 13 = 1 _

2

Since AB _ AD

= AC _ AE

and "A # "A by Reflex. Prop. of #,

'ABC , 'ADE by SAS ,.



JK = ) ********* (!3 + 1 ) 2 + (!4 ! 0 ) 2 = ) ** 20 = 2 ) * 5

JL = ) ********* (3 + 1 ) 2 + (!2 ! 0 ) 2 = ) ** 20 = 2 ) * 5

JM = ) ********* (!4 + 1 ) 2 + (!6 ! 0 ) 2 = ) ** 45 = 3 ) * 5

JN = ) ********* (5 + 1 ) 2 + (!3 ! 0 ) 2 = ) ** 45 = 3 ) * 5

Step 3 Find similarity ratio.

JK _ JM

= 2 ) * 5 _ 3 ) * 5

= 2 _ 3

JL _ JN

= 2 ) * 5 _ 3 ) * 5

= 2 _ 3

Since JK _ JM

= JL _ JN

and "J # "J by Reflex. Prop.

of #, 'JKL , 'JMN by SAS ,.


8. Step 1 Multiply each coord. by 2 to find coords of vertices of 'A9B9C9.A(1, 4) 0 A9(2(1), 2(4)) = A9(2, 8)B(1, 1) 0 B9(2(1), 2(1)) = B9(2, 2)C(3, 1) 0 C9(2(3), 2(1)) = C9(6, 2)Step 2 Graph 'A9B9C9.


AB = ) ******** (1 ! 1 ) 2 + (1 ! 4 ) 2 = 3

A9B9 = ) ******** (2 ! 2 ) 2 + (2 ! 8 ) 2 = 6

BC = ) ******** (3 ! 1 ) 2 + (1 ! 1 ) 2 = 2

B9C9 = ) ******** (6 ! 2 ) 2 + (2 ! 2 ) 2 = 4

AC = ) ******** (3 ! 1 ) 2 + (1 ! 4 ) 2 = ) ** 13

A9C9 = ) ******** (6 ! 2 ) 2 + (2 ! 8 ) 2 = ) ** 52 = 2 ) ** 13 Step 4 Find similarity ratio.

A9B9 _ AB

= 6 _ 3 = 2, B9C9 _

BC = 4 _

2 = 2, A9C9 _

AC = 2 ) ** 13 _

) ** 13 = 2

Since A9B9 _ AB

= B9C9 _ BC

= A9C9 _ AC

, 'ABC , 'A9B9C9 by

SSS ,.

9. Step 1 Multiply each coord. by 3 _ 2 to find coords of

vertices of 'R9S9T9.R(!2, 2) 0 R9 ( 3 _

2 (!2), 3 _

2 (2)) = R9(!3, 3)

S(2, 4) 0 S9 ( 3 _ 2 (2), 3 _

2 (4)) = S9(3, 6)

T(0, !2) 0 T9 ( 3 _ 2 (0), 3 _

2 (!2)) = T9(0, !3)

Step 2 Graph 'R9S9T9.


RS = ) ******** (2 + 2 ) 2 + (4 ! 2 ) 2 = ) ** 20 = 2 ) * 5

R9S9 = ) ******** (3 + 3 ) 2 + (6 ! 3 ) 2 = ) ** 45 = 3 ) * 5

ST = ) ********* (0 ! 2 ) 2 + (!2 ! 4 ) 2 = ) ** 40 = 2 ) ** 10

S9T9 = ) ********* (0 ! 3 ) 2 + (!3 ! 6 ) 2 = ) ** 90 = 3 ) ** 10

RT = ) ********* (0 + 2 ) 2 + (!2 ! 2 ) 2 = ) ** 20 = 2 ) * 5

R9T9 = ) ********* (0 + 3 ) 2 + (!3 ! 3 ) 2 = ) ** 45 = 3 ) * 5 Step 4 Find similarity ratio.

R9S9 _ RS

= 3 ) * 5 _ 2 ) * 5

= 3 _ 2 , S9T9 _

ST = 3 ) ** 10 _

2 ) ** 10 = 3 _

2 ,

R9T9 _ RT

= 3 ) * 5 _ 2 ) * 5

= 3 _ 2

Since R9S9 _ RS

= S9T9 _ ST

= R9T9 _ RT

, 'RST , 'R9S9T9 by SSS ,.


10. Coords. of kite are A(4, 5), B(9, 7), C(10, 11), and

D(6, 10).Coords. of image are A(2, 2.5), B(4.5, 3.5), C(5, 5.5), and D(3, 5).

11. UO _ XO

= OV _ OY

!9 _ XO

= !3 _ !8

72 = !3XO XO = !24X on x-axis 0 X = (!24, 0)

(!9, 0) 0 ( 8 _ 3 (!9), 8 _

3 (0)) = (!24, 0), so scale

factor is 8 _ 3 .


12. MO _ KO

= ON _ OL

16 _ KO

= !24 _ !15

!240 = !24KO KO = 10K on y-axis 0 K = (0, 10)

(0, 16) 0 ( 5 _ 8 (0), 5 _

8 (16)) = (0, 10), so scale factor

is 5 _ 8 .

13. DE = ) **** 2 2 + 4 2 = 2 ) * 5 , DF = ) **** 4 2 + 4 2 = 4 ) * 2

DG = ) **** 3 2 + 6 2 = 3 ) * 5 , DH = ) **** 6 2 + 6 2 = 6 ) * 2

DE _ DG

= 2 ) * 5 _ 3 ) * 5

= 2 _ 3 , DF _

DH = 4 ) * 2 _

6 ) * 2 = 2 _

3

"D # "D by Reflex. Prop. of #. So 'DEF , 'DGH by SAS ,.

14. MN = ) **** 5 2 + 10 2 = 5 ) * 5 , MP = ) **** 15 2 + 5 2 = 5 ) ** 10

MQ = ) **** 10 2 + 20 2 = 10 ) * 5 , MR = ) **** 30 2 + 10 2 = 10 ) ** 10

MN _ MQ

= 5 ) * 5 _ 10 ) * 5

= 1 _ 2 , MP _

MR = 5 ) ** 10 _

10 ) ** 10 = 1 _

2

"M # "M by Reflex. Prop. of #. So 'MNP , 'MQR by SAS ,.

15. Step 1 Multiply each coord. by 3 to find coords of vertices of 'J9K9L9.J(!2, 0) 0 J9(3(!2), 3(0)) = J9(!6, 0)K(!1, !1) 0 K9(3(!1), 3(!1)) = K9(!3, !3)L(!3, !2) 0 K9(3(!3), 3(!2)) = L9(!9, !6)Step 2 Graph 'J9K9L9.

Step 3 Find side lengths.

JK = ) **** 1 2 + 1 2 = ) * 2 , J9K9 = ) **** 3 2 + 3 2 = 3 ) * 2

KL = ) **** 2 2 + 1 2 = ) * 5 , K9L9 = ) **** 6 2 + 3 2 = 3 ) * 5

JL = ) **** 1 2 + 2 2 = ) * 5 , J9L9 = ) **** 3 2 + 6 2 = 3 ) * 5 Step 4 Verify similarity.

Since J9K9 _ JK

= K9L9 _ KL

= J9L9 _ JL

= 3, 'JKL , 'J9K9L9 by SSS ,.

16. Step 1 Multiply each coord. by 1 _ 2 to find coords of

vertices of 'M9N9P9.

M(0, 4) 0 M9 ( 1 _ 2 (0), 1 _

2 (4)) = M9(0, 2)

N(4, 2) 0 N9 ( 1 _ 2 (4), 1 _

2 (2)) = N9(2, 1)

P(2, !2) 0 P9 ( 1 _ 2 (2), 1 _

2 (!2)) = P9(1, !1)

Step 2 Graph 'M9N9P9.

Step 3 Find side lengths.

MN = ) **** 4 2 + 2 2 = 2 ) * 5 , M9N9 = ) **** 2 2 + 1 2 = ) * 5

NP = ) **** 2 2 + 4 2 = 2 ) * 5 , N9P9 = ) **** 1 2 + 2 2 = ) * 5

MP = ) **** 2 2 + 6 2 = 2 ) ** 10 , M9P9 = ) **** 1 2 + 3 2 = ) ** 10 Step 4 Verify similarity.

Since M9N9 _ MN

= N9P9 _ NP

= M9P9 _ MP

= 1 _ 2 , 'MNP , 'M9N9P9

by SSS ,.

17. It is not a dilation; it changes shape of transformed figure.

18. Solution B is incorrect. Scale factor is ratio of a lin. measure of image to corr. lin. measure of preimage, so scale factor is UW _

RT = 3 _

2 .

19. They are reciprocals. Similarity ratio of 'ABC to 'A9B9C9 is AB _

A9B9 . Scale factor is A9B9 _

AB .

20a. Should use origin as vertex of rt. "; 1 unit reps. 60 cm 0 3 units rep. 180 cm; so coords. are J(0, 1), K(0, 0), L(3, 0).

b. J 0 J9(3(0), 3(1)) = J9(0, 3)K 0 K9(3(0), 3(0)) = K9(0, 0)L 0 L9(3(3), 3(0)) = L9(9, 0)

TEST PREP, PAGE 500

21. ACheck similarity ratio: 2.4 _

4 = 3 _

5 = !6 _ !10

22. HPerimeter is a lin. measure. So P 9 = 2P = 2(60) = 120.

23. AAB = 4, AC = BC = ) **** 2 2 + 4 2 = 2 ) * 5

DE = :3 ! 1; = 2, DF = EF = ) **** 1 2 + 2 2 = ) * 5

DE _ AB

= EF _ BC

= DF _ AC

= 1 _ 2


24. 15A 0 A9(3(3), 3(2)) = A9(9, 6)B 0 B9(3(7), 3(5)) = B9(21, 15)

A9B9 = ) **** 12 2 + 9 2 = ) ** 225 = 15


25. Possible , statements: 'XYZ , 'MNP, 'MPN, 'NMP, 'NPM, 'PMN, or 'PNM. For each , statement, Z could lie either above or below 5 67 XY. So there are 2(6) = 12 different 3. They are all different, since MN, NP, and MP are all $.

26. scale factor = XY _ MP

= 2 _ 4

= 1 _ 2

From M to N is rise of 2 and run of 1. So from X to Z is either rise of 1 and run of 1 _

2 or rise of !1 and

run of 1 _ 2 . Therefore Z = (1 ± 1 _

2 , !2 ± 1) = (1 1 _

2 , !1)

or (1 1 _ 2 , !3) .

27. All corr. % of rects. are # because they are all rt. %. Suppose 1st rect. has vertex on line y = 2x at (a, b). This pt. is a solution to the eqn., so b = 2a, and coords. of vertex are (a, 2a). Similarly, for 2nd rect., coords. of vertex on line y = 2x must be (c, 2c).

1st rect. has dimensions a and 2a, and 2nd rect. has dimensions c and 2c. So all

ratios of corr. sides = c _ a .

Therefore rects. are , by def.

28. scale factor = DE _ AB

= 6 _ 3 = 2

From A to C is rise of 2 and run of 1.2 positions for F are reflections in horiz. line 5 67 DE . So from D to F is rise of ±4 and run of 2. Therefore F = (1 + 2, !1 ± 4) = (3, 3) or (3, !5).


29. Possible answer: 2(50) + 5 + w < 250 105 + w < 250

30. Think: 'DEH # 'FEH by HL. So by CPCTC, ((

HF # ((

DF HF = DF = 6.71

31. Think: By Isosc. ' Thm., "EDH # "EFH, so by Rt. " # Thm., 3rd % Thm, and ASA, 'DFG # 'FDJ. So by CPCTC, ((

JF # ((

GD JF = GD = 5

32. Think: Use Pyth. Thm.

CF = ) ***** CH 2 + HF 2

= ) ***** 2 2 + 6.71 2 + 7.00

33. RT _ UV

= RS _ US

RT _ 9 = 6 + 2 _

6 = 4 _

3

3RT = 36 RT = 12

34. VT _ VS

= RU _ US

x _ x + 3

= 2 _ 6 = 1 _

3

3x = x + 3 2x = 3 x = 1.5VT = x = 1.5

35. ST = SV + VT= x + 3 + x= 2x + 3= 2(1.5) + 3 = 6

DIRECT VARIATION, PAGE 501

TRY THIS, PAGE 501 1. Step 1 Make a table to record data.

Scale Factor x

Side Length s = x(6)

Perimeter P = 6s

1 _ 2 3 18

2 12 72

3 18 108

4 24 144

5 30 180

Step 2 Graph pts.

Since pts. are collinear and line that contains them includes origin, relationship is a direct variation.

Step 3 Find eqn. of direct variation. y = kx180 = k(5) 36 = kThus constant of variation is 36.


2. Step 1 Make a table to record data.

Scale Factor x

Side Lengths Perimeter P = a + b + ca = x (3) b = x (6) c = x (7)

1 _ 2

1 1 _ 2 3 3 1 _

2 8

2 6 12 14 32

3 9 18 21 48

4 12 24 28 64

5 15 30 35 80

Step 2 Graph pts.


Step 3 Find eqn. of direct variation. y = kx80 = k(5) k = 16Thus constant of variation is 16.

3. Step 1 Make a table to record data.

Scale Factor x

Side Length s = x (3)

Perimeter P = 4s

1 _ 2 1 1 _

2 6

2 6 24

3 9 36

4 12 48

5 15 60

Step 2 Graph pts.


Step 3 Find eqn. of direct variation. y = kx60 = k(5) k = 12Thus constant of variation is 12.


1. EG _ FH

= GJ _ HK

= JC _ KC

= AE _ BF

= 42.2 _ 40

= 1.055

EG = 1.055FH= 1.055(40) = 42.2 cm

GJ = 1.055HK= 1.055(35) + 36.9 cm

JC = 1.055KC= 1.055(35) + 36.9 cm

2. area of 'ABC = 1 _ 2 (BC)(AB)

= 1 _ 2 (40 + 40 + 35 + 35)(50)

= 3750 cm 2 Think: Use Proportional Perimeters and Areas Thm.area of drawing = (scale factor ) 2 (area of 'ABC)

= ( 1 _ 25

) 2 (3750)

= 1 _ 625

(3750) = 6 cm 2

3.

7B READY TO GO ON?, PAGE 503

1. ST _ QT

= RT _ PT

ST _ ST + 16

= 14 _ 14 + 12

26ST = 14(ST + 16) 26ST = 14ST + 224 12ST = 224 ST = 18 2 __

3

2. AB _ AC

= BD _ CD

4y ! 1

_ 5y

= 6 _ 8

8(4y ! 1) = 6(5y) 32y ! 8 = 30y 2y = 8 y = 4AB = 4(4) ! 1 = 15AC = 5(4) = 20

3. FH _ EG

= HK _ GJ

FH _ 3.6

= 2 _ 2.4

2.4FH = 7.2 FH = 3 cm

4. plan length of

(( AB __

AB = 0.25 _

AB = 1.5 _

60

15 = 1.5AB AB = 10 ft

5. plan length of

(( BC __

BC = 0.75 _

BC = 1.5 _

60

45 = 1.5BC BC = 30 ft

6. plan length of

(( CD __

CD = 1 _

CD = 1.5 _

60

60 = 1.5CD CD= 40 ft


7. plan length of

(( EF __

EF = 0.5 _

EF = 1.5 _

60

30 = 1.5EF EF = 20 ft

8. 5 ft 3 in. = 5(12) + 3 in. = 63 in.5 ft 10 in. = 5(12) + 10 in. = 70 in.40 ft = 40(12) in. = 480 in.

h _ 63

= 480 _ 70

70h = 63(480) h = 432 in. = 36 ft

9. By the Dist. Formula:

AD = ) **** 1 2 + 2 2 = ) * 5 ; AB = ) **** 2 2 + 4 2 = 2 ) * 5

AE = ) **** 2 2 + 1 2 = ) * 5 ; AC = ) **** 4 2 + 2 2 = 2 ) * 5

"A # "A by the Reflex. Prop. of #.

By SAS ,, 'ADE , 'ABC.


RS = ) **** 2 2 + 1 2 = ) * 5 ; RU = ) **** 4 2 + 2 2 = 2 ) * 5 RT = :!3 ! 0; = 3; RV = :6 ! 0; = 6

RS _ RU

= RT _ RV

= 1 _ 2 . "SRT # "URV by the Vert. % Thm.

By SAS ,, 'RST , 'RUV.

11.

PQ = QR = 2; P9Q9 = Q9R9 = 6

PR = ) **** 2 2 + 2 2 = 2 ) * 2 ; P9R9 = ) **** 6 2 + 6 2 = 6 ) * 2

P9Q9 _ PQ

= Q9R9 _ QR

= 6 _ 2 = 3; P9R9 _

PR = 6 ) * 2 _

2 ) * 2 = 3

By SSS ,, 'P9Q9R9 , 'PQR.

12.

AB = ) **** 4 2 + 2 2 = 2 ) * 5 ; A9B9 = ) **** 6 2 + 3 2 = 3 ) * 5

BC = ) **** 2 2 + 6 2 = 2 ) ** 10 ; B9C9 = ) **** 3 2 + 9 2 = 3 ) ** 10

AC = ) **** 6 2 + 4 2 = 2 ) ** 13 ; A9C9 = ) **** 9 2 + 6 2 = 3 ) ** 13

A9B9 _ AB

= B9C9 _ BC

= A9B9 _ AB

= 3 _ 2

By SSS ,, 'A9B9C9 , 'ABC.


1. proportion 2. dilation

3. means 4. ratio


5. slope of m = 1 _ 5 6. slope of n = !3 _

6 = ! 1 _

2

7. slope of p = 6 _ 4 = 3 _

2

8. Let x, y be the largest and smallest parts respectively.

x + y

_ 84

= 6 + 3 _ 3 + 5 + 6

x + y = 84(9)

_ 14

x + y = 54The sum of the smallest and largest parts is 54.

9. ! _ w = 7 _ 12

! = 7 _ 12

w

P = 2! + 2w

= 2 ( 7 _ 12

w) + 2w

6P = 7w + 12w6(95) = 19w w = 30! = 7 _

12 (30) = 17.5

Side lengths are 17.5, 30, 17.5, 30.

10. y _

7 = 9 _

3

3y = 63 y = 21

11. 10 _ 4 = 25 _ s

10s = 100 s = 10

12. x _ 4 = 9 _ x

x 2 = 36 x = ±6

13. 4 _ z ! 1

= z ! 1 _ 36

144 = (z ! 1 ) 2 z ! 1 = ±12 z = 1 ± 12

= 13 or !11

14. 12 _ 2x

= 3x _ 32

384 = 6 x 2 x 2 = 64 x = ±8

15. y + 1

_ 24

= 2 _ 3(y + 1)

3(y + 1 ) 2 = 48 (y + 1 ) 2 = 16 y + 1 = ±4 y = !1 ± 4

= 3 or !5

LESSON 7-2, PAGE 505 16. JK _

PQ = 8 _

4.8 = 5 _

3 ; JM _

PS = 5 _

3 ; all % are rt %, so #

yes, by def. of ,; , ratio = 5 _ 3 ; JKLM , PQRS

17. yes, by AA ,; , ratio = TU _ WX

= 12 _ 6 = 2;

'TUV , 'WXY




1. JL = 1 _ 3 JN, JK = 1 _

3 JM 1. Given

2. JL _ JN

= 1 _ 3 , JK _

JM = 1 _

3 2. Div. Prop. of =

3. JL _ JN

= JK _ JM


4. "J # "J 4. Reflex. Prop. of #5. 'JKL , 'JMN 5. SAS , Steps 3, 4


1. ((

QR 1 ((

ST 1. Given2. "RQP # "STP 2. Alt. Int. % Thm.3. "RPQ # "SPT 3. Vert. % Thm.4. 'PQR , 'PTS 4. AA , Steps 2, 3


1. ((

BC 1 ((

CE 1. Given2. "ABD # "C 2. Corr. % Post.3. "ADB # "E 3. Corr. % Post.4. 'ABD , 'ACE 4. AA , Steps 2, 3

5. AB _ AC

= BD _ CE

5. Def. of , polygons

6. AB(CE) = AC(BD) 6. Cross Products Prop.


21. CE _ 15

= 8 _ 12

12CE = 120 CE = 10

22. ST _ 10

= 3 _ 9

9ST = 30 ST = 3 1 _

3

23. JK _ JM

= JL _ JN

= 1 _ 2

Since JK _ JM

= JL _ JN

((

KL 1 (((

MN by Conv. of ' Prop. Thm.

24. EC/EA = ED _ EB

= 3 _ 7

Since EC _ EA

= ED _ EB

((

AB 1 ((


25. SU _ RU

= SV _ RV

y + 1

_ 8 =

2y _

12

12(y + 1) = 8(2y) 12y + 12 = 16y 12 = 4y y = 3SU = 3 + 1 = 4SV = 2(3) = 6

26. x + 6 _ 30

= 2x _ 24

24(x + 6) = 30(2x)24x + 144 = 60x 144 = 36x x = 4AB = x + 6 + 2x

= 3x + 6= 3(4) + 6 = 18

27. P = a + b + c where b = a + x, c = 3 + 5 = 8

3 _ a = 5 _ a + x

3(a + x) = 5a 3a + 3x = 5a 2a = 3xP = a + a + x + 8

= 2a + x + 8= 4x + 8

LESSON 7-5, PAGE 507 28. 3 ft = 3(12) in. = 36 in.

5 ft 4 in. = 5(12) + 4 in. = 64 in.14 ft 3 in. = 14(12) + 3 in. = 171 in.

x _ 64

= 171 _ 36

36x = 10,944 x = 304 in. = 25 ft 4 in.

29. 6 _ x = 12 _ 3 + x

6(3 + x) = 12x 18 + 6x = 12x 18 = 6x x = 3 ft



RS = ) **** 2 2 + 2 2 = 2 ) * 2 ; RU = ) **** 4 2 + 4 2 = 4 ) * 2

RT = ) **** 1 2 + 3 2 = ) ** 10 ; RV = ) **** 2 2 + 6 2 = 2 ) ** 10

RS _ RU

= RT _ RV

= 1 _ 2 . "R # "R by the Reflex. Prop. of #.

So 'RST , 'RUV by SAS ,.


JK = ) **** 2 2 + 1 2 = ) * 5 ; JM = ) **** 8 2 + 4 2 = 4 ) * 5 JL = :2 ! 4; = 2; JN = :!4 ! 4; = 8

JK _ JM

= JL _ JN

= 1 _ 4 . "J # "J by the Reflex. Prop. of #.

So 'JKL , 'JMN by SAS ,.

32. AO _ CO

= OB _ OD

12 _ 18

= OB _ !9

!108 = 18OB OB = !6Since x-coord. of B is 0, B = (0, !6).

Scale factor = 12 _ 18

= 2 _ 3 .

33. Image vertices are K9(0, 9), L9(0, 0), M9(12, 0). By the Dist. Formula: KL = 3; K9L9 = 9; LM = 4; L9M9 = 12

KM = ) **** 3 2 + 4 2 = 5; K9M9 = ) **** 9 2 + 12 2 = 15All proportions = 3, so 'KLM , 'K9L9M9 by SSS ,.


1. slope of ! = !6 ! 4 _ 10 + 6

= ! 5 _ 8

2. 5 _ 8 = 3.5 _ w

5w = 28 w = 5.6 in.

3. "B # "N and "C # "P; yes, by AA ,; , ratio = AB _

MN = 40 _

60 = 2 _

3 ; 'ABC , 'MNP


4. DE _ HJ

= 55 _ 22

= 5 _ 2 ; DG _

HL = 40 _

16 = 5 _

2

yes; since all % are rt. % and therefore #; , ratio = 5 _

2 ; DEFG , HJKL by def.


1. RSTU is a /. 1. Given2. ((

RU 1 ((

ST 2. Def. of /3. "VRW # "TSW 3. Alt. Int. % Thm.4. "RWV # "SWT 4. Vert % Thm.5. 'RWV , 'SWT 5. AA , Steps 3, 4

6. CD _ AB

= DG _ BG

CD _ 2.5

= 6 _ 9

9CD = 2.5(6) = 15 CD + 1.7 ft

EF _ AB

= FG _ BG

EF _ 2.5

= 3 _ 9

9FG = 7.5 FG + 0.8 ft

7. PR _ 21

= 10 _ 18

18PR = 210

PR = 11 2 _ 3

8. YW _ XY

= WZ _ XZ

t __ 2 _

8 = t ! 2 _

12.8

12.8 ( t _ 2

) = 8(t ! 2)

6.4t = 8t ! 16 16 = 1.6t t = 10

YW = t _ 2 = 5

WZ = t ! 2 = 8

9. 5 ft 8 in. = 5(12) + 8 in. = 68 in.3 ft = 36 in.; 27 ft = 324 in.

h _ 68

= 324 _ 36

= 9

h = 68(9) = 612 in. = 51 ft

10. plan length of

(( AB __

AB = 1.5 _

30

1.25 _ AB

= 1.5 _ 30

37.5 = 1.5AB

(( AB = 25 ft


AB = ) **** 3 2 + 1 2 = ) ** 10 ; AD = ) **** 9 2 + 3 2 = 3 ) ** 10 AC = :3 ! 5; = 2; AE = :!1 ! 5; = 6

AB _ AD

= AC _ AE

= 1 _ 3

. "A # "A by the Reflex. Prop. of #.

So 'JKL , 'JMN by SAS ,.

12.


1. A BC _

CD = AB _

DE

BC _ 9 ! BC

= 4 _ 8 = 1 _

2

2BC = 9 ! BC 3BC = 9 BC = 3

2. C x _ 21

= 6 _ 14

14x = 126 x = 9

Since ((

BD is horiz., y-coord. of C is 1; so C = (1 + 3, 1) = (4, 1).

3. D; x + y + z = 750,000 and x : y : z = 4 : 5 : 6

z _ 750,000

= 6 _ 4 + 5 + 6

= 2 _ 5

5z = 1,500,000 z = 300,000

4. D

35 _ 9 = h _

1.2

42 = 9h

h = 4 2 _ 3

ft = 4 ft 8 in.

5. DIn any square, all % are rt %, so #; all sides are #.


Solutions KeyRight Triangles and Trigonometry8

CHAPTER


1. D 2. C

3. A 4. E

5. PR ___ RT

= 10 ___ 5

= 2; QR ___ RS

= 12 ___ 6

= 2

!PRQ " !TRS by Vert. # Thm.yes; $PRQ % $TRS by SAS %

6. AB ___ FE

= 6 __ 4 = 3 __

2 ; BC ___

ED = 15 ___

10 = 3 __

2

!B " !E by Rt. ! " Thm.yes; $ABC % $FED by SAS %

7. x = 5 & ' 2 8. 16 = x & ' 2 16 & ' 2 = 2x x = 8 & ' 2

9. x = 4 & ' 3 10. x = 2(3) = 6

11. 3(x ( 1) = 12 x ( 1 = 4 x = 5

12. (2(y + 5) = (1 y + 5 = 0.5 y = (4.5

13. 6 = 8(x ( 3) 6 = 8x ( 2430 = 8x x = 3.75

14. 2 = (1(z + 4)2 = (z ( 4z = (6

15. 4 __ y = 6 ___ 18

4 __ y = 1 __ 3

4(3) = y(1) y = 12

16. 5 __ 8 = x ___

32

5(32) = 8(x) 160 = 8x x = 20

17. m __ 9 = 8 ___

12 = 2 __

3

3m = 9(2) = 18 m = 6

18. y __

4 = 9 __ y

y(y) = 4(9) y 2 = 36 y = ±6

19. 13.118 ) 13.12 20. 37.91 ) 37.9

21. 15.992 ) 16.0 22. 173.05 ) 173

8-1 SIMILARITY IN RIGHT TRIANGLES, PAGES 518–523

CHECK IT OUT! PAGES 518–520 1. Sketch the 3 rt. * with # of * in corr. positions.

By Thm. 8-1-1, $LJK % $JMK % $LMJ.

2a. x 2 = (2)(8) = 16 x = 4

b. x 2 = (10)(30) = 300 x = & '' 300 = 10 & ' 3

c. x 2 = (8)(9) = 72 x = & '' 72 = 6 & ' 2

3. 9 2 = (u)(3)81 = 3uu = 27

v 2 = (3)(3 + u) v 2 = (3)(30) = 90 v = & '' 90 = 3 & '' 10

w 2 = (u)(3 + u) w 2 = (27)(30) = 810 w = & '' 810 = 9 & '' 10

4. Let x be height of cliff above eye level.(28 ) 2 = 5.5x x ) 142.5 ftCliff is about 142.5 + 5.5, or 148 ft high.


1. Set up the proportion 7 __ x = x ___ 21

, and solve for x. x 2 = 7(21) = 147x = & '' 147 = 7 & ' 3

2.


1. 8 is geometric mean of 2 and 32.

2. Sketch the 3 rt. * with # of * in corr. positions.

By Thm. 8-1-1, $RPQ % $PSQ % $RSP.



By Thm. 8-1-1, $BED % $ECD % $BCE.


By Thm. 8-1-1, $XYZ % $XWY % $YWZ.

5. x 2 = (2)(50) = 100 x = 10

6. x 2 = (4)(16) = 64 x = 8

7. x 2 = ( 1 __ 2

) (8) = 4

x = 2

8. x 2 = (9)(12) = 108 x = & '' 108 = 6 & ' 3

9. x 2 = (16)(25) = 400 x = 20

10. x 2 = (7)(11) = 77 x = & '' 77

11. x 2 = (10)(6) = 60 x = & '' 60 = 2 & '' 15

z 2 = (10)(4) = 40 z = & '' 40 = 2 & '' 10

y 2 = (6)(4) = 24 y = & '' 24 = 2 & ' 6

12. 10 2 = 100 = 20x x = 5 z 2 = (5)(20 + 5) = 125 z = & '' 125 = 5 & ' 5

y 2 = (20)(20 + 5) = 500 y = & '' 500 = 10 & ' 5

13. (6 & '' 13 ) 2 = (18)(18 + z) 468 = 324 + 18z 144 = 18z z = 8 y 2 = (8)(18 + 8) = 208 y = & '' 208 = 4 & '' 13

x 2 = (8)(18) = 144 x = 12

14. RS 2 = (64)(60) = 3840 RS = & '' 3840 ) 62.0 m


15. By Thm. 8-1-1, $MPN % $PQN % $MQP.

16. By Thm. 8-1-1, $CAB % $ADB % $CDA.

17. By Thm. 8-1-1, $RSU % $RTS % $STU.

18. x 2 = (5)(45) = 225 x = 15

19. x 2 = (3)(15) = 45 x = 3 & ' 5

20. x 2 = (5)(8) = 40 x = 2 & '' 10

21. x 2 = ( 1 __ 4 ) (80) = 20

x = 2 & ' 5

22. x 2 = (1.5)(12) = 18 x = 3 & ' 2

23. x 2 = ( 2 __ 3 ) ( 27 ___

40 ) = 9 ___

20

x = 3 ____ 2 & ' 5

= 3 & ' 5 ____ 10

24. 12 2 = 4(4 + x)144 = 16 + 4x128 = 4x x = 32

z 2 = 32(4 + 32) = 1152 z = 24 & ' 2

y 2 = 4(32) = 128 y = 8 & ' 2

25. x 2 = (30)(40) = 1200 x = 20 & ' 3

z 2 = (70)(40) = 2800 z = 20 & ' 7

y 2 = (30)(70) = 2100 y = 10 & '' 21

26. 9.6 2 = (z)(12.8)92.16 = 12.8z z = 7.2

y 2 = (12.8)(12.8 + 7.2) = 256 y = 16

x 2 = (7.2)(12.8 + 7.2) = 144 x = 12

27. Let h represent height of tower above eye level.91 ft 3 in. = 91.25 ft(91.25 ) 2 = 5h h ) 1665 ftTower is about 1665 + 5 = 1670 ft high.

28. 8 2 = 64 = 2x x = 32

29. (2 & ' 5 ) 2 = 20 = 6x

x = 10 ___ 3 = 3 1 __

3

30. y; x __ z = z __ y 31. x + y; x + y

_____ u = u __ x

32. y; x + y

_____ v = v __ y 33. z; y __ z = z __ x

34. v; v 2 = y(x + y) 35. x; u 2 = (x + y)x

36. BD 2 = (AD)(CD)= (12)(8) = 96

BD = 4 & ' 6

37. BC 2 = (AC)(CD)= (16)(5) = 80

BC = 4 & ' 5


38. BD 2 = (AC)(CD)

= ( & ' 2 ) ( & ' 2 ) = 2

BD = & ' 2

39. BC 2 = (AC)(CD) 5 = CD & '' 10 5 & '' 10 = 10CD

CD = & '' 10 ____

2

40. & ''''' (0.1) (0.03) X 100% ) 5.5%

41. B is incorrect; proportion should be 12 ___ EF

= EF ___ 8 .

42. a 2 = (2)(5) = 10 a = & '' 10 ) 3.2Altitude is about 3.2 cm long.

43. By Corollary 8-1-3, a 2 = x(x + y) and b 2 = y(x + y). So a 2 + b 2 = x(x + y) + y(x + y). By Distrib. Prop.,this expression simplifies to (x + y)(x + y) = (x + y ) 2 = c 2 . So a 2 + b 2 = c 2 .

44a. S W 2 = (RS)(ST ) = (4)(3) = 12 SW = & '' 12 ) 3.46 ft, or 3 ft 6 in.

b. R W 2 = (RS)(RT ) = (4)(7) = 28 RW = & '' 28 ) 5.29 ft, or 5 ft 3 in.

45. Area of rect. is ab, and area of square is s 2 . It is given that s 2 = ab, so s is geometric mean of a and b.

46. Let z be geometric mean of x and y, where x = a 2

and y = b 2 . So z = & '' a 2 b 2 = ab, which is a whole number.

TEST PREP, PAGE 523

47. D XY 2 = (8)(11) = 88 XY ) 9.4 ft

48. H BD 2 = (9)(4) = 36 BD = 6Area = 1 __

2 (BD)(AC)

= 1 __ 2 (6)(13) = 39 m 2

49. A RS 2 = (1)(y + 1) = y + 1 RS = & ''' y + 1


50. Let x be length of shorter seg. 8 2 = (x)(4x) = 4 x 2

8 = & '' 4 x 2 = 2x x = 4Lengths of segs. are 4 in. and 4(4) = 16 in.

51. (2 & '' 21 ) 2 = (x)(x + 5)

84 = x 2 + 5x 0 = x 2 + 5x ( 84 0 = (x ( 7)(x + 12) x = 7 (since x > 0)

y 2 = (7)(5) = 35 y = & '' 35

z 2 = 5(5 + 7) = 60 z = 2 & '' 15

52. Let AD = DC = a. By Corollary 8-1-3, AB 2 = (a)(2a) = 2 a 2 , and BC 2 = (a)(2a) = 2 a 2 . So AB = BC = a & ' 2 . Therefore $ABC is isosc., so it is a 45°-45°-90° $.

53. Step 1 Apply Cor. 8-1-3 in $BDE to find BF and BD. E F 2 = (BF)(FD)3. 28 2 = 4.86BF BF ) 2.214 BD ) 7.074Step 2 Apply Cor. 8-1-3 in $BDE to find BE.B E 2 = (BF)(BD) ) 15.662 BE ) 3.958Step 3 Apply Cor. 8-1-3 in $BCD to find BC. BD 2 = (BE)(BC)7. 074 2 ) 3.958BC BC ) 12.643Step 4 Apply Cor. 8-1-3 in $BCD to find CD.C D 2 = (BC)(EC) ) 109.806 CD ) 10.479Step 5 Apply Cor. 8-1-3 in $ABC to find AC. B C 2 = (AC)(CD)12. 643 2 ) 10.479AC AC ) 15.26 cmStep 6 Apply Pyth. Thm. in $ABD to find AB.A B 2 = B D 2 + A D 2 A B 2 ) 7. 07 2 + (15.26 ( 10.48 ) 2 AB ) 8.53 cm


54. at x-intercept, y = 03(0) + 4 = 4 = 6x

x = 4 __ 6 = 2 __

3

at y-intercept, x = 03y + 4 = 6(0) 3y = (4

y = ( 4 __ 3

55. at x-intercept, y = 0x + 4 = 2(0) x = (4

at y-intercept, x = 00 + 4 = 2y y = 2

56. at x-intercept, y = 03(0) ( 15 = (15 = 15x x = (1

at y-intercept, x = 03y ( 15 = 15(0) 3y = 15 y = 5


57.

c = 2(3) = 6

58.

c = 2(7) = 14

59.

c = 2(2) = 4

60. !DEC is a rt. !, so 30y = 90 + y = 3m!EDC = 8(3) + 15 = 39°

61. ,,- DB bisects !ADC, so m!EDA = m!EDC = 39°

62. AB = BC2x + 8 = 4x 8 = 2x x = 4AB = 2(4) + 8 = 16

TECHNOLOGY LAB: EXPLORE TRIGONOMETRIC RATIOS, PAGE 524

TRY THIS, PAGE 524 1. m!A stays the same. Each $ will have 2 # (!DEA

and !A) " to 2 # in every other $, so by AA % Post., these * are % to each other.

2. Values of ratios do not change, because ratios of side lengths are equal in % *.

3. As C moves, m!A changes. Once C is in a new position, moving D does not change the ratios.

4. DE ___ AE

= 1, and m!A = 45°.

If DE ___ AD

= AE ___ AD

, then DE = AE, so DE ___ AE

= 1. Since 2

sides are =, $DEA is an isosc. $. It is also a rt. $. Thus $DEA must be a 45°-45°-90° special rt. $, so m!A = 45°.

8-2 TRIGONOMETRIC RATIOS, PAGES 525–532

CHECK IT OUT! PAGES 526–528 1a. cos A = 24 ___

25 = 0.96 b. tan B = 24 ___

7 ) 3.43

c. sin B = 24 ___ 25

= 0.96 2.

tan 45° = s __ s = 1

3a.

tan 11° ) 0.19

b.

sin 62° ) 0.88

c.

cos 30° ) 0.87

4a. ..

DF is the hyp. Given: EF, opp. to given !D. Since opp. side and hyp. are involved, use a sine ratio.

sin D = opp. leg

_______ hyp.

= EF ___ DF

sin 51° = 17 ___ DF

DF = 17 ______ sin 51°

) 21.87 m

b. ..

ST is adj. to the given !. Given: TU, the hyp. Since adj. side and hyp. are involved, use a cosine ratio.

cos T = adj. leg

______ hyp.

= ST ___ TU

cos 42° = ST ___ 9.5

9.5(cos 42°) = ST ST ) 7.06 in.

c. ..

BC is adj. to the given !. Given: AC, opp. !B. Since opp. side and adj. side are involved, use a tangent ratio. tan B =

opp. leg _______

adj. leg = AC ___

BC

tan 18° = 12 ___ BC

BC = 12 ______ tan 18°

) 36.93 ft


d. ..

JL is opp. the given !. Given: KL, the hyp. Since opp. side and hyp. are involved, use a sine ratio.

sin K = opp. leg

_______ hyp.

= JL ___ KL

sin 27° = JL ____ 13.6

13.6(sin 27°) = JL JL ) 6.17 cm

5. 1 Understand the ProblemMake a sketch. The answer is AC.2 Make a Plan ..

AC is the hyp. You are given AB, the leg opp. !C. Since opp. leg and hyp. are involved, write an equation using a sine ratio.3 Solve sin C = AB ___

AC

sin 4.8° = 1.2 ___ AC

AC = 1.2 _______ sin 4.8°

) 14.34 ft

4 Look BackProblem asks for AC rounded to nearest hundredth, so round the length to 14.34. Length AC of ramp is 14.34 ft.


1. Solve sin 32° = 4 ___ AB

. 2. Solve cos 32° = 6.4 ___ AB

.

3.


1. sin J = LK ___ JL

2. tan N = MP ___ MN

3. sin C = 4 __ 5 = 0.8 4. tan A = 3 __

4 = 0.75

5. cos A = 4 __ 5 = 0.8 6. cos C = 3 __

5 = 0.6

7. tan C = 4 __ 3 ) 1.33 8. sin A = 3 __

5 = 0.6

9.

cos 60° = x ___ 2x

= 1 __ 2

10. tan 30° = x ____ x & ' 3

= & ' 3 ___ 3

11.

sin 45° = s ____ s & ' 2

= & ' 2 ___ 2

12.

tan 67° ) 2.36

13.

sin 23° ) 0.39

14.

sin 49° ) 0.75

15.

cos 88° ) 0.03

16.

cos 12° ) 0.98

17.

tan 9° ) 0.16

18. ..

BC is opp. the given !. Given: AC, the hyp. Since opp. side and hyp. are involved, use a sine ratio.

sin A = opp. leg

_______ hyp.

= BC ___ AC

sin 23° = BC ___ 4

4(sin 23°) = BC BC ) 1.56 in.

19. ..

QR is opp. the given !. Given: PQ, adj. to given !. Since opp. and adj. sides are involved, use a tangent ratio.

tan P = opp. leg

_______ adj. leg

= QR ___ PQ

tan 50° = QR ___ 8.1

8.1(tan 50°) = QR QR ) 9.65 m

20. ..

KL is adj. to the given !. Given: JL, the hyp. Since adj. side and hyp. are involved, use a cosine ratio.

cos L = adj. leg

______ hyp.

= KL ___ JL

cos 61° = KL ___ 2.5

2.5(cos 61°) = KL KL ) 1.21 cm


21. 1 Understand the ProblemThe answer is XY, opp. the given !.2 Make a PlanYou are given WZ, which is twice WY, the leg adj. to !W. First, calculate WY. Then, since opp. and adj. legs are involved, write an equation using a tangent ratio.3 SolveWY = 1 __

2 WZ

= 1 __ 2 (56) = 28 ft

tan W = XY ____ WY

tan 15° = XY ___ 28

XY = 28(tan 15°) ) 7.5028 ft4 Look BackProblem asks for XY rounded to nearest inch. Height XY of pediment is 7 ft 6 in.


22. cos D = 8 ___ 17

) 0.47 23. tan D = 15 ___ 8 ) 1.88

24. tan F = 8 ___ 15

) 0.53 25. cos F = 15 ___ 17

) 0.88

26. sin F = 8 ___ 17

) 0.47 27. sin D = 15 ___ 17

) 0.88

28.

tan 60° = x & ' 3 ____ x = & ' 3

29. sin 30° = x ___ 2x

= 1 __ 2

30. cos 45° = s ____ s & ' 2

= & ' 2 ___ 2

31. tan 51° ) 1.23 32. sin 80° ) 0.98

33. cos 77° ) 0.22 34. tan 14° ) 0.25

35. sin 55° ) 0.82 36. cos 48° ) 0.67

37. PQ = 11 sin 19° ) 3.58 cm

38. cos 46° = 19.2 ____ AC

AC = 19.2 _______ cos 46°

) 27.64 in.

39. sin 34° = 11 ___ GH

GH = 11 ______ sin 34°

) 19.67 ft

40. cos 25° = 33 ___ XZ

XZ = 33 _______ cos 25°

) 36.41 in.

41. tan 61° = 9.5 ___ KL

KL = 9.5 ______ tan 61°

) 5.27 ft

42. EF = 83.1 tan 12°) 17.66 m

43. sin 15° = 1.58 ____ !

! = 1.58 ______ sin 15°

) 6.10 m

44. If a and b are opp. and adj. leg lengths,tan (m!) = a __

b = 1

a = b$ is 45°-45°-90°, so m! = 45°

45. sin 45° = s ____ s & ' 2

= cos 45°

So sine and cosine ratios are =.

46. sin 30° = x ___ 2x

= 1 __ 2

Sine of a 30° ! is 0.5.

47. cos 30° = x & ' 3 ____ 2x

= sin 60°

cos 30° = sine of a 60° !

48. h = 10 sin 75.5° ) 9.7 ft

49. BC = AD= 3 tan (90 ( 68)°) 1.2 ft

50. SU = RS _______ cos 49°

= UT _______ cos 49°

= 9.4 _______ cos 49°

) 14.3 in.

51. The tangent ratio is < 1 for # measuring < 45° and > 1 for # measuring > 45°. In a 45°-45°-90° $, both legs have same length, so tan 45° = 1. If the acute ! measure increases, opp. leg length also increases, so tangent ratio is > 1. If the acute ! measure decreases, the opp. leg length also decreases, so tangent ratio is < 1.

52a. AC = AB _____ sin C

= 25 ______ sin 65°

) 27.58 ft ) 27 ft 7 in.

b. AD = AB sin !ABD= 25 sin (90 ( 28)°) 22.07 ft) 22 ft 1 in.

53. From diagram,

sin A = 3 __ 5

= 0.6.


54. From diagram,

tan D = 24 ___ 7 ) 3.43.

55. Let 1 face be $ABC with A the apex; let M be mdpt. of

.. BC .

BC = 2BM

= 2h ______ tan 52°

= 2(482)

______ tan 52°

) 753 ft

56a, b. Check students’ work.

c. 20° d. sin 20° = 0.34cos 20° = 0.94tan 20° = 0.36

e. Values in part d should be close to estimate-based values in part b.

57a. tan 30° = x ____ x & ' 3

= & ' 3 ___ 3

sin 30° = x ___ 2x

= 1 __ 2

cos 30° = x & ' 3 ____ 2x

= & ' 3 ___ 2

1 _ 2 ___

& ' 3 ___ 2 = 1 ___

& ' 3 =

& ' 3 ___ 3

So tan 30° = sin 30° _______ cos 30°

b. tan A = a __ b , sin A = a __ c , cos A = b __ c

c. sin A _____ cos A

= a _ c __ b _ c

= a __ c · c __ b

= a __ b = tan A

58. (sin 45° ) 2 + (cos 45° ) 2 = ( & ' 2 ___ 2 ) 2 + ( & ' 2 ___

2 ) 2

= 2 __ 4 + 2 __

4 = 1

59. (sin 30° ) 2 + (cos 30° ) 2 = ( 1 __ 2 ) 2 + ( & ' 3 ___

2 ) 2

= 1 __ 4 + 3 __

4 = 1

60. (sin 60° ) 2 + (cos 60° ) 2 = ( & ' 3 ___ 2 ) 2 + ( 1 __

2 ) 2

= 3 __ 4 + 1 __

4 = 1

61a. sin A = a __ c , cos A = b __ c

b. (sin A ) 2 + (cos A ) 2 = ( a __ c ) 2 + ( b __ c ) 2

= a 2 + b 2 _______ c 2

= c 2 __ c 2

= 1

c. Derivation of identity uses fact that in a rt. $, a 2 + b 2 = c 2 , which is Pyth. Thm.

sin A = BC ___ AB

; cos B = BC ___ AB

; sin A = cos B;

! A and ! B are comp.; the sine of an ! is equal to the cosine of its comp.

62. P = 2 + 2 tan 24° + 2/cos 24° ) 5.08 mA = 1 __

2 (2)(2 tan 24°) ) 0.89 m 2

63. P = 7.2 + 7.2 ___ 2 _______

cos 51° +

7.2 ___ 2 _______

cos 51° ) 18.64 cm

A = 1 __ 2 (7.2) ( 7.2 ___

2 tan 51°) ) 16.00 cm 2

64. P = 4 + 4 ______ sin 58°

+ 4 ______ tan 58°

) 11.22 ft

A = 1 __ 2 (4) ( 4 ______

tan 58° ) ) 5.00 ft 2

65. P = 10 + 10 sin 72° + 10 cos 72° ) 22.60 in.A = 1 __

2 (10 sin 72°)(10 cos 72°) ) 14.69 in. 2

66. sin A = BC ___ AB

; cos B = BC ___ AB

; sin A = cos B; !A and !B

are comp.; sine of an ! is = to cosine of its comp.

67. Tangent of an acute ! increases as measure of the ! increases.

TEST PREP, PAGE 532

68. A

69. H17 tan 65° ) 36 ft

70. Ccos N = NP ___

MN = sin M


71. AB tan A = BC (4x) tan 42° = 3x + 3(4 tan 42° ( 3)(x) = 3 x = 3 ___________

4 tan 42° ( 3 ) 5

AB ) 4(5) ) 20BC ) 3(5) + 3 ) 18

AC ) & '''' 2 0 2 + 1 8 2 ) 27

72. AC cos A = AB (15x) cos 21° = 5x + 27(15 cos 21° ( 5)(x) = 27 x = 27 ____________

15 cos 21° ( 5 ) 3

AB ) 5(3) + 27 ) 42AC ) 15(3) ) 45

BC ) & '''' 4 5 2 ( 4 2 2 ) 16

73. (tan A ) 2 + 1 = ( sin A _____ cos A

) 2 + 1

= (sin A ) 2 + (cos A ) 2

________________ (cos A ) 2

= 1 _______ (cos A ) 2


74. Int. # of a reg. pentagon

measure

( 5 ( 2 _____ 5 ) (180) = 108°.

In the diagram,

m!1 = 1 __ 2 (108) = 54°.

Therefore,

r = 0.5 _______ cos 54°

) 0.85 in.

75. csc Y = 1 _____ sin Y

= 1 ___ XZ __ YZ

= YZ ___ XZ

= 5 __ 4 = 1.25

76. sec Z = 1 _____ cos Z

= 1 ___ XZ __ YZ

= YZ ___ XZ

= 5 __ 4 = 1.25

77. cot Y = 1 _____ tan Y

= 1 ___ XZ __ XY

= XY ___ XZ

= 3 __ 4 = 0.75


78.–80. Possible answers given.

78. ((3, (15), ((1, (9), (0, (6)

79. ((2, 11), (0, 10), (2, 9) 80. ((2, 14), (0, 2), (4, 2)

81. Trans. Prop. of " 82. Reflex. Prop. of "

83. Symm. Prop. of " 84. & ''' 3 · 27 = & '' 81 = 9

85. & ''' 6 · 24 = & '' 144 = 12 86. & ''' 8 · 32 = & '' 256 = 16

CONNECTING GEOMETRY TO ALGEBRA: INVERSE FUNCTIONS, PAGE 533

TRY THIS, PAGE 533 1. x = 0°

si n (1 (0) = 0° 2. x = 60°

co s (1 ( 1 __ 2

) = 60°

3. x = 45°ta n (1 (1) = 45°

4. x = 90°co s (1 (0) = 90°

5. x = 0°ta n (1 (0) = 0°

6. x = 30°

si n (1 ( 1 __ 2 ) = 30°

8-3 SOLVING RIGHT TRIANGLES, PAGES 534–541


1a. 14.4 ____ 30.6

= 8 ___ 17

= sin A

+ !A is !2

b. 27 ____ 14.4

= 1.875 = tan A

+ !A is !1

2a.

ta n (1 (0.75) ) 37°

b.

co s (1 (0.05) ) 87°

c.

si n (1 (0.67) ) 42°

3. DF = DE ____ sin F

= 14 ______ sin 58°

) 16.51

EF = DE _____ tan F

= 14 ______ tan 58°

) 8.75

Acute # of a rt. $ are comp. So,m!D = 90 ( 58 = 32°

4. Step 1 Find side lengths.Plot pts. R, S, and T.RS = 7 ST = 7RT = & ''''''''' ((3 ( 4 ) 2 + (5 + 2 ) 2

= & ''''' ((7 ) 2 + 7 2 = & '' 98 ) 9.90

Step 2 Find ! measures.m!S = 90°m!R = ta n (1 ( 7 __

7 ) = 45°

m!T = 90 ( 45 = 45°


5. 38% = 38 ____ 100

A 38% grade means Baldwin St. rises 38 ft for every 100 ft of horiz. dist.

m!D = ta n (1 ( 38 ____ 100

) ) 21°

THINK AND DISCUSS, PAGE 537 1. Find RS using Pyth. Thm. Then find m!R using

m!R = si n (1 ( 3.5 ___ 4.1

) , and find m!T using either

m!T = co s (1 ( 3.5 ___ 4.1

) or m!T = 90 ( m!R.

2. co s (1 (0.35) = m!Z

3.


1. 8 ___ 10

= 4 __ 5

= sin A

+ !A is !1

2. 8 __ 6 = 1 1 __

3 = tan A

+ !A is !1

3. 6 ___ 10

= 0.6 = cos A

+ !A is !1

4. 8 ___ 10

= 0.8 = cos A

+ !A is !2

5. 6 __ 8 = 0.75 = tan A

+ !A is !2

6. 6 ___ 10

= 0.6 = sin A

+ !A is !2

7.

ta n (1 (2.1) ) 65°

8.

co s (1 ( 1 __ 3 ) ) 71°

9.

co s (1 ( 5 __ 6 ) ) 34°

10.

si n (1 (0.5) = 30°

11.

si n (1 (0.61) ) 38°

12.

ta n (1 (0.09) ) 5°

13. tan P = 3.1 ___ 8.9

m!P = ta n (1 ( 3.1 ___ 8.9

) ) 19°

Acute # of a rt. $ are comp. Som!R ) 90 ( 19 ) 71°.

RP = 3.1 _____ sin P

= 3.1 ____________ sin (ta n (1 ( 3.1 ___

8.9 ) )

) 9.42

14. AB = 7.4 cos 32° ) 6.28BC = 7.4 sin 32° ) 3.92Acute # of a rt. $ are comp. Som!C = 90 ( 32 = 58°.

15. By Pyth. Thm.,

YZ = & ''''' 11 2 + 8.6 2 = & ''' 194.96 ) 13.96

tan Y = 8.6 ___ 11

m!Y = ta n (1 ( 8.6 ___ 11

) ) 38°

tan Z = 11 ___ 8.6

m!Z = ta n (1 ( 11 ___ 8.6

) ) 52°

16. Step 1 Find side lengths.Plot pts. D, E, and F.DE = 3 EF = 6

DF = & ''''''''' ((2 ( 4 ) 2 + ((2 ( 1 ) 2

= & '''''' ((6 ) 2 + ((3 ) 2

= & '' 45 ) 6.71

Step 2 Find ! measures.m!E = 90°m!D = ta n (1 ( 6 __

3 ) ) 63°

m!F ) 90 ( 63 ) 27°

17. Step 1 Find side lengths.Plot pts. R, S, and T.RS = 5 ST = 6

RT = & ''''''''' ((2 ( 3 ) 2 + ((3 ( 3 ) 2

= & '''''' ((5 ) 2 + ((6 ) 2

= & '' 61 ) 7.81

Step 2 Find ! measures.m!S = 90°m!R = ta n (1 ( 6 __

5 ) ) 50°

m!T ) 90 ( 50 ) 40°


18. Step 1 Find side lengths.Plot pts. X, Y, and Z.XZ = 7 YZ = 7

XY = & ''''''''' ((3 ( 4 ) 2 + (1 + 6 ) 2

= & ''''' ((7 ) 2 + 7 2 = & '' 98 ) 9.90

Step 2 Find ! measures.m!Z = 90°m!X = ta n (1 ( 7 __

7 ) = 45°

m!Y = 90 ( 45 = 45°

19. Step 1 Find side lengths.Plot pts. A, B, and C.AB = 2 BC = 4

AC = & '''''''' (1 + 1 ) 2 + (5 ( 1 ) 2

= & '''' 2 2 + 4 2 = & '' 20 ) 4.47

Step 2 Find ! measures.m!B = 90°m!A = ta n (1 ( 4 __

2 ) ) 63°

m!C = 90 ( 63 ) 27°

20. 8% = 8 ____ 100

An 8% grade means hill rises 8 m for every 100 m of horiz. dist.

m! = ta n (1 ( 8 ____ 100

) ) 5°


21. 5 ___ 12

= 7.5 ___ 18

= tan!2

!A = !2

22. 2.4 = 18 ___ 7.5

= tan!1

!A = !1

23. 12 ___ 13

= 18 ____ 19.5

= sin!1

!A = !1

24. 5 ___ 13

= 7.5 ____ 19.5

= sin!2

!A = !2

24. 12 ___ 13

= 18 ____ 19.5

= cos!2

!A = !2

26. 5 ___ 13

= 7.5 ____ 19.5

= cos!1

!A = !1

27. si n (1 (0.31) ) 18° 28. ta n (1 (1) = 45°

29. co s (1 (0.8) ) 37° 30. co s (1 (0.72) ) 44°

31. ta n (1 (1.55) ) 57° 32. si n (1 ( 9 ___ 17

) ) 32°

33. Step 1 Find unknown side lengths.JK = 3.2 cos 26° ) 2.88LK = 3.2 sin 26° ) 1.40Step 2 Find unknown ! measure.m!L = 90 ( 26 = 64°

34. Step 1 Find unknown side length.

DF = & '''''' ( & ' 5 ) 2 + ( & ' 2 ) 2

= & ''' 5 + 2 = & ' 7 ) 2.65Step 2 Find unknown ! measures.

m!D = ta n (1 ( & ' 2 ___ & ' 5

) ) 32°

m!F = ta n (1 ( & ' 5 ___ & ' 2

) ) 58°

35. Step 1 Find unknown ! measures.

m!P = co s (1 ( 6.7 ___ 8.3

) ) 36°

m!R = si n (1 ( 6.7 ___ 8.3

) ) 54°

Step 2 Find unknown side length.QR = 8.3 sin P

= 8.3 sin (co s (1 ( 6.7 ___ 8.3

) ) ) 4.90

36. Step 1 Find side lengths. ..

AB is vert., AB = 5; ..

BC is horiz., BC = 1

By Pyth. Thm., AC = & '''' 5 2 + 1 2 = & '' 26 ) 5.10 Step 2 Find ! measures.m!B = 90°m!A = ta n (1 ( BC ___

AB ) = si n (1 ( 1 __

5 ) ) 11°

m!C ) 90 ( 11 ) 79°

37. Step 1 Find side lengths.

...

MN is vert., MN = 4; ..

NP is horiz., NP = 4

By Pyth. Thm., MP = & '''' 4 2 + 4 2 = & '' 32 ) 5.66Step 2 Find ! measures.m!N = 90°m!M = ta n (1 ( NP ___

MN ) = ta n (1 ( 4 __

4 ) = 45°

m!P = 90 ( 45 = 45°

38. Range of ! measures is between ta n (1 ( 1 ___ 20

) ) 3°

and ta n (1 ( 1 ___ 16

) ) 4°.

39. ta n (1 (3.5) ) 74°tan 74° ) 3.5

40. si n (1 ( 2 __ 3 ) ) 42°

sin 42° ) 2 __ 3

41. cos 42° ) 0.74 42. cos 12° ) 0.98co s (1 (0.98) ) 12°

43. sin 69° ) 0.93si n (1 (0.93) ) 69°

44. cos 60° = 1 __ 2

45. Assume square has sides of length a. Then either rt. $ formed by a diag. has legs of length a. So measure of ! formed by diag. and a side is ta n (1 ( a __ a ) = ta n (1 (1) = 45°.

46a. Possible answer: m!P ) 40°

b. RQ ) 2.2 cm, PQ ) 3.1 cm

c. m!P = ta n (1 ( RQ ___ PQ

) ) ta n (1 ( 2.2 ___

3.1 ) ) 35°


d. Possible answer: Answer in part c is likely more accurate, since it is easier to measure lengths to the nearest tenth than to measure # to the nearest degree.

47a. m!1 = ta n (1 ( 8 ____ 100

) ) 5°

b. m!1 ) 90 ( 5 ) 85°

c. h = 31 _________________ sin (90 ( ta n (1 ( 8 ___

100 ) )

) 31.10 ft, or 31 ft 1 in.

48. si n (1 ( 3 __ 5 ) ) 37°, si n (1 ( 4 __

5 ) ) 53°

49. si n (1 ( 5 ___ 13

) ) 23°, si n (1 ( 12 ___ 13

) ) 67°

50. si n (1 ( 8 ___ 17

) ) 28°, si n (1 ( 15 ___ 17

) ) 62°

51. ta n (1

( 45 ___ 28

) ) 58°, ta n (1

( 90 ___ 28

) ) 73°

Acute ! measure changes from about 58° to about 73°, an increase by a factor of 1.26.

52. m! = ta n (1 ( 28 ____ 100

) ) 16°

53a. AB = & '''''''' (6 + 1 ) 2 + (1 ( 0 ) 2

= & '''' 7 2 + 1 2 = & '' 50 = 5 & ' 2

BC = & '''''''' (0 ( 6 ) 2 + (3 ( 1 ) 2

= & '''' 6 2 + 2 2 = & '' 40 = 2 & '' 10

AC = & '''''''' (0 + 1 ) 2 + (3 ( 0 ) 2

= & '''' 1 2 + 3 2 = & '' 10

b. A C 2 + B C 2 = 10 + 40 = 50 = A B 2

So $ABC is a rt. $, and C is the rt. !.

c. m!A = si n (1 ( BC ___ AB

) = si n (1 ( 2 & '' 10 _____

5 & ' 2 )

= si n (1 ( 2 & ' 5 ____ 5

) ) 63°

m!B = 90 ( m!A ) 27°

54. m!BDC = ta n (1 ( 2 __ 7 ) ) 16°

55. m!STV = ta n (1 ( 3.2 ___ 4.5

) ) 35°

56. m!DGF = 2m!DGH = 2 si n (1 ( 2.4 ___ 4.4

) ) 66°

57. m!LKN = ta n (1 ( 9 ___ 4.8

) ) 62°

58. tan 70° > tan 60°; possible answer: consider 2 rt. *, 1 with a 60° ! and 1 with a 70° !. Suppose that legs adj. to these # have length 1 unit. Leg opp. 70° ! will be longer than leg opp. 60° !. So tan 70° is greaater than tan 60°.

59. ta n (1 (m) = ta n (1 (3) ) 72°

60. ta n (1 (m) = ta n (1 ( 2 __ 3 ) ) 34°

61. 5y = 4x + 3 y = 4 __

5 x + 3 __

5

ta n (1 (m) = ta n (1 ( 4 __ 5 ) ) 39°

62. Since $ is not a rt. $., trig. ratios do not apply.

63. No; possible answer: you only need to know 2 side lengths. You can use Pyth. Thm. to find 3rd side length or use trig. ratios to find acute ! measures.

64. AD = AC cos A

= 10 cos (ta n (1 ( 6 ___ 10

) ) ) 8.57

BD = BC cos B

= 6 cos (ta n (1 ( 10 ___ 6 ) ) ) 3.09

CD = BC sin B

= 6 sin (ta n (1 ( 10 ___ 6 ) ) ) 5.14

(8.57)(3.09) ) 26 ) (5.14 ) 2 (8.57)(8.57 + 3.09) ) 100 = (10 ) 2 (3.09)(8.57 + 3.09) ) 36 = (6 ) 2

TEST PREP, PAGE 540

65. D 66. J

67. Ata n (1 ( 1.4 ___

2.7 ) ) 27°

68. 9°ta n (1 ( 3 ___

20 ) ) 9°

CHALLENGE AND EXTEND, PAGES 540–541

69. LH = 10 sin J = 20 sin 25° sin J = 2 sin 25° m!J = si n (1 (2 sin 25°) ) 58°

70. BD = 3.2 tan A = 8 cos 64° tan A = 2.5 cos 64° m!A = ta n (1 (2.5 cos 64°) ) 48°

71. Let !A be an acute ! with m!A = co s (1 (cos 34°).Then cos A = cos 34°. Since cos is a 1-to-1 function on acute ! measures, m!A = 34°.

72. Since tan is a 1-to-1 function on acute ! measures, x = tan [ta n (1 (1.5)] + x = 1.5

73. Since sin is a 1-to-1 function on acute ! measures,y = sin (si n (1 x) + y = x

74. y = 40 sin (ta n (1 ( 6 ____ 100

) ) ) 2.40 ft

75. Possible answer: The expression sin (1 (1.5) represents an ! measure that has a sine of 1.5. The sine of an acute ! of a rt. $ must be between 0 and 1, so the expression sin (1 (1.5) is undefined.

76. Let ..

BD be altitude. ThenArea = 1 __

2 (base)(height)

= 1 __ 2 (AC)(BD)

= 1 __ 2 (b)(c sin A)

= 1 __ 2 bc sin A.



77. false; 6.8 / 2 + 2.5 + 3.3 = 7.8

78. true; 2 + 2.5 + 3.3 + 6.8 + 3.6 _____________________ 5

) 3.5 in.

79. False; rainfall decreased from April to May.

80. !B " !E 37 = 2x + 11 26 = 2x x = 13

81. AB ___ DE

= AC ___ DF

3y + 7

______ 2y + 6

= 1.4 + 1.6 ________ 1.4 + 1.6

= 1

3y + 7 = 2y + 6 y = (1

82. DF = 1.4 + 1.6 = 3 83. sin 63° ) 0.89

84. cos 27° ) 0.89 85. tan 64° ) 2.05

USING TECHNOLOGY, PAGE 541 1.–5. Check students’ work.


1. (DB ) 2 = (DA)(DC)= (30 ( 6)(6) = 144

DB = & '' 144 = 12 ft

2. (AB ) 2 = (AD)(AC)= (24)(30)= 720

AB = & '' 720 ) 26.83 ft or 26 ft 10 in.

3. m!ABD = ta n (1 ( AD ___ DB

) = ta n (1 ( 24 ___

12 )

= ta n (1 (2) = 63°

4. !A and !ABD are comp.m!A = 90 ( m!ABD

) 90 ( 63 ) 27°

5. grade = DB ___ DC

· 100% = 6 ___ 12

· 100% = 50%


1. x = & ''' (5)(12) = & '' 60 = 2 & '' 15

2. x = & '''' (2.75)(44) = & '' 121 = 11

3. x = & ''''

( 5 __ 2 ) ( 15 ___

8 )

= & '' 75 ___ 16

= 5 & ' 3 ____ 4

4. x 2 = (4)(8) = 32 x = & '' 32 = 4 & ' 2 y 2 = (4)(12) = 48 y = & '' 48 = 4 & ' 3 z 2 = (8)(12) = 96 z = & '' 96 = 4 & ' 6

5. (12 & ' 5 ) 2 = (24)(x + 24) 720 = 24x + 576 144 = 24x x = 6 y 2 = 24x

= 24(6) = 144 y = 12

z 2 = (24 + x)(x)= (30)(6) = 180

z = & '' 180 = 6 & ' 5

6. 6 2 = 12x36 = 12x x = 3 y 2 = 12(12 + x)

= (12)(15) = 180 y = & '' 180 = 6 & ' 5

z 2 = (12 + x)x= (15)(3) = 45

z = & '' 45 = 3 & ' 5

7. (AB ) 2 = (BC)(BD)= (22)(30) = 660

AB = & '' 660 ) 25.7 m

8. Let legs of 45°-45°-90° $ have length x.tan 45° = x __ x = 1

9. Let 30°-60°-90° $ have side lengths x, x & ' 3 , 2x.sin 30° = x ___

2x = 1 __

2

10. cos 30° = x & ' 3 ____ 2x

= & ' 3 ___ 2 11. sin 16° ) 0.28

12. cos 79° ) 0.19 13. tan 27° ) 0.51

14. QR = 14 ______ tan 31°

) 23.30 in.

15. AB = 6 cos 50° ) 3.86 m

16. LM = 4.2 sin 62° ) 3.71 cm

17. m!A = 90 ( 32 = 58°

BC = 22 ______ tan 32°

) 35.21

AC = 22 ______ sin 32°

) 41.52

18. HJ = & ''''' 7 2 + 10. 5 2 = & ''' 159.25 ) 12.62

m!H = ta n (1 ( 10.5 ____ 7 ) ) 56°

m!J = ta n (1 ( 7 ____ 10.5

) ) 34°

19. m!Z = 90 ( 28 = 62°XY = 5.1 cos 28° ) 4.50YZ = 5.1 sin 28° ) 2.39

20. ta n (1 ( 1 ___ 18

) ) 3°


8-4 ANGLES OF ELEVATION AND DEPRESSION, PAGES 544–549

CHECK IT OUT! PAGES 544–546 1a. !5 is formed by a horiz. line and a line of sight to a

pt. below the line. It is an ! of depression.

b. !6 is formed by a horiz. line and a line of sight to a pt. above the line. It is an ! of elevation.

2. Let A represent airport and P represent plane. Let x be horiz. distance between plane and airport.

tan 29° = 3500 _____ x

x = 3500 ______ tan 29°

) 6314 ft

3. Let T represent top of tower and F represent fire. Let x be horiz. distance between tower and fire.By Alt. Int. # Thm., m!F = 3°.

tan 3° = 90 ___ x

x = 90 _____ tan 3°

) 1717 ft

4. Step 1 Let P represent plane, and A and B represent two airports. Let x be distance between airports.Step 2 Find y.By Alt. Int. # Thm., m!CAP = 78°. In $APC,tan 78° =

12,000 ______ y

y = 12,000

______ tan 78°

) 2550.7 ft

Step 3 Find z.By Alt. Int. # Thm., m!CBP = 19°. In $BPC,

tan 19° = 12,000

______ z

z = 12,000

______ tan 19°

) 34,850.6 ft

Step 4 Find x.x = z ( y) 34,850.6 ( 2550.7 ) 32,300 ft

THINK AND DISCUSS, PAGE 546 1. It increases, because height of skyscraper is

constant and horiz. dist. is decreasing.

2.


1. elevation 2. depression

3. !1 is formed by a horiz. line and a line of sight to a pt. above the line. It is an ! of elevation.

4. !2 is formed by a horiz. line and a line of sight to a pt. below the line. It is an ! of depression.

5. !3 is formed by a horiz. line and a line of sight to a pt. above the line. It is an ! of elevation.

6. !4 is formed by a horiz. line and a line of sight to a pt. below the line. It is an ! of depression.

7. Let h be height of flagpole.

tan 37° = h ____ 24.2

h = 24 tan 37° ) 18 ft

8. Let H represent helicopter and A represent accident. Let x be horiz. dist. between helicopter and accident.By Alt. Int. # Thm., m!A = 18°.

tan 18° = 1560 _____ x

x = 1560 ______ tan 18°

) 4801 ft


9. Step 1 Let T represent top of canyon, and A and B represent near and far sides of river. Let w be width of river.Step 2 Find y.By Alt. Int. # Thm., m!CAT = 74°. In $ATC,

tan 74° = 191 ____ y

y = 191 ______ tan 74°

) 54.77 m

Step 3 Find z.By Alt. Int. # Thm., m!CBT = 58°. In $BTC,

tan 58° = 191 ____ z

z = 191 ______ tan 58°

) 119.35 m

Step 4 Find w.w = z ( y

) 119.35 ( 54.77 ) 64.6 m


10. !1: ! of depression 11. !2: ! of elevation

12. !3: ! of elevation 13. !4: ! of depression

14. h = 1.5 + 100 tan 67° ) 237 m

15. x = 120 _______ tan 3.5°

) 1962 ft 16. z = y ( x= 1 tan 74° ( 1 tan 16°) 3.2 mi

17. true 18. true

19. false; ! of elevation gets closer to 90°

20. true 21. !1 and !3

22. m!2 = 30° (given)m!1 = 90 ( m!2

= 90 ( 30 = 60° (comp. #)m!3 = m!1 = 60°,m!4 = m!2 = 30° (Alt. Int. # Thm.)

23. Possible answer: As a hot air ballon descends vertically, ! of depression to an object on the ground decreases.

24. By Alt. Int. # Thm.,! of depression = ta n (1 ( 165 ____

50 ) ) 73°

25a. x = 1000 ______ tan 67°

) 424 ft b. z = y ( x

= 1000 ______ tan 55°

( 1000 ______ tan 67°

) 276 ft

26. When the ! of elevation is exactly 45°, the length of the shadow will be the same as the length of telephone pole, since a rt. isosc. $ is formed and tan 45° = 1.

27a. x = 1250 ______ tan 31°

) 2080 ft b. v = s __ t

t = s __ v

) 2080 _____ 150

) 14 s

TEST PREP, PAGE 549

28. Ddist. = 1600 ______

tan 35° ) 2285 ft

29. Jheight = 93 tan 60° ) 161 ft

30. The ! of elevation increases as Jim moves closer to trail marker.


31. Let x and y be dists. from Jorge and from Susan to foot of Big Ben; let h be height of Big Ben.Jorge: h = x tan 49.5°Susan: h = y tan 65° y = x ( 38h = x tan 49.5° = (x ( 38) tan 65° 38 tan 65° = x(tan 65° ( tan 49.5°)

x = 38 tan 65° _________________ (tan 65° ( tan 49.5°)

h = x tan 49.5°

= 38 tan 65° _________________ (tan 65° ( tan 49.5°)

(tan 49.5°) ) 98 m

32. Speed = 500 mi ___ h · 1 h ______

60 min · 5280 ft ______

1 mi = 44,000 ft/min

Let time until over lake be t. Then horiz. dist to lake is

s = 44,000t = 14,000

______ tan 6°

t = 14,000 ___________

44,000 tan 6° ) 3 min.

33. h = x tan 5° = (10 ( x) tan 2°x(tan 5° + tan 2°) = 10 tan 2°

x = 10 tan 2° ____________ tan 5° + tan 2°

h = x tan 5°

= 10 tan 2° ____________ tan 5° + tan 2°

(tan 5°)

) 0.2496 mi ) 1318 ft

34. h = y ( x= 46 tan 42° ( 46 tan 18°) 26.47 ft or 26 ft 6 in.


35. Let x and y be dists. run by Emma and mother in time t. When they meet, x + y = 16t + 4t = 1 10t = 1 t = 0.1 h or 6 min

36. Let p be original price.discounted price = 0.7pprice after coupon = 0.85(0.7p) = 17.85

p = 17.85 ________ 0.85(0.7)

= 30

Original price was $30.

37. rhombus, square 38. rectangle, square


39. rectangle, rhombus, square

40. rectangle, rhombus, square

41. x 2 + 3 2 = 5 2 x 2 + 9 = 25 x 2 = 16 x = 4

42. y __ x = 5 __

3

y = 5x ___ 3

= 5(4)

____ 3 = 20 ___

3

43. x 2 = 3z 4 2 = 3z

z = 16 ___ 3

GEOMETRY LAB: INDIRECT MEASUREMENT USING TRIGONOMETRY, PAGE 550

TRY THIS, PAGE 550 1. The ! reading from clinometer is comp. of ! of

elevation.


3. Check students’ work. Results should be similar.

4. Possible answers: Measuring the distance between observer and object, measuring height of observer’s eyes, and reading the ! measure from clinometer.

5. It can be used to measure height of tall objects that cannot be measured directly.

8-5 LAW OF SINES AND LAW OF COSINES, PAGES 551–558

CHECK IT OUT! PAGES 551–554 1a.

tan 175° ) (0.09

b.

cos 92° ) (0.03

c.

sin 160° ) 0.34

2a. sin N ____ MP

= sin M _____ NP

sin 39° ______ 22

= sin 88° ______ NP

NP sin 39° = 22 sin 88°

NP = 22 sin 88° ________ sin 39°

) 34.9

b. sin L ____ JK

= sin K ____ JL

sin L ____ 6 = sin 125° _______

10

sin L = 6 sin 125° ________ 10

m!L = si n (1 ( 6 sin 125° ________ 10

) ) 29°

c. sin X ____ YZ

= sin Y ____ XZ

sin X ____ 4.3

= sin 50° ______ 7.6

sin X = 4.3 sin 50° _________ 7.6

m!X = si n (1 ( 4.3 sin 50° ________ 7.6

) ) 26°

d. m!A = 180 ( 67° ( 44° = 69°

sin A ____ BC

= sin B ____ AC

sin 69° ______ 18

= sin 67° ______ AC

AC sin 69° = 18 sin 67°

AC = 18 sin 67° ________ sin 69°

) 17.7

3a. D E 2 = D F 2 + E F 2 ( 2(DF)(EF) cos F = 1 6 2 + 1 8 2 ( 2(16)(18) cos 21°D E 2 ) 42.2577 DE ) 6.5

b. J L 2 = J K 2 + K L 2 ( 2(JK)(KL) cos K 8 2 = 1 5 2 + 1 0 2 ( 2(15)(10) cos K 64 = 325 ( 300 cos K(261 = (300 cos K

cos K = 261 ____ 300

m!K = co s (1 ( 261 ____ 300

) ) 30°

c. Y Z 2 = X Y 2 + X Z 2 ( 2(XY)(XZ) cos X = 1 0 2 + 4 2 ( 2(10)(4) cos 34°Y Z 2 ) 49.6770 YZ ) 7.0

d. P Q 2 = Q R 2 + P R 2 ( 2(QR)(PR) cos R 9. 6 2 = 10. 5

2 + 5. 9

2 ( 2(10.5)(5.9) cos R

92.16 = 145.06 ( 123.9 cos R(52.9 = (123.9 cos R

cos R = 52.9 _____ 123.9

m!R = co s (1 ( 52.9 _____ 123.9

) ) 65°

4. Step 1 Find length of cable.A C 2 = A B 2 + B C 2 ( 2(AB)(BC) cos B = 3 1 2 + 5 6 2 ( 2(31)(56) cos 100°A C 2 = 4699.9065 AC = 68.6 mStep 2 Find angle measure between cable and ground.

sin A ____ BC

= sin B ____ AC

sin A ____ 56

= sin 100° _______ 68.56

sin A = 56 sin 100° _________ 68.56

m!A = si n (1 ( 56 sin 100° _________ 68.56

) ) 54°


THINK AND DISCUSS, PAGE 554 1. m!A.

2.


1.

sin 100° ) 0.98

2.

cos 167° ) (0.97

3.

tan 92° ) (28.64

4.

tan 141° ) (0.81

5.

cos 133° ) (0.68

6.

sin 150° = 0.5

7.

sin 147° ) 0.54

8.

tan 164° ) (0.29

9.

cos 156° ) (0.91

10. sin R ____ ST

= sin S ____ RT

sin 36° ______ 15

= sin 70° ______ RT

RT sin 36° = 15 sin 70°

RT = 15 sin 70° ________ sin 36°

) 24.0

11. sin B ____ AC

= sin C ____ AB

sin B ____ 14

= sin 101° _______ 20

sin B = 14 sin 101° _________ 20

B = si n (1 ( 14 sin 101° _________ 20

) ) 43°

12. sin F ____ DE

= sin D ____ EF

sin F ____ 20

= sin 84° ______ 31

sin F = 20 sin 84° ________ 31

F = si n (1 ( 20 sin 84° ________ 31

) ) 40°

13. P R 2 = P Q 2 + Q R 2 ( 2(PQ)(QR) cos Q 7 2 = 6 2 + 10 2 ( 2(6)(10) cos Q 49 = 136 ( 120 cos Q (87 = (120 cos Q

cos Q = 87 ____ 120

m!Q = co s (1 ( 87 ____ 120

) ) 44°

14. M N 2 = L M 2 + L N 2 ( 2(LM)(LN) cos L = 3 0 2 + 2 5 2 ( 2(30)(25) cos77°M N 2 ) 1187.5734 MN ) 34.5


15. A B 2 = A C 2 + B C 2 ( 2(AC)(BC) cos C = 8 2 + 1 1 2 ( 2(8)(11) cos 131A B 2 ) 300.4664 AB ) 17.3

16. Think: Find each ! using Law of Cosines. 2 0 2 = 2 4 2 + 3 0 2 ( 2(24)(30) cos!1 400 = 1476 ( 1440 cos!1(1076 = (1440 cos!1

m!1 = co s (1 ( 1076 _____ 1440

) ) 42°

2 4 2 = 2 0 2 + 3 0 2 ( 2(20)(30) cos!2 576 = 1300 ( 1200 cos!2(724 = (1200 cos!2

m!2 = co s (1 ( 724 _____ 1200

) ) 53°

3 0 2 = 2 4 2 + 2 0 2 ( 2(24)(20) cos!3 900 = 976 ( 960 cos!3 76 = (960 cos!3

m!3 = co s (1 ( 76 ____ 960

) ) 85°


17. cos 95° ) (0.09 18. tan 178° ) (0.03

19. tan 118° ) (1.88 20. sin 132° ) 0.74

21. sin 98° ) 0.99 22. cos 124° ) (0.56

23. tan 139° ) (0.87 24. cos 145° ) (0.82

25. sin 128° ) 0.79

26. sin C ____ 6.8

= sin 122° _______ 10.2

m!C = si n (1 ( 6.8 sin 122° __________ 10.2

) ) 34°

27. sin 17° ______ 8.5

= sin 135° _______ PR

PR = 8.5 sin 135° __________ sin 17°

) 20.6

28. sin 140° _______ 9 = sin 20° ______

JL

JL = 9 sin 20° _______ sin 140°

) 4.8

29. sin 56° ______ 11.7

= sin 47° ______ EF

EF = 11.7 sin 47° _________ sin 56°

) 10.4

30. sin J ____ 61

= sin 80° ______ 100

m!J = s in (1 ( 61 sin 80° ________ 100

) ) 37°

31. sin X ____ 3.6

= sin 78° ______ 3.9

m!X = si n (1 ( 3.6 sin 78° _________ 3.9

) ) 65°

32. A B 2 = 1 3 2 + 5. 8 2 ( 2(13)(5.8) cos 67°) 143.7177

AB ) 12.0

33. 9. 7 2 = 14. 7 2 + 6. 8 2 ( 2(14.7)(6.8) cos Z 94.09 = 262.33 ( 199.92 cos Z(168.24 = (199.92 cos Z

m!Z = co s (1 ( 168.24 ______ 199.92

) ) 33°

34. 5 2 = 1 2 2 + 1 3 2 ( 2(12)(13) cos R 25 = 313 ( 312 cos R(288 = (312 cos R

m!R = co s (1 ( 288 ____ 312

) ) 23°

35. E F 2 = 8. 4 2 + 10. 6 2 ( 2(8.4)(10.6) cos 51°) 70.8506

EF ) 8.4

36. L M 2 = 10. 1 2 + 12. 9 2 ( 2(10.1)(12.9) cos 112°) 366.0350

LM ) 19.1

37. 5 2 = 1 3 2 + 1 4 2 ( 2(13)(14) cos G 25 = 365 ( 364 cos G(340 = (364 cos G

m!G = co s (1 ( 340 ____ 364

) ) 21°

38. A B 2 = 1 08 2 + 5 5 2 ( 2(108)(55) cos 59°) 8570.3477

AB ) 92.6

sin B ____ 55

) sin 59° ______ 92.576

m!B ) si n (1 ( 55 sin 59° ________ 92.576

) ) 31°

39. sin B ____ b = sin A ____ a

sin 22° ______ 3.2

= sin 74° ______ a

a = 3.2 sin 74° _________ sin 22°

) 8.2 cm

40. c 2 = a 2 + b 2 ( 2ab cos C= 9. 5 2 + 7. 1 2 ( 2(9.5)(7.1) cos 100°) 164.0851

c ) 12.8 in.

41. b 2 = a 2 + c 2 ( 2ac cos B 3. 1 2 = 2. 2 2 + 4 2 ( 2(2.2)(4) cos B 9.61 = 20.84 ( 17.6 cos B(11.23 = (17.6 cos B

m!B = co s (1 ( 11.23 _____ 17.6

) ) 50°

42. sin C ____ c = sin A ____ a

sin C ____ 8.4

= sin 45° ______ 10.3

m!C = si n (1 ( 8.4 sin 45° _________ 10.3

) ) 35°

43. No; 3 ! measures do not uniquely determine a $. There is not enough information to use either Law of Sines or Law of Cosines.

44. c 2 = a 2 + b 2 ( 2ab cos C= a 2 + b 2 ( 2ab cos 90°= a 2 + b 2

Law of Cosines simplifies to Pyth. Thm.

45. Let ! of turn be !1 and let !2 be opp. 6-km side. 6 2 = 3 2 + 4 2 ( 2(3)(4) cos !236 = 25 ( 24 cos !2 11 = (24 cos !2

m!2 = co s (1 (( 11 ___ 24

) m!1 = 180 ( m!2

= 180 ( co s (1 (( 11 ___ 24

) ) 63°


46. Step 1 Find 3rd side length. Think: Use Law of Cosines. x 2 = 5 2 + 9 2 ( 2(5)(9) cos 109° ) 135.3011x ) 11.6 cmStep 2 Find perimeter.P ) 5 + 9 + 11.6 ) 25.6 cm

47. Step 1 Find 2nd side length. Think: Use $ ! Sum Thm., Law of Sines.m!3 = 180 ( (93 + 24) = 63°

sin 63° ______ 16

= sin 24° ______ x

x = 16 sin 24° ________ sin 63°

) 7.30 ft

Step 2 Find 3rd side length. Think: Use Law of Sines.

sin 63° ______ 16

= sin 93° ______ y

y = 16 sin 93° ________ sin 63°

) 17.93 ft

Step 3 Find perimeter.P ) 16 + 7.30 + 17.93 ) 41.2 ft

48. Step 1 Find 2nd side length. Think: Use Law of Sines.

sin 45° ______ 7.3

= sin 115° _______ x

x = 7.3 sin 115° _________ sin 45°

) 9.36 in.

Step 2 Find 3rd side length. Think: Use $ ! Sum Thm., Law of Sines.m!3 = 180 ( (45 + 115) = 20°

sin 45° ______ 7.3

= sin 20° ______ y

y = 7.3 sin 20° _________ sin 45°

) 3.53 in.

Step 3 Find perimeter.P = 7.3 + 9.36 + 3.53 ) 20.2

49.

Figure shows two possible positions for C. Since ..

BC " ...

BC' , !C " !BC'C, so !C and !BC'A are supp.

sin C ____ 12

= sin 30° ______ 9

m!C = si n (1 ( 12sin 30° ________ 9

) ) 42°

m!BC'A = 180 ( m!C) 180 ( 42 ) 138°

50a. Think: Use $ ! Sum Thm.m!F + 51 + 38 = 180 m!F + 89 = 180 m!F = 91°

b. sin 91° ______ 18.3

= sin 38° ______ AF

AF = 18.3sin 38° _________ sin 91°

) 11 mi

sin 91° ______ 18.3

= sin 51° ______ BF

BF = 18.3sin 51° _________ sin 91°

) 14 mi

c. Dist. = speed · time AF = 150 t 1 BF = 150 t 2 BF ( AF = 150( t 1 ( t 2 )

t 1 ( t 2 = BF ( AF ________ 150

) 14.2 ( 11.3 __________ 150

) 0.193 h or 1.2 min

51. Given ! is opp. one of given sides, so use Law of Sines.

52. Given ! is included by given sides, so use Law of Cosines.

53. One of given # is opp. given side, so use Law of Sines.

54a. RS = & '''' 3 2 + 2 2 = & '' 13 ) 3.6

ST = & '''' 6 2 + 2 2 = & '' 40 = 2 & '' 10 ) 6.3

RT = & '''' 3 2 + 4 2 = 5

b. !R, because it is opp. the longest side.

c. S T 2 = R S 2 + R T 2 ( 2(RS)(RT) cos R

40 = 13 + 25 ( 2 ( & '' 13 ) (5) cos R 2 = (10 & '' 13 (cos R)

R = co s (1 (( 2 ______ 10 & '' 13

) ) 93°

55. B C 2 = 6.4 6 2 + 7.1 4 2 ( 2(6.46)(7.14) cos 104° ) 115.12197 BC ) 10.73 cmA B 2 = 3.8 6 2 + 7.1 4 2 ( 2(3.86)(7.14) cos 138° ) 106.84194 AB ) 10.34 cm

sin ABE _______ 3.86

) sin 138° _______ 10.34

m!ABE ) si n (1 ( 3.86 sin 138° ___________ 10.34

) ) 14.47°

sin EBC _______ 6.46

) sin 104° _______ 10.73

m!EBC ) si n (1 ( 6.46 sin 104° ___________ 10.73

) ) 35.74°

m!ABC = m!ABE + m!EBC) 14.47 + 35.74 ) 50°

56. A is incorrect; possible answer: the fraction on the right side of the proportion is incorrect.

It should be sin 70° ______ 12

= sin 85° ______ x , as in B.

57a. y 2 + h 2 b. b 2

c. a 2 = c 2 ( 2cx + x 2 + h 2

d. a 2 = c 2 + b 2 ( 2cx

e. b cos A f. Subst.

58. No; possible answer: to use Law of Sines, you need to know at least 1 side length and ! measure opp. that side.


TEST PREP, PAGE 558

59. AA B 2 = 1 2 2 + 1 4 2 ( 2(12)(14) cos 23° ) 32.71 AB ) 5.7 cmNearest given value is 5.5 cm.

60. H

61. C m!Y = 180 ( (25 + 135) = 20°

sin 20° ______ 100

= sin 25° ______ XY

XY = 100 sin 25° _________ sin 20°

) 124 m


62. A B 2 = A C 2 + B C 2 ( 2(AC)(BC) cos ACB(2 + 3 ) 2 = (2 + 4 ) 2 + (3 + 4 ) 2

( 2(2 + 4)(3 + 4) cos ACB 25 = 85 ( 84 cos ACB (60 = (84 cos ACB

m!ACB = co s (1 ( 60 ___ 84

) ) 44°

63. Let pts. be A((1, 1), B(1, 3), and C(3, 2)

AB = & '''' 2 2 + 2 2 = & ' 8 ; AC = & '''' 4 2 + 1 2 = & '' 17 ;

BC = & '''' 2 2 + 1 2 = & ' 5 B C 2 = A B 2 + A C 2 ( 2(AB)(AC) cos A 5 = 8 + 17 ( 2 ( & ' 8 ) ( & '' 17 ) cos A (20 = (4 & '' 34 (cos A)

m!A = co s (1 ( 20 _____ 4 & '' 34

) ) 31°

64. Let P be position of boat after 45 min = 0.75 h.Given information: AB = 5 mi, AP = (6 mi/h)(0.75 h) = 4.5 mi, !A = 180 ( 32 = 148°B P 2 = A B 2 + A P 2 ( 2(AB)(AP) cos A = 5 2 + 4. 5 2 ( 2(5)(4.5) cos 148° ) 83.4122 BP ) 9.1 mi


65. 3n 66. 2n + 1

67. 2n + 2 68. Alt. Ext. # Thm.

69. Alt. Int. # Thm. 70. Same-Side Int. # Thm.

71. Alt. Ext. # Thm. 72. !2

73. !1 74. !1

8-6 VECTORS, PAGES 559–567

CHECK IT OUT! PAGES 559–562 1a. Horiz. change along

0 u is (3 units.

Vert. change along 0 u is (4 units.

So component form of 0 u is 1(3, (42.

b. Horiz. change from L to M is 7 units.Vert. change from L to M is 1 unit.So component form of

3330 LM is 17, 12.

2. Step 1 Draw vector on a coord. plane. Use origin as initial pt. Then ((3, 1) is terminal pt.Step 2 Find magnitude. Use Dist. Formula.

41(3, 125 = & ''''''''' ((3 ( 0 ) 2 + (1 ( 0 ) 2 = & '' 10 ) 3.2

3. Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step 2 Find direction. Draw rt. $ABC as shown. !A is ! formed by vector and x-axis, and tan A = 3 __

7 . So

m!A = ta n (1 ( 3 __ 7 ) ) 23°

4a. 3330 PQ =

3330 RS (same magnitude and direction)

b. 3330 PQ 6

3330 RS and

3330 XY 6

3330 MN (same or opp. direction)

5. Step 1 Sketch vectors for kayaker and current.

Step 2 Write vector for kayaker in component form. It has magn. 4 mi/h and makes ! of 70° with x-axis. cos 70° = x __

4 , so x = 4 cos 70° ) 1.37

sin 70° = y __

4 , so y = 4 sin 70° ) 3.76

Kayaker’s vector is 11.37, 3.762.Step 3 Write vector for current in component form: 11, 02.Step 4 Find and sketch resultant vector

3330 AB .

Add components of kayaker’s vector and current’s vector.11.37, 3.762 + 11, 02 = 12.37, 3.762

Step 5 Find magn. and direction of resultant vector.Magn. of resultant vector is kayak’s actual speed.

412.37, 3.7625 = & '''''''''' (2.37 ( 0 ) 2 + (3.76 ( 0 ) 2 ) 4.4 mi/h! measure formed by resultant vector gives kayak’s actual direction.

tan A ) 3.76 ____ 2.37

, so m!A = ta n (1 ( 3.76 ____ 2.37

) ) 58°, or N 32° E.


THINK AND DISCUSS, PAGE 563 1. It does not have a direction. 2. Pyth. Thm. 3. Possible answer: Write each vector in component

form and add horizontal and vertical components. 4.


1. equal 2. parallel 3. magnitude 4. Horiz. change from A to C is 5 units.

Vert. change from A to C is 3 units.So component form of

3330 AC is 15, 32.

5. Horiz. change from M to N is 8 units.Vert. change from M to N is (8 units.So component form of

3330 MN is 18, (82.

6. Horiz. change from P to Q is 2 units.Vert. change from P to Q is 5 units.So component form of

3330 PQ is 12, 52.

7. Step 1 Draw vector on a coord. plane. Use origin as initial pt. Then (1, 4) is terminal pt.Step 2 Find magnitude. Use Dist. Formula.

411, 425 = & '''''''' (1 ( 0 ) 2 + (4 ( 0 ) 2 = & '' 17 ) 4.1

8. Step 1 Draw vector on a coord. plane. Use origin as initial pt. Then ((3, (2) is terminal pt.Step 2 Find magnitude. Use Dist. Formula.

41(3, (225 = & ''''''''' ((3 ( 0 ) 2 + ((2 ( 0 ) 2

= & '' 13 ) 3.6 9. Step 1 Draw vector

on a coord. plane. Use origin as initial pt. Then (5, (3) is terminal pt.Step 2 Find magnitude. Use Dist. Formula.

415, (325 = & ''''''''' (5 ( 0 ) 2 + ((3 ( 0 ) 2 = & '' 34 ) 5.8


4 . So

m!A = ta n (1 ( 6 __ 4 ) ) 56°


5 . So

m!A = ta n (1 ( 1 __ 5 ) ) 11°


6 . So

m!A = ta n (1 ( 3 __ 6 ) ) 27°.

13. 3330 CD =

3330 EF (same magn. and direction)

14. 3330 CD 6

3330 EF and

3330 AB 6

3330 GH (same or opp. direction)

15. 3330 RS =

3330 XY (same magn. and direction)

16. 3330 RS 6

3330 XY and

3330 MN 6

3330 PQ (same or opp. direction)

17. Step 1 Sketch vectors for 2 stages of hike.

Step 2 Write vector for 1st stage in component form. It has magn. 2 mi and makes ! of 50° with

x-axis. cos 50° = x __

2 , so x = 2 cos 50° ) 1.29

sin 50° = y __

2 , so y = 2 sin 50° ) 1.53

Vector for 1st stage is 11.29, 1.532.Step 3 Write vector for 2nd stage in component form: 13, 02.Step 4 Find and sketch resultant vector

3330 AB .

Add components of 1st- and 2nd-stage vectors.11.29, 1.532 + 13, 02 = 14.29, 1.532

Step 5 Find magn. and direction of resultant vector.Magn. of resultant vector is straight-line dist. to campsite. 414.29, 1.5325 = & '''''''''' (4.29 ( 0 ) 2 + (1.53 ( 0 ) 2 ) 4.6 mi! measure formed by resultant vector gives direction of hike.

tan A ) 1.53 ____ 4.29

, so m!A = ta n (1 ( 1.53 ____ 4.29

) ) 20°, or N 70° E.



18. 19, 22 19. 1(3.5, 5.52

20. 1(4, (42

21. Step 1 Draw vector on a coord. plane. Use origin as initial pt.Step 2 Find magnitude. Since vector is horiz.,

41(2, 025 = 4(25 = 2.0

22. Step 1 Draw vector on a coord. plane. Use origin as initial pt.Step 2 Find magnitude. Use Pyth. Thm.

411.5, 1.525 = & ''''' 1. 5 2 + 1. 5 2 = & '' 4.5 ) 2.1

23. Step 1 Draw vector on a coord. plane. Use origin as initial pt.Step 2 Find magnitude. Use Pyth. Thm.

412.5, (3.525 = & ''''' 2. 5 2 + 3. 5 2 = & '' 18.5 ) 4.3

24. Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step 2 Find direction.

m!A = ta n (1 ( 1.5 ___ 4 ) ) 21°


m!A = ta n (1 ( 2.5 ___ 3.5

) ) 36°


m!A = ta n (1 ( 5 __ 2 ) ) 68°

27. ,,- DE =

,,- LM 28. All 4 vectors are 6.

29. ,,- RS =

,,- UV

30. ,,- RS 6

,,- UV 6

,,- AB and

,,- CD 6

,,- XY

31. Step 1 Write airplane’s vector in component form. x = 200 cos 65° ) 84.524; y = 200 sin 65° ) 181.262Airplane’s vector is 184.524, 181.2622.Step 2 Write windspeed vector in component form. x = 40 cos 45° ) 28.284; y = (40 sin 45° ) (28.284Windspeed vector is 128.284, (28.2842.Step 3 Find resultant vector

3330 AB . Add components

of airplane’s and windspeed vectors.184.524, 181.2622 + 128.284, (28.2842 ) 1112.81, 152.982Step 4 Find magn. and direction of resultant vector.

41112.81, 152.9825 = & ''''''' 112.8 1 2 + 152.9 8 2 ) 190.1 km/h

m!A ) ta n (1 ( 152.98 ______ 112.81

) ) 54°, or N 36° E.

32. 11, 22 + 10, 62 = 11 + 0, 2 + 62 = 11, 82

33. 1(3, 42 + 15, (22 = 1(3 + 5, 4 + ((2)2 = 12, 22

34. 10, 12 + 17, 02 = 10 + 7, 1 + 02 = 17, 12

35. 18, 32 + 1(2, (12 = 18 + ((2), 3 + ((1)2 = 16, 22

36. Yes; possible answer: if you use head-to-tail method in both orders, you end up with a 7 and its diag. Resultant vector is the diag. See figures below.

37a. Let 8 ,- FG be a vert. line. Use Alt. Int # Thm.m!F = m!GFH + m!GFX

= 45 + 53 = 98°

b. H X 2 = 5 0 2 + 4 1 2 ( 2(50)(41) cos 98° ) 4751.6097 HX ) 68.9 mi/h

c. sin FHX _______ 41

) sin 98° ______ 68.9

m!FHX ) si n (1 ( 41 sin 98° ________ 68.9

) ) 36°

d. direction ) 45 + 36 = 81° E of N, or N 81° E

38. 115 cos 42°, 15 sin 42°2 ) 111.1, 10.02

39. 17.2 cos 9°, 7.2 sin 9°2 ) 17.1, 1.12

40. direction relative to x-axis = 90 ( 57 = 33°112.1 cos 33°, 12.1 sin 33°2 ) 110.1, 6.62

41. direction relative to x-axis = 90 ( 22 = 68°15.8 cos 68°, 5.8 sin 68°2 ) 12.2, 5.42

42a. 10 sin 45° ) 7.1 lb b. 10 sin 75° ) 9.7 lb

c. Taneka; she applies more vert. force.

43a. Prob. of 1 on 1st draw is 1 __ 4 ; prob. of then drawing

2 is 1 __ 3 . So Prob. (11, 22) = 1 __

4 · 1 __

3 = 1 ___

12 .

b. Prob. (vector 6 to 11, 22) = Prob. (11, 22 or 12, 42) = 1 ___

12 + 1 ___

12 = 1 __

6


44a. b. Estimates should be between 50° and 60°.

c. Answers should be close to 56°.

d. direction = ta n (1 ( 6 __ 4 ) ) 56°

e. Possible answer: Estimate is accurate only to within ) 5°; measurement is accurate to within 1 or 2°; calculation is accurate to nearest degree.

45. 4 0 u 5 = 445 = 4

direction of 0 u = 0°

46. 4 0 v 5 = 435 = 3

direction of 0 v = 90°

47. 4 0 w 5 = & '''' 2 2 + 3 2 = & '' 13 ) 3.6

direction of 0 w = ta n (1 ( 3 __

2 ) ) 56°

48. 4 0 z 5 = & '''' 4 2 + 1 2 = & '' 17 ) 4.1

direction of 0 z = ta n (1 ( 1 __

4 ) ) 14°

49. Pass pattern vectors are 10, 102 and 110, 02.Resultant vector is 10, 102 + 110, 02 = 110, 102.

Magn. of resultant is & '''' 1 0 2 + 1 0 2 = 10 & ' 2 ;

Direction of resultant is ta n (1 ( 10 ___ 10

) = 45°.

Jason’s move is equivalent to resultant.

50.–52. Possible answers given.

50. Think: Change sign of one component only.13, 62 has same magn. but different direction.Think: Multiply both components by the same factor.1(6, 122 has same direction but different magn.

51. 1(12, (52 has same magn. but different (opp.) direction.124, 102 has same direction but different magn.

52. 1(8, 112 has same magn. but different (opp.) direction.14, (5.52 has same direction but different magn.

53. 0 u +

0 v = 11 + 2.5, 2 + ((1)2 = 13.5, 12

4 0 u +

0 v 5 = & '''' 3. 5 2 + 1 2 = & ''' 13.25 ) 3.6

direction of 0 u +

0 v = ta n (1 ( 1 ___

3.5 ) ) 16°

54. 0 u +

0 v = 1(2 + 4.8, 7 + ((3.1)2 = 12.8, 3.92

4 0 u +

0 v 5 = & ''''' 2. 8 2 + 3. 9 2 = & ''' 23.05 ) 4.8

direction of 0 u +

0 v = ta n (1 ( 3.9 ___

2.8 ) ) 54°

55. 0 u +

0 v = 16 + ((2), 0 + 42 = 14, 42

4 0 u +

0 v 5 = & '''' 4 2 + 4 2 = 4 & ' 2 ) 5.7

direction of 0 u +

0 v = ta n (1 ( 4 __

4 ) = 45°

56. 0 u +

0 v = 1(1.2 + 5.2, 8 + ((2.1)2 = 14, 5.92

4 0 u +

0 v 5 = & '''' 4 2 + 5. 9 2 = & ''' 50.81 ) 7.1

direction of 0 u +

0 v = ta n (1 ( 5.9 ___

4 ) ) 56°

57a. 0 v = 11, 32; 2

0 v = 12, 62

b. 4 0 v 5 = & '''' 1 2 + 3 2 = & '' 10 ; 42

0 v 5 = & '''' 2 2 + 6 2 = 2 & '' 10

2 0 v is twice the magnitude of

0 v .

c. direction of 0 v = ta n (1 ( 3 __

1 ) = 72°

direction of 2 0 v = ta n (1 ( 6 __

2 ) = ta n (1 ( 3 __

1 ) = 72°

Directions are the same.

d. Multiply each component by k.

e. ( ,,- v = ((1)

,,- v = ((1)1x, y2

= 1((1)x, ((1)y2 = 1(x, (y2

58. If u > v, resultant points due west, with magn. u ( v. If v > u, resultant points due east, with magn. v ( u. If u = v, resultant is 10, 02.

59. A line seg. has magnitude (or length), but no direction. A ray is a part of a line that continues indefinitely in one direction. Thus it has direction and infinite magnitude. A vector has both direction and magnitude.

TEST PREP, PAGE 567

60. C 0 w = ((2)12, 12

61. Gta n (1 ( 9 __

7 ) ) 52°

62. C & '''' 5 2 + 1 1 2 = & '' 146 ) 12

63. 8.2 4 3330 AB 5 = & ''''''''' ((5 + 3 ) 2 + ((2 ( 6 ) 2 = & '' 68 ) 8.2


64. 1(2, 32 is in 2nd quadrant, so direction is between 90° and 180°.direction = 180 + ta n (1 ( 3 ___

(2 ) ) 124°

65. 1(4, 02 lies along negative x-axis, so direction = 180°.

66. 1(5, (32 is in 3rd quadrant, so direction is between 180° and 270°.direction = 180 + ta n (1 ( (3 ___

(5 ) ) 211°

67. Let 0 v = 1x, y2 be required velocity vector. Then

1x, y2 + 14, 02 = 110 cos 20°, 10 sin 20°2x + 4 = 10 cos 20° x = 10 cos 20° ( 4 ) 5.40 mi/hy = 10 sin 20° ) 3.42 mi/h

4 0 v 5 = & '''''' 5.4 0 2 + 3.4 2 2 ) 6.4 mi/h

bearing = ta n (1 ( 3.42 ____ 5.40

) ) 32°, or N 58° E

68. 0 v = 1x, y2 = 13 cos 60°, 3 cos 60°2 + 16, 02

+ 14 cos 40°, 4 sin 40°2x = 3 cos 60° + 6 + 4 cos 40° ) 10.56 kmy = 3 sin 60° + 0 + 4 sin 40° ) 5.17 km

4 0 v 5 = & '''''' 10.5 6 2 + 5.1 7 2 ) 11.8 km

bearing = ta n (1 ( 5.17 _____ 10.56

) = 26°, or N 64° E



69. 9

: ; x ( y = (5

y = 3x + 1

+ 9

: ; y = x + 5

y = 3x + 1

Lines intersect at (2, 7).

70. 9

: ; x ( 2y = 0

2y + x = 8

+ 9

: ; y = 0.5x

y = (0.5x + 4

Lines intersect at (4, 2).

71. 9

: ; x + y = 5

3y + 15 = 2x

+ 9

: ; y = (x + 5

y = 2 __

3 x ( 5

Lines intersect at (6, (1).

72. NP = 3JLPerim. = 3(12) = 36 cm

73. Area = (3 ) 2 (6) = 54 c m 2

74. B C 2 = 3. 5 2 + 4 2 ( 2(3.5)(4) cos 50° ) 10.2519 BC ) 3.2

75. sin B ____ 4

) sin 50° ______ 3.20

m!B ) si n (1 ( 4 sin 50° _______ 3.20

) ) 73°

76. m!C ) 180 ( (50 + 73) ) 57°


1. By Alt. Int. # Thm., dist. = 1500 ______ tan 14°

) 6016 ft.

2. By Alt. Int. # Thm., m!S = 57°H A 2 = 10 0 2 + 3 0 2 ( 2(100)(30) cos 57° ) 7632.1658 HA ) 87 mi/h

3. sin H ____ 30

) sin 57° ______ 87.4

m!H = si n (1 ( 30 sin 57° ________ 87.4

) )73°

So direction of 3330 HA is N 17° E.


1. By Alt. Int. # Thm., dist. = 1600 ______ tan 34°

) 2372 ft.

2. height = 6 tan 78° ) 28.2 m

3. sin A ____ 14

= sin 118° _______ 20

A = si n (1 ( 14 sin 118° _________ 20

) ) 38°

4. sin 41° ______ 7 = sin 84° ______

GH

GH = 7 sin 84° _______ sin 41°

) 10.6

5. m!X = 180 ( (92 + 62) = 26°

sin 26° ______ 8 = sin 92° ______

XZ

XZ = 8 sin 92° _______ sin 26°

) 18.2

6. U V 2 = 1 2 2 + 9 2 ( 2(12)(9) cos 35° ) 48.0632 UV ) 6.9

7. 4 2 = 5 2 + 6 2 ( 2(5)(6) cos F 16 = 61 ( 60 cos F (45 = (60 cos F

m!F = co s (1 ( 45 ___ 60

) ) 41°

8. Q S 2 = 10. 5 2 + 6 2 ( 2(10.5)(6) cos 39° ) 48.3296 QS ) 7.0

9. 413, 125 = & '''' 3 2 + 1 2 = & '' 10 ) 3.2

10. 41(2, (425 = & '''' 2 2 + 4 2 = & '' 20 ) 4.5

11. 410, 525 = 455 = 5

12. direction = ta n (1 ( 1 __ 2

) ) 27°


13. direction = ta n (1 ( 3 __ 5 )

) 31°

14. direction = ta n (1 ( 4 __ 4 )

= 45°

15. Let 1x, y2 be resultant vector.1x, y2 = 16 sin 32°, 6 cos 32°2 + 18, 02x = 6 sin 32° + 8 ) 11.18 kmy = 6 cos 32° ) 5.09 km

dist. = & '''''' 11.1 8 2 + 5.0 9 2 ) 12.3 km

direction = ta n (1 ( 5.09 _____ 11.18

) ) 24°, or N 66° E


VOCABULARY, PAGE 572 1. component form 2. equal vectors

3. geometric mean 4. angle of elevation

5. trigonometric ratio

LESSON 8-1, PAGE 572 6. $PRQ % $RSQ % $PSR

7. x 2 = ( 1 __ 4

) (100) = 25

x = & '' 25 = 5

8. x 2 = (3)(17) = 51 x = & '' 51

9. x 2 = (5)(7) = 35 x = & '' 35

y 2 = 5(5 + 7) = 60 y = & '' 60 = 2 & '' 15

z 2 = 7(5 + 7) = 84 z = & '' 84 = 2 & '' 21

10. 6 2 = (x)(12)36 = 12x x = 3

y 2 = (3)(3 + 12) = 45 y = & '' 45 = 3 & ' 5

z 2 = (12)(3 + 12) = 180 z = & '' 180 = 6 & ' 5

11. ( & ' 6 ) 2 = (1)(1 + x) 6 = 1 + x x = 5

z 2 = (5)(1 + 5) = 30 z = & '' 30

y 2 = (1)(5) = 5 y = & ' 5


12. sin 80° = 11 ___ UV

UV = 11 ______ sin 80°

) 11.17 m

13. cos 29° = PR ___ 7.2

PR = 7.2 cos 29° ) 6.30 m

14. cos 33° = XY ____ 12.3

XY = 12.3 cos 33° ) 10.32 cm

15. tan 47° = 1.4 ___ JL

JL = 1.4 ______ tan 47°

) 1.31 cm

LESSON 8-3, PAGE 573 16. m!C = 90 ( 22 = 68°

AB = 5.2 cos 22° ) 4.82AC = 5.2 sin 22° ) 1.95

17. m!H = ta n (1 ( 4.7 ___ 3.5

) ) 53°

m!G ) 90 ( 53 ) 37°

HG = & ''''' 3. 5 2 + 4. 7 2 ) 5.86

18. m!S = 90 ( 50 = 40°

RS = 32.5 ______ sin 50°

) 42.43

RT = 32.5 ______ tan 50°

) 27.27

19. m!Q = ta n (1 ( 8.6 ___ 9.9

) ) 41°

m!N ) 90 ( 41 ) 49°

QN = & ''''' 9. 9 2 + 8. 6 2 ) 13.11

LESSON 8-4, PAGE 574 20. ! of depression 21. ! of elevation

22. height = 5.1 tan 82° ) 36 ft

23. horiz. dist. = 32 _____ tan 4°

) 458 m


24. sin Z ____ 4 = sin 40° ______

7

m!Z = si n (1 ( 4 sin 40° _______ 7 )

) 22°

25. sin 23° ______ 16

= sin 130° _______ MN

MN = 16 sin 130° _________ sin 23°

) 31.4

26. E F 2 = 1 4 2 + 1 2 2 + 2(14)(12) cos 101° ) 404.1118 EF ) 20.1

27. 1 0 2 = 6 2 + 1 2 2 ( 2(6)(12) cos Q 100 = 180 ( 144 cos Q (80 = (144 cos Q

m!Q = co s (1 ( 80 ____ 144

) ) 56°


3330 AB = 1(2 ( 5, 3 ( 12 = 1(7, 22

29. 3330 MN = 1(1 ( ((2), (2 ( 42 = 11, (62

30. 3330 RS = 1(2, (52

31. 41(5, (325 = & ''' 5 2 + 3 2 = & '' 34 ) 5.8


32. 41(2, 025 = 425 = 2

33. 414, (425 = & ''' 4 2 + 4 2 = & '' 32 ) 5.7

34. direction = ta n (1 ( 5 __ 4 )

= 51°

35. direction = ta n (1 ( 2 __ 7 )

= 16°

36. plane’s vector = 1600 cos 35°, 600 sin 35°2crosswind vector = 150, 02resultant vector = 1600 cos 35° + 50, 600 sin 35°2

) 1541.49, 344.152

speed ) & '''''''' 541.4 9 2 + 344.1 5 2 ) 641.6 mi/h

direction ) ta n (1 ( 344.15 ______ 541.49

) ) 32°, or N 58° E


1. 4 2 = (x)(8)16 = 8x x = 2

z 2 = 2(2 + 8) = 20 z = & '' 20 = 2 & ' 5

y 2 = 8(2 + 8) = 80 y = & '' 80 = 4 & ' 5

2. x 2 = (6)(12) = 72x = & '' 72 = 6 & ' 2

z 2 = 6(6 + 12) = 108 z = & '' 108 = 6 & ' 3

y 2 = 12(6 + 12) = 216 y = & '' 216 = 6 & ' 6

3. (2 & '' 30 ) 2 = 10(10 + x) 120 = 100 + 10x 20 = 10x x = 2

z 2 = 2(2 + 10) = 24 z = & '' 24 = 2 & ' 6

y 2 = (2)(10) = 20 y = & '' 20 = 2 & ' 5

4. Let 30°-60°-90° $ have sides x, x & ' 3 , 2x.cos 60° = x ___

2x = 1 __

2

5. Let 45°-45°-90° $ have sides s, s, s & ' 2 .sin 45° = s ____

s & ' 2 =

& ' 2 ___ 2

6. tan 60° = x & ' 3 ____ x = & ' 3 7. PR = 4.5 sin 18° ) 1.39 m

8. AB = 9 ______ cos 51°

) 14.30 cm

9. FG = 6.1 ______ tan 34°

) 9.04 in.

10. m! = ta n (1 ( 3.5 ___ 10

) ) 19°

11. horiz. dist. = 910 ______ tan 61°

) 504 ft

12. sin B ____ 4 = sin 85° ______

9

m!B = si n (1 ( 4 sin 85° _______ 9 )

) 26°

13. sin 35° ______ 11

= sin 108° _______ RS

RS = 11 sin 108° _________ sin 35°

) 18.2

14. 7 2 = 1 0 2 + 1 5 2 ( 2(10)(15) cos M 49 = 325 ( 300 cos M(276 = (300 cos M

m!M = co s (1 ( 276 ____ 300

) ) 23°

15. 411, 325 = & ''' 1 2 + 3 2 = & '' 10 ) 3.2

16. 41(4, 125 = & ''' 4 2 + 1 2 = & '' 17 ) 4.1

17. 412, (325 = & ''' 2 2 + 3 2 = & '' 13 ) 3.6

18. direction = ta n (1 ( 5 __ 3

) = 59°

19. direction = ta n (1 ( 1 __ 4

) = 14°

20. boat’s vector = 13.5 cos 50°, 3.5 sin 50°2current’s vector = 12, 02resultant vector = 13.5 cos 50° + 2, 3.5 sin 50°2

) 14.25, 2.682

speed ) & '''''' 4.2 5 2 + 2.6 8 2 ) 5.0 mi/h

direction ) ta n (1 ( 2.68 ____ 4.25

) ) 32°, or N 58° E



1. A(x + 4y = 12 4y = x + 12 y = 1 __

4 x + 3

tan P = slope = 1 __ 4

m!P = ta n (1 ( 1 __ 4 ) ) 14°

2. Dcos F = EF ___

DF = 36 ___

39 = 12 ___

13

3. C

sin 61° ______ 8 = sin 100° _______ x

x = 8 sin 100° ________ sin 61°

) 9 cm

4. Dswimmer’s vector = 10, 22current vector = 17, 02resultant vector = 17, 22

speed = & '''' 7 2 + 2 2 = & '' 53 ) 7.3 m/s

5. B 1 6 2 = 1 7 2 + 1 7 2 ( 2(17)(17) cos A 256 = 578 ( 578 cos A(322 = (578 cos A

m!A = co s (1 ( 322 ____ 578

) ) 56.1°


Solutions KeyExtending Perimeter, Circumference, and Area9

CHAPTER


1. C 2. D

3. E 4. A

5. 12 mi = 12 · 1760 yd = 21,120 yd

6. 7.3 km = 7.3 · 1000 m = 7300 m

7. 6 in = (6 ÷ 12) ft = 0.5 ft

8. 15 m = 15 · 1000 mm = 15,000 mm

9. x 2 = 3. 1 2 + 5. 8 2 x 2 = 43.25 x = ! """ 43.25 # 6.6 in.

10. 1 0 2 = x 2 + 8 2 x 2 = 100 $ 64

x 2 = 36 x = 6 cm

11. 9. 9 2 = x 2 + 4. 3 2 x 2 = 98.01 $ 18.49

x 2 = 79.52 x = ! """ 79.52 # 8.9 m

12. 5 __ 8 in.; 1.5 cm 13. 1 1 __

8 in.; 3 cm

14. 1 3 __ 4 in.; 4.5 cm

15. A = 1 __ 2 bh

2A = bh

b = 2A ___ h

16. P = 2b + 2hP $ 2b = 2h h = P $ 2b ______

2

17. A = 1 __ 2 ( b 1 + b 2 )h

2A ___ h = b 1 + b 2

b 1 = 2A ___ h

$ b 2

18. A = 1 __ 2 d 1 d 2

2A = d 1 d 2

d 1 = 2A ___ d 2

CONNECTING GEOMETRY TO ALGEBRA: LITERAL EQUATIONS, PAGE 588

TRY THIS, PAGE 588 1. P = 2! + 2w

P $ 2! = 2w P $ 2! ______

2 = w

w = P $ 2! ______ 2

= 24 $ 2(2)

________ 2 = 10 cm

w = P $ 2! ______ 2

= 24 $ 2(3)

________ 2 = 9 cm

w = P $ 2! ______ 2

= 24 $ 2(4)

________ 2 = 8 cm

w = P $ 2! ______ 2

= 24 $ 2(6)

________ 2 = 6 cm

w = P $ 2! ______ 2

= 24 $ 2(8)

________ 2 = 4 cm

2. a 2 + b 2 = c 2 a 2 = c 2 $ b 2

a = ! """ c 2 $ b 2

a = ! """ c 2 $ b 2 = ! """" 65 2 $ 16 2 = ! "" 3969 = 63 ft

a = ! """ c 2 $ b 2 = ! """" 65 2 $ 25 2 = ! "" 3600 = 60 ft

a = ! """ c 2 $ b 2 = ! """" 65 2 $ 33 2 = ! "" 3136 = 56 ft

a = ! """ c 2 $ b 2 = ! """" 65 2 $ 39 2 = ! "" 2704 = 52 ft

3. P = a + b + c112 = a + b + c a = 112 $ b $ c

a b c

112 $ 48 $ 35 = 29 48 35

112 $ 36 $ 36 = 40 36 36

112 $ 14 $ 50 = 48 14 50

9-1 DEVELOPING FORMULAS FOR TRIANGLES AND QUADRILATERALS, PAGES 589–597


1. A = bh28 = 56b b = 0.5 yd

2. b 2 + h 2 = c 2 b 2 + 12 2 = 20 2 b 2 = 256 = 1 6 2 b = 16 mA = 1 __

2 bh

= 1 __ 2 (16)(12) = 96 m 2

3. A = 1 __ 2 d 1 d 2

12xy = 1 __ 2 (3x) d 2

24xy = (3x) d 2 d 2 = 8y m

4. P = 4 + 2 ! " 2 + 2 ! " 2 = (4 + 4 ! " 2 ) cm

A = 1 __ 2 bh

= 1 __ 2 (4)(2) = 4 cm 2

THINK AND DISCUSS, PAGE 593 1. Because 2 congruent copies of the triangle fit

together to form a parallelogram with same base and height as the triangle

2. The area of a rectangle is the base times the height, and area of a trapezoid is the average of the bases times the height.

3.




1. A = bh= (12)(10) = 120 c m 2

2. A = bh10 x 2 = (2x)h h = 5x ft

3. A = s 2 169 = s 2 s = 13P = 4s

= 4(13) = 52 cm

4. A = 1 __ 2 ( b 1 + b 2 )h

= 1 __ 2 (9 + 15)(20)

= 240 m 2

5. A = 1 __ 2

bh

58.5 = 1 __ 2

b(9) = 4.5b

b = 13 in.

6. A = 1 __ 2

( b 1 + b 2 )h

48x + 68 = 1 __ 2

( b 1 + 9x + 12)(8)

48x + 68 = 4 b 1 + 36x + 4812x + 20 = 4 b 1 b 1 = (3x + 5) in.

7. ( d 1 __

2 ) 2 + ( d 2

__ 2 ) 2 = c 2

d 1 2 + d 2 2 = 4 c 2

d 1 2 + 1 4 2 = 4(25 ) 2 d 1 2 = 2500 $ 196 = 2304 d 1 = 48 in.A = 1 __

2 d 1 d 2

= 1 __ 2 (48)(14) = 336 i n. 2

8. A = 1 __ 2 d 1 d 2

187.5 = 1 __ 2 (15) d 2

= 7.5 d 2 d 2 = 25 m

9. A = 1 __ 2 d 1 d 2

12 x 2 y 3 = 1 __ 2 (3xy) d 2

24 x 2 y 3 = (3xy) d 2

d 2 = 8x y 2 cm

10. Parallelogram: A = bh = ! " 2 (1.5 ! " 2 ) ft 2 = 3 ft 2 Rectangles: A = bh = 1(2) = 2 ft 2 Triangles: A = 1 __

2 bh = 1 __

2 (1)(2) = 1 ft 2

Trapezoids: A = 1 __ 2 ( b 1 + b 2 )h = 1 __

2 (1 + 2)(1) = 1.5 ft 2


11. A = bh7.5 = 6h h = 1.25 m

12. P = 2! + 2w= 2(x + 2) + 2(x $ 1)= 2x + 4 + 2x $ 2= (4x + 2) in.

13. A = bh= (3x + 5)(7x $ 1)

= (21 x 2 + 32x $ 5) f t 2

15. A = 1 __ 2 ( b 1 + b 2 )h

280 = 1 __ 2 (8 + 20)h = 14h

h = 20 cm

14. Let b = x + y x 2 + 1 5 2 = 1 7 2 x 2 = 64 x = 8 in. y 2 + 1 5 2 = 2 5 2 y 2 = 400 y = 20 in.A = 1 __

2 bh

= 1 __ 2 (8 + 20)(15)

= 210 i n. 2

16. A = 1 __ 2 bh

= 1 __ 2 (x + 1)(8x)

= (4 x 2 + 4x) f t 2

17. Right triangles are 30°-60°-90° triangles; so d 1 = 7 + 21 = 28 in.A = 1 __

2 d 1 d 2

= 1 __ 2 (28) (2 (7 ! " 3 ) )

= 196 ! " 3 i n. 2

18. A = 1 __ 2 d 1 d 2

3 x 2 + 6x = 1 __ 2 (x + 2) d 2

3x(x + 2) = 1 __ 2 (x + 2) d 2

d 2 = 6x m

19. A = 1 __ 2 d 1 d 2

= 1 __ 2 (6x + 5)(4x + 8)

= (12 x 2 + 34x + 20) ft

20. A = bh = 6(3) = 18 in. 2

21. A = 1 __ 2 bh = 1 __

2 (3)(3) = 4.5 i n. 2

22. A = 1 __ 2 ( b 1 + b 2 )h = 1 __

2 (3 + 6)(3) = 13.5 i n. 2

23. h = 6 ! " 3 ____ 2 = 3 ! " 3

A = bh= (10)3 ! " 3 = 30 ! " 3 c m 2

24. b 2 + 5 2 = 13 2 b = 12 mA = 1 __

2 bh

= 1 __ 2 (5)(12) = 30 m 2

25. h 2 + 7 2 = 2 5 2 h = 24(b $ 7 ) 2 + 2 4 2 = 3 0 2 b $ 7 = 18 b = 25A = 1 __

2 bh

= 1 __ 2 (25)(24) = 300 i n. 2

26. b = x, h = x ! " 3 ____ 2

A = 1 __ 2 (x) x ! " 3 ____

2 = x 2 ! " 3 _____

4

27. b = x ! " 3 , h = x

A = 1 __ 2 (x ! " 3 ) x = x 2 ! " 3 _____

2

28. b = h = xA = 1 __

2 (x)(x) = x 2 __

2

29a. h = 36 ! " 3 _____ 2 # 31.2 in.


b. A # 1 __ 2 (36)(31.18)

# 561.2 i n. 2

c. Material left is# 3 6 2 $ 561.2 # 734.8 i n. 2

Base b Height h Area A Perimeter P

30. 12 16 12(16) = 192

2(12) + 2(16) = 56

31. 17 8 136 = 17(8)

2(17) + 2(8)= 50

32. 14 11 14(11)= 154

50 = 2(14) + 2(11)

33. 9 24 216 = 9(24)

66= 2(9) + 2(24)

34. P = 2b + 2h72 = 2(3h) + 2h = 8h h = 9 in.A = bh

= (3(9))9 = 243 i n. 2

35. A = 1 __ 2 bh

50 = 1 __ 2 (4h)h = 2 h 2

25 = h 2 h = 5 cm

36. P = s 1 + b 1 + s 2 + b 2 40 = s + 11 + s + 1910 = 2s s = 5 ftLet 19 = 11 + 2x, so x = 4 ft. x 2 + h 2 = s 2 16 + h 2 = 25 h = 3 ftA = 1 __

2 ( b 1 + b 2 )h

= 1 __ 2 (11 + 19)(3) = 45 f t 2

37. 1 y d 2 = (3 ft ) 2 = 9 f t 2

38. 1 m 2 = (100 cm ) 2 = 10,000 c m 2

39. 1 c m 2 = (10 mm ) 2 = 100 m m 2

40. 1 m i 2 = (5280(12) in. ) 2 = 4,014,489,600 i n. 2

41. A = 1 __ 2 (3)(8) = 12 y d 2

= 12(9 f t 2 ) = 108 f t 2

42. A = 1 __ 2 (500)(800) = 200,000 y d 2

= 200,000 y d 2

___________ (1760 yd/mi ) 2

# 0.065 m i 2

43a. A = (a + b)

______ 2 (a + b)

= 1 __ 2 (a + b ) 2

b. 1 __ 2 ab; 1 __

2 ab; 1 __

2 c 2

c. 1 __ 2 (a + b ) 2 = 1 __

2 ab + 1 __

2 ab + 1 __

2 c 2

(a + b ) 2 = 2ab + c 2 a 2 + 2ab + b 2 = 2ab + c 2 a 2 + b 2 = c 2

44. The area of the large square is (b + h ) 2 . The area of the medium square is b 2 and the area of the small square is h 2 . The total area is the sum of the areas. Let A represent area of the rectangle.

(b + h ) 2 = b 2 + h 2 + 2A b 2 + 2bh + h 2 = b 2 + h 2 + 2A 2bh = 2A A = bh

45. Opposite sides of a parallelogram are congruent, so the diagonal divides the parallelogram into 2 congruent triangles. Let A represent the area of each triangle. The sum of the triangles’ areas is the area of the parallelogram.

2A = bh A = 1 __

2 bh

46. Both triangles have height h. The area of the upper triangle is 1 _

2 b 1 h and the area of the lower triangle

is 1 _ 2 b 2 h. The area of the trapezoid is the sum of

the areas of the triangles.

A = 1 __ 2 b 1 h + 1 __

2 b 2 h

= 1 __ 2 ( b 1 + b 2 )h

47a. Possible answers:

A: (2.1)(2.0) = 4.2 c m 2 B: (1.2)(3.2) = 3.8 c m 2 C: (2.7)(1.6) = 4.3 c m 2

b. C has greatest area.

48. w = 40A + 20 d 1 + 20 d 1 = 40 1 __

2 d 1 d 2 + 20( d 1 + d 2 )

= 20( d 1 d 2 + d 1 + d 2 )= 20(0.90(0.80) + 0.90 + 0.80) = 48.4 g

49. There are 9 tiles per 1-ft square. So Tom needs 1.15[12(18)(9)] # 2236 tiles, or 23 cases.

50. From the given measurements, the area is 12 c m 2 . If the actual measurements were 5.9 cm and 1.9 cm, the area would be 11.21 c m 2 . If the actual measurements were 6.1 cm and 2.1 cm, the area would be 12.81 c m 2 . The maximum error is 0.81 c m 2 .

51. A square is a parallelogram and a rectangle in

which b = h = s, so A = bh = (s) (s) = s 2 . A square is a rhombus in which d 1 = d 2 = s ! " 2 ,

so A = 1 _ 2 (s ! " 2 ) (s ! " 2 ) = 1 _

2 s 2 (2) = s 2 .


52. B 53. H

54. C 1 __ 2 (16) 1 __

2 (18) = 8(9)

= 72

55. HJK = ! """" 6 2 + 1 0 2

= ! "" 136 # 11.7 cm

56. $1309C = 2.75A

= 2.75bh= 2.75(28)(17) = 1309



57. A = 3h = 6(2) 3h = 12 h = 4 in.

58. A = 25h = 15(20) 25h = 300 h = 12 m

59. A = 42 x 2 + 51x + 15 = (7x + 5)(6x + 3)

P = 26x + 16 = 2(13x + 8) = 2(7x + 5) + 2(6x + 3)

b = (7x + 5) cm

h = (6x + 3) cm

60. Let ABCD be a quadrilateral with perpendicular diagonals.

%% AC and

%% BD that intersect at E. Let

d 1 = AC, d 2 = BD, and x = BE. &ABC has b = d 1 and h = x, so A = 1 _

2 d 1 x. &ADC

has b = d 1 and h = d 2 $ x, so A = 1 _ 2 d 1 ( d 2 $ x).

ABCD has area A = 1 _ 2 d 1 x + 1 _

2 d 1 ( d 2 $ x)

= 1 _ 2 d 1 (x + d 2 $ x) = 1 _

2 d 1 d 2 .

61a. 2x + 2y = 24 2y = 24 $ 2x y = 12 $ xA = xy = x(12 $ x)

b.

D: {x | 0 < x < 12}R: {y | 0 < y ' 36}

c. Area is maximized when x = 6; therefore, the dimensions are 6 ft by 6 ft.

d. Solve the area formula for y and substitute the expression into the perimeter formula. Graph, and find the minimum value.


62. $4 ' x ' 6$4 $ 3 ' x $ 3 ' 6 $ 3 $7 ' y ' 3

63. $2 ' x ' 2

0 ' x 2 ' 4

$4 ' $ x 2 ' 0

$4 + 2 ' $ x 2 + 2 ' 0 + 2 $2 ' y ' 2

64. P = 2(x + 2) + 2(2x)= 2x + 4 + 4x= 6x + 4

A = (x + 2)(2x)= 2 x 2 + 4x

65. P = x + 7 + (x + 1)= 2x + 8

A = 1 __ 2 (x)(7)

= 7x ___ 2

66. ((() LM = (5 $ 4, 10 $ 3) = *1, 7+

67. ((() ST = (4 $ ($2), 6 $ ($2)) = *6, 8+

9-2 GEOMETRY LAB: DEVELOP !, PAGES 598–599

ACTIVITY 1, TRY THIS, PAGE 598 1. No; possible answer: all circles are similar, so

the ratio of circumference to diameter is always the same.

2. Solving the relationship for C gives a formula in terms of d and ".

3. If the circumference is n", then the diameter is n. Check students’ measurements.

ACTIVITY 2, TRY THIS, PAGE 599 4. Outer hexagon: let r and s represent the radius and

hexagon side length respectively. Then r = s ! " 3 ___ 2 = 1,

so s = 2 ! " 3 ___ 3 and P = 6s = 4 ! " 3 .

Inner hexagon: s = r = 1, so P = 66 < C < 4 ! " 3 6 < "(2) < 4 ! " 3 3 < " < 3.46

5. Possible answer: The second inequality values are closer together. With more sides, the values would be even closer together. You can estimate " by averaging the upper and lower values.

6. Possible answer: Average the areas of the inscribed and circumscribed polygons.

9-2 DEVELOPING FORMULAS FOR CIRCLES AND REGULAR POLYGONS, PAGES 600–605

CHECK IT OUT! PAGES 601–602 1. Step 1 Use given circumference to solve for r.

C = 2"r(4x $ 6)" = 2"r 2x $ 3 = rStep 2 Use expression for r to find area.A = " r 2 A = "(2x $ 3 ) 2 A = (4 x 2 $ 12x + 9)" m 2

2. C = "(10) # 31.4 in.; C = "(12) # 37.7 in.; C = "(14) # 44.0 in.

3. Step 1 Find perimeter.P = 8(4) = 32 cmStep 2 Use tangent ratio to find apothem.tan 22.5° = 2 __ a

a = 2 ________ tan 22.5°

Step 3 Use apothem and perimeter to find area.

A = 1 __ 2 aP

A = 1 __ 2 ( 2 ________

tan 22.5° ) (32)

A # 77.3 c m 2


THINK AND DISCUSS, PAGE 602 1. Circumference of a circle is " times diameter.

2. Divide 360° by n.

3.



1. Draw a segment perpendicular to a side with one endpoint at the center. The apothem is 1 _

2 s.

2. C = "d = " ( 10 ___ " ) = 10 cm

3. A = " r 2 = "(3x ) 2 = 9 x 2 " i n. 2

4. Step 1 Use given area to solve for r. A = " r 2 36" = " r 2 36 = r 2 r = 6 in.Step 2 Use value of r to find circumference.C = 2"rC = 2"(6) = 12" in.

5. A = " ( 8 __ 2 ) 2 # 50.3 i n. 2 ; A = " ( 10 ___

2 ) 2 # 78.5 i n. 2 ;

A = " ( 12 ___ 2 ) 2 # 113.1 i n. 2

6. Step 1 Find perimeter.P = 6(10) = 60 in.Step 2 Use properties of 30°-60°-90° & to find apothem.a = 5 ! " 3 in.Step 3 Use apothem and perimeter to find area.A = 1 __

2 aP

A = 1 __ 2 (5 ! " 3 ) (60) # 259.8 in. 2

7. Step 1 Find perimeter.P = 7(3) = 21 cmStep 2 Use tangent ratio to find apothem.

tan ( 360 ____ 14

) ° = 1.5 ___ a

a = 1.5 ________ tan ( 360 ___

14 ) °

Step 3 Use apothem and perimeter to find area.A = 1 __

2 aP

A = 1 __ 2 ( 1.5 ________

tan ( 360 ___ 14

) ° ) (21) # 32.7 c m 2

8. Step 1 Find side length. s = 2a ! " 3 = 4 ! " 3 ftStep 2 Find perimeter.P = 3s = 34 ! " 3 = 12 ! " 3 ftStep 3 Find area.A = 1 __

2 aP

A = 1 __ 2 (2) (12 ! " 3 ) # 20.8 ft 2

9. Step 1 Find perimeter.P = 12(5) = 60 mStep 2 Use tangent ratio to find apothem.

tan 15° = 2.5 ___ a

a = 2.5 ______ tan 15°

Step 3 Use apothem and perimeter to find area.A = 1 __

2 aP

A = 1 __ 2 ( 2.5 ______

tan 15° ) (60) # 279.9 m 2


10. A = "(7 ) 2 = 49" y d 2 11. C = "(5) = 5" m

12. C = "d10 = "d

d = 10 ___ " ft

13. A = " ( 35 ___ 2 ) 2 # 962.1 f t 2 ;

A = " ( 50 ___ 2 ) 2 # 1963.5 f t 2 ;

A = " ( 66 ___ 2 ) 2 # 3421.2 f t 2

14. A = (2(12) ) 2 = 576 c m 2

15. Step 1 Find side length and perimeter.

tan 22.5° = s _ 2 __ a = s __

4

s = 4 tan 22.5°P = 8s = 32 tan 22.5°Step 2 Find area.A = 1 __

2 aP

= 1 __ 2 (2)(32 tan 22.5°) # 13.3 f t 2

16. Step 1 Find side length and apothem. P = 9s 144 = 9s s = 16

tan 20° = 8 __ a

a = 8 ______ tan 20°

Step 2 Find area.A = 1 __

2 aP

= 1 __ 2 ( 8 ______

tan 20° ) (144) # 1582.5 i n. 2


17. Step 1 Find side length and perimeter.

tan 36° = s _ 2 __ a = s __

4

s = 4 tan 36°P = 5s = 20 tan 36°Step 2 Find area.A = 1 __

2 aP

= 1 __ 2 (2)(20 tan 36°) # 14.5 f t 2

18. 360° ____ 3 = 120° 19. 360° ____

4 = 90°

20. 360° ____ 5 = 72° 21. 360° ____

6 = 60°

22. 360° ____ 7 # 51.4° 23. 360° ____

8 = 45°

24. 360° ____ 9 = 40° 25. 360° ____

10 = 36°

26. s = 2a tan 30° = 28 tan 30°P = 6s = 168 tan 30°A = 1 __

2 (14)168 tan 30° # 679.0 in. 2

27. s = 2a tan 25.71° = 10 tan 25.71°P = 7s = 70 tan 25.71°A = 1 __

2 (5)70 tan 25.71° # 84.3 cm 2

28. a = s _______ 2 tan 20°

= 3 ______ tan 20°

P = 9s = 54

A = 1 __ 2 ( 3 ______

tan 20° ) (54) # 222.5 in. 2

29. s = 2a tan 60° = 6 tan 60°P = 3s = 18 tan 60°A = 1 __

2 (3)18 tan 60° # 46.8 m 2

30. a = r cos 22.5° = 2 cos 22.5°s = 2r sin 22.5° = 4 sin 22.5°P = 8s = 32 sin 22.5°A = 1 __

2 (2 cos 22.5°)(32 sin 22.5°) # 11.3 cm 2

31. a = s __________ 2 tan 25.71°

= 2.5 _________ tan 25.71°

P = 7s = 35

A = 1 __ 2 ( 2.5 _________

tan 25.71° ) (35) # 90.8 ft 2

32. C = 100 = 2"(r + 0.5)

r = 50 ___ " $ 0.5

a = r __ w = 50 __ " $ 0.5

_______ 0.2

# 77 yr

33. A is incorrect because the diameter, instead of the radius, is used to find the area.

Diam. d Radius r Area A Circ. C

34. 6 3 "(3 ) 2 = 9! 6!

35. 20 " # ! ______ ! ! "" 100 ____ "

= 10 " # ! ______ !

100 20 " # !

36. 34 17 289! 34!

37. 36 18 324! 36"

38. A garden = A circle $ A gazebo

= " r 2 $ 1 __ 2 aP

= "(10 + 6 ) 2 $ 1 __ 2 ( 6 ! " 3 ____

2 ) (36) # 711 f t 2

39a. Step 1 Find side length and perimeter.

30 = s ! " 2 ____ 2 + s + s ! " 2 ____

2

s = 30 _______ 1 + ! " 2

= 30 ( ! " 2 $ 1) P = 8s = 240 ( ! " 2 $ 1)

Step 2 Find apothem.2a = 30 a = 15Step 3 Find area.

A = 1 __ 2 aP

= 1 __ 2 (15)240 ( ! " 2 $ 1) # 745.6 in. 2

b. Step 1 Find side length and perimeter. s = 36 _______

1 + ! " 2 = 36 ( ! " 2 $ 1)

P = 8s = 288 ( ! " 2 $ 1) Step 2 Find apothem.2a = 36 a = 18Step 3 Find area.

A = 1 __ 2 aP

= 1 __ 2 (18)288 ( ! " 2 $ 1) # 1073.6 in. 2

c. s, P, and a are all proportional to the given sign height. Therefore the area is proportional to the square of the height.

Percent increase = A(36)

_____ A(30)

$ 100%

= ( 36 ___ 30

) 2 $ 100% = 44%

40. C = 1 = "d d = 1 __ " # 0.318 m

41. Possible answer: The circular table would fit at least as many people as the rectangle table. At the rectangle table, 2 people would fit at each of the 4-ft sides and 3 people would fit at each of the 6-ft sides, for a total of 10 people. Each person would have 2 ft of space. Between 10 and 12 people would fit around the circular table, with about 1 ft 9 in of space per person.

42. The circumference is proportional to the diameter. So, the circumference of the largest circle is the sum of the circumferences of the 4 smaller circles.

TEST PREP, PAGE 605

43. Bs (1 + ! " 2 ) = 2(6)

s = 12 ( ! " 2 $ 1) P = 8s

= 96 ( ! " 2 $ 1) # 40 cm

44. F 45. B



46. C large $ C small = 2" r large $ 2" r small

= 2"( r large $ r small )

= 2"(5) = 10" units

47. C = 2"r r = C ___

2"

A = " r 2

= " ( C ___ 2"

) 2 = C 2 ___ 4"

48. As n gets very large, the regular n-gon begins to look like a circle with C # P and r # a. The area

of the polygon is A = 1 __ 2 aP, which is close to 1 __

2 rC

or 1 __ 2 r(2"r) = " r 2 .


49. m = 17 $ 2 ______ 10 $ 5

= 3

y = 3x $ 13

50. m = $1 $ 2 ________ 0 $ ($3)

= $1

y = $x $ 1

51. By Isosceles Triangle Theorem, ,A - ,Cm,B = 180 $ (m,A + m,C)

= 180 $ 2(28) = 124°

52. AB = BC 6x = 3x + 15 3x = 15 x = 5AB = 6(5) = 30

53. A = 1 __ 2 d 1 d 2

14 = 1 __ 2 (20) d 2

d 2 = 1.4 cm

54. A = 1 __ 2 ( b 1 + b 2 )h

= 1 __ 2 (3 + 6)(4) = 18 yd 2

9-3 COMPOSITE FIGURES, PAGES 606–612

CHECK IT OUT! PAGES 606–608 1. Divide the figure into a rectangle and a triangle.

Area of rectangle (on left): A = bh = (37.5)(22.5) = 843.75 m 2

Area of triangle (on right): b = 75 $ 37.5 = 37.5 m

h = ! """""" 62. 5 2 $ 37. 5 2 = 50 mA = 1 __

2 bh = 1 __

2 (37.5)(50) = 937.5 m 2

Shaded area:A = 843.75 + 937.5 = 1781.25 m 2

2. Area of circle:A = " r 2 = 9" in. 2 Area of square:A = s 2 = (3 ! " 2 ) 2 = 18 in. 2 Shaded area:A = 9" $ 18 # 10.3 in. 2

3. Xeriscape garden will save 375.75(79 $ 17) = 23,296.5 gal per year.

4. Area of triangle a: A = 1 _

2 bh = 1 _

2 (4)(1) = 2

Area of trapezoid b:A = 1 _

2 ( b 1 + b 2 )h

= 1 _ 2 (2 + 4)(1) = 3

Area of trapezoid c:A = 1 _

2 ( b 1 + b 2 )h

= 1 _ 2 (3 + 2)(1) = 2.5

Area of rectangle d:A = bh = (3)(1) = 3Area of triangle e:A = 1 _

2 bh = 1 _

2 (3)(1) = 1.5

Shaded area of composite figure is = 2 + 3 + 2.5 + 3 + 1.5 = 12 ft 2 So the shaded area is : A # 12 ft 2 .

THINK AND DISCUSS, PAGE 608 1. Possible answer: figure with a hole in the middle

2. Draw a composite figure with an area close to the area of the irregular shape. Divide the composite figure into simpler shapes, such as triangles, rectangles, and trapezoids. Find the sum of the areas of the simpler figures.

3.



1. Possible answer:

2. Area of top rectangle:A = bh = (12 $ 5 $ 3)(4) = 16 cm 2 Area of bottom rectangle:A = bh = (12)(2) = 24 cm 2 Shaded area:A = 16 + 24 = 40 cm

2

3. Area of semicircle:A = 1 _

2 " r 2 = 1 _

2 "(2 ) 2 = 2" ft 2

Area of triangle:A = 1 _

2 bh = 1 _

2 (4)(5) = 10 ft 2

Shaded area:A = 2" + 10 # 16.3 ft 2

4. Area of rectangle:A = bh = (18)(8) = 144 in. 2 Area of circle:A = " r 2 = "(3 ) 2 = 9" in. 2 Shaded area:A = 144 $ 9" # 115.7 in. 2


5. Area of rectangle:A = bh = (6)(3) = 18 m 2 Area of triangle at corner:A = 1 _

2 bh = 1 _

2 (6 $ 5)(3 $ 2) = 0.5 m 2

Shaded area: A = 18 $ 0.5 = 17.5 m 2

6. Area of rectangle:A = bh = (4.5)(7) = 31.5 yd 2 Area of middle rectangle:A = bh = (2)(7 $ 5.5) = 3 yd 2 Area of right rectangle:A = bh = (1.5)(7) = 10.5 yd 2 Area of carpet:A = 31.5 + 3 + 10.5 = 45 yd 2 Cost to install: 45(6) = $270

7. Area of triangle a: A = 1 _

2 (2)(1) = 1

Area of trapezoid b:A = 1 _

2 (2 + 3)(1) = 2.5

Area of triangle c:A = 1 _

2 (2)(1) = 1

Shaded area is: A = 1 + 2.5 + 1 = 4.5 in. 2

8. Area of triangle a:A = 1 _

2 (2)(1) = 1

Area of triangle b:A = 1 _

2 (2)(1) = 1

Area of triangle c:A = 1 _

2 (4)(2) = 4

Shaded area is: A = 1 + 1 + 4 = 6 in. 2


9. Area of rectangle (on left):A = (7)(6) = 42 m m 2 Area of triangle (on right):A = 1 _

2 (12 $ 7)(6 $ 3) = 7.5 m m 2

Shaded area:A = 42 + 7.5 = 49.5 m m 2

10. Area of each semicircle:A = 1 _

2 "(10 ) 2 = 50" yd 2

Area of rectangle:A = (40)(20) = 800 yd 2 Shaded area:A = 800 + 2(50") # 1114.2 yd 2

11. Area of square:A = 2 2 = 4 m 2 Area of missing triangle:

A = 1 _ 2 (2) ! " 3 = ! " 3 m 2

Shaded area:A = 4 $ ! " 3 # 2.3 m 2

12. Area of trapzoid:A = 1 _

2 (51 + 24)(18) = 675 in. 2

Area of missing triangle:A = 1 _

2 (9)(12) = 54 in. 2

Shaded area:A = 675 $ 54 = 621 in. 2

13. Area of each triangle:A = 1 _

2 (10)(22 $ 15) = 35 ft 2

Area of rectangle:A = (30)(15) = 450 ft 2 Area of backdrop:A = 450 + 3(35) = 555 ft 2 Paint required = 555 ÷ 90 # 6.2 qtPat must buy 7 qt of paint.

14. A # 1 _ 2 (3)(1) + (3)(1)

+ 1 _ 2 (3)(2)

# 1.5 + 3 + 3 # 7.5 m 2

15. A # 1 _ 4 "(2 ) 2 + 1 _

2 (3)(2)

+ 1 _ 2 (2)(2) + 1 _

2 (2)(1)

# 9 m 2

16. Adding:Upper rectangle: A = (16)(3) = 48 cm 2 Middle rectangle: A = (16 $ 8)(3) = 24 cm 2 Lower rectangle: A = (16)(4) = 64 cm 2 Composite:A = 48 + 24 + 64 = 136 cm 2 Subtracting:Outer rectangle: A = (16)(10) = 160 cm 2 Missing rectangle: A = (8)(3) = 24 cm 2 Composite:A = (16)(10) $ (8)(3) = 136 cm 2 ;

the answers are the same.

17. Adding:Left trapezoid: A = 1 _

2 (21 + 9)(18) = 270 in. 2

Right trapezoid: A = 1 _ 2 (21 + 9)(18) = 270 in. 2

Composite: A = 270 + 270 = 540 in. 2 Subtracting:Outer rectangle: A = (18 + 18)(21) = 756 in. 2 Missing triangle: A = 1 _

2 (18 + 18)(21 $ 9) = 216 in. 2

Composite:A = 756 $ 216 = 540 in. 2 ;the answers are the same.

18. Divide composite into kite with diagonals 16 + 32 = 48 m and 2(30) = 60 m, and semicircles with

radius 1 _ 2 ! """" 16 2 + 30 2 = 17 m. Area of composite is

A = 1 _ 2 (48)(60) + 2 ( 1 _

2 "(17 ) 2 ) = (1440 + 289") m 2 .

19. Area of triangle:A = 1 _

2 (10) (5 ! " 3 ) = 25 ! " 3 in. 2

Area of each semicircle:

A = 3 ( 1 _ 2 "(5 ) 2 ) = 75" ____

2 in. 2

Area of composite:

A = (25 ! " 3 + 75" ____ 2 ) in. 2


20. Outer circle: A = "(8 ) 2 = 64" cm 2 Inner circle: A = "(5 ) 2 = 25" cm 2 Composite: A = 64" $ 25" = 39" cm 2

21. Possible answer: 35,000 mi 2

22. Let b 1 and b 2 be the bases of the trapezoid; let h be the height of the trapezoid, triangles, and rectangle; let x and y be the bases of the triangles. Then x + b 1 + y = b 2 . The area of the trapezoid is:A = 1 _

2 xh + b 1 h + 1 _

2 yh

= 1 _ 2 h(x + 2 b 1 + y)

= 1 _ 2 h( b 1 + x + b 1 + y)

= 1 _ 2 h( b 1 + b 2 ).

23a. Lower rectangle:A = (30)(15) = 450 in. 2 Upper triangle:A = 1 _

2 (30)(30 $ 15) = 225 in. 2

Composite:A = 450 + 225 = 675 in. 2

b.

c. Area of metal left:A = (105)(45) $ 6(675) = 675 in. 2

24. A = (3)(4) $ "(1 ) 2 = (12 $ ") cm 2

Possible drawing:

25. A = (4 ) 2 + 1 _ 2 (4)(5)

+ 1 _ 2 "(2 ) 2

= (26 + 2") in. 2

Possible drawing:

26. A = "(5 ) 2 $ 1 _ 2 (8)(6)

= (25" $ 24) cm 2 Possible drawing:

27. A 1 = 1 _ 4 "(2 ) 2 = "

A 2 = 1 _ 2 (2)(2) = 2

A 3 = 1 _ 2 " ( ! " 2 ) 2 = "

A 4 = A 3 $ ( A 2 $ A 1 )= " $ (" $ 2) = 2

28. Possible answer: A # 13.4 cm 2

29. Possible answer: A # 10 cm 2

30. Possible answer: Use addition to find the area of a figure that can be divided into triangles, rectangles, trapezoids, and semicircles. Use subtraction to find the area of a figure that has a shape removed from its interior.


31. A

32. GUpper trapezoid:A = 1 _

2 (7.8 + 5.4)(2.5) = 16.5 cm 2

Lower triangle:A = 1 _

2 (2.2)(2) = 2.2 cm 2

Composite:A = 16.5 + 2.2 # 19 cm 2

33. CA = 105(45) $ 45(30) $ 1 _

2 (20)(45) = 2925 m 2


34. A = " R 2 $ " r 2 = " ( R 2 $ r 2 ) 35. Possible answer:



37. Sale price = (100% $ 20%)(19.95) = 0.8(19.95) = $15.96

38. Sale price = (100% $ 15%)34.60= 0.85(34.60) = $29.41

39. BC ___ AB

= GF ___ AG

BC ___ 2.8

= 1 __ 2

2BC = 2.8 %%

BC = 1.4

40. CD ___ AB

= FE ___ AG

CD ___ 2.8

= 0.5 ___ 2

2CD = 0.5(2.8) = 1.4 %%

CD = 0.7

41. b = 3 cm, h = 3 ! " 3 ____ 2

cm

A = 1 __ 2 (3) ( 3 ! " 3 ____

2 ) # 3.9 cm 2


42. s ! " 3 ____ 2

= a = 4 ! " 3

s ! " 3 = 8 ! " 3 s = 8 mP = 6(8) = 48 mA = 1 _

2 aP = 1 _

2 (4 ! " 3 ) (48) # 166.3 m 2

9-3 GEOMETRY LAB: DEVELOP PICK’S THEOREM FOR AREA OF LATTICE POLYGONS, PAGE 613

TRY THIS, PAGE 613

1. Possible answer: A = 1 _ 2 B + I $ 1


3. Possible answer: A # 1 _ 2 (6) + 3 $ 1 = 5 unit s 2

4. Split outer figure into 2 trapezoids and 2 triangles

Upper left trapezoid: A = 1 _ 2 (2 + 1)(2) = 3

Upper right trapezoid: A = 1 _ 2 (2 + 1)(1) = 1.5

Lower left triangle: A = 1 _ 2 (2)(1) = 1

Lower right triangle: A = 1 _ 2 (2)(2) = 2

Missing square: A = 1 2 = 1Shaded area: A = 3 + 1.5 + 1 + 2 $ 1 = 6.5 units 2 A formula = 1 _

2 B + I $ 1 = 1 _

2 (11) + (1) $ 1 = 5.5; no


1. Area of each circle: A = "(15 ) 2 = 225" in. 2 Area of sheet: A = (90)(60) = 5400 in. 2 Area left over: A = 5400 $ 6(225") # 1159 in. 2

2. For stop sign, 30 = 2a = s( ! " 2 + 1)a = 15; s = 30( ! " 2 $ 1); P = 8s = 240( ! " 2 $ 1)Area of each sign: A = 1 _

2 aP = 1 _

2 (15)(240( ! " 2 $ 1)) = 1800( ! " 2 $ 1) in. 2

Area left over: A = 5400 $ 1800( ! " 2 $ 1) # 926 in. 2

3. For yield sign, b = 30, h = 15 ! " 3 Area of each sign: A = 1 _

2 (30)15 ! " 3 = 225 ! " 3 in. 2

Area left over: A = 5400 $ 10(225R3) # 1503 in. 2

4. Stop sign results in least amount of waste.


1. A = bh = (10)(5) = 50 ft 2

2. A = bh(24 x 2 + 8x) = b(4x) 4x(6x + 2) = b(4x) b = (6x + 2) m

3. A = 1 _ 2 d 1 d 2

126 = 1 _ 2 d 1 (12) = 6 d 1

d 1 = 21 ft

4. d 1 = 18 cm, d 2 = 2 ! """" 1 5 2 $ 9 2 = 24 cm

A = 1 _ 2 d 1 d 2 = 1 _

2 (18)(24) = 216 cm 2

5. Green triangle:

P = 2 + 1 + ! """" 2 2 + 1 2 = 3 + ! " 5 # 5.2 cm A = 1 _

2 (2)(1) = 1 cm 2

Blue trapezoid: P = 2 + 1 + 1 + ! " 2 = 4 + ! " 2 # 5.4 cmA = 1 _

2 (2 + 1)(1) = 1.5 cm 2

Yellow parallelogram:P = ! " 2 + ! " 5 + ! " 2 + ! " 5 = 2 ( ! " 2 + ! " 5 ) # 7.3 cmA = ( ! " 2 ) (1.5 ! " 2 ) = 3 cm 2

6. C = "d = 18" in.

7. A = " r 2 = "(6x ) 2 = 36 x 2 " ft 2

8. s ! " 3 ____ 2 = a = 6

3s = 12 ! " 3 s = 4 ! " 3 P = 6s = 24 ! " 3 A = 1 _

2 aP = 1 _

2 (6)24 ! " 3 # 124.7 ft 2

9. tan 36° = s _ 2 __ a = 6 __ a

a = 6 ______ tan 36°

P = 5(12) = 60A = 1 _

2 aP = 1 __

2 ( 6 ______

tan 36° ) (60) # 247.7 m 2

10. Triangle: A = 1 _ 2 (12)(12) = 72 cm 2

Square: A = 12 2 = 144 cm 2 Missing semicircle: A = 1 _

2 "(6 )

2 = 18" cm 2

Shaded area: A = 72 + 144 $ 18" # 159.5 cm 2

11. Outer rectangle: A = (16)(12) = 192 ft 2 Missing rectangle: A = (12)(4) = 48 ft 2 Shaded area: A = 192 $ 48 = 144 ft 2

12. Triangle a: 1 _ 2 (2)(1) = 1

Trapezoid b: 1 _ 2 (3 + 2)(1) = 2.5

Parallelogram c: (3)(1) = 3Trapezoid d: 1 _ 2 (1 + 3)(1) = 2

Area of garden:A # 1 + 2.5 + 3 + 2 = 8.5 yd 2 Cost of grass:8.5($6.50) # $55


9-4 PERIMETER AND AREA IN THE COORDINATE PLANE, PAGES 616–621

CHECK IT OUT! PAGES 616–618 1. Method 1

The area is about 1.5 + 2.5 + 4.5 + 5.5 + 5 + 6 + 8 + 3.5 + 1.5 # 38 units 2 .Method 2There are about 32 whole squares and 8 half-squares, so the area is about 32 + 1 _

2 (11) # 38 units 2 .

2. Step 1 Draw the polygon.

Step 2 HJKL appears to be a parallelogram.

%%

HL and %%

JK are vertical, therefore parallel.slope of

%% HJ = 6 $ 4 ________

2 $ ($3) = 2 __

5

slope of %%

LK = 1 $ ($1)

________ 2 $ ($3)

= 2 __ 5

%%

HJ and %%

LK are also ., so HJKL is a parallelogram. Step 3 Let

%% HL be the base; let c = HJ

b = 4 $ ($1) = 5; h = 2 $ ($3) = 5;

c = ! """"""""" (2 $ ($3) ) 2 + (6 $ 4 ) 2 = ! "" 29 P = 2b + 2c = (20 + 2 ! "" 29 ) # 20.8 unitsA = bh = (5)(5) = 25 units 2 .

3.

Area of rectangle: A = bh = (12)(8) = 96 units 2 Area of triangles:

a: A = 1 _ 2 bh = 1 _

2 (4)(6) = 12 units 2

b: A = 1 _ 2 bh = 1 _

2 (8)(6) = 24 units 2

c: A = 1 _ 2 bh = 1 _

2 (2)(2) = 2 units 2

d: A = 1 _ 2 bh = 1 _

2 (10)(2) = 10 units 2

Area of polygon: 96 $ 12 $ 24 $ 2 $ 10 = 48 units 2


THINK AND DISCUSS, PAGE 619 1. One way: draw a composite figure that approximates

the irregular shape and then find its area. Another way: count grid squares, estimating half squares.

2. If the quadrilateral is a parallelogram, rectangle, or trapezoid, use the Distance Formula to find the height and base or bases. If it is a rhombus or kite, use the Distance Formula to find the lengths of the diagonals.

3.



1. Method 1

The area is about 3.25 + 4.75 + 4 + 5 + 6.5 + 7 +

7.5 + 1.5 + 1 # 40.5 units 2 .Method 2There are about 33 whole squares and 15 half-squares, so the area is about 33 + 1 _

2 (15) # 40.5 units 2 .

2. Method 1

The area is about 1 + 1 + 5 + 3.5 + 4 + 1 + 5.5 +

4.5 + 4 + 4.5 + 5.5 + 3.5 # 43 units 2 . Method 2

There are about 32 whole squares and 22 half-squares, so the area is about 32 + 1 _

2 (22) # 43 units 2 .



Step 2 VWX appears to be an isosceles triangle.

VX = ! """"""""" (0 $ ($3) ) 2 + (3 $ 0 ) 2 = ! "" 18 = 3 ! " 2

WX = ! """""""" (3 $ 0 ) 2 + (0 $ 3 ) 2 = ! "" 18 = 3 ! " 2 VX = WX, so VWX is an isosceles triangle.Step 3 Let

%%% VW be base; let a = VX, c = WX.

b = 3 $ ($3) = 6; h = 3 $ 0 = 3; a = c = 3 ! " 2 P = a + b + c = 3 ! " 2 + 6 + 3 ! " 2 = (6 + 6 ! " 2 ) unitsA = 1 _

2 bh = 1 _

2 (6)(3) = 9 units 2


Step 2 FGH appears to be an isosceles triangle.

FG = ! """""""" (4 $ 2 ) 2 + (4 $ 8 ) 2 = ! "" 20 = 2 ! " 5

GH = ! """""""" (4 $ 2 ) 2 + (4 $ 0 ) 2 = ! "" 20 = 2 ! " 5 FG = GH, so FGH is an isosceles triangle.Step 3 Let

%% FH be the base; let a = FG, c = GH.

b = 8 $ 0 = 8; h = 4 $ 2 = 2; a = c = 2 ! " 5 P = a + b + c = 2 ! " 5 + 8 + 2 ! " 5 = (8 + 4 ! " 5 ) unitsA = 1 _

2 bh = 1 _

2 (8)(2) = 8 units 2


Step 2 PQRS appears to be a rectangle.

%%

PQ and %%

RS are horizontal; %%

QR and %%

SP are vertical. Consecutive sides are perpendicular, so PQRS is a rectangle.Step 3 Let

%% PQ be the base; let

%% QR be height.

b = 8 $ ($2) = 10; h = 5 $ 1 = 4P = 2b + 2h = 2(10) + 2(4) = 28 unitsA = bh = (10)(4) = 40 units 2


Step 2 ABCD appears to be an isosceles trapezoid. %%

AD and %%

BC are horizontal, so ABCD is a trapezoid.

AB = ! """""""""" ($2 $ ($4) ) 2 + (6 $ 2 ) 2 = ! "" 20 = 2 ! " 5

CD = ! """""""" (8 $ 6 ) 2 + (2 $ 6 ) 2 = ! "" 20 = 2 ! " 5 AB = CD, so trapezoid ABCD is isosceles.Step 3 Let

%% AD and

%% BC be the bases;

let a = AB, c = CD b 1 = 8 $ ($4) = 12; b 2 = 6 $ ($2) = 8; h = 6 $ 2 = 4; a = c = 2 ! " 5 P = a + b 1 + b 2 + c

= 2 ! " 5 + 8 + 12 + 2 ! " 5 = (20 + 4 ! " 5 ) unitsA = 1 _

2 ( b 1 + b 2 )h = 1 _

2 (12 + 8)(4) = 40 units 2

7.

Area of rectangle: A = bh = (6)(7) = 42 units 2 Area of triangles:a: A = 1 _

2 bh = 1 _

2 (1)(7) = 3.5 units 2

b: A = 1 _ 2 bh = 1 _

2 (5)(5) = 12.5 units 2

c: A = 1 _ 2 bh = 1 _

2 (6)(2) = 6 units 2

Area of &STU: 42 $ 3.5 $ 12.5 $ 6 = 20 units 2

8.


2 bh = 1 _

2 (3)(3) = 4.5 units 2

b: A = 1 _ 2 bh = 1 _

2 (3)(2) = 3 units 2

c: A = 1 _ 2 bh = 1 _

2 (4)(6) = 12 units 2

d: A = 1 _ 2 bh = 1 _

2 (2)(5) = 5 units 2

Area of polygon: 48 $ 4.5 $ 3 $ 12 $ 5 = 23.5 units 2


9. First polygon: A = 1 _

2 bh = 1 _

2 (4)(3) = 6 units 2

P = 4 + 3 + ! """" 4 2 + 3 2 = 12 unitsSecond polygon:The area is reduced by 1 unit 2 : A = 6 $ 1 = 5 units 2 The perimeter is unchanged: P = 12 unitsThird Polygon: remove one square at edge:

Fourth Polygon: remove one more square at edge:


10. Method 1

The area is about 4 + 1.5 + 5.5 + 4.5 + 5.5 + 5.5

+ 6.5 + 3.5 + 1.5 + 0.5 # 38.5 units 2 .Method 2There are about 28 whole squares and 21 half-squares, so the area is about 28 + 1 _

2 (21) # 38.5 units 2 .

11. Method 1

The area is about 2.5 + 15.5 + 2 + 2 + 1.5 + 4.5

+ 4 + 4 + 4.5+ 3 # 43.5 units 2 .Method 2There are about 35 whole squares and 17 half-squares, so the area is about 35 + 1 _

2 (17) # 43.5 units 2 .


Step 2 HJK is a right triangle.

Step 3 Let %%

HJ be the height and %%

JK be the base; let c = HK.h = 3 $ ($3) = 6; b = 5 $ ($3) = 8;

c = ! """" 6 2 + 8 2 = 10P = b + h + c = 6 + 8 + 10 = 24 unitsA = 1 _

2 bh = 1 _

2 (6)(8) = 24 units 2


Step 2 LMNP appears to be a rhombus.

LM = ! """""""" (7 $ 5 ) 2 + (5 $ 0 ) 2 = ! "" 29

MN = ! """""""" (3 $ 5 ) 2 + (5 $ 0 ) 2 = ! "" 29

NP = ! """""""" (5 $ 3 ) 2 + (10 $ 5 ) 2 = ! "" 29

LP = ! """""""" (5 $ 7 ) 2 + (10 $ 5 ) 2 = ! "" 29 All 4 sides are congruent, so LMNP is a rhombus.Step 3 Let d 1 = LN and d 2 = MP. d 1 = 7 $ 3 = 4; d 2 = 10 $ 0 = 10P = 4 ! " 29 unitsA = 1 _

2 d 1 d 2 = 1 _

2 (4)(10) = 20 units 2


Step 2 XYZ appears to be a scalene triangle.

XY = ! """""""" (5 $ 2 ) 2 + (3 $ 1 ) 2 = ! "" 13

YZ = ! """""""" (5 $ 7 ) 2 + (3 $ 1 ) 2 = ! " 8 = 2 ! " 2 XZ = 7 $ 2 = 5The 3 sides are different in length, so XYZ is a scalene triangle.Step 3 b = XZ = 5; h = 3 $ 1 = 2P = (5 + 2 ! " 2 + ! " 13 ) unitsA = 1 _

2 bh = 1 _

2 (5)(2) = 5 units 2



Step 2 TUVW appears to be an isosceles trapezoid.

AB = ! """"""""" ($3 $ 2) 2 + (5 $ 7) 2 = ! """ 25+4 = ! "" 29

CD = ! """"""""" (2 $ ($3)) 2 + (1 $ 3) 2 = ! """ 25+4 = ! "" 29 AB = CD, so trapezoid ABCD is isosceles.Step 3 Let

%% AD and

%% BC be the bases;

let a = AB, c = CD. b 1 = 5 $ 3 = 2; b 2 = 7 $ 1 = 6; h = 2 $ ($3) = 5; a = c = ! "" 29 P = a + b 1 + b 2 + c

= ! "" 29 + 2 + 6 + ! "" 29 = (8 + 2 ! "" 29 ) unitsA = 1 _

2 ( b 1 + b 2 )h = 1 _

2 (2 + 6)(5) = 20 units 2

16.


2 bh = 1 _

2 (13)(8) = 52 units 2

b: A = 1 _ 2 bh = 1 _

2 (5)(13) = 32.5 units 2

c: A = 1 _ 2 bh = 1 _

2 (8)(5) = 20 units 2

Area of &ABC: 169 $ 52 $ 32.5 $ 20 = 64.5 units 2

17.


2 bh = 1 _

2 (1)(3) = 1.5 units 2

b: A = 1 _ 2 bh = 1 _

2 (9)(1) = 4.5 units 2

c: A = 1 _ 2 bh = 1 _

2 (1)(8) = 4 units 2

d: A = 1 _ 2 bh = 1 _

2 (9)(6) = 27 units 2

Area of polygon: 90 $ 1.5 $ 4.5 $ 4 $ 27 = 53 units 2

18. Figure A: A = 1(2) + 3(2) = 8 units 2 Figure B: A = (2 ! " 2 ) ( ! " 2 ) + 1(1) + 4(1) = 9 units 2 Figure C: A = 1(3) + 4(1) + ( ! " 2 ) ( ! " 2 ) = 9 units 2 Figures B and C have same area.

19.

b = 3, h = 3, c = ! "" 18 = 3 ! " 2 P = 3 + 3 + 3 ! " 2 = (6 + 3 ! " 2 ) unitsA = 1 _

2 (3)(3) = 4.5 units 2

20.

b = 4, h = 8, c = ! """" 4 2 + 8 2 = ! "" 80 = 4 ! " 5 P = 4 + 8 + 4 ! " 5 = (12 + 4 ! " 5 ) unitsA = 1 _

2 (4)(8) = 16 units 2

21a. A = bh = (1 h)(20 mi/h) = 20 mi 2

b. Upper trapezoid: A = 1 _ 2 (4 + 2)(1)(20 mi) = 60 mi

Lower trapezoid: A = 1 _ 2 (5 + 4)(1)(20 mi) = 90 mi

Shaded area: A # 60 + 90 # 150 mi 2

c. The area represents the distance the boat traveled in 5 h.

22. Possible answer: Draw polygon ABCDE. Draw a rectangle with base 6 and height 5 around polygon. The rectangle has area 30 units 2 , and regions not included in ABCDE have areas 6, 3, 1, and 3.5 units 2 ; so the area of ABCDE is 30 $ 6 $ 3 $ 1 $ 3.5 = 16.5 units 2 .

23a. A = bh = (3)(2) = 6 units 2

b. Possible answer: ABC: A = 1 _

2 bh

6 = 1 _ 2 (3)h = 1.5h

h= 4y-coordinate of A, B is 5, so y-coordinate of C is

5 $ 4 = 1 or 5 + 4 = 9. Therefore, let the y-coordinate of C be 1 or 9. A possible coordinate for C is C = (2, 1). DEFG: the x-coordinate of D must be 8.

A = 1 _ 2 d 1 d 2

6 = 1 _ 2 (2) d 2

d 2 = 6 The y-coordinate of F is 8, so the y-coordinate of D

is 8 $ 6 = 2; D = (8, 2).


TEST PREP, PAGE 621

24. Dr = ! """""""" (3 $ 0 ) 2 = (4 $ 0 ) 2 = 5A = " r 2 = "(5 ) 2 # 78.5 units 2

25. JThe area of ABC would be: 1 _ 2 bh = 1 _

2 (3 $ 1)(5 $ ($3)) = 1 _

2 (2)(8) = 8 units 2 .

26a. Mike estimated the area by using a square with vertices at ($4, 4), (4, 4), (4, $4), and ($4, $4). This does not include the area at corners of the graph.

b. The composite figure is made of a square with area 64 units 2 , 4 triangles each with area 2.5 units 2 , and 4 other triangles each with area 2 units 2 . The area is 64 + 4(2.5) + 4(2) = 82 units 2 .

c. The irregular shape encloses a square with area 64 units 2 and is enclosed in a square with area 100 units 2 . The average of the areas

is 64 + 100 ______ 2 = 82 units 2 .


27. Split the area into a triangle to the left of the y-axis and 3 trapezoids to the right of the y-axis.A # 1 _

2 (2)(1) + 1 _

2 (1 + 2)(1) + 1 _

2 (2 + 4)(1) + 1 _

2 (1 + 2)(1)

= 1 + 1.5 + 3 + 6 # 10.5 units 2

28. Split the area into a triangle and 2 trapezoids.A # 1 _

2 (1)(1) + 1 _

2 (1 + 4)(1) + 1 _

2 (4 + 9)(1)

= 0.5 + 2.5 + 6.5 # 9.5 units 2

29. Split the area into a triangle and 2 trapezoids.A # 1 _

2 (1)(1) + 1 _

2 (1 + 2)(3) + 1 _

2 (2 + 3)(5)

= 0.5 + 4.5 + 12.5 # 17.5 units 2

30.

Starting at ($2, 3),P # ! " 5 + 2 + ! " 2 + ! "" 17 + 1 + ! " 5 + ! "" 17 + 1

+ ! " 2 + ! " 2 + ! "" 13 + ! "" 10 + 1 # 28.7 units 2


31.

The vertices of the octagon are (0, 1), ( ! " 2 ___ 2 ,

! " 2 ___ 2 ) ,

(1, 0),

( ! " 2 ___ 2 ,

! " 2 ___ 2 ) , (0, $1), ( ! " 2 ___

2 ,

! " 2 ___ 2 ) , ($1, 0), ( ! " 2 ___

2 ,

! " 2 ___ 2 ) .

s = ! """""""""

(0 $ ! " 2 ___ 2 ) 2 + (1 $

! " 2 ___ 2 ) 2

= ! """"""" 1 _ 2 + 1 $ ! " 2 + 1 _

2 = ! """ 2 $ ! " 2

P = 8s = 8 ! """ 2 $ ! " 2 unitsThe coordinates of the midpoint of the upper-right

side are ( ! " 2 ___ 4 , 1 _

2 +

! " 2 ___ 4 ) . Therefore,

a = ! """"""""

( ! " 2 ___ 4 ) 2 + ( 1 _

2 +

! " 2 ___ 4 ) 2

= ! """""""

1 _ 8 + 1 _

4 +

! " 2 ___ 4 + 1 _

8

= ! """

1 _ 2 +

! " 2 ___ 4 = 1 _

2 ! """ 2 + ! " 2

A = 1 _ 2 aP

= 1 _ 2 ( 1 _

2 ! """ 2 + ! " 2 ) (8 ! """ 2 $ ! " 2 )

= 2 ! """""""" (2 $ ! " 2 ) (2 + ! " 2 ) = 2 ! """"" 2 2 $ ( ! " 2 ) 2 = 2 ! " 2 units 2

32. $4 < x + 4 < 7$4 $ 3 < x < 7 $ 3 $7 < x < 4

33. 0 < 2a + 4 < 100 $ 4 < 2a < 10 $ 4 $4 < 2a < 6 $2 < a < 3

34. 12 ' $2m + 10 ' 20 2 ' $2m ' 10$10 ' 2m ' $2 $5 ' m ' $1


1. %%

DC - %%

BC , ,DCA - ,ACB

1. Given

2. %%

AC - %%

AC 2. Reflex Prop. of -

3. &DCA - &BCA 3. SAS

4. ,DAC - ,BAC 4. CPCTC


36. C = 2"r16" = 2"r r = 8 cmA = " r 2

= "(8 ) 2 = 64" cm 2

37. A = " r 2 121" = " r 2 121 = r 2 r = 11 ftd = 2r

= 2(11) = 22 ft

9-5 EFFECTS OF CHANGING DIMENSIONS PROPORTIONALLY, PAGES 622–627

CHECK IT OUT! PAGES 622–624 1. Original dimensions:

A = bh = 7(4) = 28 ft 2

Triple the height:A = bh

= 7(12) = 84 ft 2 84 = 3(28); if height is tripled, area is also tripled.

2. Original dimensions:b = 7 $ 2 = 5, h = 5 $ 1 = 4, c = ! """" 5 2 + 4 2 = ! "" 41 P = 5 + 4 + ! "" 41 = (9 + ! "" 41 ) unitsA = 1 _

2 (5)(4) = 10 units 2

Dimensions multiplied by 3:b = 3(5) = 15, h = 3(4) = 12, c = 3 ! "" 41 P = 15 + 12 + 3 ! "" 41 = (27 + 3 ! "" 41 ) unitsA = 1 _

2 (15)(12) = 90 units 2

Perimeter is multiplied by 3. Area is multiplied by 3 2 , or 9.

3. Original perimeter is P = 4s = 36 mm; side length is 9 mm, and area is A = 9 2 = 81 mm 2 . If the area is multiplied by 1 _

2 , the new area is 40.5 mm.

s 2 = 40.5 s = ! "" 40.5 = 9 ___

! " 2

4.5 = 1 ___ ! " 2

(9); side length is multiplied by 1 ___ ! " 2

.

4. Possible answer:

THINK AND DISCUSS, PAGE 624 1. If one dimension of a rectangle is multiplied

by a, the area is also multiplied by a. If both dimensions of a rectangle are multiplied by a the perimeter is multiplied by a.

2.



1. Original dimensions:A = 1 _

2 bh

= 1 _ 2 (21)(12) = 126 m 2

Double the height:A = 1 _

2 bh

= 1 _ 2 (21)(24) = 252 m 2

252 = 2(126); if the height is doubled, the area is also doubled.

2. Original dimensions:A = 1 _

2 ( b 1 + b 2 )h

= 1 _ 2 (12 + 18)(5)

= 75 cm 2

Multiply height by 1 _ 3 :

A = 1 _ 2 ( b 1 + b 2 )h

= 1 _ 2 (12 + 18) ( 5 _

3 )

= 25 cm 2 25 = 1 _

3 (75); if the height is multiplied by 1 _

3 , the area

is also multiplied by 1 _ 3 .

3. Original dimensions:

b = 12, h = 6, c = ! """" 1 2 2 + 6 2 = ! "" 180 = 6 ! " 5 P = 12 + 6 + 6 ! " 5 = (18 + 6 ! " 5 ) in.A = 1 _

2 (12)(6) = 36 in. 2

Dimensions multiplied by 3:b = 3(12) = 36, h = 3(6) = 18, c = 18 ! " 5 P = 36 + 18 + 18 ! " 5 = (54 + 18 ! " 5 ) in.A = 1 _

2 (36)(18) = 324 in. 2

Perimeter is multiplied by 3. Area is multiplied by 3 2 ,or 9.

4. Original dimensions:P = 2(18) + 2(6) = 48 ftA = (18)(6) = 108 ft 2 Dimensions multiplied by 1 _

2 :

P = 2(9) + 2(3) = 24 ftA = (9)(3) = 27 ft 2 Perimeter is multiplied by 1 _

2 . Area is multiplied

by ( 1 _ 2 ) 2 , or 1 _

4 .

5. The original area is A = s 2 = 36 m 2 ; the side length is 6 m. If the area is doubled, the new area is 72 m. s 2 = 72 s = ! "" 72 = 6 ! " 2 mThe side length is multiplied by ! " 2 .

6. The original diameter is d = 2r = 14 ft; the radius is 7 ft, area is A = "(7 ) 2 = 49" ft 2 , and circumference is 14" ft. If the area is tripled, the new area is 147" ft 2 ." r 2 = 147" r 2 = 147 r = ! "" 147 =7 ! " 3 C = 2"7 ! " 3 = 14" ! " 3 ftThe circumference is multiplied by ! " 3 .


7. Old area = (2)(4) = 8 in. 2 New area = (4)(8) = 32 in. 2 32 = 4(8), so the area is multiplied by 4. Therefore the cost is multiplied by 4: Cost of new ad = 4($36.75) = $147


8. Original dimensions: A original = 1 _

2 bh

Multiply height by 4: A new = 1 _

2 b(4h) = 2bh

2bh = 4 ( 1 _ 2 bh) ; if the height is multiplied by 4,

the area is also multiplied by 4.

9. Original dimensions:A = bh

= (24)(9) = 216 in. 2

Double the height:A = bh

= (16)(9) = 144 in. 2 144 = 2 _

3 (216); if the base is multiplied by 2 _

3 ,

the area is also multiplied by 2 _ 3 .

10. Original dimensions:a = b = c = 10, h = 5 ! " 3 P = 10 + 10 + 10 = 30 cmA = 1 _

2 (10)5 ! " 3 = 25 ! " 3 cm 2

Dimensions doubled:a = b = c = 20, h = 10 ! " 3 P = 20 + 20 + 20 = 60 cmA = 1 _

2 (20)10 ! " 3 = 1000 ! " 3 cm 2

Perimeter is doubled. Area is multiplied by 2 2 , or 4.

11. Original dimensions:r = 5 $ 0 = 5C = 2"(5) = 10" unitsA = "(5 ) 2 = 25" units 2 Dimensions multiplied by 3 _

5 :

r = 3 _ 5 (5) = 3

C = 2"(3) = 6" unitsA = "(3 ) 2 = 9" units 2 Circumference is multiplied by 3 _

5 . Area is multiplied

by ( 3 _

5 ) 2 , or 9 __

25 .

12. Original circumference is C = 2"r = 16" mm; radius is 8 mm, and area is A = "(8 ) 2 = 64" ft 2 .

If the area is multiplied by 1 _ 3 , the new area is 64" ___

3 ft 2 .

" r 2 = 64" ___ 3

r 2 = 64 __ 3

r = ! " 64 __ 3 = 8 ! " 3 ___

3

8 ! " 3 ___ 3 = 81 ___

! " 3 ; radius is multiplied by 1 ___

! " 3 .

13. The original side length is 8 $ 3 = 5 units, and the original area is 5 2 = 25 units 2 . The new area

is 3(25) = 75 units 2 . s 2 = 75 s = ! "" 75 = 5 ! " 3 unitsThe side length is multiplied by ! " 3 .

14a. Smaller screen: 32 2 = b 2 + h 2 Larger screen:

3 6 2 = (kb ) 2 + (kh ) 2 = k 2 ( b 2 + h 2 ) = k 2 (32 ) 2

36 = k(32)kh : h = 36 : 32 = 9 : 8

b. Ratio of areas: (kb)(kh) : bh = k 2 (bh) : bh= k 2 : 1

= ( 9 _ 8 ) 2 : 1

= 9 2 : 8 2 = 81 : 64

15. Original dimensions: A = 1 _ 2 d 1 d 2

New dimensions: A = 1 _ 2 (8 d 1 )(8 d 2 ) = 32 d 1 d 2

32 d 1 d 2 = 64 ( 1 _ 2 d 1 d 2 ) , so area is multiplied by 64.

16. Original dimensions: C = 2"r, A = "r^2New dimensions: C = 2.4(2"r) = 2"(2.4r);Therefore the new radius is 2.4r, and A = "(2.4r ) 2 = 5.76(" r 2 ). So area is multiplied by 5.76.

17. Original dimensions: A = bhNew dimensions: A = (4b)(7h) = 28(bh)The area is multiplied by 28.

18. Original dimensions: s = 2a tan 22.5°, P = 8s = 16a tan 22.5°A = 1 _

2 aP

= 1 _ 2 a(16a tan 22.5°)

= 8 a 2 tan 22.5°New dimensions: P = 16(3a) tan 22.5° = 48a tan 22.5°A = 1 _

2 aP

= 1 _ 2 (3a)(48a tan 22.5°)

= 72 a 2 tan 22.5°72 a 2 tan 22.5° = 9(8 a 2 tan 22.5°). So the area is multiplied by 9.

19. Original dimensions: d = s ! " 2 , A = s 2 New dimensions: d = s ! " 2 ÷ 4 = ( s _

4 ) ! " 2 ;

Therefore the new side length is s _ 4 , and

A = ( s _ 4 ) 2 = s 2 ÷ 16. So the area is divided by 16.

20. Original dimensions: A = 1 _ 2 d 1 d 2

New dimensions: A = 1 _ 2 ( 1 _

7 d 1 ) (8 d 2 ) = 1 __

14 d 1 d 2

1 __ 14

d 1 d 2 = 1 _ 7 ( 1 _

2 d 1 d 2 ) , so the area is multiplied by 1 _

7 .

21. Original dimensions: P = 3s, h = s ! " 3 ___ 2 ,

A = 1 _ 2 s (s

! " 3 ___ 2 ) = s 2

! " 3 ___ 4

New dimensions: P = 2(3s) = 3(2s);Therefore the new side length is 2s, and A = ( s _

2 ) 2

! " 3 ___ 4

= s 2 ! " 3 ___ 16

= 1 _ 4 ( s 2

! " 3 ___ 4 ) . So the area is multiplied by 4.


22a. A = 1 _ 2 (42 + 24)(15) = 495 cm 2

top base doubled: A = 1 _

2 (42 + 2(24))(15) = 675 cm 2

675 # 1.4(495), so the area is multiplied by about 1.4.

b. A = 1 _ 2 (2(42) + 2(24))(15) = 990 cm 2

990 = 2(495), so the area is doubled.

c. A = 1 _ 2 (42 + 24)(2(15)) = 990 cm 2

990 = 2(495), so the area is doubled.

d. A = 1 _ 2 (2(42) + 2(24))(2(15)) = 1980 cm 2

1980 = 4(495), so the area is multiplied by 4.

23. 1 square inch = 1 0 2 mi 2 = 100 mi 2 = 100(640 acres) = 64,000 acres

12.5 sq. in. = 12.5(64,000 acres) = 800,000 acres

24. If the dimensions are multiplied by x, the area is multiplied by x 2 . x 2 = 50% = 1 _

2

x = ! " 1 _ 2 = 1 ___

! " 2

25a. Original dimensions:b = 2 $ ($2) = 4, h = 3 $ ($2) = 5A = 1 _

2 (4)(5) = 10 units 2

New dimensions:b = 6 $ ($6) = 12, h = 3 $ ($2) = 5A = 1 _

2 (12)(5) = 30 units 2


b. New dimensions:b = 2 $ ($2) = 4, h = 9 $ ($6) = 15A = 1 _

2 (4)(15) = 30 units 2


c. New dimensions:b = 6 $ ($6) = 12, h = 9 $ ($6) = 15A = 1 _

2 (12)(15) = 90 units 2


26a. Original dimensions:Outer rectangle: A = (4 $ ($4))(4 $ ($3)) = 48Missing upper left, lower right triangles:A = 1 _

2 ($1 $ ($4))(4 $ 0) = 6

Missing lower left, upper right triangles:A = 1 _

2 (4 $ ($1))(4 $ 1) = 7.5

Area of figure:A = 48 $ 2(6) $ 2(7.5) = 21 units 2 New dimensions:Outer rectangle: A = (12 $ ($12))(4 $ ($3)) = 144Missing upper left, lower right triangles:A = 1 _

2 ($3 $ ($12))(4 $ 0) = 18


2 (12 $ ($3))(4 $ 1) = 22.5

Area of figure:A = 144 $ 2(18) $ 2(22.5) = 63 units 2 63 = 3(21), so the area is multiplied by 3.

b. New dimensions:Outer rectangle: A = (4 $ ($4))(12 $ ($9)) = 144Missing upper left, lower right triangles:A = 1 _

2 ($1 $ ($4))(12 $ 0) = 18


2 (4 $ ($1))(12 $ 3) = 22.5


c. New dimensions:Outer rectangle: A = (12 $ ($12))(12 $ ($9)) = 432Missing upper left, lower right triangles:A = 1 _

2 ($3 $ ($12))(12 $ 0) = 54


2 (12 $ ($3))(12 $ 3) = 67.5


27a. Original dimensions:Left rectangle: A = ($1 $ ($3))(3 $ ($2)) = 10Middle square: A = (0 $ ($1))($1 $ ($2)) = 1Right rectangle: A = (2 $ (0))(1 $ ($2)) = 6Area of figure: A = 10 + 1 + 6 = 17 units 2 New dimensions:Left rectangle: A = ($3 $ ($9))(3 $ ($2)) = 30Middle square: A = (0 $ ($3))($1 $ ($2)) = 3Right rectangle: A = (6 $ (0))(1 $ ($2)) = 18Area of figure: A = 30 + 3 + 18 = 51 units 2 51 = 3(17), so the area is multiplied by 3.

b. New dimensions:Left rectangle: A = ($1 $ ($3))(9 $ ($6)) = 30Middle square: A = (0 $ ($1))($3 $ ($6)) = 3Right rectangle: A = (2 $ (0))(3 $ ($6)) = 18Area of figure: A = 30 + 3 + 18 = 51 units 2 51 = 3(17), so the area is multiplied by 3.

c. New dimensions:Left rectangle: A = ($3 $ ($9))(9 $ ($6)) = 90Middle square: A = (0 $ ($3))($3 $ ($6)) = 9Right rectangle: A = (6 $ (0))(3 $ ($6)) = 54Area of figure: A = 90 + 9 + 54 = 153 units 2 153 = 9(17), so the area is multiplied by 9.

28. Possible answers:Multiply the base or height by 5.Multiply the base and height by ! " 5 .

29a. Original area is " ( 8 __ 2 ) 2 = 16 "

Now we want 2 (16)" = 32 ", which means the new diameter = (2) ! "" 32 = 8 ! " 2 in.

b. Now we want the area to be 0.5 (16 ") = 8 ".

So now the new diameter = 2 ( ! " 8 ) = 4 ! " 2 in.

TEST PREP, PAGE 627

30. DA = (2s ) 2 = 4( s 2 )

31. GA = 4(" r 2 ) = "(2r ) 2 ;d = 2(2r)


32. Cbh = 60 ft 2 A = (1.5b)(1.5h) = 2.25bh = 2.25(60) = 135 ft 2

33. 36Old dimensions: P = a + b + c = 18 in.New dimensions: P = 2a + 2b + 2c = 2(a + b + c) = 2(18) = 36 in.


34. A = (5(2x + 5) ) 2 = 25(4 x 2 + 20x + 25)= (100 x 2 + 500x + 625) cm 2

35. Old dimensions: C = 6" in.New dimensions: C = 6"(x + 3) = 2"(3(x + 3)) in.r = 3(x + 3) = 3x + 9 in.A = " (3x + 9) 2 = (9" x 2 + 54"x + 81") in. 2

36. Possible answers:Multiply all lengths of the horizontal segments by 2.Multiply all side lengths by ! " 2 .


37. t tortillas ___________ 2 tortillas/min

= 36 min

t __ 2 = 36

38. m mi ________ 25 mi/gal

= (13 $ 8) gal

m ___ 25

= 13 $ 8

39. 7m(int. ,) = (7 $ 2)180° = 900° m(int. ,) # 128.6°7m(ext. ,) = 360° m(ext. ,) # 51.4°

40. 10m(int. ,) = (10 $ 2)180° = 1440° m(int. ,) = 144°10m(ext. ,) = 360° m(ext. ,) = 36°

41. 14m(int. ,) = (14 $ 2)180° = 2160° m(int. ,) # 154.3°14m(ext. ,) = 360° m(ext. ,) # 25.7°

42. Outer rectangle: A = (6)(7) = 42Missing triangles: A = 1 _

2 (6)(1) = 3

A = 1 _ 2 (4)(7) = 14

A = 1 _ 2 (2)(6) = 6

Area of figure: A = 42 $ 3 $ 14 $ 6 = 19 units 2

43. Outer rectangle: A = (8)(8) = 64Missing triangles: A = 1 _

2 (2)(2) = 2

A = 1 _ 2 (6)(2) = 6

A = 1 _ 2 (2)(6) = 6

A = 1 _ 2 (6)(6) = 18

Area of figure: A = 64 $ 2 $ 6 $ 6 $ 18 = 32 units 2

CONNECTING GEOMETRY TO PROBABILITY, PAGES 628–629

TRY THIS, PAGE 629 1. The event “choosing a circle” contains only 1

outcome. The probability is:

P(circle) = # outcomes in event _________________ # possible outcomes

= 1 __ 6

2. The event “choosing a shape with area 36 cm 2 ” contains 2 outcomes: the square has an area 6 2 = 36 cm 2 , and the rectangle has an area (9)(4) = 36 cm 2 . The probability is:


= 2 __ 6 = 1 __

3

3. The event “choosing a triangle or quadrilateral” contains 5 outcomes. The probability is:


= 5 __ 6

4. The event “not choosing a triangle” contains 4 outcomes. The probability is:


= 4 __ 6 = 2 __

3

9-6 GEOMETRIC PROBABILITY, PAGES 630–636


1. P ( %% BD ) = P ( %% BC ) + P ( %% CD ) = 3 ___ 12

+ 5 ___ 12

= 8 ___ 12

= 2 __ 3

2. P(not red) = P(green or yellow)= P(green) + P(yellow)

= 25 ___ 60

+ 5 ___ 60

= 30 ___ 60

= 1 __ 2

3. P = 80 + 100 ________ 360

= 180 ____ 360

= 1 __ 2

4. Area of the triangle (which contains the circle) is # 187 m 2 .Area of the trapezoid is 75 m 2 .Area of the rectangle is 900 m 2

P # 900 $ (187 + 75)

_______________ 900

= 638 ____ 900

# 0.71

THINK AND DISCUSS, PAGE 633 1. In a geometric model, there are an infinite number

of outcomes in each event.

2. Subtract 1 _ 2 and 1 _

3 from 1 to find what part of the

spinner is yellow.

3.




1. Possible answer: a spinner

2. P ( %% XZ ) = P ( %% XY ) + P ( %% YZ ) = 5 ___ 10

+ 3 ___ 10

= 8 ___ 10

= 4 __ 5

3. P (not %%

XY ) = P ( %%% WX ) + P ( %% YZ ) = 2 ___ 10

+ 3 ___ 10

= 5 ___ 10

= 1 __ 2

4. P ( %%% WX or %%

YZ ) = P ( %%% WX ) + P ( %% YZ ) = 1 __ 2

5. P ( %%% WY ) = P ( %%% WX ) + P ( %% XY ) = 2 ___ 10

+ 5 ___ 10

= 7 ___ 10

6. P = 1.5 min _______ 10 min

= 0.15

7. P(wait < 3 min, once) = 3 + 1.5 _______ 10

= 0.45

In 20 times, expect to wait < 3 min, 0.45(20) = 9 times.

8. P = 45 ____ 360

= 1 __ 8

9. P = 45 + 90 _______ 360

= 135 ____ 360

= 3 __ 8

10. P = 360 $ 120 _________ 360

= 240 ____ 360

= 2 __ 3

11. P = 60 + 90 _______ 360

= 150 ____ 360

= 5 ___ 12

12. Area of the triangle is A = 1 _ 2 (10)(10) = 50 ft 2

Area of the rectangle is A = (48)(24) = 1152 ft 2

P = 50 _____ 1152

# 0.04

13. Area of the trapezoid is A = 1 _ 2 (18 + 12)(6) = 90 ft 2

P = 90 _____ 1152

# 0.08

14. Area of the square is A = (10 ) 2 = 100 ft 2 P = 100 _____

1152 # 0.09

15. The combined area of the smaller shapes is A = 50 + 90 + 100 = 240 ft 2 .

P = 1152 $ 240 __________ 1152

# 0.79


16. HM = 16.4 + 21.9 + 15.3 + 14.8 = 68.4

P ( %% JK ) = 21.9 ____ 68.4

# 0.32

17. P (not %%

LM ) = 68.4 $ 14.8 __________ 68.4

# 0.78

18. P ( %% HJ or %%

KL ) = P ( %% HJ ) + P ( %% KL ) = 16.4 ____

68.4 + 15.3 ____

68.4 # 0.46

19. P (not %%

JK or %%

LM ) = P ( %% HJ or %%

KL ) # 0.46

20. P = 0.75 min ________ 15 min

= 0.05

21. P = 15 $ (5 + 0.75)

_____________ 15

# 0.62

22. Assume that if the report has started, you must wait to hear the whole next one.P = 1 ___

15

In 50 times, expect to wait < 1 min, 1 ___ 15

(50) # 3 times.

23. P(red) = 180 ____ 360

= 1 __ 2

24. P(yellow or blue) = 45 + 45 _______ 360

= 90 ____ 360

= 1 __ 4

25. P(not green) = 360 $ 90 ________ 360

= 270 ____ 360

= 3 __ 4

26. P(red or green) = 180 + 90 ________ 360

= 270 ____ 360

= 3 __ 4

27. Area of the triangle is

A = 1 __ 2 ( 30 _____

! " 3 /2 ) (30) = 1 __

2 (20 ! " 3 ) (30) = 300 ! " 3 m 2

Area of the rectangle is A = (20 ! " 3 ) (30) = 600 ! " 3 m 2

P = 300 ! " 3 ______ 600 ! " 3

= 1 __ 2 or 0.5

28. Area of the square is A = (10 ! " 2 ) 2 = 200 m 2

P = 200 ______ 600 ! " 3

# 0.19

29. Area of the circle is A = "(10 ) 2 = 100" m 2 P = 100" $ 200 __________

600 ! " 3 # 0.11

30. The circle and square are inside the triangle, so the remaining area is A = 600 ! " 3 $ 300 ! " 3 = 300 ! " 3 m 2

P = 300 ! " 3 ______ 600 ! " 3

= 1 __ 2 or 0.5

31. The value in A is incorrect because the sectors have different angle measures, so they are not equally likely outcomes.

32. Area of rectangle: A = (15 $ 2)(8 $ 1) = 91 units 2 Area of triangle: A = (5)(4) $ 1 _

2 (1)(4) $ 1 _

2 (4)(2) $ 1 _

2 (5)(2)

= 20 $ 2 $ 4 $ 5 = 9 units 2 P = 9 ___

91 # 0.10

33. Only half of / P lies inside ABCD. The area of the semicircle is A = 1 _

2 "(3 ) 2 = 4.5" units 2 .

P = 91 $ 4.5" _________ 91

# 0.84

34. Area of outer square: A = (10 ) 2 = 100 units 2 Area of parallelogram: A = (2)(3) = 6 units 2 P = 6 ____

100 = 0.06

35. Area of circle: A = "(2 ) 2 = 4" units 2 P = 4" ____

100 # 0.13

36. Area of triangle is A = 1 _ 2 (3)(3) = 4.5 units 2

P = 4.5 + 4" ________ 100

# 0.17

37. P = 100 $ (4.5 + 6 + 4")

__________________ 100

# 0.77


38a. Area of central region is A = "(6.1 ) 2 cm 2 Area of target is A = "(61 ) 2 cm 2

P = "(6.1 ) 2

______ "(61 ) 2

= ( 6.1 ___ 61

) 2 = (0.1 ) 2 = 0.01

b. Inner radius of blue rings: r = 4(6.1) cmOuter radius of black rings: r = 8(6.1) cmArea of blue and black rings: A = "(8(6.1) ) 2 $ "(4(6.1) ) 2

= "(6.1 ) 2 (64 $ 16) = (48)"(6.1 ) 2 cm 2

P = 48"(6.1 ) 2

________ "(61 ) 2

= 48 ____ 100

= 0.48

c. Area of 5 inner rings: A = "(5(6.1) ) 2 = 25"(6.1 ) 2 cm 2

P = 25"(6.1 ) 2

________ "(61 ) 2

= 25 ____ 100

= 0.25

d. The probabilities might be different because an archer would be aiming for the center, not a random point.

39. Possible answer: The point lies on %%

AC .

40. Possible answer: The point lies in the red or yellow region.

41. Possible answer: The point lies in the blue or green triangle.

42a. Area of blue parallelogram: A = (2)(1) = 2 units 2 Area of tangram: A = (4 ) 2 = 16 units 2

P = 2 __ 16

= 1 _ 8

b. Area of purple triangle: A = 1 _ 2 (2)(2) = 2 units 2

P = 2 __ 16

= 1 _ 8

c. Area of large yellow triangle: A = 1 _

2 (4)(2) = 4 units 2

P = 4 __ 16

= 1 _ 4

d. No, because areas are the same.

43. P = 4 _ 8 = 1 _

2 ; it does not matter which regions are

shaded because they all have the same area.

44a. Area of each balloon: A = "(1.5 ) 2 = 2.25" in. 2 Area of board: A = (50)(30) = 1500 in. 2 For 40 balloons,

P = 40(2.25")

_________ 1500

# 0.19

b. For n balloons, if probability is 0 0.25,

P = n(2.25")

________ 1500

0 0.25

n 0 1500 _____ 2.25"

(0.25) # 53.1

n 0 54 balloons

TEST PREP, PAGE 636

45. AP =

2(1.5) _____

6(3.5) = 3 ___

21 # 0.14

46. GP ( %% AB ) = 18 _______

18 + 24 = 18 ___

42 # 0.43

47. DArea of triangle: A = 1 _

2 (10)(20) = 100 m 2

Area of circle: A = "(20 ) 2 = 400" m 2 Area of square: A = (25 ) 2 = 625 m 2 Area of field: A = 100(70) = 7000 m 2

P = 7000 $ (100 + 400" + 625)

_______________________ 7000

# 0.717

48a. Let P(r), P(b), P(g) be the probabilities of each color. From the given info:P(r) = 2P(b), P(g) = P(b)Substitute into this eqation: P(r) + P(b) + P(g) = 12P(b) + P(b) + P(b) = 1 4P(b) = 1 P(b) = 1 _

4

P(g) = P(b) = 1 _ 4 or 0.25

b. 3; the probability of landing on green is 0.25, so the number of green regions is 0.25(12) = 3.


49. Area of each red region: A = 1 2 $ 4 ( 1 _

4 "(0.5 ) 2 ) = 1 $ 0.25"

P = 1 $ 0.25" _________ 1 = 4 $ " _____

4 # 0.21

50. P = s 2 _______ (18)(24)

= s 2 ____ 432

= 1 __ 3

s 2 = 144 s = 12 ft The square will be 12 ft by 12 ft.

P = s 2 ____ 432

= 3 __ 4

s 2 = 324 s = 18 ft

The square will be 18 ft by 18 ft.

51. Possible answer: The probabilities must add to 1, so P(yellow) + P(blue) + P(red) = 1. I would make the regions different sizes, and I would want each region to be worth more points the smaller it is. The point value for red is 6 times the point value for yellow, so I would make 6 · P(red) = P(yellow).

The point value for blue is 3 times the point value for yellow, so I would make 3 · P(blue) = P(yellow).

Then P(yellow) + 1 __ 3 P(yellow) + 1 __

6 P(yellow) = 1.

This means P(yellow) = 2 __ 3 , P(blue) = 2 __

9 ,

and P(red) = 1 __ 9 .

The angle measure for the yellow region would be 240°, for the blue region would be 80°, and for the red region would be 40°.


52. (3 x 2 y) (4 x 3 y 2 ) = 3(4) x (2 + 3) y (1 + 2) = 12 x 5 y 3

53. (2 m 5 ) 2 = 2 2 m 5(2) = 4 m 10


54. $8 a 4 b 6 _______ 2a b 3

= $ 8 __ 2 a (4 $ 1) b (6 $ 3)

= $4 a 3 b 3

55. By Distribution Formula, AB = AC = 2 ! " 5 , BC = 4, AD = AE = 4 ! " 5 , DE = 8.

AB ___ AD

= AC ___ AE

= BC ___ DE

= 1 __ 2 , so &ABC 1 &ADE by SSS .

56. Each circle: A = "(2 ) 2 = 4" cm 2 Square: A = (8 ) 2 = 64 cm 2 Shaded area: A = 64 $ 2(4") # 38.9 cm 2

57. Triangle: A = 1 _ 2 (2)(2) = 2 in. 2

Circle: A = "(2 ) 2 = 4" in. 2 Shaded area: A = 4" $ 2 # 10.6 in. 2

9-6 GEOMETRY LAB: USE GEOMETRIC PROBABILITY TO ESTIMATE !, PAGE 637

TRY THIS, PAGE 637 1. Check students’ work.

2a. A = 4 ( 1 _ 4 " r 2 ) = " r 2 2b. A = (2r ) 2 = 4 r 2

2c. P = " r 2 ___ 4 r 2

= " __ 4

3. The probability is " __ 4

, so 4 times the probability is ".

9B MULTI-STEP TEST PREP, PAGE 638

1. Area of each balloon: A = "(2 ) 2 = 4" in. 2 Area of board: A = (48)(24) = 1152 in. 2 For 15 balloons,

P = 15(4")

______ 1152

# 0.16

2. New area of each balloon: A = "(4 ) 2 = 16" in. 2

P = 15(16")

_______ 1152

= 4 ( 15(4") ______

1152 )

The probability is 4 times as great.

3. Area of board: A = (100)(60) = 6000 units 2 Area of missing triangles: A = 1 _

2 (40)(30) = 600

units 2 A = 1 _

2 (60)(30) = 900 units 2

A = 1 _ 2 (40)(20) = 400 units 2

A = 1 _ 2 (60)(40) =1200 units 2

Area of ABCD: A = 6000 $ (200 + 900 + 400 + 1200) = 2900 units 2

P = 2900 _____ 1600

# 0.48

4. 0.16 < 0.48 < 4(0.16) = 0.64, so balloon game in Problem 2 gives the best chance.

9B READY TO GO ON? PAGE 639

1.

The figure is a trapezoid.

AB = ! """""""" (2 + 2 ) 2 + (4 $ 2 ) 2 = ! "" 20 = 2 ! " 5

BC = 4 + 4 = 8

CD = ! """"""""" ($2 $ 2 ) 2 + ($2 + 4 ) 2 = ! "" 20 = 2 ! " 5

AD = 2 + 2 = 4

P = 2 ! " 5 + 8 + 2 ! " 5 + 4 = (12 + 4 ! " 5 ) units

A = 1 _ 2 (8 + 4)(2 + 2) = 24 units 2

2.

The figure is a rectangle.EF = GH = 3 + 1 = 4, FG = EH = 3 + 5 = 8P = 4 + 8 + 4 + 8 = 24 unitsA = (8)(4) = 32 units 2

3. Outer rectangle: A = (6)(6) = 36 units 2 Missing rectangles: A = 1 _

2 (5)(1) = 2.5 units 2 ,

A = 1 _ 2 (3)(5) = 7.5 units 2 , A = 1 _

2 (3)(2) = 3 units 2 ,

A = 1 _ 2 (1)(4) = 2 units 2

Area of JKLM: A = 36 $ (2.5 + 7.5 + 3 + 2) = 21 units 2

4. Outer rectangle: A = (8)(7) = 56 units 2 Missing rectangles: A = 1 _

2 (6)(2) = 6 units 2 ,

A = 1 _ 2 (2)(2) = 2 units 2 , A = 1 _

2 (3)(5) = 7.5 units 2 ,

A = 1 _ 2 (5)(5) = 12.5 units 2

Area of NPQR: A = 56 $ (6 + 2 + 7.5 + 12.5) = 28 units 2

5. Old dimensions: P = 4(7) = 28 mA = 7 2 = 49 m 2 New dimensions: P = 4(21) = 84 mA = 2 1 2 = 441 m 2 84 = 3(21), so the perimeter is tripled.441 = 9(49), so the area is multiplied by 9.


6. Old dimensions:Side length of rhombus: s 2 = 1. 5 2 + 4. 5 2 = 22.5s = ! "" 22.5 = 1.5 ! "" 10 ftP = 4s = 6 ! "" 10 ftA = 1 _

2 d 1 d 2 = 1 _

2 (3)(9) = 13.5 ft 2

New dimensions: s 2 = 0. 5 2 + 1. 5 2 = 2.5s = ! "" 2.5 = 0.5 ! "" 10 ftP = 4s = 2 ! "" 10 ftA = 1 _

2 d 1 d 2 = 1 _

2 (1)(3) = 1.5 ft 2

2 ! "" 10 = 1 _ 3 (6 ! "" 10 ) ,

so the perimeter is multiplied by 1 _ 3 .

1.5 = 1 _ 9 (13.5), so the area is multiplied by 1 _

9 .

7. Old dimensions: P = 2(15) + 2(9) = 48 cmA = (15)(9) = 135 cm 2 New dimensions: P = 2(30) + 2(18) = 96 cmA = (30)(18) = 540 cm 2 96 = 2(48), so the perimeter is doubled.540 = 4(135), so the area is multiplied by 4.

8. Old dimensions: c = ! """" 1 5 2 + 8 2 = 17 in.P = 15 + 8 + 17 = 40 in.A = 1 _

2 (15)(8) = 60 in. 2

New dimensions: c = ! """"

3 2 + ( 8 _ 3 ) 2 = 17 __

3 in.

P = 5 + 8 _ 3 + 17 __

3 = 40 __

3 in.

A = 1 _ 2 (5) ( 8 _

3 ) = 20 __

3 in. 2

40 __ 3 = 1 _

3 (40), so the perimeter is multiplied by 1 _

3 .

20 __ 3 = 1 _

9 (60), so the area is multiplied by 1 _

9 .

9. Old dimensions: s = 4 units, A = 4^2 = 16 units 2 new dimensions: A = s 2 4(16) = s 2 64 = s 2 s = 8 units8 = 2(4), so the side length is doubled.

10. Assume the batter required, B, is proportional to area.

B reg

_____ B silver

= A reg

_____ A silver

B reg

____ 1/8

= "(2.5(4) ) 2

_________ "(4 ) 2

= 100 ____ 16

= 25 ___ 4

B reg = 25 ___ 4

( 1 __ 8 ) # 0.78 cup

11. P = 120 ____ 360

= 1 __ 3

12. P = 120 + 100 _________ 360

= 220 ____ 360

= 11 ___ 18

13. P = 360 $ 95 ________ 360

= 265 ____ 360

= 53 ___ 72

14. P = 100 + 45 ________ 360

= 145 ____ 360

= 29 ___ 72

15. Commercials play for 12 min out of every 60 min.P = 12 ___

60 = 1 __

5 or 0.2


VOCABULARY, PAGE 640 1. apothem 2. center of the circle

3. geometric probability

LESSON 9-1, PAGE 640 4. P = 36 = 4s

s = 9A = 9 2 = 81 in. 2

5. A = bh28 = (4)h h = 7P = 2(4) + 2(7) = 22 cm

6. A = 1 _ 2 bh

6 x 3 y = 1 _ 2 (4xy)h

6 x 2 = 2h h = 3 x 2 in.

7. A = 1 _ 2 ( b 1 + b 2 )h

48xy = 1 _ 2 (9xy + 3xy)h

48 = 6h h = 8 ft

8. A = 1 _ 2 d 1 d 2 = 1 _

2 (21)(24) = 252 yd 2

9. A = 1 _ 2 d 1 d 2

630 x 3 y 7 = 1 _ 2 (30 x 2 y 3 ) d 2

630x y 4 = 15 d 2 d 2 = 42x y 4 in.

10. A = 1 _ 2 d 1 d 2 = 1 _

2 (32)(18) = 288 m 2


11. C = "d = " ( 2 __ " ) = 2 ft

12. C = 2"r14" = 2"r r = 7A = " r 2

= "(7 ) 2 = 49" yd 2 # 153.9 yd 2

13. A = " r 2 64 x 2 " = " r 2 64 x 2 = r 2 r = 8x md = 2r = 16x m

14. tan 36° = s _ 2 __ a = 5 __ a

a = 5 ______ tan 36°

ft

P = 5s = 5(10) = 50 ft

A = 1 __ 2 aP = 1 __

2 ( 5 ______

tan 36° ) (50) # 172.0 ft 2

15. b = 4 = 2(2), so h = 2 ! " 3 A = 1 _

2 bh = 1 _

2 (4) (2 ! " 3 ) = 4 ! " 3 in. 2 # 6.9 in. 2

16. tan 22.5° = s/2 ___ a = 4 __ a

a = 4 ________ tan 22.5°

cm

P = 8s = 8(8) = 64 cm

A = 1 __ 2 aP = 1 __

2 ( 4 ________

tan 22.5° ) (64) # 309.0 cm 2

17. d 2 = s 2 + s 2 = 2 s 2 12 2 = 2A144 = 2A A = 72 m 2



18. Area of triangle: A = 1 _ 2 (15)(15) = 112.5 ft 2

Area of semicircle: A = 1 _ 2 "(7.5 ) 2 = 28.125" ft 2

Shaded area: A = 112.5 + 28.125" # 200.9 ft 2

19. Left rectangle: A = (8)(6) = 48 cm 2 Middle rectangle: A = (6)(6 + 6) = 72 cm 2 Right rectangle: A = (4)(18) = 72 cm 2 Shaded area: A = 48 + 72 + 72 = 192 cm 2

20. Triangle: b = 8 = 2(4), so h = 4 ! " 3 A = 1 _

2 (8)4 ! " 3 = 16 ! " 3 mm 2

Missing semicircle: A = 1 _ 2 "(2 ) 2 = 2" mm 2

Shaded area: A = 16 ! " 3 $ 2" # 21.4 mm 2

LESSON 9-4, PAGE 642 21. The shape has approximately 41 whole squares

and 17 half squares. Total area is # 41 + 1 _

2 (17) = 49.5 units 2 .

22. The shape has approximately 35 whole squares and 18 half squares. Total area is # 35 + 1 _

2 (18) = 44 units 2 .

23.

The figure is a square.

s = ! """" 3 2 + 3 2 = ! "" 18 = 3 ! " 2 unitsP = 4 (3 ! " 2 ) = 12 ! " 2 units

A = (3 ! " 2 ) 2 = 9(2) = 18 units 2

24.

The figure is a right triangle.

b = 5, h = 7, so c = ! """" 5 2 + 7 2 = ! "" 74 unitsP = 5 + 7 + ! "" 74 = (12 + ! "" 74 ) unitsA = 1 _

2 (5)(7) = 17.5 units 2

25.

The figure is an isosceles trapezoid. b 1 = CD = 4 + 4 = 8, b 2 = AB = 2 + 2 = 4,

h = 3 + 1 = 4, AD = BC = ! """" 2 2 + 4 2 = ! "" 20 = 2 ! " 5 P = 8 + 2 ! " 5 + 4 + 2 ! " 5 = (12 + 4 ! " 5 ) unitsA = 1 _

2 (8 + 4)(4) = 24 units 2

26.

The figure is a parallelogram.b = EF = GH = 3 + 1 = 4, h = 3 $ 0 = 3,

EH = FG = ! """" 2 2 + 3 2 = ! "" 13 P = 4 + ! "" 13 + 4 + ! "" 13 = (8 + 2 ! "" 13 ) unitsA = (4)(3) = 12 units 2

27. Outer rectangle: A = (7)(8) = 56 units 2 Missing triangles: A = 1 _

2 (3)(1) = 1.5 units 2 ,

A = 1 _ 2 (2)(7) = 7 units 2 , A = 1 _

2 (5)(2) = 5 units 2 ,

A = 1 _ 2 (4)(6) = 12 units 2

Area of QRST: A = 56 $ (1.5 + 7 + 5 + 12) = 30.5 units 2


2 (6)(2) = 6 units 2 ,

A = 1 _ 2 (2)(3) = 3 units 2 , A = 1 _

2 (5)(3) = 7.5 units 2 ,

A = 1 _ 2 (1)(2) = 1 unit 2

Area of VWXY: A = 35 $ (6 + 3 + 7.5 + 1) = 17.5 units 2


2 (1)(1) = 0.5 units 2 ,

A = 1 _ 2 (2)(6) = 6 units 2 , A = 1 _

2 (2)(2) = 2 units 2 ,

A = 1 _ 2 (3)(5) = 7.5 units 2

Area of ABCD: A = 28 $ (0.5 + 6 + 2 + 7.5) = 12 units 2


2 (3)(2) = 3 units 2 ,

A = 1 _ 2 (1)(3) = 1.5 units 2 , A = 1 _

2 (5)(2) = 5 units 2 ,

A = 1 _ 2 (3)(3) = 4.5 units 2

Area of EFGH: A = 30 $ (3 + 1.5 + 5 + 4.5) = 16 units 2



31. Original: P = 5 + ! """" 2 2 + 5 2 + ! """" 3 2 + 5 2 = (5 + ! "" 29 + ! "" 34 ) unitsA = 1 _

2 bh = 1 _

2 (5)(5) = 12.5 units 2

Tripled: P = 15 + ! """" 6 2 + 1 5 2 + ! """" 9 2 + 1 5 2 = (15 + 3 ! "" 29 + 3 ! "" 34 ) unitsA = 1 _

2 bh = 1 _

2 (15)(15) = 112.5 units 2

15 + 3 ! "" 29 + 3 ! "" 34 = 3 (15 + 3 ! "" 29 + 3 ! "" 34 ) , so the perimeter is tripled. 112.5 = 9(12.5), so the area is multiplied by 9.

32. Original: P = 4s = 4(4) = 16 unitsA = s 2 = (4 ) 2 = 16 units 2 Doubled: P = 4(8) = 32 unitsA = (8 ) 2 = 64 units 2 32 = 2(16), so the perimeter is doubled. 64 = 4(16), so the area is multiplied by 4.

33. Original: C = 2"r = 2"(11) = 22" mA = " r 2 = "(11 ) 2 = 121" m 2 Halved: C = 2"(5.5) = 11" mA = "(5.5 ) 2 = 30.25" m 2 11" = 1 _

2 (22"), so the circumference is multiplied

by 1 _ 2 .

30.25" = 1 _ 4 (121"), so the area is multiplied by 1 _

4 .

34. Let the other 2 sides of the triangle (besides its base) have lengths x and y. Assume these side lengths are also multiplied by 4.Original: P = b + x + y = (8 + x + y) ftA = 1 _

2 bh = 1 _

2 (8)(20) = 80 ft 2

New: P = (32 + 4x + 4y) ftA = 1 _

2 (32)(80) = 1280 ft 2

32 + 4x + 4y = 4(8 + x + y), so the perimeter is multiplied by 4. 1280 = 16(80), so the area is multiplied by 16.

LESSON 9-6, PAGE 643 35. AD = 7 + 1 + 5 = 13

P ( %% AB ) = 7 __ 13

36. P (not %%

CD ) = P ( %% AB or %%

BC ) = P ( %% AB ) + P ( %% BC ) = 7 __

13 + 1 __

13 = 8 __

13

37. P ( %% AB or %%

CD ) = P ( %% AB ) + P ( %% CD ) = 7 __ 13

+ 5 __ 13

= 12 __ 13

38. P ( %% BC or %%

CD ) = P ( %% BC ) + P ( %% CD ) = 1 __ 13

+ 5 __ 13

= 6 __ 13

39. Outer rectangle: A = (40)(24) = 960 m 2 Hexagon: s = 8 = 2(4), so a = 4 ! " 3 ; P = 6(8) = 48 mA = 1 _

2 aP = 1 _

2 (4 ! " 3 ) (48) = 96 ! " 3 m 2

P = 96 ! " 3 _____ 960

# 0.17

40. Triangle: A = 1 _ 2 (10)(10) = 50 m 2

P = 50 ____ 960

# 0.05

41. Circle: A = "(6 ) 2 = 36" m 2

P = 36" + 50 ________ 960

# 0.17

42. P = 960 $ (96 ! " 3 + 50 + 36")

______________________ 960

# 0.66


1. A = 1 _ 2 bh

12 x 2 y = 1 _ 2 (3x)h

24xy = 3h h = 8xy ft

2. A = 1 _ 2 ( b 1 + b 2 )h

161.5 = 1 _ 2 ( b 1 + 13)(17)

323 = 17( b 1 + 13) 19 = b 1 + 13 b 1 = 6 cm

3. A = 1 _ 2 d 1 d 2 = 1 _

2 (25)(12) = 150 in. 2

4. C = "d = 12" in.r = 1 _

2 d = 1 _

2 (12) = 6 in.

A = " r 2 = "(6 ) 2 = 36" in. 2

5. s = 14 = 2(7), so a = 7 ! " 3 mP = 6(14) = 84 mA = 1 _

2 aP

= 1 _ 2 (7 ! " 3 ) (84)

= 294 ! " 3 m 2 # 509.2 m 2

6. Rectangle: A = (15)(8) = 120 cm 2 Missing triangle: A = 1 _

2 (6)(8) = 24 cm 2

Missing semicircle: A = 1 _ 2 "(4 ) 2 = 8" cm 2

Shaded area: A = 120 $ (24 + 8") = (96 $ 8") cm 2 # 70.9 cm 2

7. Lower rectangle: A = (26)(10) = 260 in. 2 Upper triangle: A = 1 _

2 (26 $ 16)(16 $ 10) = 1 _

2 10)(6) = 30 in. 2

Shaded area: A = 260 + 30 = 290 in. 2

8. Triangle (row 1): A = 1 _ 2 (2)(1) = 1 yd 2

Parallelogram (row 2): A = (2)(1) = 2 yd 2 Trapezoid (row 3): A = 1 _

2 (2 + 4)(1) = 3 yd 2

Triangle (row 4): A = 1 _ 2 (4)(1) = 2 yd 2

Pond: A # 1 + 2 + 3 + 2 = 8 yd 2

9.

The figure is a rectangle.

AB = CD = ! """" 1 2 + 2 2 = ! " 5

AD = BC = ! """" 4 2 + 2 2 = ! "" 20 = 2 ! " 5 P = 2 ( ! " 5 ) + 2 (2 ! " 5 ) = 6 ! " 5 units

A = (2 ! " 5 ) ( ! " 5 ) = 2(5) = 10 units 2



2 (4)(3) = 6 units 2 ,

A = 1 _ 2 (1)(5) = 2.5 units 2 , A = 1 _

2 (4)(5) = 10 units 2 ,

A = 1 _ 2 (1)(3) = 1.5 units 2

Area of EFGH: A = 40 $ (6 + 2.5 + 10 + 1.5) = 20 units 2


2 (1)(5) = 2.5 units 2 ,

A = 1 _ 2 (6)(3) = 9 units 2 , A = 1 _

2 (1)(7) = 3.5 units 2 ,

A = 1 _ 2 (6)(1) = 3 units 2

Area of JKLM: A = 56 $ (2.5 + 9 + 3.5 + 3) = 38 units 2

12. Let the other 2 sides of the triangle (besides its base) have lengths x and y. Assume these side lengths are also multiplied by 3.Original: P = b + x + y = (10 + x + y) cmA = 1 _

2 bh = 1 _

2 (10)(12) = 60 cm 2

New: P = (30 + 3x + 3y) cmA = 1 _

2 (30)(36) = 540 cm 2

30 + 3x + 3y = 3(10 + x + y), so the perimeter is multiplied by 3; 540 = 9(60), so the area is multiplied by 9.

13. Original: C = 2"r = 2"(12) = 24" mA = " r 2 = "(12 ) 2 = 144" m 2 New: C = 2"(6) = 12" mA = "(6 ) 2 = 36" m 2 12" = 1 _

2 (24") so the circumference is multiplied

by 1 _ 2 .

36" = 1 _ 4 (144"), so the area is multiplied by 1 _

4 .

14. Original: C = 9" ft, A = "(4.5 ) 2 = 20.25" ft 2 New: A = 1 _

9 (20.25")

" r 2 = 2.25" r 2 = 2.25 r = 1.5 ftC = 2"(1.5) = 3" ft3" = 1 _

3 (9"), so the circumference will be 1 _

3 as long.

15. NS = 12 + 6 + 8 = 26P ( %% NQ ) = 12 __

26 = 6 __

13

16. P (not %%

QR ) = 26 $ 6 _____ 26

= 10 __ 13

17. P ( %% NQ or %%

RS ) = P ( %% NQ ) + P ( %% NS ) = 12 __ 26

+ 8 __ 26

= 10 __ 13

18. P = 2 min ______ 18 min

= 1 __ 9


1. 36The 3rd angle measures 60°, so the triangle is equiangular and therefore equilateral. The remaining side lengths are also 12, so P = 12 + 12 + 12 = 36.

2. 36Let the shaded square have side length s.2 + s + 2 = 10 s = 6A = 6 2 = 36

3. 101m,U = m,R = 180 $ (m,P + m,R) x = 180 $ (22 + 57) = 101

4. 10Points (0 + 5, 5) = (5, 5) and (2 + 4, 4) = (6, 4) lie on !.

Slope of ! = 4 $ 5 _____ 6 $ 5

= $1

Equation of !: y $ 5 = $1(x $ 5) y $ 5 = $ x + 5 y = $x + 10y-intercept = 10

5. 135By Linear Pair Post., x + 3x = 180 4x = 180 x = 45By Vertical Angles Theorem, y = 3x = 3(45) = 135.

6. 512x + 3x + 7y = 18015x = 180 $ 7y7y > 60, and y is an integer, so y 0 9.180 $ 7(9) = 117; not divisible by 15180 $ 7(10) = 110; not divisible by 15 2180 $ 7(15) = 75 = 5(15) 2180 $ 7(25) = 5; not divisible by 15Only solution is x = 5, y = 15.


Solutions KeySpatial Reasoning10

CHAPTER


1. D 2. C

3. A 4. E

5. b = AB = 5 ! 0 = 5; h = 3 ! (!1) = 4A = 1 _

2 bh = 1 _

2 (5)(4) = 10 units 2

6. b = LM = 6 ! (!2) = 8, h = KL = 7 ! 3 = 4A = bh = (8)(4) = 32 units 2

7. r = PQ = 2 ! (!6) = 8A = ! r 2 = !(8 ) 2 = 64! units 2

8. C = 2!(8) = 16! cmA = !(8 ) 2 = 64! cm 2

9. C = !(21) = 21! ftA = !(10.5 ) 2 = 110.25! ft 2

10. C = ! ( 32 ___ ! ) = 32 in.

A = ! ( 16 ___ ! ) 2 = 256 ____ ! in 2

11. AB = " ######### (5 ! (!3) ) 2 + (6 ! 2 ) 2 = " ## 80 = 4 " # 5 $ 8.9 units

M = ( !3 + 5 _______ 2 , 2 + 6 _____

2 ) = ( 2 __

2 , 8 __

2 ) = (1, 4)

12. CD = " ########### (2 ! (!4) ) 2 + (!3 ! (!4) ) 2 = " ## 37 $ 6.1 units

M = ( !4 + 2 _______ 2

, !4 + (!3)

_________ 2 ) = ( !2 ___

2 , !7 ___

2 ) = (!1, !3.5)

13. EF = " ######### (!3 ! 0 ) 2 + (4 ! 1 ) 2 = " ## 18 = 3 " # 2 $ 4.2 units

M = ( 0 + (!3) ________

2 , 1 + 4 _____

2 ) = ( !3 ___

2 , 5 __

2 ) = (!1.5, 2.5)

14. GH = " ########### (!2 ! 2 ) 2 + (!2 ! (!5) ) 2 = " ## 25 = 5 units

M = ( 2 + (!2) ________

2 ,

!5 + (!2) _________

2 ) = ( 0 __

2 , !7 ___

2 ) = (0, !3.5)

15. r = " # A __ ! = " ### 121! _____ ! = " ## 121 = 11 cm

16. a = 2A ___ P

= 2(128)

______ 32

= 8 ft

17. b = " ### c 2 ! a 2 = " ##### (17 ) 2 ! (8 ) 2 = " ## 225 = 15 m

18. b 2 = 2A ___ h ! b 1 =

2(60) _____

6 ! 8 = 20 ! 8 = 12 in.

10-1 SOLID GEOMETRY, PAGES 654–660

CHECK IT OUT! PAGES 655–656 1a. cone

vertex: Nedges: nonebase: circle M

b. triangular prismvertices: T, U, V, W, X, Yedges:

%% TU , %%

TV , %%

UV , %%%

TW, %%

UX , %%

VY , %%%

WX , %%%

WY , %%

XY bases: &TUV, &WXY

2a. The net has a triangular central face and 3 other triangular faces. So, it forms a triangular pyramid.

b. The net has a rectangular face between 2 circular faces. So, it forms a cylinder.

3a. The cross section is a hexagon.

b. The cross section is a triangle.

4.

Cut through mdpts. of three edges that meet at one vertex.

THINK AND DISCUSS, PAGE 656 1. Both prisms and cylinders have 2 congruent parallel

bases. The bases of a prism are polygons, and the bases of a cylinder are circles. The bases of a prism are connected by ', and the bases of a cylinder are connected by a curved surface.

2.



1. cylinder 2. conevertex: Aedges: nonebase: circle B

3. rect. prismvertices: C, D, E, F, G, H, J, Kedges:

%% GH , %%

GK , %%

HJ , %%

JK , %%

GF , %%

HE , %%

JD , %%

KC , %%

FC , %%

CD ,

%% DE , %%

EF bases: rect. CDEF, rect. GHJK


4. triangular pyramidvertices: L, M, N, Pedges:

%% LM , %%

LN , %%

LP , %%%

MN , %%

MP , %%

NP base: triangle LMP

5. The net has 2 (, nonadjacent rect. faces, and remaining faces are '. So, the net forms a rectangular prism.

6. The net has 1 circular face and 1 other curved face that is a sector of a circle. So, the net forms a cone.

7. The net has 6 ( square faces and folds up without overlapping. So, the net forms a cube.

8. cross section is a circle

9. cross section is a pentagon

10. cross section is a rectangle

11. cut ) to bases 12. cut * to bases


13. cubevertices: S, T, U, V, W, X, Y, Zedges:

%% ST , %%

TU , %%

UV , %%

VS , %%%

SW , %%

TX , %%

UY , %%

VZ , %%%

WX ,

%% XY , %%

YZ , %%%

ZW bases: STUV, WXYZ

14. rect, pyramidvertices: A, B, C, D, Eedges:

%% AB , %%

BC , %%

CD , %%

AD , %%

AE , %%

BE , %%

CE , %%

DE base: ABCD

15. cylindervertices: noneedges: nonebases: circle Q, circle R

16. The net has 2 ( pentagonal faces and 5 + faces. So, it forms a pentagonal prism.

17. The net has a central triangular face, surrounded by 3 other triangular faces. So, net forms a triangular pyramid.

18. The net has a rectangular face between 2 circular faces. So, it forms a cylinder.

19. square 20. rectangle

21. rectangle

22. cut ) to ground 23. cut * to ground

24–27. Possible answers:

24. cube 25. rectangular prism

26. cylinder 27. hexagonal prism

28. Possible answer: The figure is a hexagonal prism whose bases are regular hexagons with 7-in. sides. Height of the prism is 13 in.

29. Possible answer: The figure is a cylinder whose bases each have radius 12 ft. Height of the cylinder is 9 ft.

30. Possible answer: The figure is a square prism with 36 cm by 36 cm bases and a height of 108 cm.

31.

32.

33.

34.

35.

36.

37a. pentagonal prism

c.

one

38. B is incorrect; bases are ( reg. hexagons. So, opposite sides of cross section must be congruent.

39.

40. Figure b; when figure is folded, shaded faces will overlap.

TEST PREP, PAGE 659

41. D 42. F

43. B 44. G


45. 46.


47. 48.

49. 50.

51a. A and B, C and F, D and G, E and H

b.

One


52. y = ! x 2 53. y = x 2 + 6

54. Possible answer: y = 1 _ 2 x 2

55. largest , is opp. longest side: ,Bsmallest , is opp. shortest side: ,C

56. largest: ,Esmallest: ,D

57. largest: ,Ismallest: ,H

58. no; 2 : 4 - 8 : 12

59. yes; 7 : 11.9 = 24 : 40.8 = 25 : 42.5 = 10 : 17

10-2 REPRESENTATIONS OF THREE-DIMENSIONAL FIGURES, PAGES 661–668

CHEC K IT OUT! PAGES 661–664 1. 2.

3a.

b.

4. no

THINK AND DISCUSS, PAGE 664 1. All 6 views are squares.

2. No; vertical lines do not meet at a vanishing point.

3.



1. perspective

2. 3.


4. 5.

6. 7.

8. one-point perspective:

two-point perspective:



10. yes 11. no

12. no 13. no


14.

15.

16.

17.


18.

19.





22. yes 23. yes

24. no 25. no

26.

27.

28.

29a.

b.

c. 3 at front + 3 along sides + 3 at rear = 9

30.

31.

32.

33. Possible answer: a right cylinder and a right square prism in which the diameter of the cylinder is equal to the side length of the square base of the prism and the heights are the same.

34a. two-point perspective


b. Extend a pair of ) lines to meet at each vanishing point.

TEST PREP, PAGE 667

35. B 36. H


Some edges that are ) on the 3-dimensional object are not ) in perspective drawing. If they were extended, they would meet at the vanishing point of drawing. All the ) edges of prism are also ) in the isometric drawing.


38.

39.

40.

41a.

b.

c.

42. Check students’ drawings.


43. x + y = 30 (1)2x ! 2y = 20 (2) x ! y = 10 (2) ÷ 2 x + y = 30 + (1) 2x = 40 x = 20y = 30 ! x = 30 ! 20 = 10

44. x ! y = 74x + y = 38 5x = 45 x = 9(9) ! y = 7 y = 2

45. y = x + 5 x + y = 5 x + 2y = x + 10 2y = 10 y = 5x = y ! 5 = 5 ! 5 = 0

46. slope of %%

AB = 1 ! 2 _____ 6 ! 4

= ! 1 __ 2

47. slope of %%

AC = 0 ! 2 _____ 3 ! 4

= !2 ___ !1

= 2

48. slope of %%

AD = 0 ! 2 _____ 2 ! 4

= !2 ___ !2

= 1

49. 2 pentagons and 5 parallelograms

50. 6 squares

51. 4 triangles

USING TECHNOLOGY, PAGE 668 1. Check students’ drawings.


2. Possible answer: First, draw a horiz. line to represent horizon and locate 2 vanishing points A and B on the line. Then, draw a vert. seg.

%% CD and

draw segs. %%

CA , %%

CB , %%

DA , and %%

DB .Draw two more vert. segs., one with endpoints on

%% CA and

%% DA , and other with endpoints on

%% CB

and %%

DB . Connect each endpoint of these segs. to both vanishing points to form a rectangular prism.

GEOMETRY LAB: USE NETS TO CREATE POLYHEDRONS, PAGE 669

TRY THIS, PAGE 669

1. Polyhedron V E F V - E + F

Tetrahedron 4 6 4 2

Octahedron 6 12 8 2

Icosahedron 12 30 20 2

Cube 8 12 6 2

Dodecahedron 20 30 12 2

2. V ! E + F is always equal to 2.

10-3 FORMULAS IN THREE DIMENSIONS, PAGES 670–677

CHECK IT OUT! PAGES 670–673 1a. V = 6, E = 12, F = 8

6 ! 12 + 8 . 2 2 = 2

b. V = 7, E = 12, F = 77 ! 12 + 7 . 2 2 = 2

2. d = " ###### 5 2 + 5 2 + 5 2 = " ###### 25 + 25 + 25 = " ## 75 = 5 " # 3 $ 8.7 cm

3. Graph center of base at (0, 0, 0). Since height is 7, graph vertex at (0, 0, 7). Radius is 5, so, base will cross x-axis at (0, 5, 0) and y-axis at (5, 0, 0). Connect vertex to base.

4a. d = " ############## ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 + ( z 2 ! z 1 ) 2

= " ############# (6 ! 0) 2 + (0 ! 9 ) 2 + (12 ! 5 ) 2 = / ###### 36 + 81 + 49 = / ## 166 $ 12.9 units

M ( x 1 + x 2 _______

2 ,

y 1 + y 2 _______

2 ,

z 1 + z 2 _______

2 )

M ( 0 + 6 _____ 2 , 9 + 0 _____

2 , 5 + 12 ______

2 )

M(3, 4.5, 8.5)

b. d = " ############## ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 + ( z 2 ! z 1 ) 2

= " ############## (12 ! 5 ) 2 + (16 ! 8 ) 2 + (20 ! 16 ) 2 = " ###### 49 + 64 + 16 = " ## 129 $ 11.4 units

M ( x 1 + x 2 _______

2 ,

y 1 + y 2 _______

2 ,

z 1 + z 2 _______

2 )

M ( 5 + 12 ______ 2 , 8 + 16 ______

2 , 16 + 20 _______

2 )

M(8.5, 12, 18)

5. Locations of divers on surface can be represented by ordered triples (18, 9, 0) and (!15, !6, 0).

d = " ############## ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 + ( z 2 ! z 1 ) 2

= " ############### (!15 ! 18 ) 2 + (!6 ! 9 ) 2 + (0 ! 0 ) 2 = " ## 1314 $ 36.2 ft

THINK AND DISCUSS, PAGE 673 1. Find the difference of x-coordinates, the

difference of y-coordinates, and the difference of z-coordinates. Square each result, and add. The distance is the square root of the sum.

2.



1. because the bases are circles, which are not polygons

2. V = 6, E = 9, F = 56 ! 9 + 5 . 2 2 = 2

3. V = 6, E = 10, F = 66 ! 10 + 6 . 2 2 = 2

4. V = 10, E = 20, F = 1210 ! 20 + 12 . 2 2 = 2

5. d = " ###### 4 2 + 8 2 + 12 2 = " ###### 16 + 64 + 144 = " ## 224 = 4 " ## 16 $ 15.0 ft

6. 13 2 = 6 2 + 10 2 + h 2 h 2 = 13 2 ! 6 2 ! 10 2

h = " ###### 13 2 ! 6 2 ! 10 2 = " ####### 169 ! 36 ! 100 = " # 33 $ 5.7 in.

7. d = " ####### 12 2 + 12 2 + 1 2 = " ###### 144 + 144 + 1 = " ## 289 = 17 in.

8. Graph the center of base at (0, 0, 0).Since the height is 4, graph the vertex at (0, 0, 4). The radius is 8, so, the base will cross the x-axis at (0, 8, 0) and the y-axis at (8, 0, 0). Connect the vertex to base.


9. Graph the center of the bottom base at (0, 0, 0).Since the height is 4, graph the center of the top base at (0, 0, 4). The radius is 3, so the bottom base will cross the x-axis at (0, 3, 0) and the y-axis at (3, 0, 0). Draw the top base parallel to the bottom base and connect the bases.

10. cube has 8 vertices:(0, 0, 0), (7, 0, 0), (7, 7, 0), (0, 7, 0), (0, 0, 7), (7, 0, 7), (7, 7, 7), (0, 7, 7)

11. d = " ############## ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 + ( z 2 ! z 1 ) 2

= " ############# (5 ! 0 ) 2 + (9 ! 0 ) 2 + (10 ! 0 ) 2 = " ###### 25 + 81 + 100 = " ## 206 $ 14.4 units

M ( x 1 + x 2 _______

2 ,

y 1 + y 2 _______

2 ,

z 1 + z 2 _______

2 )

M ( 0 + 5 _____ 2

, 0 + 9 _____ 2 , 0 + 10 ______

2 )

M (2.5, 4.5, 5)

12. d = " ############## ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 + ( z 2 ! z 1 ) 2

= " ############# (7 ! 0 ) 2 + (0 ! 3 ) 2 + (14 ! 8 ) 2 = " ##### 49 + 9 + 36 = " ## 94 $ 9.7 units

M ( x 1 + x 2 _______

2 ,

y 1 + y 2 _______

2 ,

z 1 + z 2 _______

2 )

M ( 0 + 7 _____ 2

, 3 + 0 _____ 2 , 8 + 14 ______

2 )

M (3.5, 1.5, 11)

13. d = " ############## ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 + ( z 2 ! z 1 ) 2

= " ############## (9 ! 4 ) 2 + (12 ! 6 ) 2 + (15 ! 10 ) 2 = " ###### 25 + 36 + 25 = " ## 86 $ 9.3 units

M ( x 1 + x 2 _______

2 ,

y 1 + y 2 _______

2 ,

z 1 + z 2 _______

2 )

M ( 4 + 9 _____ 2

, 6 + 12 ______ 2 , 10 + 15 _______

2 )

M (6.5, 9, 12.5)

14. Represent the locations of the starting point and the camp by ordered triples (0, 0, 0) and (3, 7, 0.6).

d = " ############## ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 + ( z 2 ! z 1 ) 2

= " ############# (3 ! 0 ) 2 + (7 ! 0 ) 2 + (0.6 ! 0 ) 2 = " ### 58.36 $ 7.6 km


15. V = 8, E = 12, F = 68 ! 12 + 6 . 22 = 2

16. V = 8, E = 18, F = 128 ! 18 + 12 . 2 2 = 2

17. V = 11, E = 20, F = 1111 ! 20 ! 11 . 2 2 = 2

18. d = " ###### 7 2 + 8

2 + 16

2

= " ###### 49 + 64 + 256 = " ## 369 = 3 " ## 41 $ 19.2 yd

19. 17 2 = 15 2 + 6 2 + h 2 h 2 = 17 2 ! 15 2 ! 6 2

h = " ###### 17 2 ! 15 2 ! 6 2 = " ###### 289 ! 225 ! 36 = " ## 28 = 2 " # 7 $ 5.3 m

20. 8 2 = s 2 + s 2 + s 2 64 = 3 s 2 8 = s " # 3

s = 8 ____ " # 3

= 8 " # 3 ____ 3

$ 4.6 cm

21. Graph the center of the bottom base at (0, 0, 0).Since the height is 3, graph the center of the top base at (0, 0, 3).The radius is 5, so, the bottom base will cross the x-axis at (0, 5, 0) and the y-axis at (5, 0, 0). Draw the top base ) to the bottom base and connect bases.

22. Graph the center of the base at (0, 0, 0). Since the height is 4, graph the vertex at (0, 0, 4). The radius is 2, so, the base will cross the x-axis at (0, 2, 0) and the y-axis at (2, 0, 0). Connect the vertex to the base.

23. prism has 8 vertices:(0, 0, 0), (5, 0, 0), (5, 5, 0), (0, 5, 0), (0, 0, 3), (5, 0, 3), (5, 5, 3), (0, 5, 3),

24. d = " ############ (4 ! 0 ) 2 + (4 ! 0 ) 2 + (4 ! 0 ) 2 = " ###### 16 + 16 + 16 = " ## 48 = 4 " # 3 $ 6.9 units

M ( 0 + 4 _____ 2 , 0 + 4 _____

2 , 0 + 4 _____

2 ) = M(2, 2, 2)

25. d = " ############# (9 ! 2 ) 2 + (10 ! 3 ) 2 + (10 ! 7 ) 2 = " ##### 49 + 49 + 9 = " ## 107 $ 10.3 units

M ( 2 + 9 _____ 2 , 3 + 10 ______

2 , 10 + 7 ______

2 ) = M(5.5, 6.5, 8.5)


26. d = " ############# (8 ! 2 ) 2 + (8 ! 5 ) 2 + (10 ! 3 ) 2 = " ##### 36 + 9 + 49 = " ## 94 $ 9.7 units

M ( 2 + 8 _____ 2 , 5 + 8 _____

2 , 3 + 10 ______

2 ) = M(5, 6.5, 6.5)

27. Let the cloud and the raindrop hitting ground have the coordinates (0, 0, 6500) and (!700, 500, 0). The distance traveled by the raindrop is

d = " ################ (!700 ! 0 ) 2 + (500 ! 0 ) 2 + (0 ! 6500 ) 2

= " ##### 42,990,000 $ 6557 ft.

28. original:

d = " ###### 12 2 + 3 2 + 4 2 = " ###### 144 + 9 + 16 = " ## 169 = 13 ftdoubled:

d = " ###### 24 2 + 6 2 + 8 2 = " ###### 576 + 36 + 64 = " ## 676 = 36 ftThe diagonal is doubled.

29. V ! E + F = 28 ! 12 + F = 2 !4 + F = 2 F = 6

30. V ! E + F = 2 V ! 9 + 5 = 2 V ! 4 = 2 V = 6

31. V ! E + F = 2 7 ! E + 7 = 2 14 ! E = 2 E = 12

32. V = n + n = 2nE = n + n + n = 3nF = 1 + n + 1 = n + 2V ! E + F = 2n ! 3n + (n + 2) = 2

33. V = n + 1E = n + n = 2nF = n + 1V ! E + F = (n + 1) ! 2n + (n + 1) = 2

34a. K(0, 2.5, 6), L(0, 5, 0), N(7, 2.5, 6), P(7, 5, 0)

b. KP = " ############# (7 ! 0 ) 2 + (5 ! 2.5 ) 2 + (0 ! 6 ) 2 = " ### 91.25 $ 9.6 ft

35. d = " ###### 6 2 + 6 2 + 6 2 = " ## 108 = 6 " # 3

36. w = " ####### 65 2 ! 24 2 ! 60 2 = " ## 49 = 7

37. h = " ####### 24 2 ! 12 2 ! 18 2 = " ## 108 = 6 " # 3

38. " = " ###### 4 2 ! 2 2 + 3 2 = " # 3

39. bottom base extends forward to (5, 2, 5)center of top base: (1, 2, 10)Draw the top base ) to bottom base.

40. base extends forward to (6, 2, 6)vertex: (3, 2, 13)

41. vertices of cube:(4, 2, 3), (10, 2, 3), (10, 8, 3), (4, 8, 3),(4, 2, 9), (10, 2, 9), (10, 8, 9), (4, 8, 9)

42.

43. base extends forward to (8, 7, 1)


44. bottom base extends forward to (7, 3, 7)Draw the top base parallel to bottom base.

45.

d = " ############ (3 ! 1 ) 2 + (2 ! 2 ) 2 + (1 ! 3 ) 2

= " # 8 = 2 " # 2 $ 2.8 units

M ( 1 + 3 _____ 2

, 2 + 2 _____ 2 , 3 + 1 _____

2 ) = M (2, 2, 2)

46.

d = " ############ (7 ! 4 ) 2 + (4 ! 3 ) 2 + (4 ! 3 ) 2

= " # 11 $ 3.3 units

M ( 4 + 7 _____ 2

, 3 + 4 _____ 2 , 3 + 4 _____

2 ) = M (5.5, 3.5, 3.5)

47.

d = " ############ (3 ! 4 ) 2 + (1 ! 7 ) 2 + (5 ! 8 ) 2

= " ## 46 $ 6.8 units

M ( 4 + 3 _____ 2

, 7 + 1 _____ 2 , 8 + 5 _____

2 ) = M (3.5, 4, 6.5)

48.

d = " ############ (8 ! 0 ) 2 + (3 ! 0 ) 2 + (6 ! 0 ) 2

= " ## 109 $ 10.4 units

M ( 0 + 8 _____ 2 , 0 + 3 _____

2 , 0 + 6 _____

2 ) = M (4, 1.5, 3)

49.

d = " ############ (2 ! 6 ) 2 + (2 ! 1 ) 2 + (6 ! 8 ) 2

= " ## 21 $ 4.6 units

M ( 6 + 2 _____ 2 , 1 + 2 _____

2 , 8 + 6 _____

2 ) = M (4, 1.5, 7)

50.

d = " ############ (3 ! 2 ) 2 + (6 ! 8 ) 2 + (3 ! 5 ) 2

= " # 9 = 3 units

M ( 2 + 3 _____ 2 , 8 + 6 _____

2 , 5 + 3 _____

2 ) = M(2.5, 7, 4)

51. 13 = " ############### (3 ! 6 ) 2 + (3 ! (!1) ) 2 + (z ! (!3) ) 2 1 3 2 = (3 ! 6 ) 2 + (3 ! (!1) ) 2 + (z ! (!3) ) 2 169 = 9 + 16 + z 2 + 6z + 9 0 = z 2 + 6z ! 135 0 = (z + 15)(z ! 9) z = 9 or !15




d $ " ######## 0. 4 2 + 0. 9 2 + 1. 5 2 $ " ## 32.2 $ 1.8 in.

54. Possible answer: a segment that connects a vertex of one base to the opposite vertex of the other base

AB = " ############ (1 ! 0 ) 2 + (0 ! 0 ) 2 + (0 ! 0 ) 2 = 1 unit

AC = " ############ (1 ! 0 ) 2 + (2 ! 0 ) 2 + (0 ! 0 ) 2 = " # 5

$ 2.2 units

AG = " ############ (1 ! 0 ) 2 + (2 ! 0 ) 2 + (2 ! 0 ) 2 = 3 units

55. AB = " ############## (5 ! 3 ) 2 + (8 ! 2 ) 2 + (6 ! (!3) ) 2

= " ## 121 = 11

AC = " ############### (!3 ! 3 ) 2 + (!5 ! 2 ) 2 + (3 ! (!3) ) 2

= " ## 121 = 11

BC = " ############## (!3 ! 5 ) 2 + (!5 ! 8 ) 2 + (3 ! 6 ) 2

= " ## 242 = 11 " # 2 AB = AC and A B 2 + AC 2 = BC 2 &ABC is a right isoceles &.

56. 10 cm; the segment is the hypotenuse of a right & in which one leg is a diameter of one base, and the opposite vertex is on the other base. Legs measure 6 cm and 2(4) = 8 cm, so, segment length is 10 cm. Segment is longest because a diameter is longest possible segment in a circle.

TEST PREP, PAGE 677

57. C 58. Hd = " ###### 1 2 2 + 8 2 + 6 2

= " ## 244 $ 15.6 m

59. Bd = " ############# (9 ! 7 ) 2 + (3 ! 14 ) 2 + (12 ! 8 ) 2 = " ## 189 = 11.9 units


60. legs of rt. & measure h and a __ 2

+ a + a __ 2 = 2a

d = " ##### (2a ) 2 + h 2 = " #### 4 a 2 + h 2

61. AB = " ############## (1 ! (!1)) 2 + (!2 ! 2 ) 2 + (6 ! 4 ) 2 = " ## 24 = 2 " # 6

AC = " ############## (3 ! (!1)) 2 + (!6 ! 2 ) 2 + (8 ! 4 ) 2 = " ## 96 = 4 " # 6

BC = " ############## (3 ! 1) 2 + (!6 ! (!2) ) 2 + (8 ! 6 ) 2 = " ## 24 = 2 " # 6 AB + BC = AC. So, points are collinear.

62. AB

= " ############# ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 + ( z 2 ! z 1 ) 2

AM

= " ####################

( x 1 + x 2 ______

2 ! x 1 ) 2 + ( y 1 + y 2

______ 2 ! y 1 ) 2 + ( z 1 + z 2

______ 2 ! z 1 ) 2

= " ################

( x 2 __

2 !

x 1 __

2 ) 2 + ( y 2

__ 2 !

y 1 __

2 ) 2 + ( z 2

__ 2 !

z 1 __

2 ) 2 , and

MB

= " #####################

( x 2 ! x 1 + x 2

_______ 2 ) 2 + ( y 2 !

y 1 + y 2 _______

2 ) 2 + ( z 2 !

z 1 + z 2 _______

2 ) 2

= " ################

( x 2 __

2 !

x 1 __

2 ) 2 + ( y 2

__ 2 !

y 1 __

2 ) 2 + ( z 2

__ 2 !

z 1 __

2 ) 2

So AM = MB. Also, AM + MB

= 2 " ################

( x 2 __

2 !

x 1 __

2 ) 2 + ( y 2

__ 2 !

y 1 __

2 ) 2 + ( z 2

__ 2 !

z 1 __

2 ) 2

= 2 " ################# 1 __ 4 ( x 2 ! x 1 ) 2 + 1 __

4 ( y 2 ! y 1 ) 2 + 1 __

4 ( z 2 ! z 1 ) 2

= " ############## ( x 2 ! x 1 ) 2 + ( y 2 ! y 1 ) 2 + ( z 2 ! z 1 ) 2 = AB

So A, M, and B are collinear. Since M is on %%

AB and AM = MB, M is the midpoint of

%% AB by def. midpt.

63. AG = " ############ (a ! 0 ) 2 + (b ! 0 ) 2 + (c ! 0 ) 2

= " ###### a 2 + b 2 + c 2

BH = " ############ (0 ! a ) 2 + (b ! 0 ) 2 + (c ! 0 ) 2

= " ###### a 2 + b 2 + c 2 AG = BH, so

%% AG (

%% BH . By the definition of ( segs.

Let M and N be mdpts. of %%

AG and %%

BH .

M = ( 0 + a _____ 2 , 0 + b _____

2 , 0 + c _____

2 ) = ( a __

2 , b __

2 , c __

2 )

N = ( 0 + a _____ 2 , 0 + b _____

2 , 0 + c _____

2 ) = ( a __

2 , b __

2 , c __

2 )

M = N. These segs. have the same midpoint, so they bisect each other..


64. 30 + 25 = 55 65. 0–9 yr old

66. A = b(2h) = 2bh 67. A = 1 _ 2 h ( 1 _

2 b 1 + b 2 )

68. A = !(3r ) 2 = 9! r 2 69. cone

70. none 71. 0C



1. A-frame tent:

pyramid tent:

2. A-frame: floor is a rectangle ABED; sleeping area = 8(7) = 56 ftpyramid: floor is a square PQRS;sleeping area = 8 2 = 56 ftThe pyramid tent has greater sleeping area.

3. A-frame: vertical distance from A to C is 7 ! 0 = 7 ftpyramid: vertical distance from P to T is 8 ! 0 = 8 ftThe pyramid tent has more headroom.

4. EF = " ############# (8 ! 8 ) 2 + (3.5 ! 7 ) 2 + (7 ! 0 ) 2 = " ### 61.25 $ 7.8 ft

TR = " ############ (8 ! 4 ) 2 + (8 ! 4 ) 2 + (8 ! 0 ) 2 = " ## 96 $ 9.8 ftThe camper should purchase the A-frame tent.


1. hexagonal pyramidvertices: A, B, C, D, E, F, Gedges:

%% AB , %%

AC , %%

AD , %%

AE , %%

AF , %%

AG , %%

BC , %%

CD , %%

DE ,

%% EF , %%

FG , %%

BG base: hexagon BCDEFG

2. conevertices: Hedges: nonebase: circle J

3. rectangular prismvertices: K, L, M, N, P, Q, R, Sedges:

%% KL , %%

LM , %%%

MN , %%

NK , %%

PQ , %%

QR , %%

RS , %%

PS , %%

KP ,

%% LQ ,

%%% MR , %%

NS ;bases: KLMN, PQRS

4. cube 5. cone

6. pentagonal pyramid 7. square

8. rectangle 9. circle

10.

11.

12.

13.

14. V = 8, E = 12, F = 6V ! E + F = 8 ! 12 + 6 = 2

15. V = 7, E = 12, F = 7V ! E + F = 7 ! 12 + 7 = 2

16. V = 4, E = 6, F = 4V ! E + F = 4 ! 6 + 4 = 2


17. Let the coordinates of the nest and the bird be (0, 0, 0) and (6, ! 7, 6).

d = " ############# (6 ! 0 ) 2 + (!7 ! 0 ) 2 + (6 ! 0 ) 2

= " ## 121 = 11 ft

18. d = " ############# (4 ! 0 ) 2 + (6 ! 0 ) 2 + (12 ! 0 ) 2

= " ## 196 = 14 units

M ( 0 + 4 _____ 2 , 0 + 6 _____

2 , 0 + 12 ______

2 ) = M(2, 3, 6)

19. d = " ############## (5 ! 3 ) 2 + (!5 ! 1 ) 2 + (7 ! (!2) ) 2

= " ## 121 = 11 units

M ( 3 + 5 _____ 2 ,

1 + (!5) ________

2 , !2 + 7 _______

2 ) = M(4, !2, 2.5)

20. d = " ############ (7 ! 3 ) 2 + (2 ! 5 ) 2 + (0 ! 9 ) 2

= " ## 106 $ 10.3 units

M ( 3 + 7 _____ 2 , 5 + 2 _____

2 , 9 + 0 _____

2 ) = M(5, 3.5, 4.5)

10-4 SURFACE AREA OF PRISMS AND CYLINDERS, PAGES 680–687

CHECK IT OUT! PAGES 681–683 1. L = Ph

= (4s)(s) = 4(8)(8) = 256 cm 2

S = L + 2B = L + 2 s 2 = 256 + 2(8 ) 2 = 384 cm 2

2. Step 1 Use the base area to find the radius. A = ! r 2 49! = ! r 2 49 = r 2 r = 7 in.Step 2 Use the radius to find the lateral area and the base area.height is 2 times radius, or 14 in.L = 2!rh = 2!(7)(14) = 196! in 2 S = 2!rh + 2! r 2 = 196! + 2(49!) = 294! in 2

3. Surface area of right rectangular prism isS = Ph + 2B

= 26(5) + 2(9)(4) = 202 cm 2 .A right cylinder is added to a rectangular prism.Lateral area of cylinder isL = 2!rh

= 2!(2)(3) = 12! cm 2 .The base area of the cylinder is not added, just raised through 3 cm.The surface area of the composite figure is the sum of the areas of all surfaces on the exterior of figure.S = (prism surface area) + (cylinder lateral area)

= 202 + 12! $ 239.7 cm 2

4. original dimensions:S = 2!rh + 2! r 2

= 2!(11)(14) + 2!(11 ) 2 = 550! cm 2

height, diameter multiplied by 1 _ 2 :

S = 2!rh + 2! r 2 = 2!(5.5)(7) + 2!(5.5 ) 2 = 137.5! cm 2

Notice 137.5! = 1 _ 4 (550!). If the height and

diameter are multiplied by 1 _ 2 , the surface area is

multiplied by 1 _ 4 .

5. The 5 cm by 5 cm by 1 cm prism has a surface area of S = Ph + 2B = 20(1) + 2(5)(5) = 70 cm 2 , which is greater than the 2 cm by 3 cm by 4 cm prism and about the same as the half cylinder. It will melt faster than 2 cm by 3 cm by 4 cm prism and at about same rate as the half cylinder.

THINK AND DISCUSS, PAGE 683 1. Use the radius of the base to find the base area,

then add twice the base area to the lateral area.

2. An oblique prism has at least one lateral face that is not a rectangle. All the lateral faces of a right prism are rectangles.

3.



1. 5 lateral faces

2. L = Ph= 24(3) = 72 ft 2

S = L + 2B= 72 + 2(5)(7) = 142 ft 2

3. The base is a rt. &, because 3, 4, 5 is a Pythag. triple.L = Ph

= 12(2) = 24 cm 2 S = L + 2B

= 24 + 2 ( 1 _ 2 (3)(4))

= 36 cm 2

4. L = Ph= (4s)(s)= 4(9)(9) = 324 in 2

S = L + 2B= L + 2 s 2 = 324 + 2(9 ) 2 = 486 in 2

5. L = 2!rh = 2!(3)(4) = 24! ft 2 S = 2!rh + 2! r 2 = 24! + 2!(3 ) 2 = 42! ft 2

6. L = 2!rh = 2!(7.5)(12) = 180! yd 2 S = 2!rh + 2! r 2 = 180! + 2!(7.5 ) 2 = 292.5! yd 2

7. Step 1 Use the base area to find the radius. A = ! r 2 64! = ! r 2 64 = r 2 r = 8 mStep 2 Use the radius to find the lateral area and base area. The height is 3 less than the radius, or 5 m.L = 2!rh = 2!(8)(5) = 80! m 2 S = 2!rh + 2! r 2 = 80! + 2!(8 ) 2 = 208! m 2


8. surface area of the right rectangular prism isS = Ph + 2B

= 44(12) + 2(14)(8) = 752 ft 2 .A right cylinder is added to right rectangular prism.The lateral area of the cylinder isL = 2!rh

= 2!(4)(8) = 64! ft 2 .The base area of the cylinder is not added, just raised through 8 ft.The surface area of the composite figure is the sum of the areas of all the surfaces on the exterior of the figure.S = (prism surface area) + (cylinder lateral area)

= 752 + 64! $ 953.1 ft 2

9. surface area of cylinder isS = 2!rh + 2! r 2

= 2!(14)(14) + 2!(14 ) 2 = 784! ft 2 A right rectangular prism is removed from the cylinder. The lateral area is L = Ph = 40(14) = 560 ft 2 .The base area is B = 14(6) = 84 ft 2 .The surface area of the composite figure is the sum of the areas of all the surfaces on the exterior of the figure.S = (cylinder surface area) + (prism rectangle area)

! (prism base area)= 784! + 560 ! 2(84)= 784! + 392 $ 2855.0 ft 2

10. original dimensions:S = 2!rh + 2! r 2

= 2!(8)(4) + 2!(8 ) 2 = 192! yd 2

dimensions multiplied by 1 _ 2 :

S = 2!rh + 2! r 2 = 2!(4)(2) + 2!(4 ) 2 = 48! yd 2

Notice 48! = 1 _ 4 (192!). If the dimensions are

multiplied by 1 _ 2 , the surface area is multiplied by 1 _

4 .

11. original dimensions:S = Ph + 2B

= 32(6) + 2(8)(8) = 320 yd 2 dimensions multiplied by 2 _

3 :

S = Ph + 2B= 64 __

3 (4) + 2 ( 16 __

3 ) ( 16 __

3 ) = 1280 ____

9 yd 2

Notice 1280 ____ 9 = 4 _

9 (320). If the dimensions are multiplied

by 2 __ 3

, the surface area is multiplied by 4 __ 9 .

12. 16-in. bulb: L = 2!rh = 2!(1)(16) = 32! in 2 23-in. bulb: L = 2!rh = 2!3/4(23) = 34.5! in 2 23-in. bulb will produce more light


13. L = Ph= 20(10) = 200 cm 2

S = L + 2B= 200 + 2(5)(5) = 250 cm 2

14. L = Ph= 6(12)(15) = 1080 m 2

apothem of base is 6 " # 3 mS = L + 2B

= 1080 + 2 ( 1 _ 2 (6 " # 3 ) (72))

= 1080 + 432 " # 3 $ 1828.2 m 2

15. L = Ph= 3(8)(14) = 336 ft 2

altitude of base is 4 " # 3 ftS = L + 2B

= 336 + 2 ( 1 _ 2 (8) (4 " # 3 ) )

= 336 + 32 " # 3 $ 391.4 ft 2

16. L = 2!rh = 2!(5.5)(7) = 77! in 2 S = 2!rh + 2! r 2 = 77! + 2!(5.5 ) 2 = 137.5! in 2

17. L = 2!rh = 2!(4)(23) = 184! cm 2 S = 2!rh + 2! r 2 = 184! + 2!(4 ) 2 = 216! cm 2

18. Step 1 Use the base circumference to find the radius. C = 2!r16! = 2!r r = 8 ydStep 2 Use the radius to find the lateral area and the base area. The height is 3 times the radius, or 24 yd.L = 2!rh = 2!(8)(24) = 384! yd 2 S = 2!rh + 2! r 2 = 384! + 2!(8 ) 2 = 512! yd 2

19. The base of the right triangular prism is a rt. &, because 6, 8, 10 is a Pythagorean triple.The surface area of the right triangular prism isS = Ph + 2B

= 24(9) + 2 ( 1 _ 2 (6)(8)) = 264 cm 2 .

A right cylinder is removed from the triangular prism.The lateral area of cylinder isL = 2!rh

= 2!(2)(9) = 36! cm 2 .The base area of cylinder is B = ! r 2 = !(2 ) 2 = 4! cm 2 .The surface area of the composite figure is the sum of areas of all surfaces on the exterior of the figure.S = (prism surface area) + (cylinder lateral area)

! (cylinder base area)= 264 + 36! ! 2(4!) = 264 + 28! $ 352.0 cm 2

20. The surface area of right rectangular prism isS = Ph + 2B

= 8(0.5) + 2(2)(2) = 12 ft 2 .A right cylinder is added to the rectangular prism.The lateral area of cylinder isL = 2!rh

= 2!(0.5)(2) = 2! ft 2 .The base area of cylinder is not added, just lowered through 2 ft.The surface area of the composite figure is the sum of the areas of all surfaces on the exterior of figure.S = (prism surface area) + (cylinder lateral area)

= 12 + 2! $ 18.3 ft 2


21. original: S = 2!rh + 2! r 2

= 2!(4.5)(11) + 2! (4.5) 2 = 139.5! ft 2

dimensions tripled:S = 2!rh + 2! r 2

= 2!(13.5)(33) + 2! (13.5) 2 = 1255.5! ft 2

1255.5! = 9(139.5!). So, surface area is multiplied by 9.

22. original: S = Ph + 2B

= 42(3) + 2(12)(9) = 342 ft 2 dimensions doubled:S = Ph + 2B

= 84(6) + 2(24)(18) = 1368 ft 2 1368 = 4(342). So, surface area is multiplied by 4.

23. left cell:S = Ph + 2B

= 90(7) + 2(35)(10) = 1330 µm 2 right cell:S = Ph + 2B

= 52(15) + 2(15)(11) = 1110 µm 2 The cell that measures 35 µm by 7 µm by 10 µm should absorb at a greater rate.

24. S = 2!rh + 2! r 2 160! = 2!(5)h + 2!(5 ) 2 160! = 10!h + 50!110! = 10!h h = 11 ft

25. S = Ph + 2B286 = 36h + 2(10)(8)286 = 36h + 160126 = 36h h = 3.5 m

26. L = Ph1368 = 6(12)h1368 = 72h h = 19 m

27. The bases are rt. 1 with leg lengths 2 and 5 units,

and hypotenuse length " #### 2 2 + 5 2 = " ## 29 units; height is 9 units.S = Ph + 2B

= (2 + 5 + " ## 29 ) (9) + 2 ( 1 _ 2 (2)(5))

= 63 + 9 " ## 29 + 10= 73 + 9 " ## 29 $ 121.5 units 2

28. S = 2!rh + 2! r 2 = 2!(9.525)(1.55) + 2!(9.525 ) 2 $ 662.81 mm 2

29. S = 2!rh + 2! r 2 = 2!(10.605)(1.95) + 2!(10.605 ) 2 $ 836.58 mm 2

30. S = 2!rh + 2! r 2 = 2!(8.955)(1.35) + 2!(8.955 ) 2 $ 579.82 mm 2

31. S = 2!rh + 2! r 2 = 2!(12.13)(1.75) + 2!(12.13 ) 2 $ 1057.86 mm 2

32. Possible answer: triple edge lengthsLet ", w, and h represent the original dimensions.original: S = Ph + 2B

= (2" + 2w)h + 2"w= 2"h + 2wh + 2"w

tripled: S = Ph + 2B

= (2(3") + 2(3w))(3h) + 2(3")(3w)= 18"h + 18wh + 18"w= 9(2"h + 2wh + 2"w)

33. Possible answer: Multiply the radius and the height by 1 _

2 .

original: S = 2!rh + 2! r 2

halved: S = 2! ( 1 _ 2 r) ( 1 _

2 h) + 2! ( 1 _

2 r) 2

= 1 _ 2 !rh + 1 _

2 ! r 2

= 1 _ 4 (2!rh + 2! r 2 )

34. the triangular prism shaped frameS = (area of 2 sides) + (area of 2 ends) = 2(10)(10) + 2 ( 1 _

2 (10)5 " # 3 )

= 200 + 50 " # 3 $ 286.6 ft 2 half cylinder:S = (area of curved panel) + (area of 2 ends) = (10) ( 1 _

2 !(10)) + 2 ( 1 _

2 !(5 ) 2 )

= 50! + 25! = 75! $ 235.6 ft 2 The triangular-prism-shaped frame requires more plastic.

35. area with given measurements:S = Ph + 2B

= 12(6) + 2(3)(3) = 90 cm 2 least possible area:S = 4"w + 2w

= 4(5.95)(2.95) + 2(2.95)(2.95) = 87.615 cm 2 greatest possible area:S = 4"w + 2w

= 4(6.05)(3.05) + 2(3.05)(3.05) = 92.415 cm 2 max. error below 90 cm 2 is 90 ! 87.615 = 2.385 cm 2 max. error above 90 cm 2 is 92.415 ! 90 = 2.415 cm 2 max. error < 2.415 cm 2

36. Find the area of each part of the net and add the areas.

37a. AB = 8 ! 1 = 7 in. BC = " #### 4 2 + 4 2 = " ## 32 = 4 " # 2 in. $ 5.7 in.

b. AC 2 + BC 2 = AB 2 AC 2 + (4 " # 2 ) 2 = 7 2 AC 2 + 32 = 49

AC 2 = 17 AC = " ## 17 in. $ 4.1 in.

c. S = Ph + 2B= 16 (4.1) + 2(16) $ 97.6 in 2


TEST PREP, PAGE 687

38. AS = (area of rectangle) + (area of circles)

= (3.0)(8.1) + 2!(1.35 ) 2 $ 35.8 cm 2

39. FL = Ph = 24(5) = 120 in 2

40. 414.5S = 2(3.14)rh + 2(3.14) r 2

= 2(3.14)(6)(5) + 2(3.14)(6 ) 2 $ 414.5 in 2


41. 1st cylinder:S = 2!rh + 2! r 2

= 2!(8)(3) + 2!(8 ) 2 = 176! cm 2 2nd cylinder: S = 2!rh + 2! r 2 176! = 2!(4)h + 2!(4 ) 2 176! = 2!(4)h + 32!144! = 8!h h = 18 cm

42. S = (area of 2 sides) + (area of 2 ends) = 2(12)(12) + 2 (2 ( 1 _

2 (12 + 18)(9)) )

= 288 + 540 = 828 ft 2 amount of paint = 828 ____

250 $ 3.3

4 gal of paint are needed, costing 4($25) = $100.

43. L = Ph144 = (2" + 2w)h144 = (2(3w) + 2w)(2w)144 = 16 w 2 9 = w 2 w = 3 cm" = 3w = 3(3) = 9 cm, h = 2w = 2(3) = 6 cmS = L + 2B

= 144 + 2(9)(3) = 198 cm 2


44. 154 + m 2 250 45. 70 2 s 2 110

46. B C 2 = AB 2 + AC 2 ! 2(AB)(AC)cos A a 2 = 7 2 + 8 2 ! 2(7)(8)cos 45° a 2 $ 33.804 a $ 5.8

47. sin B ____ AC

= sin A ____ BC

sin B ____ 8

$ sin 45° ______ 5.81

m,B $ si n !1 ( 8 sin 45° _______ 5.81

) $ 77°

48.

49.

50.

GEOMETRY LAB: MODEL RIGHT AND OBLIQUE CYLINDERS, PAGE 688

ACTIVITY 1, TRY THIS, PAGE 688 1. Each cross section is a circle.

2. The cross section is an ellipse.

ACTIVITY 2, TRY THIS, PAGE 688 3. The base of rectangle is the distance around the

cylinder, the height of the rectangle is the slant height of the cylinder, and the area of the rectangle is the lateral area of the cylinder. Check students’ estimates.

10-5 SURFACE AREA OF PYRAMIDS AND CONES, PAGES 689–696

CHECK IT OUT! PAGES 690–692 1. Step 1 Find the base perimeter and the altitude.

The base perimeter is 3(6) = 18 ft.The altitude is 3 " # 3 ft, so, the base area is 1 _

2 bh =

1 _ 2 (6) (3 " # 3 ) = 9 " # 3 ft 2.

Step 2 Find the lateral area.L = 1 _

2 P"

= 1 _ 2 (18)(10) = 90 ft 2

Step 3 Find the surface area.S = 1 _

2 P" + B

= 90 + 9 " # 3 $ 105.6 ft 2

2. Step 1 Use the Pythag. Thm. to find ".

" = " #### 8 2 + 6 2 = 10 cmStep 2 Find the lateral area and the surface area.L = !r"

= !(8)(10) = 80! cm 2 S = !r" + ! r 2

= 80! + !(8 ) 2 = 144! cm 2

3. original:S = 1 _

2 P" + B

= 1 _ 2 (60)(12) + (15 ) 2 = 585 ft 2

base edge length and slant height multiplied by 2/3:S = 1 _

2 P" + B

= 1 _ 2 (40)(8) + (10 ) 2 = 260 ft 2

Notice that 260 = 4 _ 9 (585). If the base edge length

and the slant height are multiplied by 2 _ 3 , the surface

area is multiplied by 4 _ 9 .


4. The base area of the cube is B = (2 ) 2 = 4 yd 2 .The lateral area of the cube is L = Ph = (8)(2) = 16 yd 2 .

By the Pythag. Thm., " = " #### 1 2 + 2 2 = " # 5 yd.The lateral area of the pyramid is L = 1 _

2 P" =

1 _ 2 (8) " # 5 = 4 " # 5 yd 2 .

A = (base area of cube) + (lateral area of cube) + (lateral area of pyramid)= 4 + 16 + 4 " # 5 = 20 + 4 " # 5 $ 28.9 yd 2

5. The radius of the large circle used to create the pattern is the slant height of the cone.The area of the pattern is the lateral area of the cone. The area of the pattern is also 3 _

4 of the area of

the large circle. So, !r" = 3 _ 4 ! " 2 .

!r(12) = 3 _ 4 !(12 ) 2

r = 9 in.

THINK AND DISCUSS, PAGE 692 1. The lateral faces are all 1 with an area of 1 _

2 the

base edge length times the slant height, the perimeter P is the sum of the base edge lengths. So, L = 1 _

2 P".

2. A radius and the axis of a right cone form the legs of a right &. Length of the hypotenuse is the slant height of the cone. The hypotenuse of a right & is always the longest side, so slant height is greater than the height.

3.



1. vertex and center of base

2. Step 1 Find the base perimeter and the apothem.The base perimeter is 6(8) = 48 cm.The apothem is 4 " # 3 cm. So, base area is 1 _

2 aP =

1 _ 2 (4 " # 3 ) (48) = 96 " # 3 cm 2 .


2 P"

= 1 _ 2 (48)(12) = 288 cm 2


2 P" + B

= 288 + 96 " # 3 $ 454.3 cm 2

3. Step 1 Find the base perimeter and the area.The base perimeter is 4(16) = 64 ft.The base area is (16 ) 2 = 256 ft 2 .Step 2 Find the lateral area.

The slant height is " #### 8 2 + 1 5 2 = 17 ft.L = 1 _

2 P"

= 1 _ 2 (64)(17) = 544 ft 2


2 P" + B

= 544 + 256 = 800 ft 2

4. Step 1 Find the base perimeter and the altitude.The base perimeter is 3(15) = 45 in.The altitude of base is 7.5 " # 3 in. So, base area is

1 _ 2 bh = 1 _

2 (15) (7.5 " # 3 ) = 56.25 " # 3 in 2 .


2 P"

= 1 _ 2 (45)(20) = 450 in 2


2 P" + B

= 450 + 56.25 " # 3 $ 547.4 in 2

5. Step 1 Use the Pythogorean Theorem to find the base radius r.

r = " #### 2 5 2 ! 2 4 2 = 7 in.Step 2 Find the lateral area and the surface area.L = !r"

= !(7)(25) = 175! in 2 S = !r" + ! r 2

= 175! + !(7 ) 2 = 224! in 2

6. L = !r"= !(14)(22) = 308! m 2

S = !r" + ! r 2 = 308! + !(14 ) 2 = 504! m 2

7. Step 1 Use the base area to find the base radius r.A = 36! = ! r 2 36 = r 2 r = 6 ftStep 2 Find the lateral area and the surface area.L = !r"

= !(6)(8) = 48! ft 2 S = !r" + ! r 2

= 48! + !(6 ) 2 = 84! ft 2

8. original:S = 1 _

2 P" + B

= 1 _ 2 (24)(10) + (6 ) 2 = 156 in 2

dimensions halved:S = 1 _

2 P" + B

= 1 _ 2 (12)(5) + (3 ) 2 = 39 in 2

Notice that 39 = 1 _ 4 (156). If the dimensions are cut in

half, the surface area is multiplied by 1 _ 4 .


9. original:S = !r" + ! r 2 = !(9)(15) + !(9 ) 2 = 216 cm 2 dimensions tripled:S = !r" + ! r 2 = !(27)(45) + !(27 ) 2 = 1944 cm 2 Notice that 1944 = 9(216). If the dimensions are tripled, the surface area is multiplied by 9.

10. The lateral area of the upper pyramid is L = 1 _

2 Ph = 1 _

2 (32)(15) = 240 ft 2 .

The lateral area of the upper pyramid is L = 1 _

2 Ph = 1 _

2 (32)(18) = 288 ft 2 .

S = (lateral area of upper pyramid) + (lateral area of lower pyramid)= 240 + 288 = 528 ft 2

11. The lateral area of upper cone is L = !r" = !(12)(26) = 312! m 2 .The lateral area of cylinder isL = 2!rh = 2!(12)(15) = 360! m 2 .The lateral area of upper cone is L = !r" = !(12)(32) = 384! m 2 .S = (lateral area of upper cone)

+ (lateral area of cylinder) + (lateral area of lower cone)= 312! + 360! + 384! = 1056! m 2

12. The radius of the large circle used to create the hat is the slant height of the cone.The area of the hat is the lateral area of the cone. The area of the hat is also 3 _

4 of the area of the large

circle. So, !r" = 3 _ 4 ! " 2 .

!r(6) = 3 _ 4 !(6 ) 2

r = 4.5d = 2(4.5) = 9 in. So, the hat will be too large.


13. Step 1 Find the base perimeter and the area.The base perimeter is 4(6) = 24 ft.The base area is (6 ) 2 = 36 ft 2 .Step 2 Find the lateral area.

The slant height is " #### 3 2 + 4 2 = 5 ft.L = 1 _

2 P"

= 1 _ 2 (24)(5) = 60 ft 2


2 P" + B

= 60 + 36 = 96 ft 2

14. Step 1 Find the base perimeter and the altitude.The base perimeter is 3(40) = 120 cm.The altitude of base is 20 " # 3 ft. So, the base area

is 1 _ 2 bh = 1 _

2 (40) (20 " # 3 ) = 400 " # 3 cm 2 .

Step 2 Find the lateral area.

The slant height is " #### 25 2 ! 20 2 = 15 cm.L = 1 _

2 P"

= 1 _ 2 (120)(15) = 900 cm 2


2 P" + B

= 900 + 400 " # 3 $ 1592.8 cm 2

15. Step 1 Find the base perimeter and the apothem.The base perimeter is 6(7) = 42 ft.The apothem is 3.5 " # 3 ft. So, the base area is

1 _ 2 aP = 1 _

2 (3.5 " # 3 ) (42) = 73.5 " # 3 ft 2 .


2 P"

= 1 _ 2 (42)(15) = 315 ft 2


2 P" + B

= 315 + 73.5 " # 3 $ 442.3 ft 2

16. L = !r" = !(11.5)(23) = 264.5! cm 2 S = !r" + ! r 2 = 264.5! + !(11.5 ) 2 = 396.75! cm 2

17. " = " #### 1 2 2 + 3 5 2 = 37 in.L = !r" = !(12)(37) = 444! in 2 S = !r" + ! r 2 = 444! + !(12 ) 2 = 588! in 2

18. h = 2(8) ! 1 = 15 m; " = " #### 8 2 + 1 5 2 = 17 mL = !r" = !(8)(17) = 136! m 2 S = !r" + ! r 2 = 136! + !(8 ) 2 = 200! m 2

19. original:S = 1 _

2 P" + 1 _

2 aP

= 1 _ 2 (24)(12) + 1 _

2 (2 " # 3 ) (24)

= (144 + 24 " # 3 ) ft 2 dimensions divided by 3:S = 1 _

2 P" + 1 _

2 aP

= 1 _ 2 (8)(4) + 1 _

2 ( 2 " # 3 ___

3 ) (8)

= (16 + 8 " # 3 ___ 3 ) ft 2

(16 + 8 " # 3 ___ 3 ) = (144 + 24 " # 3 ) ÷ 9. So, the surface

area is divided by 9.

20. original:S = !r" + ! r 2 = !(2)(5) + !(2 ) 2 = 14! m 2 dimensions doubled:S = !r" + ! r 2 = !(4)(10) + !(4 ) 2 = 56! m 2 56! = 4(14!). So, the surface area is multiplied by 4.

21. lateral area of left cone = !(7)(24) = 168! in 2 lateral area of right cone = !(7)(17) = 119! in 2 S = (lateral area of left cone)

+ (lateral area of right cone)= 168! + 119! = 287! in 2

22. lateral area of left pyramid = 1 _

2 P" = 1 _

2 (36)(15) = 270 cm 2

lateral area of cube = Ph = (36)(9) = 324 cm 2 lateral area of right pyramid = 1 _

2 P" = 1 _

2 (36)(19) = 342 cm 2

S = (lateral area of left pyramid) + (lateral area of cube) + (lateral area of right pyramid)= 270 + 324 + 342 = 936 cm 2

23. S = !r" = 1 _ 2 ! " 2

!r(6) = 1 _ 2 !(6 ) 2

r = 3 d = 2(3) = 6 in.


24. B = s 2 = 36. So, s = 6 cm and P = 24 cmS = 1 _

2 P" + B

= 1 _ 2 (24)(5) + 36 = 96 cm 2

25. B = 1 _ 2 as

" # 3 = 1 _ 2 ( s " # 3 ___

2 ) s = 1 _

4 s 2 " # 3

4 = s 2 s = 2 m P = 3(2) = 6 mS = 1 _

2 P" + B

= 1 _ 2 (6) ( " # 3 ) + " # 3 = 4 " # 3 m 2

26. B = ! r 2 = 16!. So, r = 4 in.S = !(4)(7) + !(4 ) 2 = 44! in 2

27. B = ! r 2 = !, so r = 1 ftS = !(1)(2) + !(1 ) 2 = 3! ft 2

28a. " = " #### 4 2 + 10 2 = " ## 116 = 2 " ## 29 cm S = 1 _

2 P" + s 2

= 1 _ 2 (32) (2 " ## 29 ) + (8 ) 2

= 32 " ## 29 + 64 $ 236.3 cm 2

b. S = !r" + ! r 2 32 " ## 29 + 64 = !(4.5)" + !(4.5 ) 2 32 " ## 29 + 64 ! 20.25! = 4.5!"

" = 32 " ## 29 + 64 ! 20.25! ___________________ 4.5!

$ 12.2 cm

29. S = !r" + ! r 2 232! = !r(21) + ! r 2 232 = 21r + r 2 0 = r 2 + 21r ! 232 0 = (r ! 8)(r + 29) r = 8 m (since r is positive)

30. P = 4s = 32. So, s = 8 ft S = 1 _

2 P" + s 2

256 = 1 _ 2 (32)" + (8 ) 2

256 = 16" + 64192 = 16" " = 12 ft

31. L = 1 _ 2 P"

120 = 1 _ 2 P(10) = 5P

P = 24 cm

32. r 2 + h 2 = " 2 r 2 + 7 2 = 25 2

r = " #### 25 2 ! 7 2 = 24 unitsS = !r" + ! r 2 = !(24)(25) + !(24 ) 2 = 1176! units 2

33. left pyramid: L left = 1 _

2 P" = 1 _

2 (30)(14) = 210 cm 2

right pyramid: L right = 1 _

2 P" = 1 _

2 (30)(8) = 120 cm 2

S = L left + L right = 210 + 210 = 330 cm 2

34. left cone: " = " #### 6 2 + 8 2 = 10 mS = !r" + ! r 2 = !(6)(10) + !(6 ) 2 = 96! m 2

right cone: " = " #### 1 0 2 + 2 4 2 = 26 mS = !r" + ! r 2 = !(10)(26) + !(10 ) 2 = 360! m 2 S = 96! + 360! = 456! m 2

35. s = 200(3 ft) = 600 ft and h = 32(10) = 320 ft. So,

" = " ##### 30 0 2 + 32 0 2 = " #### 192,400 $ 438.6 ftL = 1 _

2 P" $ 1 _

2 (2400)(438.6) $ 526,000 ft 2

36. A triangle is formed with 2 vertices at the midpoints of opposite sides of the square base and the third vertex at the vertex of the pyramid. The side lengths of the triangle are ", ", and s, the edge length of the base. By the Triangle Inequality Theorem, " + " >s, so 2 ">s.Therefore, " > 1 _

2 s.

37. In an oblique cone, the distance from a point on edge of the base to the vertex is not the same for each point on the edge of the base.

TEST PREP, PAGE 695

38. DA = ( 1 _

4 t) 2 + 1 _

2 t"

= t 2 ___

16 + t" __

2 (II)

= t __ 2 ( t __

8 + ") (III)

39. F L = 1 _

2 P"

216 = 1 _ 2 (4s)(18)

432 = 72s s = 6 cmS = L + B = 216 + (6 ) 2 = 252 cm 2

40. B" = " #### 9 2 + 40 2 = 41 cmL = !r" = !(9)(41) = 369! cm 2


41a. " ___ 10

= " ! 20 ______ 5

5" = 10(" ! 20) 5" = 10" ! 200200 = 5" " = 40 cmS = !r" + ! r 2 = !(10)(40) + !(10 ) 2 = 500! cm 2

b. " = 40 ! 20 = 20 cmL = !r" = !(5)(20) = 100! cm 2

c. B = !(5 ) 2 = 25! cm 2

d. S = (surface area of cone) + (area of top base) ! (lateral area of top of cone)= 500! + 25! ! 100! = 425! cm 2

42. L = 4 ( 1 _ 2 ( b 1 + b 2 )") = 2( b 1 + b 2 )"

43a. c = 2!r b. C = 2!"

c. c __ C

= 2!r ____ 2!"

= r __ "

d. A = ! " 2 L = c __

C (A) = ! " 2 ( r __

" ) =

!r"



44. Since the area of a circle depends on the square of its radius, the surface area of a cone cannot be described by a linear function.

45. Since perimeter is 1-dimensional, the perimeter of a rectangle can be described by a linear function.

46. Since the area of a circle depends on the square of its radius, it cannot be described by a linear function.

47. area of ACEF = 4 2 = 16 cm 2 area of &BDG = 1 _

2 (4)(2) = 4 cm 2

P = 4 ___ 16

= 0.25

48. area of circle H = !(2 ) 2 = 4! cm 2 P = 4! ___

16 = ! __

4 $ 0.79

49. area of shaded region = (16 ! 4!) cm 2 P = 16 ! 4! _______

16 = 4 ! ! _____

4 $ 0.21

50. S = L + B = 1 _

2 (8 + 15 + 17)(10) + 2 ( 1 _

2 (8)(15))

= 400 + 120 = 520 in 2

51. S = L + B = (2(8) + 2(10))(15) + 2((8)(10)) = 540 + 160 = 700 cm 2

52. S = 2!rh + 2! r 2 = 2!(2)(3) + 2!(2 ) 2 = 20! $ 62.8 cm 2

10-6 VOLUME OF PRISMS AND CYLINDERS, PAGES 697–704

CHECK IT OUT! PAGES 698–700 1. V = Bh

= ( 1 _ 2 (5)(7)) (9) = 157.5 yd 3

2. Step 1 Find the volume of the aquarium in cubic feet. V = "wh = (120)(60)(16) = 115,200 ft 3

Step 2 Use the conversion factor 1 gal _______

0.134 ft 3 to

estimate the volume in gallons.

115,200 ft 3 · 1 gal _______

0.134 ft 3 $ 859,702 gal

Step 3 Use the conversion factor 8.33 lb ______ 1 gal

to estimate the weight of water.

859,702 gal · 8.33 lb ______ 1 gal

$ 7,161,318 lb

3. V = ! r 2 h= !(8 ) 2 (17) = 1088! in 3 $ 3418.1 in 3

4. original dimensions:V = "wh = (4)(3)(1.5) = 18 ft 3 length, width, and height doubled:V = "wh = (8)(6)(3) = 144 ft 3 Notice that 144 = 8(18). If the length, width, and height are doubled, the volume is multiplied by 8.

5. The volume of cylinder is V = ! r 2 h = !(3 ) 2 (5) = 45! cm 3 .The volume of prism is V = Bh = (3 " # 2 ) 2 (5) = 90 cm 3 .The net volume of figure is difference of volumes.V = 45! ! 90 $ 51.4 cm 3

THINK AND DISCUSS, PAGE 700 1. In both formulas, the volume equals the base

area times the height. The base area of a cylinder = ! r 2 , and the base area of a prism is given by the area formula for that polygon type.

2. An oblique prism has the same cross-sectional area at every level as a right prism with the same base area and height. By Cavalieri’s Principle, the oblique prism and the right prism have the same volume.

3.



1. the same length as

2. V = "wh= (9)(4)(6) = 216 cm 3

3. V = Bh = 1 _

2 aP · h

= 1 _ 2 (3 " # 3 ) (36)(8)

= 432 " # 3 $ 748.2 m 3

4. V = s 3 = (8 ) 3 = 512 ft 3

5. Step 1 Find the volume of the ice cream cake in cubic feet.V = "wh = (19)(9)(2) = 342 ft 3

Step 2 Use the conversion factor 1 gal _______

0.134 ft 3 to

estimate the volume in gallons.

342 ft 3 · 1 gal _______

0.134 ft 3 $ 2552 gal

Step 3 Use the conversion factor 8.33 lb ______ 1 gal

to estimate the weight of water.

2552 gal · 4.73 lb ______ 1 gal

$ 12,071 lb

6. V = ! r 2 h= !(6 ) 2 (10) = 360! ft 3

$ 1131.0 ft 3

7. V = ! r 2 h= !(3 ) 2 (5) = 45! m 3

$ 141.4 m 3


8. B = ! r 2 25! = ! r 2 25 = r 2 r = 5 cmh = 5 + 3 = 8 cmV = ! r 2 h

= !(5 ) 2 (8) = 200! cm 3 $ 628.3 cm 3

9. original dimensions:V = "wh = (12)(4)(8) = 384 ft 3 dimensions multiplied by 1 _

4 :

V = "wh = (3)(1)(2) = 6 ft 3 Notice 6 = 1 __

64 (384). If the dimensions are multiplied

by 1 _ 4 , the volume is multiplied by 1 __

64 .

10. original dimensions:V = ! r 2 h = !(2 ) 2 (7) = 28! in 3 dimensions tripled:V = ! r 2 h = !(6 ) 2 (21) = 756! in 3 Notice 756! = 27(28!). If the dimensions are tripled, the volume is multiplied by 27.

11. The volume of rect. prism isV = "wh = (12)(14)(6) = 1008 ft 3 .The volume of cylinder is V = ! r 2 h = !(4 ) 2 (4) = 624! ft 3 .The total volume of the figure is the sum of its volumes.V = 1008 + 64! $ 1209.1 ft 3

12. The volume of outside cylinder isV = ! r 2 h = !(10 ) 2 (15) = 1500! in 3 .The volume of inside cylinder isV = ! r 2 h = !(5 ) 2 (15) = 375! in 3 .The net volume of the figure is the difference of the volumes.V = 1500! ! 375! = 1125! $ 3534.3 in 3


13. V = Bh = ( 1 _ 2 (9)(15)) (12) = 810 yd 3

14. V = Bh = 1 __

2 aP · h

= 1 __ 2 ( 5 ______

tan 36° ) (50)(15) $ 2580.7 m 3

15. B = s 2 49 = s 2 s = 7 fth = 7 ! 2 = 5 ftV = Bh = (49)(5) = 245 ft 3

16. Step 1 Find the volume in cubic feet.V = "wh = (9)(16) ( 1 _

3 ) = 48 ft 3

Step 2 Use the conversion factor 1 yd 3

_____ 27 ft 3

to find the

volume in cubic yards. Then round up to nearest

cubic yard.

V = 48 · 1 yd 3

_____ 27 ft 3

$ 1.78 yd 3

Colin must buy 2 yd 3 of dirt.Step 3 Use the cost per yard to find the cost of dirt.Cost = 2($25) = $50

17. V = ! r 2 h= !(14 ) 2 (9) = 1764! cm 3

$ 5541.8 cm 3

18. V = ! r 2 h= !(6 ) 2 (3) = 108! in 3

$ 339.3 in 3

19. V = Bh= 24!(16) = 384! cm 3

$ 1206.4 cm 3

20. original dimensions:V = ! r 2 h = !(2 ) 2 (3) = 12! yd 3 dimensions multiplied by 5:V = ! r 2 h = !(10 ) 2 (15) = 1500! yd 3 1500! = 125(12!). So, the volume is multiplied by 125.

21. original dimensions:V = Bh = (5 ) 2 (10) = 250! m 3 dimensions multiplied by 3 _

5 :

V = ! r 2 h = !(3 ) 2 (6) = 54! m 3 54! = 27 ___

125 (250!). So, the volume is multiplied

by 27 ___ 125

.

22. volume of lower cube: V = s 3 = (8 ) 3 = 512 cm 3 volume of middle cube: V = s 3 = (6 ) 3 = 216 cm 3 volume of upper cube: V = s 3 = (4 ) 3 = 64 cm 3 total volume: V = 512 + 216 + 64 = 792 cm 3

23. volume of square-based prism: V = Bh = (4 ) 2 (12) = 192 ft 3 volume of each half-cylinder:V = 1 _

2 ! r 2 h = 1 _

2 !(2 ) 2 (4) = 8! ft 3

total volume: V = 8! + 192 + 8! = 192 + 16! $ 242.3 ft 3

24. radius r = 2 in.: V = ! r 2 h14.4375 = !(2 ) 2 h = 4!h

h = 14.4375 _______ 4!

$ 1.1489 in.

radius r = 1.5 in.: V = ! r 2 h14.4375 = !(1.5 ) 2 h = 2.25!h

h = 14.4375 _______ 2.25!

$ 2.0425 in.

25a. V = ! r 2 h = !(5 ) 2 (3) = 75! $ 235.6 in 2


b. V = "wh = (3)(1)(3) = 9 in 3 P = 9 ____

75! $ 0.04

26a. (3 ! 1 ) 2 + h 2 = 5 2 4 + h 2 = 25 h 2 = 21 h = " ## 21 $ 4.6 in.

b. V = ! r 2 h = !(1.5 ) 2 ( " ## 21 ) $ 32.5 in 3

c. 32.5 in 3 · 0.55 oz _______ 1 in 3

$ 17.9 oz

27. V = "wh495 = (5)(9)h = 45h h = 11 ft.

28. V = Bh360 = B(9) B = 40 in 2

29. S = 2!rh + 2! r 2 210! = 2!r(8) + 2! r 2 210 = 16r + 2 r 2 105 = 8r + r 2 0 = r 2 + 8r ! 105 0 = (r ! 7)(r + 15) r = 7 mV = ! r 2 h = !(7 ) 2 (8) = 392! m 3

30. " = 7 units, w = 3 units, h = 6 unitsV = "wh = (7)(3)(6) = 126 units 3

31. V = "wh = (1)(2) ( 1 _ 6 ) = 1 _

3 ft

3

orV = (12)(24)(2) = 576 in

3

32. V = Bh = (4 ) 2 ( 1 _ 4 ) = 4 in 3 = 4 (1 in 3 )

4 servings

33. Step 1 Find the volume in cubic feet.V = ! r 2 h = !(45 ) 2 (52) = 105,300! ft 3 Step 2 Convert the volume to gallons.

105,300! ft 3 · 1 gal _______

0.134 ft 3 $ 2,468,729 gal

34. Step 1 Find the volume in cubic feet.V = "wh = (50)(100) ( 1 _

4 ) = 1250 ft 3

Step 2 Convert the volume to gallons.

1250 ft 3 · 1 gal _______

0.134 ft 3 $ 9328 gal

Step 3 Convert to weight in pounds.

9328 gal · 8.33 lb ______ 1 gal

$ 77,705 lb

35. For a scale factor of k, the ratio of surface

area to volume of the new prism is 1 __ k times the

ratio of the surface area to the volume of the old prism. A cube with edge length 1 has a surface area-to-volume ratio of 6:1. If each edge length is multiplied by 2, the surface area-to-volume ratio is 24:8 = 3:1.

36. V new = 2 V old

s new 3 = 2 ( s old 3 ) s new = 3 " # 2 ( s old ) Multiply edge length by 3 " # 2 .


37. A V candle = ! r 2 h = !(3.4 ) 2 (6.0) = 69.36! cm 3 V wax = "wh = (15)(12)(18) = 3240 cm 3

# candles 2 3240 ______ 69.36!

$ 14.9

# candles = 14

38. Fwire length = (# edges)(edge length) 96 = 12s s = 8 in.V = s 3 = (8 ) 3 = 512 in 3

39. B V prism = Bh = (3 ) 2 (9) = 81 in 3 V cylinder = ! r 2 h = !(1.75 ) 2 (9) $ 86.6 in 3

40. Hupper rect. prism: V = (10 ! 4)(10)(4) = 240 cm 3 lower rect. prism: V = (10)(10)(10 ! 4) = 600 cm 3 total: V = 240 + 600 = 800 cm 3


41. V = "wh = (x + 2)(x ! 1)(x)= x 3 + x 2 ! 2x

42. V = ! r 2 h= !(x + 1 ) 2 (x)= ! x 3 + 2! x 2 + !x

43. B = 1 __ 2 (x) ( x " # 3 ____

2 ) = x 2 " # 3 _____

4

V = Bh = ( x 2 " # 3 _____ 4 ) (x + 1) = x 3 " # 3 + x 2 " # 3 ___________

4

44. The volume is equal to the surface area,

so ! r 2 h = 2! r 2 + 2!rh. Solve for r to get r = 2h _____ h ! 2

.

If h < 2, then r < 0, so h must be greater than 2.

Similarly, if you solve for h, you get h = 2r _____ r ! 2

.

If r < 2, then h < 0, so r must be greater than 2.


45. 3

4

5

4 6

m = 2r !100 (1)

t = r + 40 (2)

r > m (3)

substitute (1) in (3): r > 2r ! 100 100 > r (4)substitute (4) in (2): t = r + 40 < 140 t 2 139So t = 139

46. typing time is

5000 words __________ 40 wpm

= 125 min = (45 + 45 + 35) min

So, he takes 2 15-min breaks. Total time is 125 + 2(15) = 155 min or 2 h 35 min.

47. ,ABC ( ,CDA m,ABC = m,CDA30x ! 10 = 23x + 4 7x = 14 x = 2m,ABC = 30(2) ! 10 = 50°


48. %%

BC ( %%

AD BC = AD 1 _ 4 z + 11 = 3 _

4 z + 3

8 = 1 _ 2 z

z = 16BC = 1 _

4 (16) + 11 = 15

49. %%

AB ( %%

DC AB = DCy + 6 = 4y 6 = 3y y = 2AB = (2) + 6 = 8

50. S = L + B = 1 _

2 P" + s 2

= 1 _ 2 (32)(10) + (8 ) 2 = 224 in 2

51. S = L + B = 1 __

2 P" + 1 __

2 aP

= 1 __ 2 (30)(8) + 1 __

2 ( 3 ______

tan 36° ) (30)

= 120 + 45 ______ tan 36°

$ 181.9 cm 2

52. C = ! = 2!r r = 0.5 ftS = !rh + ! r 2

= !(0.5)(2) + !(0.5 ) 2 = 1.25! ft 2 $ 3.9 ft 2

10-7 VOLUME OF PYRAMIDS AND CONES, PAGES 705–712

CHECK IT OUT! PAGES 706–708 1. Step 1 Find the area of the base.

B = 1 _ 2 aP

= 1 _ 2 ( " # 3 ) (12) = 6 " # 3 cm 2

Step 2 Use the base and the height to find the volume. The height is equal to the base.V = 1 _

3 Bh

= 1 _ 3 (6 " # 3 ) (6 " # 3 ) = 36 cm 3

2. First find the volume in cubic yards.V = 1 _

3 Bh

= 1 _ 3 (7 0 2 )(66) = 107,800 yd 3

Then, convert the answer to find the volume in cubic feet.

Use the conversion factor 27 ft 3 _____ 1 yd 3

to find the volume in cubic feet. 107,800 y d 3 · 27 ft 3 _____

1 yd 3 = 2,910,600 f t 3

3. V = 1 _ 3 ! r 2 h

= 1 _ 3 !(9 ) 2 (8) = 216! $ 678.6 m 3

4. original dimensions:V = 1 _

3 ! r 2 h

= 1 _ 3 !(9 ) 2 (18) = 486! cm 3

radius and height doubled:V = 1 _

3 ! r 2 h

= 1 _ 3 !(18 ) 2 (36) = 3888! cm 3

Notice that 3888! = 8(486!). If the radius and height are doubled, the volume is multiplied by 8.

5. The volume of the rectangular prism isV = "wh = (25)(12)(15) = 4500 ft 3 .The volume of the rectangular pyramid isV = 1 _

3 Bh = 1 _

3 ((25)(12))(15) = 1500 ft 3 .

The volume of the composite figure is the difference of the volumes.V = 4500 ! 1500 = 3000 ft 3


1. The volume of a pyramid is 1 _ 3 the volume of a prism

with the same base and height.

2.



1. perpendicular

2. Step 1 Find the area of base.B = "w

= (6)(4) = 24 in. 2 Step 2 Use the base and the height to find the volume.V = 1 _

3 Bh

= 1 _ 3 (24)(17) = 136 in 3

3. Step 1 Find the area of base.B = 1 _

2 aP

= 1 _ 2 (2 " # 3 ) (24) = 24 " # 3 cm 2

Step 2 Use the base and the height to find the volume.V = 1 _

3 Bh

= 1 _ 3 (24 " # 3 ) (4 " # 3 ) = 96 cm 3

4. V = 1 _ 3 Bh

= 1 _ 3 (25)(9) = 75 ft 3

5. V = 2 ( 1 _ 3 Bh)

= 2 ( 1 _ 3 (5. 7 2 )(3)) $ 65 mm 3

6. V = 1 _ 3 ! r 2 h

= 1 _ 3 !(9 ) 2 (14) = 378! cm 3

$ 1187.5 cm 3

7. V = 1 _ 3 ! r 2 h

= 1 _ 3 !(12 ) 2 (30) = 1440! in 3

$ 4523.9 in 3


8. V = 1 _ 3 ! r 2 h

= 1 _ 3 !(12 ) 2 (20) = 960! m 3

$ 3015.9 m 3


3 ! r 2 h

= 1 _ 3 !(5 ) 2 (3) = 25! cm 3

dimensions tripled:V = 1 _

3 ! r 2 h

= 1 _ 3 !(15 ) 2 (9) = 675! cm 3

Notice that 675! = 27(25!). If the dimensions are tripled, the volume is multiplied by 27.


3 Bh

= 1 _ 3 ( 9 2 )(15) = 405 cm 3


V = 1 _ 3 Bh

= 1 _ 3 ( 4.5 2 )(7.5) = 50.625 cm 3

Notice that 405 = 1 _ 8 (50.625). If the dimensions are

multiplied by 1 _ 2 , the volume is multiplied by 1 _

8 .

11. The volume of the cube isV = s 3 = (12 ) 3 = 1728 cm 3 .The volume of the rectangular pyramid isV = 1 _

3 Bh = 1 _

3 (1 2 2 )(18) = 864 cm 3 .

The volume of the composite figure is the sum of the volumes.V = 1728 + 864 = 2592 cm 3

12. The volume of the outer cone isV = 1 _

3 ! r 2 h = 1 _

3 !(8 ) 2 (12) = 256! in 3 .

The volume of the inner cone isV = 1 _

3 ! r 2 h = 1 _

3 !(4 ) 2 (6) = 32! in 3 .

The volume of the composite figure is the difference of the volumes.V = 256! ! 32! = 224! in 3

$ 703.7 in 3


13. Step 1 Find the area of the base.B = "w

= (8)(6) = 48 ft 2 Step 2 Use the base and height to find the volume.V = 1 _

3 Bh

= 1 _ 3 (48)(10) = 160 ft 3

14. Step 1 Find the area of the base. 5 2 + 12 2 = 13 2 . So, the base is a right triangle with b = 12 m and h = 5 m.B = 1 _

2 bh

= 1 _ 2 (12)(5) = 30 m 2

Step 2 Use the base and height to find the volume.V = 1 _

3 Bh

= 1 _ 3 (30)(9) = 90 m 2

15. Step 1 Find the height. 6 2 + h 2 = 10 2 h = 8 ftStep 2 Use the height and base edge length to find the volume.V = 1 _

3 Bh

= 1 _ 3 (1 2 2 )(8) = 384 ft 3

16. Step 1 Find the volume in cubic feet. The height is 5(3) = 15 ft.V = 1 _

3 Bh

= 1 _ 3 (4 5 2 )(15) = 10,125 ft 3

Step 2 Convert the volume to cubic yards.

Use the conversion factor 1 yd 3

_____ 27 ft 3

.

10,125 ft 3 · 1 yd 3

_____ 27 ft 3

$ 375 yd 3

17. V = 1 _ 3 ! r 2 h

= 1 _ 3 !(9 ) 2 (41) = 1107! m 3

$ 3477.7 m 3

18. V = 1 _ 3 ! r 2 h

= 1 _ 3 !(2 ) 2 (4) = 16 __

3 ! in 3

$ 16.8 in 3

19. B = ! r 2 36! = ! r 2 r = 6 fth = 2r = 2(6) = 12 ftV = 1 _

3 Bh

= 1 _ 3 (36!)(12) = 144! ft 3

$ 452.4 ft 3


3 ! r 2 h

= 1 _ 3 !(15 ) 2 (21) = 1575! in 3


V = 1 _ 3 ! r 2 h

= 1 _ 3 !(5 ) 2 (7) = 175 ___

3 ! in 3

Notice that 175 ___ 3 ! = 1 __

27 (1575!). If the dimensions

are multiplied by 1 _ 3 , the volume is multiplied by 1 __

27 .


3 Bh

= 1 _ 3 ( 7 2 )(4) = 196 ___

3 ft 3

dimensions multiplied by 6:V = 1 _

3 Bh

= 1 _ 3 ( 42 2 )(24) = 14,112 cm 3

Notice that 14,112 = 216 ( 196 ___ 3 ) . If the dimensions are

multiplied by 6, the volume is multiplied by 216.


22. The volume of the cylinder is V = ! r 2 h = !(6 ) 2 (10) = 360! ft 3 .The volume of the cone is V = 1 _

3 ! r 2 h = 1 _

3 !(6 ) 2 (10) = 120! ft 3 .

The volume of the composite figure is the difference of the volumes.V = 360! ! 120! = 240! ft 3

$ 754.0 ft 3

23. The volume of the rectangular prism isV = "wh = (10)(5)(2) = 100 ft 3 .The volume of each square-based pyramid isV = 1 _

3 Bh = 1 _

3 ( 5 2 )(3) = 25 ft 3 .

The volume of the composite figure is the sum of the volumes.V = 100 + 25 + 25 = 150 ft 3

24. V = 1 _ 3 ! r 2 h = 1 _

3 !(3 ) 2 (7) = 21! in 3

25. r = 1 _ 2 d = 5 _

2 m

V = 1 _ 3 ! r 2 h = 1 _

3 ! ( 5 _

2 ) 2 (2) = 25 __

6 ! m 3

26. r 2 + h 2 = " 2 28 2 + h 2 = 53 2 h = 45 ftV = 1 _

3 ! r 2 h = 1 _

3 !(28 ) 2 (45) = 11,760! ft 3

27. r = 1 _ 2 d = 1 _

2 (6) = 12 cm

r 2 + h 2 = " 2 12 2 + h 2 = 13 2 h = 5 cmV = 1 _

3 ! r 2 h = 1 _

3 !(12 ) 2 (5) = 240! ft 3

28. B = 1 _ 2 b h base = 1 _

2 (10) (5 " # 3 ) = 25 " # 3 ft 2

V = 1 _ 3 Bh = 1 _

3 (25 " # 3 ) (6) $ 86.6 ft 3

29. V = 1 _ 3 Bh = 1 _

3 (1 5 2 )(18) = 1350 m 3

30. Step 1 Find the apothem of the base.tan 36° = 4.5 ___ a

a = 4.5 ______ tan 36°

in.

Step 2 Find the area of the base.

B = 1 __ 2 aP = 1 __

2 ( 4.5 ______

tan 36° ) (45) = 101.25 ______

tan 36° in 2

Step 3 Find the volume of the pyramid.

V = 1 __ 3 Bh = 1 __

3 ( 101.25 ______

tan 36° ) (12) $ 557.4 in 3

31. B = 1 _ 2 aP = 1 _

2 (4 " # 3 ) (48) = 96 " # 3 cm 2

V = 1 _ 3 Bh = 1 _

3 (96 " # 3 ) (3) $ 166.3 cm 3

32. V = 1 _ 3 Bh

V = 1 _ 3 lwh

112 = 1 _ 3 (3)(8)h

112 = 8h h = 14 m

33. V = 1 _ 3 ! r 2 h

125! = 1 _ 3 ! r 2 (5)

125! = 5 _ 3 ! r 2

75 = r 2 r = 5 " # 3 cmC = 2!r = 2! (5 " # 3 ) = 10! " # 3 cm

34. r 2 + h 2 = " 2 r 2 + 8 2 = 10 2 r = 6 ftV = 1 _

3 ! r 2 h = 1 _

3 !(6 ) 2 (8) = 96! ft 3

35. S = 1 _ 2 P" + B

800 = 1 _ 2 (4s)(17) + s 2

800 = 34s + s 2 0 = s 2 + 34s ! 800 0 = (s ! 16)(s + 50) s = 16 in. ( 1 _

2 s) 2 + h 2 = " 2

8 2 + h 2 = 17 2 h = 15 in.V = 1 _

3 Bh = 1 _

3 (1 6 2 )(15) = 1280 in 3

36. V = 1 _ 3 ! r 2 h

1500! = 1 _ 3 ! r 2 (20)

4500! = 20! r 2 225 = r 2 r = 15 yd r 2 + h 2 = " 2 15 2 + 20 2 = " 2 " = 25 ydS = !r" + ! r 2

= !(15)(25) + !(15 ) 2 = 600! yd 2

37. The base is a right triangle with a base length of 5 units and a height of 3 units. The height of the pyramid is 7 units.B = 1 _

2 (5)(3) = 7.5 units 2

V = 1 _ 3 Bh = 1 _

3 (7.5)(7) = 17.5 units 3

38. A is incorrect because it uses the slant height of the cone instead of the height.

39. 3 : 2; The base areas are the same for both figures The volume of the prism is By, and the volume of

the figure formed by 2 pyramids is 1 __ 3 B(2y).

The ratio of the volumes is By : 1 __ 3 B(2y), which is

equivalent to 3:2.

40. Possible answer: Substitute the given values for r and S into the surface area formula and solve for ". Then, use the Pythagorean Theorem and the values for r and " to solve for h. Substitute the values for r and h into the volume formula.

41a. V = 1 _ 3 ! r 2 h = 1 _

3 !(2 ) 2 (8) = 32 __

3 ! $ 33.5 in 3

b. V = 1 _ 3 ! r 2 h = 1 _

3 !(4 ) 2 (8) = 128 ___

3 ! $ 134.0 in 3


c. The large size holds 4 times as much. So, the price should be 4($1.25) = $5.

TEST PREP, PAGE 712

42. A1 2 2 + h 2 = 15 2 h = 9 cmV = 1 _

3 ! r 2 h = 1 _

3 !(12 ) 2 (9) = 432! cm 3

43. H L = 1 _

2 P"

350 = 1 _ 2 (4s)(25)

350 = 50s s = 7 m3. 5 2 + h 2 = 2 5 2 h = " ### 612.75 $ 24.75V = 1 _

3 Bh $ 1 _

3 ( 7 2 )(24.75) $ 404 m 3

44. BV = 1 _

3 ! r 2 h = 1 _

3 !(3 ) 2 (6) = 18! in 3

45. 9 V = 1 _

3 Bh

243 = 1 _ 3 ( h 2 )h

729 = h 3 h = 9 cm


46. radius = apothem of regular triangle

1 _ 2 s = 1. So, r = 1 ___

" # 3 =

" # 3 ___ 3 ft

V = 1 _ 3 ! r 2 h = 1 _

3 ! ( " # 3 ___

3 ) 2 (2) = 2 _

9 ! ft 3

47. r = 1 _ 2 s = 1 ft

V = 1 _ 3 ! r 2 h = 1 _

3 !(1 ) 2 (2) = 2 _

3 ! ft 3

48. radius = apothem of regular hexagons = 2. So, r = " # 3 ftV = 1 _

3 ! r 2 h = 1 _

3 ! ( " # 3 ) 2 (2) = 2! ft 3

49. slant height is altitude of a face; s = 10. So, " = 5 " # 3 cm.

( 1 _ 2 s) 2 + h 2 = " 2

5 2 + h 2 = (5 " # 3 ) 2 h = " ## 50 = 5 " # 2 cm

V = 2 ( 1 _ 3 Bh) = 2 _

3 (1 0 2 ) (5 " # 2 ) = 1000 " # 2 _______

3 cm 3

50. h = 9 in.; the volume of a cone with the same base and height as the cylinder is 1 _

3 the volume of the

cylinder. For a cone to have the same volume as the cylinder, the height of cone must be 3 times the height of the cylinder.


51. 3

5 6 x ! y = 24 (1)

x = 3y ! 4 (2)

Substitute (2) into (1):3y ! 4 ! y = 24 2y = 28 y = 14x = 3(14) ! 4 = 38The numbers are 38 and 14.

52. 3

5 6 3x + y = 88 (1)

10x = 4y (2)

Solve (2) for y and substitute into (1):

y = 10x ____ 4 = 5x ___

2

3x + 5x ___ 2 = 88

6x + 5x = 176 11x = 176 x = 16

y = 10(16)

______ 4 = 40

The numbers are 16 and 40.

53. 3

5 6 x + y = 197 (1)

x = 1 _

2 y + 20 (2)

Substitute (2) into (1): 1 _ 2 y + 20 + y = 197

y + 40 + 2y = 394 3y = 354 y = 118x = 1 _

2 (118) + 20 = 79

The numbers are 79 and 118.

54. DF ___ AC

= 25.5 ____ 17

= 3 __ 2 ; EF ___

BC = 31.5 ____

21 = 3 __

2 ; ,C ( ,F

triangles are 7 by 7 SAS

AB ___ 10

= 3 __ 2

2AB = 3(10) = 30 AB = 15

55. %%

LM and %%

PQ are both perpendicular to %%

LN . So, %%

LM parallel

%% PQ . By Corr. 8 Post., ,M ( ,NQP; by

Reflex. Prop. of (, ,N ( ,N. So 1 are 7 by AA 7.

PQ ___ LM

= PN ___ LN

PQ ___ 9 = 16 ___

24

24PQ = 9(16) = 144 PQ = 6

56. AB = " ############# (8 ! 1 ) 2 + (9 ! 1 ) 2 + (10 ! 2 ) 2

= " ###### 49 + 64 + 64 = " ## 177 $ 13.3

M ( 1 + 8 _____ 2 , 1 + 9 _____

2 , 2 + 10 ______

2 ) = M(4.5, 5, 6)

57. AB = " ################ (5 ! (!4) ) 2 + (1 ! (!1) ) 2 + (!4 ! 0 ) 2

= " ##### 81 + 4 + 16 = " ## 102 $ 10.0

M ( !4 + 5 _______ 2 , !1 + 1 _______

2 , 0 + !4 _______

2 ) = M(0.5, 0 , !2)

58. AB = " ############### (!2 ! 2 ) 2 + (2 ! (!2) ) 2 + (!4 ! 4 ) 2

= " ###### 16 + 16 + 64 = " ## 96 $ 9.8

M ( 2 + (!2) ________

2 , !2 + 2 _______

2 ,

4 + (!4) ________

2 ) = M(0, 0, 0)

59. AB = " ################ (!1 ! (!3) ) 2 + ( 5 ! (!1)) 2 + ( 5 ! 2) 2

= " ##### 4 + 36 + 9 = " ## 49 = 7

M ( !3 + (!1) _________

2 , !1 + 5 _______

2 , 2 + 5 _____

2 ) = M(!2, 2, 3.5)


CONNECTING GEOMETRY TO ALGEBRA: FUNCTIONAL RELATIONSHIPS IN FORMULAS, PAGE 713

TRY THIS, PAGE 713 1. First use the surface area formula to write an

equation.S = !r" + ! r 2 S = !(10)" + !(10 ) 2 S = 10!" + 100!Then, solve for " to get an equation for " in terms of S.S ! 100! = 10!" " = S ____

10! ! 10

Since the equation is linear, use slope and y-intercept to graph it.

The slant height increases as the surface area increases.

2. First use the volume formula to write an equation.V = ! r 2 hV = ! r 2 (5)Then, solve for " to get an equation for " in terms of S.

V ___ 5!

= r 2

r = " ## V ___ 5!

Make a table of V- and r-values. Then, plot points and draw a smooth curve through them.

V r

0 0

5 0.46

10 0.80

15 0.98

20 1.13

The radius increases as the volume increases.

10-8 SPHERES, PAGES 714–721


1. V = 4 _ 3 ! r 3

2304! = 4 _ 3 ! r 3

1728 = r 3 r = 12 ft

2. hummingbird eyeball:V = 4 _

3 ! r 3

= 4 _ 3 !(1.25 ) 3 $ 2.604! c m 3

human eyeball:V = 4 _

3 ! r 3

= 4 _ 3 !(0.3 ) 3 = 0.036! c m 3

A human eyeball is about 2.604! ______ 0.036!

$ 72.3 times as

great in volume as a hummingbird eyeball.

3. S = 4! r 2 = 4!(25 ) 2 = 2500! cm 2

4. original dimensions:S = 4! r 2

= 4!(3 ) 2 = 36! m 3

radius multiplied by 1 _ 3 :

S = 4! r 2 = 4!(1 ) 2 = 4! m 3

Notice that 4! = 1 _ 9 (36!). If the radius is multiplied

by 1 _ 3 , the surface area is divided by 9.

5. Step 1 Find the surface area of the composite figure. The surface area of the composite figure is the sum of the surface area of the hemisphere, the lateral area of cylinder, and the area of the lower base of cylinder. S hemisphere = 1 _

2 (4! r 2 ) = 2!(3 ) 2 = 18! ft 2

L cylinder = 2!rh = 2!(3)(5) = 30! ft 2 B cylinder = ! r 2 = !(3 ) 2 = 9! ft 2 The surface area of composite figure is 18! + 30! + 9! = 57! ft 2 .Step 2 Find the volume of the composite figure.The volume of composite figure is difference of volume of cylinder and volume of hemisphere. V cylinder = ! r 2 h = !(3 ) 2 (5) = 45! ft 3

V hemisphere = 1 _ 2 ( 4 _

3 ! r 3 ) = 2 _

3 !(3 ) 3 = 18! ft 3

The volume of composite figure is 45! ! 18! = 27! ft 3 .

THINK AND DISCUSS, PAGE 717 1. The surface area is 4 times the area of the

great circle.

2. Both the area of sphere and the area of the composite figure have a volume of 4 _

3 ! r 3 .

3.



1. One endpoint is the center of the sphere, and the other is a point on the sphere.

2. V = 1 _ 2 ( 4 _

3 ! r 3 ) = 2 _

3 !(11 ) 3 = 2662 ____

3 ! in 3


3. V = 4 _ 3 ! r 3 = 4 _

3 !(1 ) 3 = 4 _

3 ! m 3

4. V = 4 _ 3 ! r 3

288! = 4 _ 3 ! r 3

216 = r 3 r = 6 cm

5. grapefruit: V = 1 _

2 ( 4 _

3 ! r 3 ) = 2 _

3 !(5 ) 3 = 50 __

3 ! cm 3

lime:V = 1 _

2 ( 4 _

3 ! r 3 ) = 2 _

3 ! ( 5 _

2 ) 3 = 25 __

12 ! cm 3

The grapefruit is 8 times as great in volume as the lime.

6. S = 4! r 2 = 4!(8 ) 2 = 256! yd 3

7. A = ! r 2 S = 4! r 2 = 4A = 4(49!) = 196! cm 2

8. S = 4! r 2 6724! = 4! r 2 1681 = r 2 r = 41 ftV = 4 _

3 ! r 3 = 4 _

3 !(41 ) 3 = 275,684 ______

3 ! ft 3


= 4!(15 ) 2 = 900! in 2

dimensions doubled:S = 4! r 2

= 4!(30 ) 2 = 3600! in 2

Notice that 3600! = 4(900!). If dimensions are doubled, the surface area is multiplied by 4.


3 ! r 3

= 4 _ 3 !(8 ) 3

= 2048 ____ 3 ! cm 3


V = 4 _ 3 ! r 3

= 4 _ 3 !(8 ) 3

= 32 __ 3 ! cm 3

Notice that 32 __ 3 ! = 1 __

64 ( 2048 ____

3 !) . If the radius is multiplied

by 1 _ 4 , the volume is multiplied by 1 __

64 .

11. Step 1 Find the surface area of the composite figure. The surface area of the composite figure is the sum of the surface area of two hemispheres and the lateral area of cylinder. S hemisphere = 1 _

2 (4! r 2 ) = 2!(2 ) 2 = 8! ft 2

L cylinder = 2!rh = 2!(2)(5) = 20! ft 2 The surface area of the composite figure is 2(8!) + 20! = 36! ft 2 .Step 2 Find the volume of the composite figure.The volume of the composite figure is the sum of the volume of the cylinder and the volume of the two hemispheres. V cylinder = ! r 2 h = !(2 ) 2 (5) = 20! ft 3


3 ! r 3 ) = 2 _

3 !(2 ) 3 = 16 __

3 ! ft 3

The volume of the composite figure is 20! + 2 ( 16 __

3 !) = 92 __

3 ! ft 3 .

12. Step 1 Find the surface area of the composite figure. The surface area of the composite figure is the sum of the lateral area of the hemisphere and the surface area of the cylinder, less the area of the base of the hemisphere. S hemisphere = 1 _

2 (4! r 2 ) = 2!(2 ) 2 = 8! in 2

S cylinder = 2!rh + 2! r 2 = 2!(8)(3) + 2!(8 ) 2 = 176! in 2

B hemisphere = ! r 2 = !(2 ) 2 = 4! in 2 The surface area of the composite figure is

8! + 176! ! 4! = 180! in 2 .Step 2 Find the volume of the composite figure.The volume of the composite figure is the difference of the volume of the cylinder and the volume of the hemisphere. V cylinder = ! r 2 h = !(8 ) 2 (3) = 192! ft 3


3 ! r 3 ) = 2 _

3 !(2 ) 3 = 16 __

3 ! in 3

The volume of the composite figure is 192! ! 16 __

3 ! = 560 ___

3 ! in 3 .


13. V = 4 _ 3 ! r 3 = 4 _

3 !(9 ) 3 = 972! cm 3

14. V = 1 _ 2 ( 4 _

3 ! r 3 ) = 2 _

3 !(7 ) 3 = 686 ___

3 ! ft 3

15. V = 4 _ 3 ! r 3

7776! = 4 _ 3 ! r 3

5832 = r 3 r = 18 d = 36 in.

16. 9-mm pearl: V = 4 _

3 ! r 3 = 4 _

3 !(4.5 ) 3 = 121.5! mm 3

6-mm pearl:V = 4 _

3 ! r 3 = 4 _

3 !(3 ) 3 = 108! mm 3

The 9-mm pearl is 3.375 times as great in volume.

17. S = 4! r 2 = 4!(21 ) 2 = 1764! in 2

18. A = ! r 2 S = 4! r 2 = 4A = 4(81!) = 324! in 2

19. S = 4! r 2 625! = 4! r 2 156.25 = r 2 r = 12.5 mV = 4 _

3 ! r 3 = 4 _

3 !(12.5 ) 3 = 15,625 _____

6 ! m 3


= 4!(0.6 ) 2 = 1.44! ft 2


S = 4! r 2 = 4!(0.12 ) 2 = 0.0576! in. 2

Notice that 0.0576! = 1 __ 25

(1.44!). If the dimensions are multiplied by 1 _

5 , the surface area is multiplied

by 1 __ 25

.



3 ! r 3

= 4 _ 3 !(14 ) 3

= 10,976 _____ 3 ! mm 3

dimensions multiplied by 6:V = 4 _

3 ! r 3

= 4 _ 3 !(84 ) 3

= 790,272! mm 3

Notice that 790,272! = 216 ( 10,976 _____ 3 !) . If the radius is

multiplied by 6, the volume is multiplied by 216.

22. Step 1 Find the surface area. S prism = Ph + 2"w

= 2(10)+ 2(5))(4) + 2(10)(5) = 220 cm 2 L hemisphere = 1 _

2 (4! r 2 ) = 2!(3 ) 2 = 18! cm 2

B hemisphere = ! r 2 = !(3 ) 2 = 9!S = S prism + L hemisphere ! B hemisphere

= 220 + 18! ! 9!= 220 + 9! $ 248.3 cm 2

Step 2 Find the volume. V prism = "wh = (10)(5)(4) = 200 cm 3


3 ! r 3 ) = 2 _

3 !(3 ) 3 = 18! cm 3

V = V prism + V hemisphere = 200 + 18! $ 256.5 cm 3

23. Step 1 Find the surface area. The slant height of

the cone is " #### 2 0 2 + 2 4 2 = 26 mm. S cone = !r" + ! r 2

= !(10)(26) + !(10 ) 2 = 360! mm 2 L hemisphere = 1 _

2 (4! r 2 ) = 2!(8 ) 2 = 128! mm 2

B hemisphere = ! r 2 = !(8 ) 2 = 64!S = S prism + L hemisphere ! B hemisphere

= 360! + 128! ! 64!= 424! $ 1332.0 mm 2

Step 2 Find the volume. V cone = 1 _

3 ! r 2h = 1 _

3 (10 ) 2 (24) = 800! mm 3


3 ! r 3 ) = 2 _

3 !(8 ) 3 = 1024 ____

3 ! mm 3

V = V cone ! V hemisphere

= 800! ! 1024 ____ 3 ! = 1376 ____

3 ! $ 1440.9 mm 3

24. V = 1 _ 2 ( 4 _

3 ! r 3 )

144! = 2 _ 3 ! r 3

216 = r 3 r = 6 cm

25. S = 4! r 2 60! = 4! r 2 15 = r 2 r = " ## 15 in.C = 2!r = 2! " ## 15 in.

26. C = 2!r36! = 2!r r = 18 ftV = 4 _

3 ! r 3 = 4 _

3 !(18 ) 3 = 7776! ft 3

27. r = " ############ (2 ! 0 ) 2 + (3 ! 0 ) 2 + (6 ! 0 ) 2 = " ## 49 = 7S = 4! r 2 = 4!(7 ) 2 = 196! units 2 V = 4 _

3 ! r 3 = 4 _

3 !(7 ) 3 = 1372 ____

3 ! units 3

28. Possible answer: Use 8 mm as the estimated height of the cylinder. S sphere = 4! r 2 = 4!(4 ) 2 = 64! mm 2 L cylinder = 2!rh = 2!(1)(8) = 16! mm 2 B cylinder = 2! r 2 = 2!(1 ) 2 = 2! mm 2 S $ S sphere + L cylinder ! B cylinder = 64! + 16! ! 2! = 78! $ 245 mm 2 V sphere = 4 _

3 ! r 3 = 4 _

3 !(4 ) 3 = 256 ___

3 ! mm 3

V cylinder = ! r 2 h = !(1 ) 2 (8) = 8! mm 3

V $ V sphere ! V cylinder

= 256 ___ 3 ! ! 8!

= 232 ___ 3 ! $ 243 mm 3

29. C = !d = !(1.68) $ 5.28 in.

S = 4! r 2 = 4!(0.84 ) 2 $ 8.87 in 2

V = 4 _ 3 ! r 3 = 4 _

3 !(0.84 ) 3 $ 2.48 in 3

30. C = !d 9 = !d d = 9 __ ! $ 2.86 in.

S = 4! r 2 = 4! ( 9 ___ 2!

) 2 $ 25.78 in 2

V = 4 _ 3 ! r 3 = 4 _

3 ! ( 9 ___

2! ) 3 $ 12.31 in 3

31. C = !d = !(2.5) $ 7.85 in.S = 4! r 2 = 4!(1.25 ) 2 $ 19.63 in 2 V = 4 _

3 ! r 3 = 4 _

3 !(1.25 ) 3 $ 8.18 in 3

32. C = !d = !(74) $ 232.48 mmS = 4! r 2 = 4!(37 ) 2 $ 17,203.36 mm 2 V = 4 _

3 ! r 3 = 4 _

3 !(37 ) 3 $ 212,174.79 mm 3

33. V outer = 4 _ 3 ! r 3 = 4 _

3 !(28.5 ) 3 = 30,865.5! in 3

V inner = 4 _ 3 ! r 3 = 4 _

3 !(27 ) 3 = 26,244! in 3

V window = ! r 2 h = !(4 ) 2 (1.5) = 24! in 3 V $ V outer ! V inner ! 3 V window = 30,865.5! ! 26,244! ! 3(24!)

= 4549.5! $ 14,293 in 3

34. S land = 1 _ 3 S Earth

= 4! r 2 $ 4!(4000 ) 2 $ 67,000,000 mi 2

35. V Jupiter

______ V Earth

= 4 _ 3 !(44,423 ) 3

__________ 4 _ 3 !(3963 ) 3

$ 1408

Volume of Jupiter is about 1408 times as great as the volume of Earth.

36. V Venus + V Mars = 4 _ 3 !(3760.5 ) 3 + 4 _

3 !(2111 ) 3

$ 2.62 9 10 11 mi 3 V Earth = 4 _

3 !(3963 ) 3 $ 2.61 9 10 11 mi 3

The sum of the volumes of Venus and Mars is about equal to the volume of the Earth.


37. S Uranus + S Neptune = 4!(15,881.5 ) 2 + 4!(15,387.5 ) 2 ! 6.14 " 10 9 mi 2 S Saturn = 4!(37,449 ) 2 ! 1.76 " 10 10 mi 2 The surface area of Saturn is greater.

38. S Earth

_____ S Mars

= 4!(3963 ) 2

_________ 4!(2111 ) 2

! 4

The surface area of Earth is about 4 times as great as the surface area of Mars.

39. The cross section of the hemisphere is a circle

with radius # $$$ r 2 % x 2 , so its area is A = ! ( r 2 % x 2 ). The cross section of the cylinder with the cone removed has an outer radius of r and an inner radius of x, so the area is

A = ! r 2 % ! x 2 = ! ( r 2 % x 2 ) .

40. 4! r 2 = 6 s 2

4! r 2 ____ 6 = s 2

s = 2r · # $ ! __ 6 or s ! 1.4r

41a. S = 4! r 2 50.3 ! 4! r 2

50.3 ____ 4!

! r 2

r ! # $$ 50.3 ____ 4!

in.

V ! 4 __ 3 ! ( # $$ 50.3 ____

4! ) 3 ! 33.5 in 3

b. V ! 4 __ 3 ! (1.1 # $$ 50.3 ____

4! ) 3 ! 44.6 in 3

TEST PREP, PAGE 721

42. A(16 ) 3 : 4 _

3 !(8 ) 3

4096 : 2048 ____ 3 !

2 : 1 _ 3 !

43. H V = 4 _

3 ! r 3

32 __ 3 ! = 4 _

3 ! r 3

8 = r 3 r = 2 in.S = 4!(2 ) 2 = 16! in 2

44. AV = (2r ) 3 + 1 _

2 ( 4 _

3 ! r 3 )

= 8 r 3 + 2 _ 3 ! r 3

= r 3 (8 + 2 _ 3 !)


45. 3300 gumballs take up about 57% of the volume of sphere.3300 ( 4 _

3 ! r 3 ) ! 0.57 ( 4 _

3 !(9 ) 3 )

3300 r 3 ! 0.57(9 ) 3 = 415.53

r 3 ! 415.53 ______ 3300

r ! 3 # $$$ 415.53 ______

3300 ! 0.50

d = 2(0.50) ! 1.0 in.

46a. S = 4! r 2 S ___

4! = r 2

r = # $$

S ___ 4!

= # $$ S! _____ 2!

b. V = 4 __ 3 ! r 3

= 4 __ 3 ! ( # $$ S!

_____ 2!

) 3

= 4 __ 3 ! ( S! # $$ S!

_______ 8 ! 3

) = S # $$ S! ______

6!

c.

Possible answer: The shape of the graph is similar

to half of a parabola.

47. V cylinder

_______ V sphere

= ! r 2 (2r)

______ 4 _ 3 ! r 3

= 2 __ 4 _ 3 = 1.5

The volume of the cylinder is 1.5 times the volume of the sphere.

48. L cylinder

_______ S sphere

= 2!r(2r)

______ 4! r 2

= 4! r 2 ____ 4! r 2

= 1

The surface area of the sphere is equal to the lateral area of the cylinder.


49. The graph resembles a parabola. The data fits the equation y = x 2 + 1.

50. The graph is a straight line with slope 1 and y-intercept 10. The equation is y = x + 10.

51. quarter-circle: A = 1 _ 4 !(4 ) 2 = 4! in 2

&: A = 1 _ 2 (4)(4) = 8 in 2

shaded area: A = 4! % 8 ! 4.6 in 2

52. rectangle: A = (10)(6) = 60 cm 2 trapezoid: A = 1 _

2 (1 + 5)(4) = 12 cm 2

shaded area: A = 60 % 12 = 48 cm 2

53. V = ( 3 _ 4 s) 3 = 27 __

64 s 3

The volume is multiplied by 27 __ 64

.

54. V = (5B)(5h) = 25BhThe volume is multiplied by 25.

10-8 TECHNOLOGY LAB: COMPARE SURFACE AREAS AND VOLUMES, PAGES 722–723

ACTIVITY 1, TRY THIS, PAGE 723 1. The cylinder with the minimum surface area will be

the one in which the height and diameter are the closest to each other.

2. Possible answer: No; a package such as a cereal box might be designed to have a large surface area so that it stands out on the shelf.



ACTIVITY 2, TRY THIS, PAGE 723 3. The cylinder with the maximum volume will have a

height equal to its diameter.

4. SA = 2LW + 2LH + 2WH becomes

H = ( SA ___

2 ! LW)

__________ L + W

when solved for H.

5. Possible answer: A sphere will have the least surface area and a pyramid will have the greatest surface area for a given volume.

6. Possible answer: If dimensions of a rectangular prism are doubled, the surface area is multiplied by 4 and the volume is multiplied by 8. So the ratio of the surface area to the volume is multiplied by

4 _ 8 = 1 _

2 . Set up a spreadsheet as in Activity 1. Enter

the values for L, W, and H in one row, and enter the values 2L, 2W, and 2H in the next row. Repeat for several values of L, W, and H.


1. 4 2 + h 2 = 5 2 h = 3 in.S = 2!rh + 2! r 2 = 2!(2)(3) + 2!(2 ) 2 = 20! $ 62.8 in 2

2. volume in cubic inches: V = ! r 2 h = !(2 ) 2 (3) = 12! in 3 volume in ounces:

12! in 3 · 0.55 oz _______ 1 in 3

$ 20.7 oz

3. 3 2 + 3 2 + h 2 = 5 2 h = " # 7 in.S = Ph + B = (4s)h + 2( s 2 ) = 4(3) ( " # 7 ) + 2(3 ) 2 = 18 + 12 " # 7 $ 49.7 in 2

4. volume in cubic inches: V = s 2 h = (3 ) 2 ( " # 7 ) = 9 " # 7 in 3 volume in ounces:

9 " # 7 in 3 · 0.55 oz _______ 1 in 3

$ 13.1 oz

5. Possible answer: I would recommend the cylinder, because the ratio of the surface area to the volume is about 3 in 2 per oz, and the ratio of the surface area to the volume of prism is about 3.8 in 2 per oz. Thus, the cylinder costs less to produce per oz of volume.


1. S = L + B= (2" + 2w)h + 2("w)= (2(12) + 2(8))(8) + 2(12)(8) = 512 cm 2

2. S = 2!rh + 2! r 2 = 2!(6)(10) + 2!(6 ) 2 = 192! $ 603.2 ft 2

3. S prism = Ph + 2 s 2 = (140)(10) + 2(35 ) 2 = 3850 in 2

L cylinder = 2!rh = 2!(5)(15) = 150! in 2 S = S prism ! B cylinder + L cylinder + B cylinder = S prism + L cylinder = 3850 + 150! $ 4321.2 in 2

4. original dimensions:S = (2" + 2w)h + 2("w)

= (2(12) + 2(8))(24) + 2(12)(8) = 1152 cm 2 dimensions multiplied by 3 _

4 :

S = (2" + 2w)h + 2("w)= (2(9) + 2(6))(18) + 2(9)(6) = 648 cm 2

648 = 9 __ 16

(1152). So, surface area is multiplied by 9 __ 16

.

5. Step 1 Find the apothem and the perimeter of the base.tan 36° = 9 __ a

a = 9 ______ tan 36°

yd

P = 5(18) = 90 ydStep 2 Find the base area.

B = 1 __ 2 aP = 1 __

2 ( 9 ______

tan 36° ) (90) = 405 ______

tan 36° yd 2

Step 3 Find the surface area.S = 1 __

2 P" + B

= 1 __ 2 (90)(20) + 405 ______

tan 36° $ 1457.4 yd 2

6. 15 2 + 8 2 = " 2 " = 17 in.S = !r" + ! r 2

= !(15)(17) + !(15 ) 2 = 480! $ 1508.0 in 2

7. L left = !r" = !(10)(34) = 340! ft 2 L right = !r" = !(10)(26) = 260! ft 2 S = L left + L right = 340! + 260!

= 600! $ 1885.0 ft 2

8. V = Bh = (23)(9) = 207 in 3

9. V = ! r 2 h = !(8 ) 2 (14) = 896! $ 2814.9 yd 3

10. Step 1 Find the volume of the bricks.V = "wh = (10)(12) ( 1 _

3 ) = 40 ft 3

Step 2 Use the conversion factor 130 lb ______ 1 ft 3

to find the weight of the bricks.

40 ft 3 · 130 lb ______ 1 ft 3

= 5200 lb

11. original dimensions:V = ! r 2 h = !(2 ) 2 (1) = 4! ft 3 dimensions doubled:V = ! r 2 h = !(4 ) 2 (2) = 32! ft 3 32! = 8(4!). So, volume is multiplied by 8.

12. V = 1 _ 3 ! r 2 h

= 1 _ 3 !(12 ) 2 (16) = 768! ft 3

$ 2412.7 ft 3

13. V = 1 _ 3 Bh = 1 _

3 ((39)(15))(16) = 3120 m 3


14. V prism = Bh = ( 1 _ 2 (13)(20)) (9) = 1170 yd 3

V pyramid = 1 _ 3 Bh = 1 _

3 ( 1 _

2 (13)(20)) (27 ! 9) = 780 yd 3

V = V prism + V pyramid = 1170 + 780 = 1950 yd 3

15. S = 4! r 2 = 4!(100 ) 2 = 400! in 2 V = 4 _

3 ! r 3 = 4 _

3 !(10 ) 3 = 4000 ____

3 ! in 3

16. S = 1 _ 2 (4! r 2 ) + ! r 2 = 3!(12 ) 2 = 432! in 2

V = 1 _ 2 ( 4 _

3 ! r 3 ) = 2 _

3 !(12 ) 3 = 1152! in 3

17. V softball _______ V baseball

= 4 _ 3 !(2.5 ) 3

_______ 4 _ 3 !(1.5 ) 3

= (2.5 ) 3

_____ (1.5 ) 3

$ 4.6 times as great


VOCABULARY, PAGE 730 1. oblique prism 2. cross section

LESSON 10-1, PAGE 730 3. cone; vertex: M; edges: none; base: circle L

4. rectangular pyramid: vertices: N, P, Q, R, S; edges:

%% NP ,

%%

NQ , %%

NR , %%

NS , %%

PQ , %%

QR , %%

RS , %%

SP ; base: PQRS.

5. cylinder 6. square pyramid


7.

8.

9.

10.

11. yes 12. no

LESSON 10-3, PAGE 731 13. V = 9; E = 16; F = 9

V ! E + F = 9 ! 16 + 9 = 2

14. V = 8; E = 12; F = 6V ! E + F = 8 ! 12 + 6 = 2

15. d = " ############ (7 ! 2 ) 2 + (1 ! 6 ) 2 + (1 ! 4 ) 2 = " ## 59 $ 7.7 units

M ( 2 + 7 _____ 2 , 6 + 1 _____

2 , 4 + 1 _____

2 ) = M(4.5, 3.5, 2.5)

16. d = " ############ (5 ! 0 ) 2 + (7 ! 3 ) 2 + (8 ! 0 ) 2 = " ## 105 $ 10.2 units

M ( 0 + 5 _____ 2 , 3 + 7 _____

2 , 0 + 8 _____

2 ) = M(2.5, 5, 4)

17. d = " ############ (9 ! 7 ) 2 + (1 ! 2 ) 2 + (5 ! 6 ) 2 = " # 6 $ 2.4 units

M ( 7 + 9 _____ 2 , 2 + 1 _____

2 , 6 + 5 _____

2 ) = M(8, 1.5, 5.5)

18. d = " ############ (2 ! 6 ) 2 + (7 ! 2 ) 2 + (4 ! 8 ) 2 = " ## 57 $ 7.5 units

M ( 6 + 2 _____ 2 , 2 + 7 _____

2 , 8 + 4 _____

2 ) = M(4, 4.5, 6)


19. L = 2!rh = 2!(5)(20) = 200! $ 628.3 yd 2 L = 2!rh + 2! r 2

= 200! + 2!(5 ) 2 = 250! $ 785.4 yd 2

20. L = Ph = (4s)s = 4(5 ) 2 = 100 ft 2 S = Ph + B = 100 + 2( 5 2 ) = 150 ft 2


21. L = Ph = (18)(7) = 126 m 2 S = L + B

= 126 + 2 ( 1 _ 2 (6) (3 " # 3 ) ) = 126 + 18 " # 3 $ 157.2 m 2

22. L = Ph = (20)(8) = 160 cm 2 S = Ph + 2 ( 1 _

2 aP)

= 160 + ( 2 ______ tan 36°

) (20) $ 215.1 cm 2


23. L = 1 _ 2 P" = 1 _

2 (60)(21) = 630 f t 2

S = L + B = 630 + (15 ) 2 = 855 ft 2

24. " 2 = 7 2 + 24 2 " = 25 mL = !r" = !(7)(25) = 175! m 2 S = !r" + ! r 2 = 175! + !(7 ) 2 = 224! m 2

25. L = !r" = !(10)(15) = 150! in 2 S = !r" + ! r 2 = 150! + !(10 ) 2 = 250! in 2

26. L upper = 1 _ 2 P" = 1 _

2 (32)(30) = 480 ft 2

L lower = 1 _ 2 P" = 1 _

2 (32)(20) = 320 ft 2

S = L upper + L lower = 480 + 320 = 800 ft 2

27. L cone = !r" = !(8)(12) = 96! m 2 L cylinder = 2!rh = 2!(8)(16) = 256! m 2 S = L cone + L cylinder + L cone

= 96! + 256! + 96! = 448! m 2


28. V = Bh = ("w)h = (9)(12)(10) = 1080 ft 3

29. V = Bh = ( 1 __ 2

aP) h = 1 __ 2 ( 4 ______

tan 36° ) (40)(15) $ 1651.7 cm 3

30. V = ! r 2 h = !(7.5 ) 2 (16) = 900! i n 3

31. V = ! r 2 h = !(3 ) 2 (5) = 45! m 3


32. V = 1 _ 3 Bh = 1 _

3 (42)(8) = 112 m 3

33. V = 1 _ 3 Bh = 1 _

3 ( 1 _

2 (3) (1.5 " # 3 ) ) (8) = 6 " # 3 $ 10.4 cm 3

34. V = 1 _ 3 ! r 2 h = 1 _

3 !(6 ) 2 (10) = 120! cm 3

35. V = 1 _ 3 Bh = 1 _

3 (16!)(9) = 48! ft 3

36. V cylinder = ! r 2 h = !(8 ) 2 (12) = 768! ft 3 V cone = 1 _

3 ! r 2 h = 1 _

3 !(8 ) 2 (12) = 256! ft 3

V = V cylinder ! V cone = 768! ! 256! = 512! ft 3

37. V cube = s 3 = (10 ) 3 = 1000 cm 3 V pyramid = 1 _

3 Bh = 1 _

3 (1 0 2 )(16) = 1600 ____

3 cm 3

V = V cube + V pyramid

= 1000 + 1600 ____ 3 = 4600 ____

3 $ 1533.3 cm 3


38. S = 100! = 4! r 2 25 = r 2 r = 5 mV = 4 _

3 ! r 3 = 4 _

3 !(5 ) 3 = 500 ___

3 ! m 3

39. V = 288! = 4 _ 3 ! r 3

216 = r 3 r = 6 in.S = 4! r 2 = 4!(6 ) 2 = 144! in 2

40. S = 256! = 4! r 2 64 = r 2 r = 8d = 2(8) = 16 ft

41. Step 1 Find the surface area. S prism = H + B = Ph + 2("w)

= (2(10) + 2(7))(5) + 2(10)(7) = 310 cm 2

L hemisphere = 1 _ 2 (4! r 2 ) = 2!(3 ) 2 = 18! cm 2

B hemisphere = ! r 2 = !(3 ) 2 = 9! cm 2 S = S prism + L hemisphere ! B hemisphere

= 310 + 18! ! 9! = 310 + 9! $ 338.3 cm 2 Step 2 Find the volume. V prism = Bh = ("w)h = (10)(7)(5) = 350 cm 3

V hemisphere = 1 _ 2 (4/3! r 3 ) = 2/3!(3 ) 3 = 18! cm 3

V = V prism ! V hemisphere = 350 ! 18! $ 293.5 cm 3

42. Step 1 Find the surface area. L cylinder = 2!rh = 2!(3)(7) = 42! ft 2 L hemisphere = 1 _

2 (4! r 2 ) = 2!(3 ) 2 = 18! ft 2

S = L cylinder + 2 L hemisphere = 42! + 2(18!) = 78! $ 245.0 ft 2

Step 2 Find the volume. V cylinder = ! r 2 h = !(3 ) 2 (7) = 63! ft 3


3 ! r 3 ) = 2 _

3 !(3 ) 3 = 18! ft 3

V = V cylinder ! 2 V hemisphere = 63! ! 2(18!) = 27! $ 84.8 ft 3


1. pentagonal pyramid; vertices: A, B, C, D, E, F; edges:

%% AB , %%

AC , %%

AD , %%

AE , %%

AF , %%

BC , %%

CD , %%

DE , %%

EF , %%

FB ;faces: &ABC, &ACD, &ADE, &AEF, &AFB, pentagon BCDEF

2. pentagon

3. V = 6; E = 10; F = 6V ! E + F = 6 ! 10 + 6 = 2

4.


5.

6.

7. d = " ############ (5 ! 0 ) 2 + (5 ! 0 ) 2 + (5 ! 0 ) 2 = " ## 75 = 5 " # 3 $ 8.7 units

M ( 0 + 5 _____ 2

, 0 + 5 _____ 2 , 0 + 5 _____

2 ) = M(2.5, 2.5, 2.5)

8. d = " ############ (7 ! 6) 2 + (1 ! 0 ) 2 + (4 ! 9 ) 2 = " ## 27 = 3 " # 3 $ 5.2 units

M ( 6 + 7 _____ 2

, 0 + 1 _____ 2 , 9 + 4 _____

2 ) = M(6.5, 0.5, 6.5)

9. d = " ############## (2 ! (!1) ) 2 + (!5 ! 4 ) 2 + (7 ! 3 ) 2 = " ## 106 $ 10.3 units

M ( !1 + 2 _______ 2 ,

4 + (!5) ________

2 , 3 + 7 _____

2 ) = M(0.5, !0.5, 5)

10. S = L + B = Ph + 2 ( 1 _ 2 aP )

= (48)(11) + (4 " # 3 ) (48) = 528 + 192 " # 3 $ 860.6 ft 2

11. S = 2!rh + 2! r 2 = 2!(10.5)(30) + 2!(10.5 ) 2 $ 2671.9 in 2

12. B = 36 = s 2 s = 6 cmS = L + B = s 2 + 1 _

2 P"

= 36 + 1 _ 2 (24)(14) = 204 cm 2

13. C = 16! = 2!r r = 8 m " 2 = 8 2 + 1 5 2 " = 17 mS = !r" + ! r 2

= !(8)(17) + !(8 ) 2 = 200! $ 628.3 m 2

14. S = 4! r 2 = 4!(5 ) 2 = 100! $ 314.2 yd 2

15. L cube = Ph = (4s)(s) = 4(6 ) 2 = 144 m 2 B cube = s 2 = (6 ) 2 = 36 m 2 L pyramid = 1 _

2 P" = 1 _

2 (24)(4) = 48 m 21

S = L cube + B cube + L pyramid = 144 + 36 + 48 = 228 m 2

16. V = "wh = (15)(9)(12) = 1620 ft 3

17. 7 2 + h 2 = 25 2 h = 24 mV = 1 _

3 ! r 2 h = 1 _

3 !(7 ) 2 (24) = 392! $ 1231.5 m 3

18. V = 1 _ 3 Bh = 1 _

3 (1 8 2 )(20) = 2160 ft 3

19. V = ! r 2 h = !(3 ) 2 (2) = 18! $ 56.5 in 3

20. 4 2 + h 2 = 5 2 h = 3 cm V cone = 1 _

3 ! r 2 h = 1 _

3 !(4 ) 2 (3) = 16! cm 3

V cylinder = ! r 2 h = !(4 ) 2 (7) = 112! cm 3


3 ! r 3 ) = 2 _

3 !(4 ) 3 = 128 ___

3 ! cm 3

V = V cone + V cylinder ! V hemisphere

= 16! + 112! ! 128 ___ 3 ! = 256 ___

3 ! $ 268.1 cm 3

21. V = 4 _ 3 ! r 3 = 4 _

3 !(6 ) 3 = 288! $ 904.8 cm 3

22. V Earth

_____ V Moon

= 4 _ 3 !(3965 ) 3

_________ 4 _ 3 !(1080 ) 3

= (3965 ) 3

_______ (1080 ) 3

$ 49.5

The Earth’s volume is about 49.5 times as great as the Moon’s volume.


1. DA and B must be opposite vertices.

AB = " ###### 4 2 + 4 2 + 4 2 = " ## 48 = 4 " # 3 cm

2. C L = 3B2!rh = 3! r 2 2h = 3r h = 3 _

2 r

3. C " 2 = 6 2 + 8 2 " = 10 ftL = !r" = !(6)(10) = 60! ft 2

4. AV = 1 _

3 ! r 2 h = 1 _

3 !(5 ) 2 (12) = 100! cubic units

5. DThe height of the cylinder is h = 32 ! 2(5) = 22 in. V cylinder = ! r 2 h = !(5 ) 2 (22) = 550! in 3


3 ! r 3 ) = 2 _

3 !(5 ) 3 = 250 ___

3 ! in 3

V = + 2 = 550! + 2 ( 250 ___ 3 !) = 2150 ____

3 ! in 3


Solutions KeyCircles11

CHAPTER


1. C 2. E

3. B 4. A

5. total # of students = 192 + 208 + 216 + 184 = 800

( 192 ____ 800

) · 100% = 24%

6. 216 ____ 800

· 100% = 27%

7. 208 + 216 _________ 800

· 100% = 53%

8. 11%(400,000) = 44,000

9. 27%(400,000) = 108,000

10. 19% + 13% = 32%

11. 32%(400,000) = 128,000

12. 11y ! 8 = 8y + 1 3y ! 8 = 1 3y = 9 y = 3

13. 12x + 32 = 10 + x11x + 32 = 10 11x = !22 x = !1

14. z + 30 = 10z ! 15 30 = 9z ! 15 45 = 9z z = 5

15. 4y + 18 = 10y + 15 18 = 6y + 15 3 = 6y y = 1 __

2

16. !2x ! 16 = x + 6 !16 = 3x + 6 !22 = 3x x = ! 22 ___

3

17. !2x ! 11 = !3x ! 1 x ! 11 = !1 x = 10

18. 17 = x 2 ! 3249 = x 2 x = ±7

19. 2 + y 2 = 18 y 2 = 16 y = ±4

20. 4 x 2 + 12 = 7 x 2 12 = 3 x 2 4 = x 2 x = ±2

21. 188 ! 6 x 2 = 38 !6 x 2 = !150 x 2 = 25 x = ±5

11-1 LINES THAT INTERSECT CIRCLES, PAGES 746–754


1. chords: ""

QR , ""

ST ; tangent: # $% UV ; radii: ""

PQ , ""

PS , ""

PT ; secant: # $% ST ; diameter:

"" ST

2. radius of circle C: 3 ! 2 = 1radius of circle D: 5 ! 2 = 3point of tangency: (2, !1)equation of tangent line: y = !1

3. 1 Understand the ProblemThe answer will be the length of an imaginary segment from the summit of Mt. Kilimanjaro to the Earth’s horizon.2 Make a PlanLet C be the center of the Earth, E be the summit of Mt. Kilimanjaro, and H be a point on the horizon. Find the length of

"" EH , which is tangent

to circle C at H. By Thm. 11-1-1,

"" EH &

"" CH .

So 'CHE is a right '.3 SolveED = 19,340 ft

= 19,340

______ 5280

( 3.66 mi

EC = CD + ED ( 4000 + 3.66 = 4003.66 mi E C 2 ( E H 2 + C H 2 4003.6 6 2 ( E H 2 + 400 0 2 29,293.40 ( E H 2 171 mi ( EH4 Look BackThe problem asks for the distance to the nearest mile. Check that the answer is reasonable by using the Pythagorean Thm. Is 17 1 2 + 400 0 2 ( 400 4 2 ? Yes, 16,029,241 ( 16,032,016.

4a. By Thm. 11-1-3, RS = RT x __

4 = x ! 6.3

x = 4x ! 25.2!3x = !25.2 x = 8.4

RS = (8.4)

____ 2 = 2.1

b. By Thm.11-1-3, RS = RTn + 3 = 2n ! 1 3 = n ! 1 4 = n RS = (4) + 3 = 7

THINK AND DISCUSS, PAGE 750 1. 4 lines

2. No; if line is tangent to the circle with the larger radius, it will not intersect the circle with the smaller radius. If the line is tangent tothe circle with the smaller radius, it will intersect the circle with the larger radius at 2 points.

3. No; a circle consists only of those points which are a given distance from the center.

4. By Thm. 11-1-1, m)PQR = 90°. So by Triangle Sum Theorem m)PRQ = 180 ! (90 + 59) = 31°.


5.



1. secant 2. concentric

3. congruent

4. chord: ""

EF ; tangent: m; radii: ""

DE , ""

DF ; secant: !; diameter:

"" EF

5. chord: ""

QS ; tangent: # $% ST ; radii: ""

PQ , ""

PR , ""

PS ;

secant: # $% QS ; diameter: ""

QS

6. radius of circle A: 4 ! 1 = 3radius of circle B: 4 ! 2 = 2point of tangency: (!1, 4)equation of tangent line: y = 4

7. radius of circle R: 4 ! 2 = 2radius of circle S: 4 ! 2 = 2point of tangency: (1, 2)equation of tangent line: x = 1

8. 1 Understand the ProblemThe answer will be the length of an imaginary segment from the ISS to the Earth’s horizon.2 Make a PlanLet C be the center of the Earth, let E be ISS, and let H be the point on the horizon. Find the length of

"" EH , which

is tangent to circle C at H. By Thm. 11-1-1, ""

EH & ""

CH . So CHE is a right '.3 SolveEC = CD + ED ( 4000 + 240 = 4240 mi E C 2 ( E H 2 + C H 2 424 0 2 ( E H 2 + 400 0 2 1,977,60 ( E H 2 1406 mi ( EH4 Look BackThe problem asks for the distance to the nearest mile. Check that answer is reasonable by using the Pythagorean Thm. Is 1406 2 + 400 0 2 ( 4240 2 ? Yes, 17,976,836 ( 17,977,600.

9. By Thm. 11-1-3, JK = JL4x ! 1 = 2x + 92x ! 1 = 9 2x = 10 x = 5JK = 4(5) ! 1 = 19

10. By Thm. 11-1-3, ST = SU y ! 4 = 3 _

4 y

4y ! 16 = 3y y ! 16 = 0 y = 16ST = (16) ! 4 = 12


11. chords: ""

RS , """

VW ; tangent: !; radii: ""

PV , """

PW ;

secant: # $$% VW ; diameter: """

VW

12. chords: ""

AC , ""

DE ; tangent: # $% CF ; radii: ""

BA , ""

BC ;

secant: # $% DE ; diameter: ""

AC

13. radius of circle C: 2 ! 0 = 2radius of circle D: 4 ! 0 = 4point of tangency: (!4, 0)equation of tangent line: x = !4

14. radius of circle M: 3 ! 2 = 1radius of circle N: 5 ! 2 = 3point of tangency: (2, 1)equation of tangent line: y = 1

15. 1 Understand the ProblemThe answer will be the length of an imaginary segment from the summit of Olympus Mons to Mars’ horizon.2 Make a PlanLet C be the center of Mars, let E be summit of Olympus Mons, and let H be a point on the horizon. Find the length of

"" EH , which is tangent

to circle C at H. By Thm. 11-1-1,

"" EH &

"" CH .

So triangle CHE is a right triangle.3 SolveEC = CD + ED

( 3397 + 25 = 3422 km E C 2 ( E H 2 + C H 2 3422 2 ( E H 2 + 3397 2 170,475 ( E H 2 413 km ( EH4 Look BackThe problem asks for the distance to the nearest km. Check that the answer is reasonable by using the Pythagorean Thm. Is 413 2 + 3397 2 ( 3422 2 ? Yes, 11,710,178 ( 11,710,084.

16. By Thm. 11-1-3, AB = AC2 x 2 = 8xSince x * 0, 2x = 8 x = 4AB = 2(4 ) 2 = 32

17. By Thm. 11-1-3,RS = RT

y = y 2

___ 7

7y = y 2 Since y * 0, 7 = y

RT = (7 ) 2

____ 7 = 7

18. S (true if circles are identical)


19. N 20. N

21. A

22. S (if chord passes through center)

23. ""

AC 24. ""

PA , ""

PB , ""

PC , ""

PD

25. ""

AC

26. By Thm. 11-1-1, Thm 11-1-3, the definition of a circle, and SAS, 'PQR + 'PQS; so

$$% PQ bisects

)RPS. Therefore, m)QPR = 1 _ 2 (42) = 21°.

By Thm 11-1-1, 'PQR is a right '. By the Sum, m)PQR + m)PRQ + m)QPR = 180 m)PQR + 90 + 21 = 180 m)PQR = 180 ! (90 + 21) = 69°m)PQS = 2m)PQR = 2(69) = 138°

27. By Thm 11-1-1, m)R = m)S = 90°.By Quad. Sum Thm.,m)P + m)Q + m)R + m)S = 360 x + 3x + 90 + 90 = 360 4x + 180 = 360 4x = 180 x = 45m)P = 45°

28a. The perpendicular segment from a point to a line is the shortest segment from the point to the line.

b. B c. radius

d. line ! & ""

AB

29. Let E be any point on the line m other than D. It is given that line m &

"" CD . So 'CDE is a right ' with

hypotenuse ""

CE . Therefore, CE > CD. Since ""

CD is a radius, E must lie in exterior of circle C. Thus D is only a point on the line m that is also on circle C. So the line m is tangent to circle C.

30. Since 2 points determine a line, draw auxiliary segments

"" PA , ""

PB , and ""

PC . Since ""

AB and ""

AC are tangents to circle P,

"" AB &

"" PB and

"" AC &

"" PC . So

'ABP and 'ACP are right ,. ""

PB + ""

PC since they are both radii of circle P, and

"" PA +

"" PA by Reflex.

Prop. of +. Therefore, 'ABP + ' ACP by HL + and

"" AB +

"" AC by CPCTC.

31. QR = QS = 5 QT 2 = QR 2 + RT 2 (ST + 5 ) 2 = 5 2 + 12 2 ST + 5 = 13 ST = 8

32. AB = AD 23 = x AC = AE 23 + x ! 5 = x + DE23 + 23 ! 5 = 23 + DE 41 = 23 + DE DE = 18

33. JK = JL and JL = JM, so, JK = JM JK = JM6y ! 2 = 30 ! 2y 8y = 32 y = 4JL = JM = 30 ! 2(4) = 22

34. Point of tangency must be (x, 2), where x ! 2 = ±3x = 5 or !1.Possible points of tangency are (5, 2) and (!1, 2).

35a. BCDE is a rectangle; by Thm. 11-1-1, )BCD and )EDC are right -. It is given that )DEB is a right ). )CBE must also be a right by Quad. Sum Thm. Thus, BCDE has 4 right - and is a rectangle.

b. BE = CE = 17 in.AE = AD ! DE = AD ! BC = 5 ! 3 = 2 in.

c. AB 2 = AE 2 + BE 2 = 2 2 + 17 2 AB 2 = 293 AB = . // 293 ( 17.1 in.

36. Not possible; if it were possible, 'XBC would contain 2 right -. which contradicts ' Sum Thm.

37. By Thm 11-1-1, )R and )S are right -. By Quad. Sum Thm.,)P + )Q + )R + )S = 360 )P + )Q + 90 + 90 = 360 )P + )Q = 180By definition, )P and )Q are supplementary angles.

TEST PREP, PAGE 754

38. CA D 2 = A B 2 + B D 2 = 10 2 + 3 2 = 109 AD = . // 109 ( 10.4 cm

39. G!2 ! (!4) = 2. So, (3, !4) lies on circle P; y = !4 meets circle P only at (3, !4). So it is tangent to circle P.

40. B

"(5 ) 2

_____ "(6 ) 2

= 25" ____ 36"

= 25 ___ 36


41. Since 2 points determine a line, draw auxiliary segments

"" GJ and

"" GK . It is given that

"" GH &

"" JK , so,

)GHJ and )GHK are right -. Therefore, 'GHJ and 'GHK are right -.

"" GH +

"" GH by Reflex. Prop.

of +, and ""

GJ + ""

GK because they are radii of circle G. Thus 'GHJ + 'GHK by HL, and

"" JH +

"" KH by

CPCTC.

42. By Thm. 11-1-1, )C and )D are right -. So BCDE is a rectangle, CE = DB = 2, and BE = DC = 12. Therefore, 'ABE is a right with leg lengths 5 ! 2 = 3 and 12. So

AB = . ///// A E 2 + B E 2 =

. //// 3 2 + 1 2 2 = . // 153 = 3 . // 17

43. Draw a segment from X to the center C of the wheel. )XYC is a right angle and m)YXC = 1 _

2 (70) = 35°. So

tan 35° = 13 ___ XY

XY = 13 ______ tan 35°

( 18.6 in.



44. 14 + 6.25h > 12.5 + 6.5h 1.5 > 0.25h 6 > hSince h is positive, 0 < h < 6.

45. P = LM + PR ________ LR

= 10 + (16 + 4)

_________________ 10 + 6 + 4 + 16 + 4

= 30 ___ 40

= 3 __ 4

46. P = LP ___ LR

= 10 + 6 + 4 __________ 40

= 20 ___ 40

= 1 __ 2

47. P = MN + PR ________ LR

= 6 + (16 + 4)

___________ 40

= 26 ___ 40

= 13 ___ 20

48. P = QR ___ LR

= 4 ___ 40

= 1 ___ 10

CONNECTING GEOMETRY TO DATA ANALYSIS: CIRCLE GRAPHS, PAGE 755

TRY THIS, PAGE 755 1. Step 1 Add all the amounts.

18 + 10 + 8 = 36Step 2 Write each part as a fraction of the whole.novels: 18 __

36 reference: 10 __

36 textbooks: 8 __

36

Step 3 Multiply each fraction by 360° to calculate central ! measure.novels: 18 __

36 (360) = 180° reference: 10 __

36 (360) = 100°

textbooks: 8 __ 36

(360) = 80°

Step 4 Match a circle graph to the data.The data match graph D.

2. Step 1 Add all the amounts.450 + 120 + 900 + 330 = 1800Step 2 Write each part as a fraction of the whole.travel: 450 ____

1800 meals: 120 ____

1800 lodging: 900 ____

1800

other: 330 ____ 1800

Step 3 Multiply each fraction by 360° to calculate central ! measure.travel: 450 ____

1800 (360) = 90° meals: 120 ____

1800 (360) = 24°

lodging: 900 ____ 1800

(360) = 180° other: 330 ____ 1800

(360) = 66°

Step 4 Match a circle graph to the data.The data match graph C.

3. Step 1 Add all the amounts.190 + 375 + 120 + 50 = 735Step 2 Write each part as a fraction of the whole.food: 190 ___

735 health: 375 ___

735 training: 120 ___

735 other: 50 ___

735

Step 3 Multiply each fraction by 360° to calculate central ! measure.food: 190 ___

735 (360) " 93° health: 375 ___

735 (360) " 184°

training: 120 ___ 735

(360) " 59° other: 50 ___ 735

(360) " 24°

Step 4 Match a circle graph to the data.The data match graph B.

11-2 ARCS AND CHORDS, PAGES 756–763

CHECK IT OUT! PAGES 756–759 1a. m!FMC = (0.03 + 0.09 + 0.10 + 0.11)360° = 108°

b. m! # AHB = (1 $ 0.25)360° = 270°

c. m!EMD = (0.10)360° = 36°

2a. m!JPK = 25° (Vert. % Thm.) m # JK = 25°m!KPL + m!LPM + m!MPN = 180° m!KPL + 40° + 25° = 180° m!KPL = 115° m # KL = 115°m # JKL = m # JK + m # KL

= 25° + 115° = 140°

b. m # LK = m # KL = 115°m!KPN = 180°m # KJN = 180°m # LJN = m # LK + m # KJN

= 180° + 115°

3a. m!RPT = m!SPT RT = ST 6x = 20 $ 4x 10x = 20 x = 2RT = 6(2) = 12

b. m!CAD = m!EBF (11-2-2(3)) mCD = mEF 25y = 30y $ 20 20 = 5y y = 4mCD = m!CAD = 25(4) = 100°

4. Step 1 Draw radius &&

PQ .PQ = 10 + 10 = 20Step 2 Use Pythagorean and 11-2-3.P T 2 + Q T 2 = P Q 2 1 0 2 + Q T 2 = 2 0 2 Q T 2 = 300 QT = ' (( 300 = 10 ' ( 3 Step 3 Find QR.QR = 2 (10 ' ( 3 ) = 20 ' ( 3 " 34.6



THINK AND DISCUSS, PAGE 759 1. The arc measures between 90° and 180°.

2. if arcs are on 2 different circles with different radii

3.



1. semicircle 2. Vertex is the center of the circle.

3. major arc 4. minor arc

5. m)PAQ = 0.45(360) = 162°

6. m)VAU = 0.07(360) = 25.2°

7. m)SAQ = (0.06 + 0.11)360 = 61.2°

8. m 0 UT = m)UAT = 0.1(360) = 36°

9. m 0 RQ = m)RAQ = 0.11(360) = 39.6°

10. m 0 UPT = (1 ! 0.1)360 = 324°

11. m 0 DE = m)DAE = 90°m 0 EF = m)EAF = m)BAC = 90 ! 51 = 39°m 0 DF = m 0 DE + m 0 EF = 90 + 39 = 129°

12. m 0 DEB = m)DAE + m)EAB = 90 + 180 = 270°

13. m)HGJ + m)JGL = m)HGL 72 + m)JGL = 180 m)JGL = 108° m 0 JL = 108°

14. m 0 HLK = m)HGL + m)LGK = 180 + 30 = 210°

15. QR = RS (Thm. 11-2-2(1))8y ! 8 = 6y 2y = 8 y = 4QR = 8(4) ! 8 = 24

16. m)CAD = m)EBF (Thm. 11-2-2(3)) 45 ! 6x = !9x 3x = !45 x = !15m)EBF = !9(!15) = 135°

17. Step 1 Draw radius ""

PR .PR = 5 + 8 = 13Step 2 Use the Pythagorean Thm. and Thm. 11-2-3.Let the intersection of

"" PQ and

"" RS be T.

P T 2 + R T 2 = P R 2 5 2 + R T 2 = 13 2 RT = 12Step 3 Find RS.RS = 2(12) = 24

18. Step 1 Draw radius ""

CE .CE = 50 + 20 = 70Step 2 Use the Pythagorean Thm. and Thm. 11-2-3.Let the intersection of

"" CD and

"" EF be G.

C G 2 + E G 2 = C E 2 5 0 2 + E G 2 = 7 0 2 RG = . // 2400 = 20 . / 6 Step 3 Find EF.EF = 2 (20 . / 6 ) = 40 . / 6 ( 98.0


19. m)ADB = 35 ____________ 35 + 39 + 29

(360) = 35 ____ 103

(360) ( 122.3°

20. m)ADC = 29 ____ 103

(360) = 101.4°

21. m 0 AB = m)ADB ( 122.3°

22. m 0 BC = m)BDC = 39 ____ 103

(360) ( 136.3°

23. m 0 ACB = 360 ! m)ADB ( 360 ! 122.3 = 237.7°

24. m 0 CAB = 360 ! m)BDC ( 360 ! 136.3 = 223.7°

25. m 0 MP = m)MJP = m)MJQ ! m)PJQ = 180 ! 28 = 152°

26. mQNL = mQNM + mML= m)QJM + m)MJL= 180 + 28 = 208°

27. m 0 WT = m 0 WS + m 0 ST = m)WXS + m)SXT= 55 + 100 = 155°

28. m 0 WTV = m 0 WS + m 0 STV = m)WXS + m)SXV= 55 + 180 = 235°

29. m)CAD = m)EBF (Thm. 11-2-2(3))10x ! 63 = 7x 3x = 63 x = 21m)CAD = 10(21) ! 63 = 147°

30. m 0 JK = m 0 LM (Thm. 11-2-2(2))4y + y = y + 68 y = 17m 0 JK = 4(17) + 17 = 85°


31. AC = AB = 2.4 + 1.7 = 4.1Let

!! AB and

!! CD meet at E.

A E 2 + CE 2 = AC 2 2. 4 2 + C E 2 = 4. 1 2 CE = " ### 11.05 CD = 2 " ### 11.05 $ 6.6

32. PR = PQ = 2(3) = 6Let

!! PQ and

!! RS meet at T.

P T 2 + R T 2 = P R 2 3 2 + R T 2 = 6 2 RT = " ## 27 = 3 " # 3 RS = 2 (3 " # 3 ) = 6 " # 3 $ 10.4

33. F; the % measures between 0° and 180°. So it could be right or obtuse.

34. F; Endpts. of a diameter determine 2 & arcs measuring exactly 180°.

35. T (Thm. 11-2-4)

36. Check students’ graphs.

37. Let m%AEB = 3x, m%BEC = 4x, and m%CED = 5x.m%AEB + m%BEC + m%CED = 180 3x + 4x + 5x = 180 12x = 128 x = 15m%AEB = 3(15) = 45° m%BEC = 4(15) = 60°m%CED = 5(15) = 75°

38. m ' JL + m ' JK + m ' KL = 3607x ( 18 + 4x ( 2 + 6x + 6 = 360 17x = 374 x = 22m ' JL = 7(22) ( 18 = 136°

39. m%QPR = 180 10x = 180 x = 18m%SPT = 6(18) = 108°


1. !!

BC & !!

DE 1. Given.2.

!! AB &

!! AD and

!!

AC & !!

AE 2. All radii of a circle

are &.3. )BAC & ) DAE 3. SSS4. %BAC & %DAE 4. CPCTC5. m%BAC = m%DAE 5. 6. m ' BC = m ' DE 6. Definition of arc

measures7. ' BC & ' DE 7. Definition of arcs


1. ' BC & ' DE 1. Given2. m ' BC = m ' DE 2. Definition of & arcs3. m%BAC = m%DAE 3. Definition of arc

measures4. %BAC & %DAE 4.


1. !!

CD * !!

EF 1. Given 2. Draw radii

!! CE

and !!

CF . 2. 2 points determine

a line. 3.

!! CE &

!! CF 3. All radii of a circle

are congruent. 4.

!!! CM &

!!! CM 4. Reflex. Prop. of &

5. %CMF and %CME are rt. +.

5. Def. of *

6. )CMF and )CME are rt. ,.

6. Def. of a rt. )

7. )CMF & )CME 7. HL Steps 3, 4 8.

!! FM &

!! EM 8. CPCTC

9. !!

CD bisects !!

EF 9. Def. of a bisector10. %FCD & %ECD 10. CPCTC11. m%FCD = m%ECD 11. Def. of & +12. m ' FD = m ' ED 12. Def. of arc measures13. ' FD & ' ED 13. Def. of arcs14.

!! CD bisects ' EF . 14. Def. of a bisector


1. !!

JK is the * bis. of !!

GH 1. Given

2. A is equidistant from G and H.

2. Def. of the center of circle

3. A lies on the * bis. of !!

GH .3. Perpendicular

Bisector Theorem4.

!! JK is a diameter of

circle A.4. Def. of diam.

44. The circle is divided into eight & sectors, each with central % measure 45°. So possible measures of the central congruent are multiples of 45° between 0(45) = 0° and 8(45) = 360°. So there are three different sizes of angles: 135°, 90°, and 45°.

45. Solution A is incorrect because it assumes that %BGC is a right %.

46. To make a circle graph, draw a circle and then draw central + that measure 0.4(360) = 144°, 0.35(360) = 126°, 0.15(360) = 54°, and 0.1(360) = 36°.

47a. AC = 1 _ 2 (27) = 13.5 in.

AD = AB ( DB = 13.5 ( 7 = 6.5 in.

b. C D 2 + A D 2 = A C 2 C D 2 + 6. 5 2 = 13. 5 2 CD = " ## 140 = 2 " ## 35 $ 11.8 in.

c. By Theorem 11-2-3, !!

AB bisects !!

CE . So CE = 2CD = 4 " ## 35 $ 23.7 in.

TEST PREP, PAGE 763

48. Dm ' WT = 90 + 18 = 108°m ' UW = 90°

m ' VR = 180 ( 41 = 139°m ' TV = 180 ( 18 = 162°

49. FCE = 1 _

2 (10) = 5

A E 2 + 5 2 = 6 2 AE = " # 11 $ 3.3



50. 90 !!

AP is a horizontal radius, and !!

BP is a vertical radius. So m " AB = m#APB = 90°.


51. AD = AB = 4 + 2 = 6cos BAD = 4 _

6 = 2 _

3

m " BD = m#BAD = co s $1 ( 2 _ 3 ) % 48.2°

52. 2 points determine 2 distinct arcs.3 points determine 6 arcs.4 points determine 12 arcs.5 points determine 20 arcs.&n points determine n(n $ 1) arcs.

53a. ! ' 180°. So ! __ 2 ' 90°, ! __

3 ' 60°, and ! __

4 ' 45°

b. 135° ' 135 ____ 180

(!) = 3! ___ 4

270° ' 270 ____ 180

(!) = 3! ___ 2


54. (3x ) 3 (2 y 2 ) ( 3 $2 y 2 ) (27 x 3 ) (2 y 2 ) ( 1 _

9 y 2 )

6 x 3 y 4

55. a 4 b 3 ($2a) $4

a 4 b 3 ( 1 __ 16

a $4 ) 1 __ 16

b 3

56. ($2 r 3 s 2 ) (3t s 2 ) 2

$2 t 3 s 2 (9 t 2 s 4 ) $18 t 5 s 6

57. 3 = 1 + 2 7 = 3 + 413 = 7 + 621 = 13 + 821 + 10 = 31

58. C, E, G, I, K, M 59. 6 = 1 + 515 = 6 + 915 + 13 = 28

60. #NPQ and #NMQ are right ( (Thm. 11-1-2).So #NMQ = 90°.

61. PQ = MQ (Thm. 11-1-3) 2x = 4x $ 9 9 = 2x x = 4.5MQ = 4(4.5) $ 9 = 9

CONSTRUCTION, PAGE 763 1. O is on the ) bisector of

!! PQ . So by Conv. of )

Bisector, OP = OQ. Similarly, O is on ) bisector of !!

QR . So OQ = OR. Thus, !!

OP , !!!

OQ , and !!

OR are radii of circle O, and circle O contains Q and R.

11-3 SECTOR AREA AND ARC LENGTH, PAGES 764–769


1a. A = ! r 2 ( m ____ 360

) = !(1 ) 2 ( 90 ____ 360

) = 1 __ 4

! m 2 % 0.79 m 2

b. A = ! r 2 ( m ____ 360

) = !(16 ) 2 ( 36 ____ 360

) = 25.6! in. 2 % 80.42 in. 2

2. A = ! r 2 ( m ____ 360

) = !(360 ) 2 ( 180 ____

360 )

% 203,575 ft 2

3. Step 1 Find the area of sector RST.

A = ! r 2 ( m ____ 360

) = !(4 ) 2 ( 90 ____ 360

) = 4! m 2

Step 2 Find the area of *RST.A = 1 _

2 bh = 1 _

2 (4)(4) = 8 m 2

Step 3 area of segment = area of sector RST $ area of

*RST= 4! $ 8 % 4.57 m 2

4a. L = 2!r ( m ____ 360

) = 2!(6) ( 40 ____ 360

) = 4 __ 3 ! m % 4.19 m

b. L = 2!r ( m ____ 360

) = 2!(4) ( 135 ____ 360

) = 3! cm % 9.42 cm

THINK AND DISCUSS, PAGE 766 1. An arc measure is measured in degrees. An arc

length is measured in linear units.

2. the radius and central # of the sector

3.



1. segment

2. A = ! r 2 ( m ____ 360

) = !(6 ) 2 ( 90 ____ 360

) = 9! m 2 % 28.27 m 2

3. A = ! r 2 ( m ____ 360

) = !(8 ) 2 ( 135 ____ 360

) = 24! cm 2 % 75.40 cm 2

4. A = ! r 2 ( m ____ 360

) = !(2 ) 2 ( 20 ____ 360

) = 2 __ 9

! ft 2 % 0.70 ft 2

5. A = ! r 2 ( m ____ 360

) = !(3 ) 2 ( 150 ____ 360

) = 15 ___ 4 ! mi 2 % 12 mi 2



6. Step 1 Find area of sector ABC.

A = " r 2 ( m ____ 360

) = "(3 ) 2 ( 90 ____ 360

) = 9 __ 4 " in. 2

Step 2 Find area of 'ABC.A = 1 _

2 bh = 1 _

2 (3)(3) = 9 _

2 in. 2

Step 3 area of segment = area of sector ABC ! area of

'ABC= 9 _

4 " ! 9 _

2 ( 2.57 in. 2

7. Step 1 Find the area of sector DEF.

A = " r 2 ( m ____ 360

) = "(20 ) 2 ( 60 ____ 360

) = 200 ____ 3 " m 2

Step 2 Find the area of 'DEF.A = 1 _

2 bh = 1 _

2 (20) (10 . / 3 ) = 100 . / 3 m 2

Step 3 area of segment = area of sector DEF ! area of

'DEF= 200 ___

3 " ! 100 . / 3 ( 36.23 m 2

8. Step 1 Find the area of sector ABC.

A = " r 2 ( m ____ 360

) = "(6 ) 2 ( 45 ____ 360

) = 9 __ 2 " cm 2

Step 2 Find the area of 'ABC.A = 1 _

2 bh = 1 _

2 (6 ) (3 . / 2 ) = 9 . / 2 cm 2

Step 3 area of segment = area of sector ABC ! area of

'ABC = 9 __

2 " ! 9 . / 2 ( 1.41 cm 2

9. L = 2"r ( m ____ 360

) = 2"(16) ( 45 ____ 360

) = 4" ft ( 12.57 ft

10. L = 2"r ( m ____ 360

) = 2"(9) ( 120 ____ 360

) = 6" m ( 18.85 m

11. L = 2"r ( m ____ 360

) = 2"(6) ( 20 ____ 360

) = 2 __ 3 " in. ( 2.09 in.


12. A = "(20 ) 2 ( 150 ____ 360

) = 500 ____ 3 " m 2 ( 523.60 m 2

13. A = "(9 ) 2 ( 100 ____ 360

) = 45 ___ 2

" in 2 ( 70.69 in. 2

14. A = "(2 ) 2 ( 47 ____ 360

) = 47 ___ 90

" ft 2 ( 1.64 ft 2

15. A = "(20 ) 2 ( 180 ____ 360

) = 200" ( 628 in. 2

16. 'ABC is a 45°-45°-90° triangle. So central angle measures 90°.area of segment = area of sector ABC ! area of

'ABC

= "(10 ) 2 ( 90 ____ 360

) ! 1 __ 2 (10)(10)

= 25" ! 50 ( 28.54 m 2

17. m)KLM = m 0 KM = 120°area of segment = area of sector KLM ! area of

'KLM

= "(5 ) 2 ( 120 ____ 360

) ! 1 __ 2 ( 5 __

2 ) (5 . / 3 )

= 25 ___ 3 " ! 25 ___

4 . / 3 ( 15.35 in. 2

18. area of segment = area of sector RST ! area of 'RST

= "(1 ) 2 ( 60 ____ 360

) ! 1 __ 2 (1) ( . / 3 ___

2 )

= 1 __ 6 " ! 1 __

4 . / 3 ( 0.09 in. 2

19. L = 2"(5) ( 50 ____ 360

) = 25 ___ 18

" mm ( 4.36 mm

20. L = 2"(1.5) ( 160 ____ 360

) = 4 __ 3 " m ( 4.19 m

21. L = 2"(2) ( 9 ____ 360

) = 1 ___ 10

" ft ( 0.31 ft

22. P = 2"(3) ( 180 ____ 360

) + 3 (2"(1) ( 180 ____ 360

) ) = 6" ( 18.8 in.

23. never

24. sometimes (if radii of arcs are equal)

25. always

26. A = " r 2 ( m ____ 360

) 9" = " r 2 ( 90 ____

360 )

36 = r 2 r = 6

27. L = 2"r ( m ____ 360

) 8" = 2"r ( 120 ____

360 )

24" = 2"r r = 12

28a. L ( 2 ( 22 ___ 7 ) (7) ( 90 ____

360 ) = 11 in.

b. L = 2"(7) ( 90 ____ 360

) = 7 __ 2 " ( 10.99557429 in.

c. overestimate, since L < 10.996 < 11

29a. L = 2"(2.5) ( 90 ____ 360

) = 5 __ 4 " ( 3.9 ft

b. 4.5 = 2"(2.5) ( m ____ 360

) 9 ____

10" = m ____

360

m = 324 ____ " ( 103°

30. The area of sector BAC is 45 ____ 360

= 1 __ 8 area of circle A.

So if area of circle A is 24 in 2 , the area of the

sector will automatically be 1 __ 8 (24) = 3 in. 2

So we need only to solve " r 2 = 24 for r." r 2 = 24

r 2 = 24 ___ "

r = . // 24 ___ " ( 2.76 in.

31. If the length of the arc is L and its degree measure is m, then L = 2"r ( m ____

360 )

360L = 2"rm

r = 360L _____ 2"m

= 180L _____ "m .


TEST PREP, PAGE 769

32. BA = !(8 ) 2 ( 90 ____

360 ) = 16!

33. GA = 2!(8) ( 90 ____

360 ) = 4!

34. 43.98A = !(12 ) 2 ( 35 ____

360 ) ! 43.98


35. A = !(5 ) 2 ( 40 ____ 360

) " !(2 ) 2 ( 40 ____ 360

) = 25 ___ 9 ! " 4 __

9 ! = 7 __

3 !

36a. V = Bh = (!(4 ) 2 ( 30 ____ 360

) ) (3) = 4! ! 12.6 in. 3

b. B = 2 (!(4 ) 2 ( 30 ____ 360

) ) = 8 __ 3 !

L = 2(3)(4) + 2 (!(4) ( 30 ____ 360

) ) (3) = 24 + !

A = 8 __ 3

! + 24 + 2! ! 38.7 in. 2

37a. A(#) = !(2 ) 2 = 4!

A(red) = 4!(1 ) 2 ( 45 ____ 360

) = 1 __ 2 !

P(red) = 1 _ 2 !

___ 4!

= 1 __ 8

b. A(blue) = 4 (!(2 ) 2 ( 45 ____ 360

) " !(1 ) 2 ( 45 ____ 360

) ) = 3 __ 2 !

P(blue) = 3 _ 2 !

___ 4!

= 3 __ 8

c. P(red or blue) = 1 _ 2 ! + 3 _

2 ! ________

4! = 1 __

2


38. 8x " 2y = 6 2y = 8x " 6 y = 4x " 3The line is $.

39. slope = 2 " 0 ______ 1 1 _

2 " 1 _

2 = 2

The line is neither $ nor %.

40. y = mx + 1 0 = m(4) + 1

m = " 1 __ 4

The line is %.

41. V = 4 _ 3 !(3 ) 3 = 36! cm 3

42. S = 4! = 4! r 2 r 2 = 1 r = 1 cmC = 2!(1) = 2! cm

43. m&KLJ = m&KLH " (m&GLJ " m&GLH)10x " 28 = 180 " (90 " (2x + 2))10x " 28 = 90 + 2x + 2 8x = 120 x = 15m&KLJ = 10(15) " 28 = 122°

44. m ' KJ = m&KLJ = 122°

45. m ' JFH = m ' JF + m ' FG + m ' GH = m&JLF + m&FLG + m&GLH= 180 + 90 + 2(15) + 2 = 302°


1. Let " represent the length of the stand. Radius r = 14 in. Form a right ( with leg lengths 14 " 10 = 4 in. and 1 _

2 " in., and

hypotenuse length 14 in.

4 2 + ( 1 _ 2 ") 2 = 14 2

16 + 1 _ 4 " 2 = 196

1 _ 4 " 2 = 180

" 2 = 720 " = ) ** 720 = 12 ) * 5 ! 26.8 in.

2. Let E be on ++

AD with ++

BE ++

CD and BE = CD. (ABE is a right (. So A E 2 + BE 2 = AB 2 (4 " 2 ) 2 + CD 2 = 15 2 C D 2 = 221 CD = ) ** 221 ! 14.9 in.

3. L = 2!r ( m ____ 360

) = 2!(4) ( 60 ____ 360

) = 4 __ 3 ! ! 4.2 in.

4. L = 2!r ( m ____ 360

) 4 __ 3 ! = 2!(2) ( m ____

360 )

1 __ 3 = m ____

360

m = 360 ____ 3 = 120°


1. chord: ++

PR ; tangent: m; radii: ++

QP , ++

QR , ++

QS ; secant: , -. PR ; diameter:

++ PR

2. chords: ++

BD , ++

BE ; tangent: , -. BC ; radii: ++

AB , ++

AE ; secant: , -. BD ; diameter:

++ BE

3. Let x be the distance to the horizon. The building forms a right ( with leg lengths x mi and 4000 mi, and hypotenuse length (4000 + 732 ____

5280 ) mi.

x 2 + (4000 ) 2 = (4000 + 732 ____ 5280

) 2

x 2 ! 1109.11 x ! 33 mi

4. m ' BC + m ' CD = m ' BD m ' BD + m&CAD = m&BAD m ' BD + 49 = 90 m ' BD = 41°

5. m ' BED = m ' BFE + m ' ED = m&BAE + m&EAD= 180 + 90 = 270°

6. m ' SR + m ' RQ = m ' SQ m ' SR + m&RPQ = m&SPQ m ' SR + 71 = 180 m ' SR = 109°



7. m 0 SQU + m 0 UT + m 0 TS = 360m 0 SQU + m)UPT + m)TPS = 360 m 0 SQU + 40 + 71 = 360 m 0 SQU = 249°

8. Let JK = 2x; then x 2 + 4 2 = (4 + 3 ) 2 x 2 = 33 x = . // 33 JK = 2 . // 33 ( 11.5

9. Let XY = 2x; then x 2 + 4 2 = 8 2

x 2 = 48 x = 4 . / 3 XY = 8 . / 3 ( 13.9

10. A = "(22 ) 2 ( 80 ____ 360

) = 968 ____ 9 " ( 338 cm 2

11. 0 AB = 2"(4) ( 150 ____ 360

) = 10 ___ 3 " ft ( 10.47 ft

12. 0 EF = 2"(2.4) ( 75 ____ 360

) = " cm ( 3.14 cm

13. arc length = 2"(5) ( 44 ____ 360

) = 11 ___ 9 " in. ( 3.84 in.

11. arc length = 2"(46) ( 180 ____ 360

) = 46" m ( 144.51 m

11-4 INSCRIBED ANGLES, PAGES 772–779


1a. m)ABC = 1 _ 2 m 0 ADC

135 = 1 _ 2 m 0 ADC

m 0 ADC = 270°

b. m)DAE = 1 _ 2 mDE

= 1 _ 2 (72) = 36°

2. m)ABD = 1 _ 2 m 0 AD = 1 _

2 (86) = 43°

m)BAC = 1 _ 2 m 0 BC

60 = 1 _ 2 m 0 BC

m 0 BC = 120°

3a. )ABC is a right anglem)ABC = 90 8z ! 6 = 90 8z = 96 z = 12

b. m)EDF = m)EGF 2x + 3 = 75 ! 2x 4x = 72 x = 18m)EDF = 2(18) + 3

= 39°

4. Step 1 Find the value of x. m)K + m)M = 18033 + 6x + 4x ! 13 = 180 10x = 160 x = 16Step 2 Find the measure of each ).m)K = 33 + 6(16) = 129°m)L =

9(16) ____

2 = 72°

m)M = 4(16) ! 13 = 51°m)J + m)L = 180 m)J + 72 = 180 m)J = 108°

THINK AND DISCUSS, PAGE 775 1. No; a quadrilateral can be inscribed in a circle if

and only if its opposite - are supplementary.

2. An arc that is 1 _ 4 of a circle measures 90°. If the arc

measures 90°, then the measure of the inscribed ) is 1 _ 2 (90) = 45°.

3.



1. inscribed

2. m)DEF = 1 _ 2 m 0 DF

= 1 _ 2 (78) = 39°

3. m)EFG = 1 _ 2 m 0 EG

29 = 1 _ 2 m 0 EG

m 0 EG = 58°

4. m)JNL = 1 _ 2 m 0 JKL

102 = 1 _ 2 m 0 JKL

m 0 JKL = 204°

5. m)LKM = 1 _ 2 m 0 LM

= 1 _ 2 (52) = 26°

6. m)QTR + m)Q + m)R = 180m)QTR + 1 _

2 m 0 RS + m)S = 180

m)QTR + 1 _ 2 (90) + 25 = 180

m)QTR + 70 = 180 m)QTR = 110°

7. )DEF is a right ) 4x ___ 5 = 90

4x = 450 x = 112.5

8. )FHG is a right ). So 'FGH is a 45°-45°-90° triangle.m)GFH = 45 3y + 6 = 45 3y = 39 y = 13

9. m)XYZ = m)XWZ 7y ! 3 = 4 + 6y y = 7m)XYZ = 7(7) ! 3 = 46°


10. Step 1 Find the value of x. m)P + m)R = 1805x + 20 + 7x ! 8 = 180 12x = 168 x = 14Step 2 Find the measures.m)P = 5(14) + 20 = 90°m)Q = 10(14) = 140°m)R = 7(14) ! 8 = 90°m)S + m)Q = 180 m)S + 140 = 180 m)S = 40°

11. Step 1 Find the value of z. m)A + m)C = 1804z ! 10 + 10 + 5z = 180 9z = 180 z = 20Step 2 Find the measures.m)A = 4(20) ! 10 = 70°m)B = 6(20) ! 5 = 115°m)C = 10 + 5(20) = 110°m)B + m)D = 180 115 + m)D = 180 m)D = 65°


12. m)MNL = 1 _ 2 m 0 ML

43 = 1 _ 2 m 0 ML

m 0 ML = 86°

13. m)KMN = 1 _ 2 m 0 KN

= 1 _ 2 (95)

= 47.5°

14. m)EJH = 1 _ 2 m 0 EGH

139 = 1 _ 2 m 0 EGH

m 0 EGH = 278°

15. m)GFH = 1 _ 2 m 0 GH

= 1 _ 2 (95.2)

= 47.6°

16. m)ADC = m)ADB + m)BDC = 1 _

2 mAB + m)BEC

= 1 _ 2 (44) + 40 = 62°

17. m)SRT = 903 y 2 ! 18 = 90 3 y 2 = 108 y 2 = 36 y = ±6

18. 'JKL is a 30°-60°-90°6z ! 4 = 60 6z = 64 z = 64 __

6 = 10 2 _

3

19. m)ADB = m)ACB 2 x 2 = 10x 2x = 10 (x * 0) x = 5m)ADB = 1 _

2 m 0 AB

2(5 ) 2 = 1 _ 2 m 0 AB

50 = 1 _ 2 m 0 AB

m 0 AB = 2(50) = 100°

20. )MPN and )MNP are complementary.m)MPN + m)MNP = 90 11x ___

3 + 3x ! 10 = 90

20x ! 30 = 270 20x = 300 x = 15

m)MPN = 11(15)

______ 3 = 55°

21. Step 1 Find the value of x.m)B + m)D = 180 x __

2 + x __

4 + 30 = 180

3x ___ 4 = 150

3x = 600 x = 200Step 2 Find the measures.m)B = 200 ___

2 = 100°

m)D = 200 ___ 4 + 30 = 80°

m)E = 200 ! 59 = 141°m)C + m)E = 180 m)C + 141 = 180 m)C = 39°

22. Step 1 Find the value of x. m)U + m)W = 18014 + 4x + 6x ! 14 = 180 10x = 180 x = 18Step 2 Find the value of y. m)T + m)V = 18012y ! 5 + 15y ! 4 = 180 27y = 189 y = 7Step 3 Find the measures.m)T = 12(7) ! 5 = 79°m)U = 14 + 4(18) = 86°m)V = 15(7) ! 4 = 101°m)W = 6(18) ! 14 = 94°

23. always 24. never

25. sometimes (if opposite - are supplementary)

26. m)ABC = 1 _ 2 m 0 AC = 1 _

2 m)ADC = 1 _

2 (112) = 56°

27. Let S be any point on the major arc from P to R. m)PQR + m)PSR = 180m)PQR + 1 _

2 m 0 PQR = 180

m)PQR + 1 _ 2 (130) = 180

m)PQR = 180 ! 65 = 115°

28. By the definition of an arc measure, m 0 JK = m)JHK. Also, the measure of an ) inscribed in a circle is half the measure of the intercepted arc. So m)JLK = 1 _

2 m 0 JK . Multiplying both sides of equation

by 2 gives 2m)JLK = m 0 JK . By substitution, m)JHK = 2m)JLK.

29a. m)BAC = 1 _ 2 m 0 BC = 1 _

2 ( 1 _

6 (360)) = 30°

b. m)CDE = 1 _ 2 mCAE = 1 _

2 ( 4 _

6 (360)) = 120°

c. )FBC is inscribed in a semicircle. So it must be a right ); therefore, 'FBC is a right ). (Also, m)CFD = 30°. So, 'FBC is a 30°-60°-90° '.)


30.

Since any 2 points determine a line, draw !!" BX .

Let D be a point where !!" BX intercepts # AC . By Case 1

of the Inscribed $ Thm., m$ABD = 1 _ 2 m # AD and

m$DBC = 1 _ 2 m # DC . By Add., Distrib., and Trans.

Props. of =, m$ABD + m$DBC = 1 _ 2 m # AD + 1 _

2 m # DC

= 1 _ 2 (m # AD + m # DC ). Thus, by $ Add. Post. and Arc

Add. Post., m$ABC = 1 _ 2 mAC.

31.

Since any 2 points determine a line, draw !!" BX .

Let D be pt. where !!" BX intercepts # ACB . By Case 1

of Inscribed $ Thm., m$ABD = 1 _ 2 m # AD and m$CBD

= 1 _ 2 m # CD . By Subtr., Distrib., and Trans. Props.

of =, m$ABD % m$CBD = 1 _ 2 m # AD % 1 _

2 m # CD

= 1 _ 2 (m # AD % m # CD ). Thus by $ Add. Post. and Arc

Add. Post., m$ABC = 1 _ 2 mAC.

32. By the Inscribed $ Thm., m$ACD = 1 _ 2 m # AB

and m$ADB = 1 _ 2 m # AB . By Substitution, m$ACD =

m$ADB, and therefore, $ACD & $ADB.

33. m$J = 1 _ 2 m # KLM = 1 _

2 (216) = 108°

m$J + m$L = 180 108 + m$L = 180 m$L = 72°m$M = 1 _

2 m # JKL = 1 _

2 (198) = 99°

m$K + m$M = 180 m$K + 99 = 180 m$K = 81°

34. ''

PR is a diag. of PQRS. $Q is an inscribed right angle. So its intercepted arc is a semicircle. Thus,

'' PR is a diameter of the circle.

35a. AB 2 + AC 2 = BC 2 , so by Conv. of Pythag. Thm., (ABC is a right ( with right $A. Since $A is an inscribed right $, it intercepts a semicircle. This means that

'' BC is a diameter.

b. m$ABC = sin %1 % ( 14 ___ 18

) ) 51.1°.

Since m$ABC = 1 _ 2 m # AC , m # AC = 102°.

36.

Draw a diagram through D and A. Label the intersection of

'' BC and

'' DE as F and the intersection

of !!" DA and

'' BE as G. Since

'' BC is a diameter of the

circle, it is a bisector of chord ''

DE . Thus, ''

DF & ''

EF ,and $BFD and $BFE are & right *.

'' BF &

'' BF

by Reflex. Prop. of &. Thus, (BFD & (BFE by SAS.

'' BD &

'' BE by CPCTC. By Trans. Prop. of &,

''

BE & ''

ED . Thus, by definition, (DBE is equilateral.

37. Agree; the opposite * of a quadrilateral are congruent. So the $ opposite the 30° $ also measures 30°. Since this pair of the opposite * are not supplementary, the quadrilateral cannot be inscribed in a circle.


TEST PREP, PAGE 778

39. Dm$BAC + m$ABC + m$ACB = 180 m$BAC + 1 _

2 mAC + m$ACB = 180

m$BAC + 1 _ 2 (76) + 61 = 180

m$BAC + 38 + 61 = 180 m$BAC = 81°

40. H 1 _ 2 m # XY = m$XCY = 1 _

2 m$XCZ

m # XY = m$XCZ = 60°

41. CLet m$A = 4x and m$C = 5x.m$A + m$C = 180 4x + 5x = 180 9x = 180 x = 20m$A = 4(20) = 80°

42. Fm$STR = 180 % (m$TRS + m$TSR)

= 180 % ( 1 _ 2 m # PS + 1 _

2 m # QR )

= 180 % (42 + 56) = 82°m$QPR = 1 _

2 m # QR = 56°

m$QPR = 1 _ 2 m # QR = 56°

m$PQS = 1 _ 2 m # PS = 42°


43. If an $ is inscribed in a semicircle, the measure of the intercepted arc is 180°. The measure of $ is 1 _

2 (180) = 90°. So the angle is a right $.

Conversely, if an inscribed angle is a right angle, then it measures 90° and its intercepted arc measures 2(90) = 180°. An arc that measures 180° is a semicircle.



44. Suppose the quadrilateral ABCD is inscribed in a

circle. Then m)A = 1 _ 2 m 0 BCD and m)C = 1 _

2 m 0 DAB .

By Add., Distrib., and Trans. Prop. of =, m)A +

m)C = 1 _ 2 m 0 BCD + 1 _

2 m 0 DAB = 1 _

2 (m 0 BCD + m 0 DAB ).

m 0 BCD + m 0 DAB = 360°. So by subst., m)A + m)C= 1 _

2 (360) = 180°. Thus, )A and )C are

supplementary. A similar proof shows that )B and )D are supplementary.

45. ""

RQ is a diameter. So )P is a right ). m 0 PQ = 2m)R = 2 ta n !1 ( 7 _

3 ) ( 134°

46. Draw ""

AC and ""

DE . m 1 _

2 m

"" CE = 19°

m)ACD = 1 _ 2 m 0 AD = 36°

m)ABC + 19 + 36 = 180 m)ABC = 125°m)ABD + 125 = 180 m)ABD = 55°



48. 1

2

3

2

4

a + b + c = 12 (1)

30a + 22.5b + 15c = 255 (2)

30a = 15c (3)

Substitute (3) in (2):15c + 22.5b + 15c = 255 3b + 4c = 34 (4)Substitute 1 __

15 (3) in 2(1):

c + 2b + 2c = 24 2b + 3c = 24 (5)3(5) ! 2(4):6b + 9c ! 6b ! 8c = 72 ! 68 c = 4Substitute in (3):30a = 15(4) = 60 a = 2Substitute in (5):2b + 3(4) = 24 2b = 12 b = 6

49. m = 1 _ 2 ! (!6)

________ 8 ! 4 1 _

2 =

6 1 _ 2 ___

3 1 _ 2 = 13 ___

7

50. m = !2 ! (!8)

_________ 0 ! (!9)

= 6 __ 9 = 2 __

3

51. m = 6 ! (!14)

_________ 11 ! 3

= 20 ___ 8 = 5 __

2

52. By Vert. - Thm., )RWV + )SWT. So by Thm. 11-2-2 (1), RV = ST2z + 15 = 9z + 1 14 = 7z z = 2 0 RV + 0 ST and 0 RS + 0 VT . So by substitution, 2m 0 ST + 2m 0 VT = 360 m 0 ST + m 0 VT = 180m 0 ST + 31(2) + 2 = 180 m 0 ST + 64 = 180 m 0 ST = 116°

53. Let BD = 2x; then 1. 5 2 + x 2 = (1 + 1.5 ) 2 2.25 + x 2 = 6.25 x 2 = 4 x = 2.'ABD has base 2(2) = 4 m and height 1.5 m. So A = 1 _

2 bh = 1 _

2 (4)(1.5) = 3 m 2 .


1. Yes; ""

CR is a radius of circle C. If a line is tangent to a circle, then it is & to the radius at the point of tangency.

11-5 TECHNOLOGY LAB: EXPLORE ANGLE RELATIONSHIPS IN CIRCLES, PAGES 780–781

TRY THIS, PAGE 781 1. The relationship is the same. Both types of - have

a measure equal to half the measure of the arc.

2. A tangent line must be & to the radius at the point of tangency. If construction is done without this & relationship, then tangent line created is not guaranteed to intersect circle at only one point. If a line is & to a radius of a circle at a point on the circle, then line is tangent to circle.

3. The relationship remains the same. The measure of ) is half difference of intercepted arcs.

4. An ) whose vertex is on the circle (inscribed - and - created by a tangent and a secant intersecting at the point of tangency) will have a measure equal to half its intercepted arc.An ) whose vertex is inside the circle (- created by intersecting secants or chords) will have a measure equal to half sum of its intercepted arcs.An ) whose vertex is outside the circle (- created by secants and/or tangents that intersect outside circle) will have a measure equal to half its intercepted arc.

5. No; it is a means to discover relationships and make conjectures.


11-5 ANGLE RELATIONSHIPS IN CIRCLES, PAGES 782–789


1a. m)STU = 1 _ 2 m 0 ST

= 1 _ 2 (166) = 83°

b. m)QSR = 1 _ 2 m 0 SR

71 = 1 _ 2 m 0 SR

m 0 SR = 142°

2a. m)ABD = 1 _ 2 (m 0 AD + m 0 CE )

= 1 _ 2 (65 + 37)

= 1 _ 2 (102) = 51°

b. m)RNP = 1 _ 2 (m 0 MQ + m 0 RP )

= 1 _ 2 (91 + 225)

= 1 _ 2 (316) = 158°

m)RNM + m)RNP = 180 m)RNM + 158 = 180 m)RNM = 22°

3. m)L = 1 _ 2 (m 0 JN ! m 0 KN )

25 = 1 _ 2 (83 ! x)

50 = 83 ! x x = 33

4. m)ACB = 1 _ 2 (m 0 AEB ! m 0 AB )

= 1 _ 2 (225 ! 135) = 45°

5. Step 1 Find m 0 PR .m)Q = 1 _

2 (m 0 MS ! mPR)

26 = 1 _ 2 (80 ! m 0 PR )

52 = 80 ! m 0 PR m 0 PR = 28°Step 2 Find m 0 LP .m 0 LP + m 0 PR = mLR m 0 LP + 28 = 100 m 0 LP = 72°

THINK AND DISCUSS, PAGE 786 1. For both chords and secants that intersect in the

interior of a circle, the measure of ) formed is half the sum of the measures of their intercepted arcs.

2.


GUIDED PRACTICE, PAGES 786–787

1. m)DAB = 1 _ 2 m 0 AB

= 1 _ 2 (140) = 70°

2. 27 = 1 _ 2 m 0 AC

m 0 AC = 54°

3. 61 = 1 _ 2 m 0 PN

m 0 PN = 122°

4. m)MNP = 1 _ 2 (238)

= 119°

5. m)STU = 1 _ 2 (m 0 SU + m 0 VW )

= 1 _ 2 (104 + 30)

= 1 _ 2 (134) = 67°

6. m)HFG = 1 _ 2 (m 0 EJ + m 0 GH )

= 1 _ 2 (59 + 23)

= 1 _ 2 (82) = 41°

7. m)NPL = 1 _ 2 (m 0 KM + m 0 NL )

= 1 _ 2 (61 + 111)

= 1 _ 2 (172) = 86°

m)NPK + m)NPL = 180 m)NPK + 86 = 180 m)NPK = 94°

8. x = 1 _ 2 (161 ! 67) = 1 _

2 (94) = 47

9. x = 1 _ 2 (238 ! 122) = 1 _

2 (116) = 58

10. 27 = 1 _ 2 (x ! 40)

54 = x ! 40 x = 94°

11. m)S = 1 _ 2 (m 0 ACB ! m 0 AB )

38 = 1 _ 2 ((360 ! x) ! x)

76 = 360 ! 2x 2x = 360 ! 76 = 284 x = 142°

12. m)E = 1 _ 2 (m 0 BF ! m 0 DF )

50 = 1 _ 2 (150 ! mDF)

100 = 150 ! m 0 DF m 0 DF = 50°

13. m 0 BC + m 0 CD + m 0 DF + m 0 BF = 360 64 + m 0 CD + 50 + 150 = 360 m 0 CD + 264 = 360 m 0 CD = 96°

14. m)NPQ = 1 _ 2 (m 0 JK + m 0 PN )

79 = 1 _ 2 (48 + m 0 PN )

158 = 48 + m 0 PN m 0 PN = 110°

15. m 0 KN + m 0 PN + m 0 JP + m 0 JL = 360 m 0 KN + 110 + 86 + 48 = 360 m 0 KN + 244 = 360 m 0 KN = 116°



16. m!BCD = 1 _ 2 m " BC = 1 _

2 (112) = 56°

17. m!ABC = 1 _ 2 (360 # 112) = 1 _

2 (248) = 124°

18. m!XZW = 1 _ 2 m " XZ = 1 _

2 (180) = 90°

19. m " XZV = m " XZ + m " ZV = 180 + 2m!VXZ = 180 + 2(40) = 260°

20. m!QPR = 1 _ 2 (m " QR + m " ST ) = 1 _

2 (31 + 98) = 64.5°

21. m!ABC = 180 # m!ABD = 180 # 1 _

2 (m " AD + m " CE )

= 180 # 1 _ 2 (100 + 45) = 107.5°

22. m!MKJ = 180 # m!MKL = 180 # 1 _

2 (m " JN + m " ML )

= 180 # 1 _ 2 (38.5 + 51.5) = 135°

23. x = 1 _ 2 (170 # (360 # (170 + 135)))

= 1 _ 2 (170 # 55) = 57.5°

24. x = 1 _ 2 (220 # 140) = 40°

25. x = 1 _ 2 ((180 # (20 + 104)) # 20)

= 1 _ 2 (56 # 20) = 18°

26. m!ABV = 1 _ 2 (180 # m " AB ) = 1 _

2 (180 # 48) = 66°

27. m!DJH = 1 _ 2 (m " DH + m " EG )

180 # 89 = 1 _ 2 (137 + m " EG )

91 = 1 _ 2 (137 + m " EG )

182 = 137 + m " EG m " EG = 45°

28. m!DJE = 1 _ 2 (m " DE + m " GH )

89 = 1 _ 2 (m " DE + 61)

178 = m " DE + 61 m " DE = 117°

29. Step 1 Find m!P.m!P = 1 _

2 m " LR = 1 _

2 (170) = 85°

Step 2 Find m!QPR.m!QPR + 50 + 85 = 180 m!QPR = 45°Step 3 Find m " PR .m!QPR = 1 _

2 m " PR

45 = 1 _ 2 m " PR

m " PR = 90°

30. 50 = 1 _ 2 (m " PS # m " ML )

= 1 _ 2 (m " PM # m " ML )

= 1 _ 2 m " LP

m " LP = 100°

31. m!ABC = 1 _ 2 m " AB

x = 1 _ 2 m " AB

m " AB = 2x °

32. m!ABD + m!ABC = 180 m!ABD + x = 180 m!ABD = (180 # x)°

33. m " AEB + m " AB = 360

m " AEB + 2x = 360

m " AEB = (360 # 2x)° 34.

Since 2 points determine a line, draw $$

BD . By Ext. ! Thm., m!ABD = m!C + m!BDC. So

m!C = m!ABD # m!BDC. m!ABD = 1 _ 2 m " AD by

Inscribed ! Thm., and m!BDC = 1 _ 2 m " BD by Thm.

11-5-1. By substitution, m!C = 1 _ 2 m " AD # 1 _

2 m " BD .

Thus, by Distrib. Prop. of =, m!C = 1 _ 2 (m " AD

# m " BD ).

35.


EG . By Ext. ! Thm., m!DEG = m!F + m!EGF. So

m!F = m!DEG # m!EGF. m!DEG = 1 _ 2 m " EHG

and m!EGF = 1 _ 2 m " EG by Theorem 11-5-1. By

substitution, m!F = 1 _ 2 m " EHG # 1 _

2 m " EG . Thus, by

Distrib. Prop. of =, m!F = 1 _ 2 (m " EHF # m " EG ).

36.


JM . By Ext. ! Thm., m!JMN = m!L + m!KJM. So

m!L = m!JMN # m!KJM. m!JMN = 1 _ 2 m " JN

and m!KJM = 1 _ 2 m " KM by Inscribed ! Thm. By

substitution, m!F = 1 _ 2 m " JN # 1 _

2 m " KM . Thus, by

Distrib. Prop. of =, m!F = 1 _ 2 (m " JN # m " KM ).

37. m!1 > m!2 because m!1 = 1 _ 2 (m " AB + m " CD )

and m!2 = 1 _ 2 (m " AB # m " CD ). Since m " CD > 0,

the expression for m!1 is greater.



38. When a tangent and secant intersect on the circle, the measure of ! formed is half the measure of the intercepted arc.

When 2 secants intersect inside the circle, the measure of each ! formed is half the sum of the measures of the intercepted arcs, or 1 _

2 (90° + 90°).

When 2 secants intersect inside the circle, the measure of each ! formed is half the difference of the measures of the intercepted arcs, or 1 _ 2 (270° " 90°).

39. m # AC + m # AD + m # CD = 3602x " 10 + x + 160 = 360 3x = 210 x = 70m!B = 1 _

2 (m # AC " m # AD ) = 1 _

2 (2(70) " 10 " 70) = 30°

m!C = 1 _ 2 m # AD = 1 _

2 (70) = 35°

m!A = 180 " (m!B + m!C) = 180 " (30 + 35) = 115°

40. m!B = 1 _ 2 (m # AD " m # DE )

4x = 1 _ 2 (18x " 15 " (8x + 1))

8x = 10x " 16 16 = 2x x = 8 m!B = 4(8) = 32°m # AD = 18(8) " 15 = 129°

m!C = 1 _ 2 (m # AED " m # AD ) = 1 _

2 (231 " 129) = 51°

m!A = 180 " (m!B " m!C) = 180 " (32 + 51) = 97°

41a. m!BHC = m # BC = 1 _ 6 (360) = 60°

b. m!EGD = 1 _ 2 (m # DE " m # DAE )

= 1 _ 2 (60 + 3(60)) = 120°

c. m!CED = m!EDF = 1 _ 2 (60) = 30° and

m!EGD > 90°, so $EGD is an obtuse isosceles.

TEST PREP, PAGE 789

42. Cm!DCE = 1 _

2 (m # AF + m # DE )

= 1 _ 2 (58 + 100) = 79°

43. J

44. 56°m!JLK = 1 _

2 (m # MN " m # JK )

45 = 1 _ 2 (146 " m # JK )

90 = 146 " m # JK m # JK = 56°


45. Case 1: Assume %%

AB is a diameter of the circle. Then m # AB = 180° and !ABC is a right !. Thus, m!ABC = 1 _

2 m # AB .

Case 2: Assume %%

AB is not a diameter of the circle. Let X be the center of the circle and draw radii

%% XA

and %%

XB . %%

XA & %%

XB . So $AXB is isosceles. Thus, !XAB & !XBA, and 2m!XBA + m!AXB = 180. This means that m!XBA = 90 " 1 _

2 m!AXB. By

Thm. 11-1-1, !XBC is a right !. So m!XBA + m!ABC = 90 or m!ABC = 90 " m!XBA. By substituting, m!ABC = 90 " (90 " 1 _

2 m!AXB)

= 1 _ 2 m!AXB. m!AXB = m # AB because !AXB is

a central !. Thus, m!ABC = 1 _ 2 m # AB .

46. Since m # WY = 90°, m!YXW = 90° because it is a central !. By Thm. 11-1-1, !XYZ and !XWZ are right '. The sum of measures of the ' of a quadrilateral is 360°. So m!WZY = 90°. Thus, all 4 ' of WXYZ are right '. So WXYZ is a rectangle. XY & XW because they are radii. By Thm. 6-5-3, WXYZ is a rhombus. Since WXYZ is a rectangle and a rhombus, it must also be a square by Theorem 6-5-6.

47. m!V = 1 _ 2 (x " 21) = 1 _

2 (124 " 50)

x " 21 = 124 " 50 = 74 x = 74 + 21 = 95°

48. Step 1 Find m!CED.m!DCE + m!ECJ = 180 m!DCE + 135 = 180 m!DCE = 45°m!CDE = m!FDG = 82°m!CED + m!DCE + m!CDE = 180 m!CED + 45 + 82 = 180 m!CED = 53°Step 2 Find m # GH .m!CED = 1 _

2 (m # GH + mKL)

53 = 1 _ 2 (m # GH + 27)

106 = m # GH + 27 m # GH = 79°


49. g(7) = 2(7 ) 2 " 15(7) " 1 = 98 " 105 " 1 = "8; yes

50. f(7) = 29 " 3(7) = 29 " 21 = 8; no

51. " 7 __ 8 (7) = " 49 ___

8 ; no 52. V = 1 _

3 Bh

= 1 _ 3 ( 1 _

2 aP) h

= 1 _ 3 ( 1 _

2 (2 ( ) 3 ) (24)) (7)

= 56 ( ) 3 * 97.0 m 3



53. L = "r!60" = "(6)! ! = 10 cm r 2 + h 2 = ! 2 6 2 + h 2 = 10 2 h = 8 cmV = 1 _

3 " r 2 h

= 1 _ 3 "(6 ) 2 (8) = 96" cm 3

( 310.6 cm 3

54. S = 1 _ 2 P! + s 2

1200 = 1 _ 2 (96)! + 576

624 = 48! ! = 13 in. ( 1 _

2 s) 2 + h 2 = ! 2

12 2 + h 2 = 13 2 h = 5 in.V = 1 _

3 Bh = 1 _

3 (576)(5) = 960 in. 3

55. m)BCA = 1 _ 2 m 0 BA = 1 _

2 (74) = 37°

56. m)DCB = 1 _ 2 m 0 DB = m)DAB = 67°

m)BDC = 90° ( 0 BC is a diam.)m)DBC = 180 ! (90 + 67) = 23°

57. m)ADC = 1 _ 2 m 0 AC = 1 _

2 (180 ! 74) = 53°

11-6 TECHNOLOGY LAB: EXPLORE SEGMENT RELATIONSHIPS IN CIRCLES, PAGES 790–791

ACTIVITY 1, TRY THIS, PAGE 790

1.

2. )CGF and )EGD; )FCG and )DEG; 'CFG and 'EDG are 5 , by AA 5 Post.

3. GC ___ GF

= GE ___ GD

; GC · GD = GE · GF

ACTIVITY 2, TRY THIS, PAGE 791 4. The products of the segment lengths for a tangent

and a secant are similar to the products of the segment lengths for 2 secants because for a tangent there is only 1 segment. Thus, the “whole segment” multiplied by the “external segment” becomes the square of the tangent seg.

5. 2 tangent segments from same external point will have = lengths, so the circle segments are +.

6. In diagram, the circle A is given with tangents ""

DC and

"" DE from D. Since 2 points determine a line,

draw radii ""

AC , ""

AE , and ""

AD . ""

AC + ""

AE because all radii are +.

"" AD +

"" AD by Reflex. Prop. of +. )ACD

and )AED are right - because they are each formed by a radius and a tangent intersecting at point of tangency. Thus, 'ACD and 'AED are right ,. 'ACD + 'AED by HL. Therefore,

"" DC

+ ""

DE by CPCTC.

ACTIVITY 3, TRY THIS, PAGE 791 7. )DGE and )FGC; )GDE and )GFC; )GED and

)GCF; 'DGE and 'FGC are 5 , by AA 5.

8. When 2 secants or 2 chords of a circle intersect, 4 segments will be formed, each with the point of intersection as 1 endpoint. The product of segment lengths on 1 secant/chord will equal the product of segment lengths on the other secant/chord.If a secant and a tangent of a circle intersect on an exterior point, 3 segments will be formed, each with exterior point as an endpoint. The product of the secant segment lengths will equal square of tangent segment length.

11-6 SEGMENT RELATIONSHIPS IN CIRCLES, PAGES 792–798

CHECK IT OUT! PAGES 792–794 1. AE · EB = CE · ED

6(5) = x(8) 30 = 8x x = 3.75AB = 6 + 5 = 11CD = (3.75) + 8 = 11.75

2. original disk: PR = 11 1 _ 3 in.

New disk:AQ · QB = PQ · QR 6(6) = 3(QR) 36 = 3QR QR = 12 in.PR = 12 + 3 = 15 in.change in PR = 15 ! 11 1 _

3 = 3 2 _

3 in.

3. GH · GJ = GK · GL13(13 + z) = 9(9 + 30) 169 + 13z = 81 + 270 13z = 182 z = 14GJ = 13 + (14) = 27GL = 9 + 30 = 39

4. DE · DF = DG 2 7(7 + y) = 10 2 49 + 7y = 100 7y = 51 y = 7 2 _

7


THINK AND DISCUSS, PAGE 795 1. Yes; in this case, chords intersect at center of the

circle. So segments of the chords are all radii, and theorem simplifies to r 2 = r 2 .

2. 2

3.


GUIDED PRACTICE, PAGES 795–796

1. tangent segment

2. HK · KJ = LK · KM 8(3) = 4(y) 24 = 4y y = 6LM = 4 + (6) = 10HJ = 8 + 3 = 11

3. AE · EB = CE · ED x(4) = 6(6) 4x = 36 x = 9AB = (9) + 4 = 13CD = 6 + 6 = 12

4. PS · SQ = RS · ST z(10) = 6(8) 10z = 48 z = 4.8PQ = 10 + (4.8) = 14.8RT = 6 + 8 = 14

5. Let the diameter be d ft.GF · FH = EF · (d ! EF) 25(25) = 20(d ! 20) 625 = 20d ! 400 225 = 20d d = 51 1 _

4 ft

6. AC · BC = EC · DC(x + 7.2)7.2 = (9 + 7.2)7.2 x + 7.2 = 9 + 7.2 x = 9AC = (9) + 7.2 = 16.2EC = 9 + 7.2 = 16.2

7. PQ · PR = PS · PT 5(5 + y) = 6(6 + 7) 25 + 5y = 78 5y = 53 y = 10.6PR = 5 + (10.6) = 15.6PT = 6 + 7 = 13

8. GH · GJ = GK · GL10(10 + 10.7) = 11.5(11.5 + x) 207 = 132.25 + 11.5x 74.75 = 11.5x x = 6.5GJ = 10 + 10.7 = 20.7GL = 11.5 + (6.5) = 18

9. AB · AC = AD 2 2(2 + 6) = x 2 16 = x 2 x = 4 (since x > 0)

10. MP · MQ = MN 2 3(3 + y) = 4 2 9 + 3y = 16 3y = 7 y = 2 1 _

3

11. RT · RU = RS 2 3(3 + 8) = x 2 33 = x 2 x = . // 33 (since x > 0)


12. DH · HE = FH · HG 3(4) = 2(y) 12 = 2y y = 6DE = 3 + 4 = 7FG = 2 + (6) = 8

13. JK · KL = MN · KN 10(x) = 6(7) 10x = 42 x = 4.2JL = 10 + (4.2) = 14.2MN = 6 + 7 = 13

14. UY · YV = WY · YZ 8(5) = x(11) 40 = 11x x = 3 7 __

11

UV = 8 +5 = 13WZ = 3 7 __

11 + 11 = 14 7 __

11

15. 590(590) = 225.4(d ! 225.4) 348,100 = 225.4d ! 50,805.16398,905.16 = 225.4d d ( 1770 ft

16. AB · AC = AD · AE5(5 + x) = 6(6 + 6) 25 + 5x = 72 5x = 47 x = 9.4AC = 5 + (9.4) = 14.4AE = 6 + 6 = 12

17. HL · JL = NL · ML(y + 10)10 = (18 + 9)9 10y + 100 = 243 10y = 143 y = 14.3HL = (14.3) + 10 = 24.3NL = 18 + 9 = 27

18. PQ · PR = PS · PT 3(3 + 6) = 2(2 + x) 27 = 4 + 2x 23 = 2x x = 11.5PR = 3 + 6 = 9PT = 2 + (11.5) = 13.5

19. UW · UX = UV 2 6(6 + 8) = z 2 84 = z 2 z = . // 84 = 2 . // 21 (since z > 0)


20. BC · BD = BA 2 2(2 + x) = 5 2 4 + 2x = 25 2x = 21 x = 10.5

21. GE · GF = GH 2 8(8 + 12) = y 2 160 = y 2 y = 4 . // 10 (since y > 0)

22a. RM · MS = PM · MQ 10(MS) = 12(12) = 144 MS = 14.4 cm

b. RS = RM + MS = 10 + 14.4 = 24.4 cm

23a. PM · MQ = RM · MS PM 2 = 4(13 ! 4) = 36 PM = 6 in.

b. PQ = 2PM = 2(6) = 12 in.

24. Step 1 Find x. AB · AC = AD · AE5(5 + 5.4) = 4(4 + x) 52 = 16 + 4x 36 = 4x x = 9Step 2 Find y.AB · AC = AF 2 52 = y 2 y = . // 52 = 2 . // 13

25. Step 1 Find x.NQ · QM = PQ · QR 4(x) = 3.2(10) = 32 x = 8Step 2 Find y. KM · KN = KL 2 6(6 + (8) + 4) = y 2 108 = y 2 y = . // 108 = 6 . / 3

26. SP 2 = SE · (SE + d) = 6000(14,000) = 84,000,000 SP = . ///// 84,000,000 = 2000 . // 21 ( 9165 mi

27. Solution B is incorrect. The first step should be AC · BC = DC 2 , not AB · BC = DC 2 .

28.

Since 2 points determine a line, draw ""

AC and ""

BD . )ACD + )DBA because they both intercept 0 AD . )CEA + )BED by Vert. - Thm. Therefore, 'ECA 5 'EBD by AA 5. Corresponding sides are proportional. So AE ___

ED = CE ___

EB . By Cross Products

Property, AE · EB = CE · ED.

29.

Since 2 points determine a line, draw ""

AD and ""

BD . m)CAD = 1 _

2 m 0 BD by Inscribed ) Thm. m)BDC =

1 _ 2 m 0 BD by Thm. 11-5-1. Thus, )CAD + )BDC. Also,

)C + )C by Reflex. Prop. of +. Therefore, 'CAD 5 'CDB by AA 5. Corresponding sides are proportional. So AC ___

DC = DC ___

BC . By Cross Products

Property, AC · BC = DC 2 .

30. Yes; PR · PQ = PT · PS, and it is given that PQ = PS.So PR = PT. Subtracting the + segments from each of these shows that

"" QR +

"" ST .

31. Method 1: By Secant-Tangent Product Theorem, BC 2 = 12(4) = 48, and so BC = . // 48 = 4 . / 3 .Method 2: By Thm. 11-1-1, )ABC is a right ). By Pythagorean Thm., BC 2 + 4 2 = 8 2 . Thus BC 2 = 64 ! 16 = 48, and BC = 4 . / 3 .

32a. AE · EC = BE · DE 5.2(4) = 3(DE) 20.8 = 3DE DE ( 6.9 cm

b. diameter = BD = BE + DE ( 3 + 6.9 ( 9.9 cm

c. OE = OB ! BE ( 1 _ 2 (9.933) ! 3 ( 1.97 cm

TEST PREP, PAGE 798

33. B PQ 2 = PR · PS = 6(6 + 8) = 84 PQ = . // 84 ( 9.2

34. F PT · PU = PQ 2 7(7 + UT) = 84 7 + UT = 12 UT = 5

35. CE = ED = 6, and by Chord-Chord Product Thm., 3(EF) = 6(6) = 36. So EF = 12, FB = 12 + 3 = 15, and radius AB = 1 _

2 (15) = 7.5.


36a. Step 1 Find the length y of the chord formed by

"" KM .

KL 2 = 144 = 8(8 + y) 18 = 8 + y y = 10

Step 2 Find the value of x.6(6 + x) = 8(8 + (10)) 36 + 6x = 144 6x = 108 x = 18

b. KL 2 + LM 2 6 KM 2 12 2 + (6 + 18 ) 2 6 (8 + 10 + 8 ) 2 720 > 676'KLM is acute.


37. Let R be center of the circle; then 'PQR is a right ). So

PR 2 = PQ 2 + QR 2 = 6 2 + 4 2 = 52 PR = . // 52 = 2 . // 13 in.Let S be the intersection of

"" PR with circle R,

so that SR = 4 in.; then the distance from P to the circle = PS = 2 . // 13 ! 4 ( 3.2 in.

38. By Chord-Chord Product Thm., (c + a)(c ! a) = b · b c 2 ! a 2 = b 2 c 2 = a 2 + b 2

39. GJ · HJ = FJ 2 (y + 6)y = 10 2 = 100 y 2 + 6y ! 100 = 0

y = !6 ± . /////// 6 2 ! 4(1)(!100)

____________________ 2

= !6 + . // 436 __________ 2 (since y > 0)

= !3 + . // 109 ( 7.44


40. P = # favorable outcomes __________________ # trials

0.035 = 14 ___ n

n = 14 _____ 0.035

= 400

41. P = 36 ___ 50

= 0.72 = 72%

42. $$% BA and

$$% CD 43.

$$% CD 44.

"" BC

45. A = "(12 ) 2 ( 55 ____ 360

) = 22" ft 2 ( 69.12 ft 2

46. L = 2"(12) ( 55 ____ 360

) = 3 2 __ 3 " ft ( 11.52 ft

47. The area of sector YZX is 40" ! 22" = 18" ft 2

A = " r 2 ( m ____ 360

) 18" = "(12 ) 2 ( m ____

360 )

m ____ 360

= 18 ____ 144

= 1 __ 8

m = 1 __ 8 (360) = 45°

11-7 CIRCLES IN THE COORDINATE PLANE, PAGES 799–805


1a. (x ! h ) 2 + (y ! k ) 2 = r 2 (x ! 0 ) 2 + (y ! (!3) ) 2 = 8 2 x 2 + (y + 3 ) 2 = 64

b. r = . ///////// (2 ! 2 ) 2 + (3 ! (!1) ) 2 = . // 16 = 4(x ! 2 ) 2 + (y ! (!1) ) 2 = 4 2 (x ! 2 ) 2 + (y + 1 ) 2 = 16

2a. Step 1 Make a table of values.Since the radius is . / 9 = 3, use ±3 and values in between for x-values.

x !3 !2 !1 0 !1 !2 !3

y 0 ±2.2 ±2.8 ±3 ±2.8 ±2.2 0

Step 2 Plot points and connect them to form a circle.

b. The equation of given circle can be written as (x ! 3 ) 2 + (y ! (!2) ) 2 = 2 2 So h = 3, k = !2, and r = 2. The center is (3, !2) and the radius is 2. Plot point (3, !2). Then graph a circle having this center and radius 2.

3. Step 1 Plot 3 given points.Step 2 Connect D, E, and F to form a '.

Step 3 Find a point that is equidistant from 3 points by constructing & bisectors of 2 sides of 'DEF. The & bisectors of sides of 'DEF intersect at a point that is equidistant from D, E, and F. The intersection of the & bisectors is P(2, !1). P is the center of circle passing through D, E, and F.



1. x 2 + y 2 = r 2

2. First find the center by finding the midpoint of the diameter. By the Midpoint Formula, the center of the circle is (!1, 4). The radius is half the length of the diameter. So r = 2. The equation is (x + 1 ) 2 + (y ! 4 ) 2 = 4.

3. No; a radius represents length, and length cannot be negative.

4.



1. (x ! h ) 2 + (y ! k ) 2 = r 2 (x ! 3 ) 2 + (y ! (!5) ) 2 = 12 2 (x ! 3 ) 2 + (y + 5 ) 2 = 144

2. (x ! h ) 2 + (y ! k ) 2 = r 2 (x ! (!4) ) 2 + (y ! 0 ) 2 = 7 2 (x + 4 ) 2 + y 2 = 49

3. r = " ######## (2 ! 4 ) 2 + (0 ! 0 ) 2 = " # 4 = 2(x ! 2 ) 2 + (y ! 0 ) 2 = 2 2 (x ! 2 ) 2 + y 2 = 4

4. r = " ########## (2 ! (!1) ) 2 + (!2 ! 2 ) 2 = " ## 25 = 5(x ! (!1) ) 2 + (y ! 2 ) 2 = 5 2 (x + 1 ) 2 + (y ! 2 ) 2 = 25

5. The equation of given circle can be written as (x ! 3 ) 2 + (y ! 3 ) 2 = 2 2 So h = 3, k = 3, and r = 2. The center is (3, 3) and radius is 2. Plot point (3, 3). Then graph a circle having this center and radius 2.

6. The equation of given circle can be written as (x ! 1 ) 2 + (y ! (!2) ) 2 = 3 2 So h = 1, k = !2, and r = 3. The center is (1, !2) and radius is 3. Plot point (1, !2). Then graph a circle having this center and radius 3.

7. The equation of given circle can be written as (x ! (!3) ) 2 + (y ! (!4) ) 2 = 1 2 So h = !3, k = !4, and r = 1. The center is (!3, !4) and radius is 1. Plot point (!3, !4). Then graph a circle having this center and radius 1.

8. The equation of given circle can be written as (x ! 3 ) 2 + (y ! (!4) ) 2 = 2 2 So h = 3, k = !4, and r = 2. The center is (3, !4) and radius is 4. Plot point (3, !4). Then graph a circle having this center and radius 4.

9a. Step 1 Plot the 3 given points.Step 2 Connect A, B, and C to form a $.

Step 3 Find a point that is equidistant from 3 points by constructing % bisectors of 2 sides of $ABC.The % bisectors of sides of $ABC intersect at a point that is equidistant from A, B, and C.The intersection of % bisectors is P(!2, 3). P is center of circle passing through A, B, and C.

b. There are approximately 10 units across the circle. So the diameter is approximately 10 ft.




10. (x ! h ) 2 + (y ! k ) 2 = r 2 (x ! (!12) ) 2 + (y ! (!10) ) 2 = 8 2 (x + 12 ) 2 + (y + 10 ) 2 = 64

11. (x ! h ) 2 + (y ! k ) 2 = r 2 (x ! 1.5 ) 2 + (y ! (!2.5) ) 2 = ( . / 3 ) 2 (x ! 1.5 ) 2 + (y + 2.5 ) 2 = 3

12. r = . //////// (2 ! 1 ) 2 + (2 ! 1 ) 2 = . / 2 (x ! 1 ) 2 + (y ! 1 ) 2 = ( . / 2 ) 2 (x ! 1 ) 2 + (y ! 1 ) 2 = 2

13. r = . ////////// (!5 ! 1 ) 2 + (1 ! (!2) ) 2 = . // 45 = 3 . / 5 (x ! 1 ) 2 + (y ! (!2) ) 2 = (3 . / 5 ) 2 (x ! 1 ) 2 + (y + 2 ) 2 = 45

14. The equation of given circle can be written as (x ! 0 ) 2 + (y ! 2 ) 2 = 3 2 .So h = 0, k = 2, and r = 3. The center is (0, 2)and radius is 3. Plot point (0, 2). Then graph a circle having this center and radius 3.

15. The equation of given circle can be written as (x ! (!1) ) 2 + (y ! 0 ) 2 = 4 2 .So h = !1, k = 0, and r = 4. The center is (!1, 0)and radius is 4. Plot point (!1, 0). Then graph a circle having this center and radius 4.

16. The equation of given circle can be written as (x ! 0 ) 2 + (y ! 0 ) 2 = 10 2 .So h = 0, k = 0, and r = 10. The center is (0, 0) and radius is 10. Then graph a circle having origin as center and radius 10.

17. The equation of given circle can be written as (x ! 0 ) 2 + (y ! (!2) ) 2 = 2 2 .So h = 0, k = !2, and r = 2. The center is (0, !2) and radius is 2. Plot point (0, !2). Then graph a circle having this center and radius 2.

18a. Step 1 Plot the 3 given points.Step 2 Connect A, B, and C to form a '.

Step 3 Find a point that is equidistant from the 3 points by constructing & bisectors of the 2 sides of 'ABC.

The & bisectors of sides of 'ABC intersect at a point that is equidistant from A, B, and C.

The intersection of & bisectors is P(!1, !2). P is center of the circle passing through A, B, and C.

b. There are approximately 10 units across the circle. So the diameter is approximately 10 ft.

19. The circle has the center (1, !2) and radius 2.(x ! 1 ) 2 + (y ! (!2) ) 2 = 2 2 (x ! 1 ) 2 + (y + 2 ) 2 = 4

20. The circle has the center (!1, 1) and radius 4.(x ! (!1) ) 2 + (y ! 1 ) 2 = 4 2 (x + 1 ) 2 + (y ! 1 ) 2 = 16

21a. r = . //// 24 2 + 32 2 = 40d = 2(40) = 80 units or 80 ft

b. x 2 + y 2 = 40 2 x 2 + y 2 = 1600

22. F; r = . / 7

23. T; (!1 ! 2 ) 2 + (!3 + 3 ) 2 = 9

24. F; center is (6, !4), in fourth quadrant.

25. T; (0, 4 ± . / 3 ) lie on y-axis and 7.

26. F; the equation is x 2 + y 2 = 3 2 = 9.


27a. Possible answer: 28 units 2

b. r = 3, so A = "(3 ) 2 = 9" ( 28.3 units 2

c. Check students’ work.

28. slope of radius from center (4, !6) to pt. (1, !10) is

m = !10 ! (!6)

__________ 1 ! 4

= !4 ___ !3

= 4 __ 3

tangent has slope ! 3 __ 4 , and eqn.

y ! (!10) = ! 3 __ 4 (x ! 1) or y + 10 = ! 3 __

4 (x ! 1)

29a. r = 3E(!3, 5 ! 2(3)) = E(!3, !1)G(0 ! 2(3), 2) = G(!6, 2)

b. d = 2(3) = 6 c. center is (0 ! 3, 2) = (!3, 2) (x ! (!3) ) 2 + (y ! 2 ) 2 = 3 2

(x + 3 ) 2 + (y ! 2 ) 2 = 9

30. (x ! 2 ) 2 + (x + 3 ) 2 = 81(x ! 2 ) 2 + (x ! (!3) ) 2 = 9 2 center (2, !3), radius 9

31. x 2 + (y + 15 ) 2 = 25(x ! 0 ) 2 + (y ! (!15) ) 2 = 5 2 center (0, !15), radius 5

32. (x + 1 ) 2 + y 2 = 7(x ! (!1) ) 2 + (y ! 0 ) 2 = ( . / 7 ) 2 center (!1, 0), radius . / 7

33. r = 3; A = "(3 ) 2 = 9"; C = 2"(3) = 6"

34. r = . / 7 ; A = " ( . / 7 ) 2 = 9"; C = 2" ( . / 7 ) = 2 . / 7 "

35. r = . ////////// (2 ! (!1) ) 2 + (!1 ! 3 ) 2 = 5A = "(5 ) 2 = 25"; C = 2"(5) = 10"

36. Graph is a single point, (0, 0).

37. The epicenter (x, y) is a solution of the equations of 3 circles. Let 1 unit represent 100 mi.seismograph A: (x + 2 ) 2 + (y ! 2 ) 2 = 3 2 x 2 + 4x + 4 + y 2 ! 4y + 4 = 9 x 2 + 4x + y 2 ! 4y = 1 (1)seismograph B: (x ! 4 ) 2 + (y + 1 ) 2 = 6 2 x 2 ! 8x + 16 + y 2 + 2y + 1 = 36 x 2 ! 8x + y 2 + 2y = 19 (2)seismograph C: (x ! 1 ) 2 + (y + 5 ) 2 = 5 2 x 2 ! 2x + 1 + y 2 + 10y + 25 = 25 x 2 ! 2x + y 2 + 10y = !1 (3)(1) ! (2):12x ! 6y = !18 2x + 3 = y (4)(1) ! (3):6x ! 14y = 2 3x = 7y + 1 (5)(4) in (5): 3x = 7(2x + 3) + 1 3x = 14x + 22!11x = 22 x = !2y = 2(!2) + 3 = !1The location of epicenter is (!200, !100).

38. The circle has a radius of 5. So 5 is tangent to x-axis if the center has y-coordinate k = ±5.

39. d = . /////////// (5 ! (!3) ) 2 + (!2 ! (!2) ) 2 = 8; r = 4

center = ( !3 + 5 _______ 2 ,

!2 + (!2) _________

2 ) = (1, !2)

equation is (x ! 1 ) 2 + (y + 2 ) 2 = 16

40. The locus is a circle with a radius 3 centered at (2, 2).

41. The point does not lie on circle P because it is not a solution to the equation (x ! 2 ) 2 + (y ! 1 ) 2 = 9. Since (3 ! 2 ) 2 + ((!1) ! 1 ) 2 < 9, the point lies in the interior of circle P.

TEST PREP, PAGE 804 42. C

(!2, 0) lies on the circle.

43. H(x ! (!3) ) 2 + (y ! 5 ) 2 = (1 ! (!3) ) 2 + (5 ! 5 ) 2 (x + 3 ) 2 + (y ! 5 ) 2 = 16

44. Adistances from statues to fountain:

. /////////// (4 ! (!1) ) 2 + (!2 ! (!2) ) 2 = 5

. /////////// (!1 ! (!1) ) 2 + (3 ! (!2) ) 2 = 5

. //////////// (!5 ! (!1) ) 2 + (!5 ! (!2) ) 2 = 5



45a. r = ! """"""""""""""" (1 # 2 ) 2 + (#2 # (#4) ) 2 + (#5 # 3 ) 2 = ! "" 69

(x # 2 ) 2 + (y # (#4) ) 2 + (z # 3 ) 2 = ( ! "" 69 ) 2

(x # 2 ) 2 + (y + 4 ) 2 + (z # 3 ) 2 = 69

b. 15; if 2 segments are tangent to a circle or sphere from same exterior point, then segments are $.

46. x + y = 5 y = 5 # xSubstitute in equation of a circle: x 2 + (5 # x ) 2 = 25 x 2 + 25 # 10x + x 2 = 25 2 x 2 # 10x = 0 2x(x # 5) = 0 x = 0 or 5The point of intersection are (0, 5 # (0)) = (0, 5) and (5, 5 # (5)) = (5, 0).

47. Given the line is % to a line through (3, 4) with slope #0.5. For point of tangency,y = 2x + 3 (1)and y # 4 = #0.5(x # 3)2(y # 4) = 3 # x 2y # 8 = 3 # x x = 11 # 2y (2)(2) in (1): y = 2(11 # 2y) + 3 y = 25 # 4y5y = 25 y = 5x = 11 # 2(5) = 1Point of tangency is (1, 5). So r 2 = (1 # 3 ) 2 + (5 # 4 ) 2 = 5.The equation of the circle is(x # 3 ) 2 + (y # 4 ) 2 = 5


48. 2 x 2 # 2(4 x 2 + 1)

_______________ 2

x 2 # (4 x 2 + 1) x 2 # 4 x 2 # 1#3 x 2 # 1

49. 18a + 4(9a + 3)

______________ 6

3a + 2 _ 3 (9a + 3)

3a + 6a + 29a + 2

50. 3(x + 3y) # 4(3x + 2y) # (x # 2y)3x + 9y # 12x # 8y # x + 2y#10x + 3y

51. By the Isosc. & Thm., m'D = m'F7x + 4 = 60 7x = 56 x = 8

52. DE = EF2y + 10 = 4y # 1 11 = 2y y = 5.5

53. 180 # 142 = 38 = 1 _ 2 (m ( LK + m ( JQ )

m ( LKQ = m ( LK + m ( KJ + m ( JQ = 2(38) + 88 = 164°m ( LNQ = 360 # 164 = 196°

54. m'NMP = 1 _ 2 (m ( LKQ # m ( NP )

= 1 _ 2 (164 # 50) = 57°


1. m'AGB = 1 _ 2 m ( AB = 1 _

2 ( 1 __

12 (360)) = 15°

2. m'KAE = 90°; the angle is inscribed in a semicircle. So it is a right '.

3. m'KMJ = 1 _ 2 (m ( KJ + m ( BE ) = 1 _

2 (30 + 90) = 60°

4. ME = 22 # 4.8 ME ) 17.2 KM · ME = JM · MB

4.8 · 17.2 = 6.4x 82.56 = 6.4x 12.9 ) x MB ) 12.9 ft

5. (x # 20 ) 2 + (y # 14 ) 2 = 11 2 = 121

6. L(20, 14 + 11) = L(20, 25)C(20 + 11, 14) = C(31, 14)F(20, 14 # 11) = F(20, 3)I(20 # 11, 14) = I(9, 14)


1. m'BAC = 1 _ 2 m ( BC = 1 _

2 (102) = 51°

2. m ( CD = 2m'CAD = 2(38) = 76°

3. 'FGH is inscribed in a semicircle. So m'FGH = 90°.

4. m ( JGF = 310°

5. m'RST = 1 _ 2 mST = 1 _

2 (266) = 133°

6. m'AEC = 1 _ 2 (m ( AC + m ( BD ) = 1 _

2 (130 + 22) = 76°

7. m'MPN = 1 _ 2 ((360 # 102) # 102) = 78°

8. AE · EB = CE · ED 2(6) = x(3) 12 = 3x x = 4AB = 2 + 6 = 8CD = (4) + 3 = 7

9. FH · GH = KH · JH (y + 3)3 = (3 + 4)4 3y + 9 = 28 3y = 19 y = 6 1 _

3

FH = (6 1 _ 3 ) + 3 = 9 1 _

3

KH = 3 + 4 = 7

10. RU · (d # UR) = SU · UT 3.9(d # 3.9) = 6.1(6.1) 3.9d # 15.21 = 37.21 3.9d = 52.42 d ) 13.44 m

11. (x # (#2) ) 2 + (y # (#3) ) 2 = 3 2 (x + 2 ) 2 + (y + 3 ) 2 = 9

12. r = ! """""""" (1 # 4 ) 2 + (1 # 5 ) 2 = 5(x # 4 ) 2 + (y # 5 ) 2 = 5 2 (x # 4 ) 2 + (y # 5 ) 2 = 25


ge07_SOLKEY_C11_273-300.indd 296ge07_SOLKEY_C11_273-300.indd 296 10/5/06 11:44:51 AM10/5/06 11:44:51 AM

13. Step 1 Plot the 3 given points.Step 2 Connect J, K, and L to form a '.

Step 3 Find a point that is equidistant from the 3 points by constructing & bisectors of 2 sides of 'JKL. The & bisectors of the sides of 'JKL intersect at a point that is equidistant from A, B, and C. The intersection of the & bisectors is P(!1, !2). P is the center of the circle passing through J, K, and L.


VOCABULARY, PAGE 810 1. segment of a circle 2. central angle 3. major arc 4. concentric circles

LESSON 11-1, PAGE 810 5. chords:

"" QS , ""

UV ; tangent: !; radii: ""

PQ , ""

PS ; secant: # $% UV ;diameter:

"" QS

6. chords: ""

KH , """

MN ; tangent: # $% KL ; radii: ""

JH , ""

JK , ""

JM , ""

JN ; secant: # $% MN ; diameters:

"" KH ,

""" MN

7. AB = BC9x ! 2 = 7x + 4 2x = 6 x = 3AB = 9(3) ! 2 = 25

8. EF = EG5y + 32 = 8 ! y 6y = !24 y = !4EG = 8 ! (!4) = 12

9. JK = JL8m ! 5 = 2m + 4 6m = 9 m = 1.5JK = 8(1.5) ! 5 = 7

10. WX = WY0.8x + 1.2 = 2.4x 1.2 = 1.6x x = 0.75WY = 2.4(0.75) = 1.8

LESSON 11-2, PAGE 811 11. m 0 KM = m)KGL + m)LGM

= m)KGL + m)HGJ= 30 + 51 = 81°

12. m 0 HMK = m)HGL + m)LGK= 180 + 30 = 210°

13. m 0 JK = m)JGK= m)HGL ! m)HGJ ! m)KGL= 180 ! 51 ! 30 = 99°

14. m 0 MJK = 360 ! m 0 KM = 360 ! 81 = 279°

15. Let ST = 2x.x(x) = 4(7 + 11) x 2 = 72 x = . // 72 = 6 . / 2 ST = 2 (6 . / 2 ) ( 17.0

16. Let CD = 2x. x 2 = 2.5(2.5 + 5)x(x) = 18.75 x = . /// 18.75 = 2.5 . / 3 CD = 2 (2.5 . / 3 ) ( 8.7


17. A = " r 2 ( m ____ 360

) = "(12 ) 2 ( 30 ____ 360

) = 12" in. 2 ( 37.70 in. 2

18. A = " r 2 ( m ____ 360

) = "(1 ) 2 ( 90 ____ 360

) = 1 _ 4 " m 2 = 0.79 m 2

19. L = 2"r ( m ____ 360

) = 2"(18) ( 160 ____ 360

) = 16" cm ( 50.27 cm

20. L = 2"r ( m ____ 360

) = 2"(2) ( 270 ____ 360

) = 3" ft ( 9.42 ft


21. m 0 JL = 2m)JNL = 2(82) = 164°

22. m)MKL = 1 _ 2 m 0 ML = 1 _

2 (64) = 32°

23. )B is inscribed in a semicircle is and therefore, a right ).3x + 12 = 90 3x = 78 x = 26

24. m)RSP = 1 _ 2 m 0 RP = m)RQP

3y + 3 = 5y ! 21 24 = 2y y = 12m)RSP = 3(12) + 3 = 39°


25. m 0 MR = 2m)PMR = 2(41) = 82°

26. m)QMR = 1 _ 2 m 0 QR = 1 _

2 (360 ! 120 ! 82) = 79°

27. m)GKH = 1 _ 2 (m 0 FJ + m 0 GH ) = 1 _

2 (41 + 93) = 67°

28. m)BXC = 1 _ 2 (m 0 AD + m 0 BC )

= 1 _ 2 ( 2 __

16 (360) + 6 __

16 (360)) = 90°

LESSON 11-6, PAGE 813 29. BA · AC = DA · AE

3(y) = 7(5) 3y = 35 y = 11 2 _

3

BC = 3 + (11 2 _ 3 ) = 14 2 _

3

DE = 7 + 5 = 12

30. QP · PR = SP · PT z(10) = 15(8) 10z = 120 z = 12QR = (12) + 10 = 22ST = 15 + 8 = 23


31. GJ · HJ = LJ · KJ(4 + 6)6 = (x + 5)5 60 = 5x + 25 35 = 5x x = 7GJ = 4 + 6 = 10LJ = (7) + 5 = 12

32. AB · AC = AD · AE4(4 + y) = 5(5 + 5) 16 + 4y = 50 4y = 34 y = 8 1 _

2

AC = 4 + (8 1 _ 2 ) = 12 1 _

2

AE = 5 + 5 = 10


33. (x ! (!4) ) 2 + (y ! (!3) ) 2 = 3 2 (x + 4 ) 2 + (y + 3 ) 2 = 9

34. r = . /////////// (!2 ! (!2) ) 2 + (!2 ! 0 ) 2 = 2(x ! (!2) ) 2 + (y ! 0 ) 2 = 2 2 (x + 2 ) 2 + y 2 = 4

35. (x ! 1 ) 2 + (y ! (!1) ) 2 = 4 2 (x ! 1 ) 2 + (y + 1 ) 2 = 16

36. (x + 2 ) 2 + (y ! 2 ) 2 = 1(x ! (!2) ) 2 + (y ! 2 ) 2 = 1 2 Graph a circle with center (!2, 2) and radius 1.


1. chord: ""

EC ; tangent: # $% AB ; radii: ""

DE , ""

DC ; secant: # $% EC ; diameter:

"" EC

2. Let x be distance to the horizon.(4000 ) 2 + x 2 = (4006.25 ) 2 x 2 = 50,039.0625 x ( 224 mi

3. m 0 JK = m)JPK = m)JPL ! m)KPL= m)MPN ! m)KPL= 84 ! 65 = 19°

4. Let UV = 2x.x(x) = 6(9 + 15) x 2 = 144 x = 12 (since x > 0)UV = 2(12) = 24

5. A = "(8 ) 2 ( 135 ____ 360

) = 24" cm 2 ( 75.40 cm 2

6. A = 2"(8) ( 135 ____ 360

) = 6" cm ( 18.85 cm

7. m 0 SR = 2m)SPR = 2(47) = 94°

8. m)PTQ = 1 _ 2 (m 0 PQ + m 0 SR ) = 1 _

2 (58 + 94) = 76°

m)QTR = 180 ! m)PTQ = 180 ! 76 = 104°

9. m)ABC = 180 ! 1 _ 2 m 0 AB = 180 ! 1 _

2 (128) = 116°

10. m)MKL = 1 _ 2 (m 0 JN + m 0 ML ) = 1 _

2 (118 + 58) = 88°

m)MKL = 180 ! m)MKL = 180 ! 88 = 92°

11. m)CSD = 1 _ 2 (m 0 CD ! m 0 AB )

42 = 1 _ 2 (124 ! m 0 AB )

84 = 124 ! m 0 AB m 0 AB = 124 ! 84 = 40°

12. z(2) = 6(4) 2z = 24 z = 12EF = (12) + 2 = 14GH = 4 + 6 = 10

13. 6(6 + x) = 8(8 + 4) 36 + 6x = 96 6x = 60 x = 10PR = 8 + 4 = 12PT = 6 + (10) = 16

14. 4(4) = 2(d ! 2) 16 = 2d ! 4 20 = 2d d = 10 in.

15. r = . ////////// (!2 ! 1 ) 2 + (4 ! (!2) ) 2 = . // 45 = 3 . / 5

(x ! 1 ) 2 + (y ! (!2) ) 2 = (3 . / 5 ) 2 (x ! 1 ) 2 + (y + 2 ) 2 = 45

16. Step 1 Plot the 3 given points.Step 2 Connect X, Y, and Z to form a '.

Step 3 Find a point that is equidistant from 3 points by constructing & bisectors of the 2 sides of 'XYZ.The & bisectors of sides of 'XYZ intersect at a point that is equidistant from X, Y, and Z.The intersection of the & bisectors is P(0, !2). P is center of the circle passing through X, Y, and Z.


1. EDraw

"" AD . m)ADC is inscribed in a semicircle. So is

a right ).m 0 AB = 2m)ADB

= 2(90 ! m)BDC) = 2(90 ! 30) = 120°

2. APossible coordinates of the center are (1 ± 2, 3 ± 2).(!1, 1) is a possible center. The circle with this center and radius 2 is(x ! (!1) ) 2 + (y ! 1 ) 2 = 2 2 (x + 1 ) 2 + (y ! 1 ) 2 = 4.


3. E PK · KM = LK · KN3(PM ! 3) = 6(10 ! 6) 3PM ! 9 = 24 3PM = 33 PM = 11

4. CL = 2!r ( m " AC _____

360 )

= 2!(6) ( 2m#ABC ________ 360

) = 12! ( 50 ____

360 ) = 5 __

3 !

5. BDraw $ with a base along the shaded area and the apex at the center of the circle. $ is a 45°-45°-90° $ with b = 9 % & 2 and h = 4.5 % & 2 .

A = ! r 2 ( 90 ____ 360

) ! 1 __ 2 bh

= !(9 ) 2 ( 1 __ 4 ) ! 1 __

2 (9 % & 2 ) (4.5 % & 2 )

= 81 ___ 4 ! ! 81 ___

2 ' 23.12

CHAPTERS 1–11, PAGES 818–819

1. C L = Ph + 1 _

2 P"

328 = 40(4) + 1 _ 2 (40)"

328 = 160 + 20"168 = 20" " = 8.4 ft

2. GA = (12 ! 2)(3 ! 0) ! 1 _

2 (12 ! 6)(3 ! 0)

= 10(3) ! 1 _ 2 (6)(3)

= 30 ! 9 = 21 units 2

3. Bm#EFD = m " ED = 45°m#FED = 180 ! (45 + 90) = 45°Since F is the center, $EFC and $CFA are also 45°-45°-90° (.

)) FB *

)) BC . So m " BC = m#BFC = 45°.

4. G

L = 2!r ( m " ED _____ 360

) 6! = 2!r ( 45 ____

360 ) = 1 __

4 !r

r = 24 cm

A = !(24 ) 2 ( 45 ____ 360

) = 72! cm 2

5. CF is not on the circle. So

)) AF is not a chord.

6. JKL = 18 and KJ = 24 + tan J = 18 __

24 = 3 _

4 ;

KL 2 + KJ 2 , JL 2 18 2 + 24 2 , 30 2 324 + 576 = 900

7. B m#P = 1 _

2 (m " RS ! m " QU )

= m#T ! 1 _ 2 m " QU

29 = 50 ! 1 _ 2 m " QU

1 _ 2 m " QU = 21

m " QU = 42°

8. FPS · PU = PR · PQ

PS = PR · PQ _______ PU

9. C

From the diagram, y = !3.

10. Gd = % &&&&&&&&&& (3 ! (!1) ) 2 + (!5 ! 1 ) 2 = % && 52 = 2 % && 13

r = % && 13

center = ( !1 + 3 _______ 2 ,

1 + (!5) ________

2 ) = (1, !2)

(x ! 1 ) 2 + (y ! (!2) ) 2 = ( % && 13 ) 2 (x ! 1 ) 2 + (y + 2 ) 2 = 13

11. DThe diagonals of a kite intersect at right -. So the shortest segment to Q is from the intersection T; ))

TQ is shortest segment.

12. HP = 1 _

2 (3b) = 3 ( 1 _

2 b)

A = 1 _ 2 ( 1 _

2 b) ( 1 _

4 b % & 3 ) = 1 __

16 b 2 % & 3 = 1 _

4 ( 1 _

2 b ( 1 _

2 b % & 3 ) )

The area is reduced by 1 _ 4 .

13. CLet the leg length be x m. A = 1 _

2 (x)(x)

36 = 1 _ 2 x 2

72 = x 2

x = 6 % & 2 hypotenuse = (6 % & 2 ) % & 2 = 12 m

14. Let the side lengths be 4x, 5x, and 8x cm. P = 4x + 5x + 8x38.25 = 17x x = 2.254x = 4(2.25) = 9 cm

15. 8 x 2 = 4(16) = 64 x = 8



16. If 'HGJ + 'LMK, ""

HG + ""

LM and ""

HJ + ""

LK .

"" HG +

"" LM

HG = LM4x + 5 = 13 4x = 8 x = 2Check: HJ = 5(2) ! 1 = 9 = KL

17. s = 2, so a = . / 3 and P = 6(2) = 12

A = 1 _ 2 aP = 1 _

2 ( . / 3 ) (12) = 6 . / 3 ( 10.4

18. 15.71 mmL = 1 _

2 (2"(5)) = 5" ( 15.71 mm

19. 452.39 cm 2 V = 4 _

3 " r 3

288" = 4 _ 3 " r 3

216 = r 3 r = 6 cmS = 4"(6 ) 2 = 144" ( 452.39 cm 2

20. 3x = 6 cos 60° = 3

21. Possible answer:180 ! 7x = 1 _

2 (4x + 10 + 6x + 14)

180 ! 7x = 1 _ 2 (10x + 24)

180 ! 7x = 5x + 12 !12x = !168 x = 14

22a. 1st unit:V = 10(5)(9) = 450 ft 3

cost/ ft 3 = 85 ____ 450

( $0.19

2nd unit:V = 11(4)(8) = 352 ft 3

cost/ ft 3 = 70 ____ 352

( $0.20

The first unit has the lower price per ft 3 .

b. V · $0.25/ ft 3 = $100 V = 400Possible dimensions: 10 ft by 5 ft by 8 ft

23a. x 2 + (y + 1 ) 2 = 25

(x ! 0 ) 2 + (y ! (!1) ) 2 = 5 2 Plot the circle with center (0, !1) and radius 5.

b. slope of radius to (3, 3) = 3 ! (!1)

________ 3 ! 0

= 4 __ 3

slope of tangent = ! 3 __ 4

equation of tangent: y ! 3 = ! 3 _

4 (x ! 3)

y ! 3 = ! 3 _ 4 x + 2 1 _

4

y = ! 3 _ 4 x + 5 1 _

4

24. The measure of the intercepted arc must be > 0° and < 180°.

25a. Statements Reasons

1. ""

AB 8 ""

CD 1. Given2. )BCD + )ABD 2. Alt. Int. - Thm.3. m)BCD = m)ABD 3. Def. of + -4. m)ACD = 1 _

2 m 0 AD ,

m)BDC = 1 _ 2 m 0 BC

4. Inscribed ) Thm.

5. 1 _ 2 m 0 BC = 1 _

2 m 0 AD 5. Subst. Prop.

6. m 0 BC = m 0 AD 6. Mult. Prop. of =

b. Possible answer: Since ""

AB 8 ""

CD , m 0 BC = m 0 AD from part a. Also, by Inscribed ) Thm., m)ACD

= 1 _ 2 m 0 AD and m)BDC = 1 _

2 m 0 BC . Then use Arc

Add. Post. to get m)ADC = 1 _ 2 (m 0 AB + m 0 BC ) and

m)BCD = 1 _ 2 (m 0 AB + m 0 AD ). Subst. m 0 BC for m 0 AD ,

which gives m)BCD = 1 _ 2 (m 0 AB + m 0 BC ) = m)ADC.

Since ABCD has 1 pair of base -, ABCD is isosceles.

c. Possible answer: Since ""

AB 8 ""

CD and ABCD is not a trapezoid,

"" AD 8

"" BC . So ABCD must be a

quadrilateral. Therefore, )ADC + )ABC. So by def. of +, m)ADC = m)ABC. Also, since ABCD can be inscribed in a circle, )ADC and )ABC are supplementary. So m)ADC + m)ABC = 180°. Subst. and solve for m)ABC: m)ABC + m)ABC = 180 2m)ABC = 180 m)ABC = 90°Therefore, )ABC is a right ). Since ABCD is a quadrilateral, it follows that every ) is a rt. ), so ABCD is a rectangle Therefore, if ABCD is not a trapezoid, it must be a rectangle.


Solutions KeyExtending Transformational Geometry12

CHAPTER


1. E 2. C

3. A 4. D

5.

6.

7.

8.

9.

10.

11. no

12. Yes; by Alt. Int. ! Thm., "DGE # "FEG and "DEG # "FGE. By Reflex. Prop. of #,

$$ GE # $$

GE . So by ASA, %DEG # %FGE.

13. Yes; JM ___ JK

= 10 ___ 4 = 5 __

2 , JN ___

JL = 7.5 ___

3 = 5 __

2 . so JM ___

JK = JN ___

JL ;

"J # "J by Reflex. Prop. of #; %JKL & %JMN by SAS &.

14. Yes; corr. ! are #, corr. sides are proportional. PQ ___

UV = QR ____

VW = RS ____

WX = SP ___

XU = 2

15. (n)m(int. ") = (n ' 2)180 8m(int. ") = 6(180) m(int. ") = 6 _

8 (180) = 135°

16. sum of internal " measures = (5 ' 2)180 = 540°

17. (n)m(ext. ") = 360 6m(ext. ") = 360 m(ext. ") = 60°

18. 90 + 5x = (6 ' 2)180 = 720 5x = 630 x = 126

12-1 REFLECTIONS, PAGES 824–830

CHECK IT OUT! PAGES 824–826 1a. No; the image does not appear to be flipped.

b. Yes; the image appears to be flipped across a line.

2. Step 1 Through each vertex draw a line ( to the line of reflection.Step 2 Measure the distance from each vertex to the line of reflection. Locate the image of each vertex on the opposite side of the line of reflection and the same distance from it.Step 3 Connect the images of the vertices.

3. If A and B were the same distance from the river, would have %ACX # %BDX (see diagram). So $$

AX = $$$

XB) (CPCTC) and therefore

$$ AX = $$

BX . So

$$ AX and

$$ BX would be

congruent.

4. The reflection of (x, y) is (x, 'y).S(3, 4) * S)(3, '4)T(3, 1) * T)(3, '1)U('2, 1) * U)('2, '1)V('2, 4) * V)('2, '4)


1. Possible answer: ABA)C is a kite, because $$

AB # $$

A)B and

$$ AC #

$$$ A)C . So there are exactly two pairs of #

adjacent sides.

2. ! is the ( bisector of $$$

AA) .

3.




1. They are congruent.

2. Yes; the image appears to be flipped across a line.

3. No; the image does not appear to be flipped.





8. 1 Understand the ProblemThe problem asks you to draw a diagram, locating point. P on a highway so that

$$ SP

+ $$

PT has the least possible value. 2 Make a PlanLet T) be the reflection of T across the highway. For any point P on the highway,

$$ PT) #

$$ PT .

So $$

SP + $$

PT = $$

SP + $$

PT). $$

SP + $$

PT) is the least when S, P, and T) are collinear.3 SolveReflect T across the highway to locate T). Draw

$$ ST) and locate

P at the intersection of $$

ST) and the highway.4 Look BackTo verify the answer, choose several possible locations for P and measure the total length of the access roads for each location.

T!

S

P

TSan

Pablo

Tanner

Highway 105

9. The reflection of (x, y) is (x, 'y).A('2, 1) * A)('2, '1)B(2, 3) * B)(2, '3)C(5, 2) * C)(5, '2)

10. The reflection of (x, y) is ('x, y).R(0, '1) * R)(0, '1)S(2, 2) * S)('2, 2)T(3, 0) * T)('3, 0)

11. The reflection of (x, y) is (y, x).M(2, 1) * M)(1, 2)N(3, 1) * N)(1, 3)P(2, '1) * P)('1, 2)Q(1, '1) * Q)('1, 1)


12. The reflection of (x, y) is (y, x).A('2, 2) * A)(2, '2)B('1, 3) * B)(3, '1)C(1, 2) * C)(2, 1)D('2, '2) * D)('2, '2)







18. Step 1 Through each vertex draw a line ( to the line of reflection.Step 2 Measure the distance from each vertex to the line of reflection. Locate image of each vertex on the opposite side of the line of reflection and the same distance from it.

Step 3 Connect the images of the vertices.

19. 1 Understand the ProblemThe problem asks you to draw a diagram, locating point X on the side rail so that

$$ AX and

$$ XC make the

same " with the side rail. 2 Make a PlanLet C) be the reflection of C across the side rail. For any point X on side rail, the ! of

$$ AX and

$$$ XC) with

the side rail are # when A, X, and C) are collinear.3 SolveReflect C across the side rail to locate C). Draw

$$ AC)

and locate P at the intersection of $$

AC) and the side rail.4 Look BackTo verify the answer, choose several possible locations for X and measure the ! for each location.

20. The reflection of (x, y) is ('x, y).A('3, 2) * A)(3, 2)B(0, 2) * B)(0, 2)C('2, 0) * C)(2, 0)

21. The reflection of (x, y) is (y, x).M('4, '1) * M)('1, '4)N('1, '1) * N)('1, '1)P('2, '2) * P)('2, '2)

22. The reflection of (x, y) is (x, 'y).J(1, 2) * J)(1, '2)K('2, '1) * K)('2, 1)L(3, '1) * L)(3, 1)

23. The reflection of (x, y) is (y, x).S('1, 1) * S)(1, '1)T(1, 4) * T)(4, 1)U(3, 2) * U)(2, 3)V(1, '3) * V)('3, 1)

24.

25.

26.

27.

28.

29.


30.

31. (5, 2) * (5, '2) 32. ('3, '7) * (3, '7)

33. (0, 12) * (0, '12) 34. ('3, '6) * ('6, '3)

35. (0, '5) * ('5, 0) 36. (4, 4) * (4, 4)

37a. No; $$

TH passes through (4, 3.25) which lies above E(4, 3).

b. H(5, 4) * H)(7, 4)

c. slope of $$$

TH) = 4 ' 1 _____ 7 ' 1

= 0.5

equation of $$$

TH) : y ' 1 = 0.5(x ' 1) or y = 0.5x + 0.5

equation of $$

BC : x = 6At the intersection, x = 6 and

y = 0.5(6) + 0.5 = 3.5.The intersection is at (6, 3.5).

38. The figures are formed by the numerals 1, 2, … 7 and their reflections. The figure formed by numeral 8 is:

39. The line of reflection is the line y = x. It is the only possible line of reflection because it must be the ( bisector of the segment connecting the given points.; there is only one such ( bisector.

40.

41.

42. The points on line ! remain fixed. Each of these pts. is its own image under a reflection across line !.

43–45. Check students’ constructions.


46. A('2, 4) * ('2, '4)

47. JP * S, M * D, J * G, and N * W, so PMJN * SDGW

48. C('3, 4) * ('('3), 4) = (3, 4)


49. y * 3 + (3 ' y) = 6 ' y(4, 2) * (4, 6 ' 2) = (4, 4)

50. x * 2(1) ' x = 2 ' x('3, 2) * (2 ' ('3), 2) = (5, 2)

51. y * x + 2 and x * y ' 2(3, 1) * (1 ' 2, 3 + 2) = ('1, 5)

52. Draw $$$

AA) and $$$

BB) ; let C be the point. where $$$

AA) intersects ! and let D be the point where

$$$ BB)

intersects !. By definition of reflection, ! is the ( bisector of

$$$ AA) and

$$$ BB) .

Therefore, "ACD and "A)CD are right. !, $$

AC #

$$$ A)C by definition of bisector, and

$$ CD # $$

CD by Reflex. Prop. of #. By SAS, %ACD # %A)CD. By CPCTC, "CDA # "CDA). "ADB is comp. to "CDA, and "A)DB) is comp. to "CDA). So "ADB # "A)DB).

$$ AD #

$$$ A)D by CPCTC, and

$$ BD #

$$$ B)D by

def. of bisector. Therefore, %ADB # %A)DB) by SAS. By CPCTC,

$$ AB #

$$$ A)B) .

53. Use the fact that the reflection of a segment is # to its preimage and the definition of # segs.

54. Draw $$

AC and $$$

A)C) . Use the fact that the reflection of a segment is # to its preimage to prove %ABC # %A)B)C) by SSS. By CPCTC "ABC # "A)B)C). So m"ABC = m"A)B)C) by definition of # !.

55. Use the fact that the reflection of a segment is # to its preimage to prove that %ABC # %A)B)C) by SSS.

56. Since C is between A and B, AC + BC = AB. Use the fact that the reflection of a segment is # to its preimage to prove that A)C) + B)C) = A)B). Then use the definition of betweenness to prove that C) is between A) and B).

57. Since A, B, and C are collinear, one point. is between other two. Case 1: If C is between A and B, then AC + BC = AB. Use the fact that the reflection of a segment is # to its preimage to prove that A)C) + B)C) = A)B). Then C) is between A) and B). So A), B), and C) are collinear. Prove the other two cases similarly.


58. P = ( 4 ___ 12

) ( 4 ___ 12

) = 16 ____ 144

= 1 __ 9

59. P = ( 10 ___ 12

) ( 10 ___ 12

) = 100 ____ 144

= 25 ___ 36

60. P = ( 6 ___ 12

) ( 4 ___ 12

) = 24 ____ 144

= 1 __ 6


61. dimensions of scale drawing: 1 cm _____ 30 m

(60 m) by

1 cm _____ 30 m

(105 m) or 2 cm by 3.5 cm

P = 2(2) + 2(3.5) = 11 cm

62. dimensions of scale drawing: 1.5 cm ______ 15 m

(60 m) by

1.5 cm ______ 15 m

(105 m) or 6 cm by 10.5 cmP = 2(6) + 2(10.5) = 33 cm

63. dimensions of scale drawing: 1 cm _____ 25 m

(60 m) by

1 cm _____ 25 m

(105 m) or 2.4 cm by 4.2 cm

P = 2(2.4) + 2(4.2) = 13.2 cm

64. BC 2 + AB 2 = AC 2 BC 2 + 2 2 = ( + , 7 ) 2 BC 2 = 3 BC = + , 3 - 1.73

65. cos A = AB ___ AC

= 2 ___ + , 7

m"A = co s '1 ( 2 ___ + , 7

) - 41°

66. m"C + m"A = 90 m"C + 41 - 90 m"C - 49°

12-2 TRANSLATIONS, PAGES 831–837

CHECK IT OUT! PAGES 831–833 1a. Yes; all the points have moved the same distance in

the same direction. b. No; not all the points have moved the same

distance.

2. Step 1 Draw a line . to the vector through each vertex of the quadrilateral.

Step 2 Measure the length of the vector. Then, from each vertex mark off this distance in the same direction as the vector, on each of the . lines.Step 3 Connect the images of the vertices.

3. The image of (x, y) is (x ' 3, y ' 3).R(2, 5) * R)(2 ' 3, 5 ' 3) = R)('1, 2)S(0, 2) * S)(0 ' 3, 2 ' 3) = S)('3, '1)T(1, '1) * T)(1 ' 3, '1 ' 3) = T)('2, '4)U(3, 1) * U)(3 ' 3, 1 ' 3) = U)(0, '2)Graph the image and the preimage.

4. The drummer’s starting coordinates are (0, 0).The second position is (0 + 16, 0) = (16, 0).The final position is (16, 0 ' 24) = (16, '24).

THINK AND DISCUSS, PAGE 833 1. Possible answer:

/ v .

$$$ AA) and 0

/ v 1 = AA)

2. Possible answer: $$$

AA) and $$$

BB) are both ( to the translation vector, so they are . to each other. They are # because their lengths equal to the length of the translation vector. So AA)B)B is a quadrilateral, because the opposite sides

$$$ AA) and

$$$ BB) are . and

#.

3.



1. No; not all the points have moved the same distance.


3. Yes; all the points have moved the same distance in the same direction.



5. Step 1 Draw a line . to the vector through each vertex of the square.

Step 2 Measure the length of the vector. Then, from each vertex mark off this distance in the same direction as the vector, on each of the . lines.Step 3 Connect the images of the vertices.

6. Step 1 Draw a line . to the vector through each vertex of the triangle.

Step 2 Measure the length of vector. Then, from each vertex mark off this distance in the same direction as the vector, on each of the . lines.Step 3 Connect the images of the vertices.

7. The image of (x, y) is (x + 5, y).A('4, '4) * A)('4 + 5, '4) = A)(1, '4)B('2, '3) * B)('2 + 5, '3) = S)(3, '3)C('1, 3) * C)('1 + 5, 3) = C)(4, 3)Graph image and preimage.

8. The image of (x, y) is (x, y ' 4).R('3, 1) * R)('3, 1 ' 4) = R)('3, '3)S('2, 3) * S)('2, 3 ' 4) = S)('2, '1)T(2, 3) * T)(2, 3 ' 4) = T)(2, '1)U(3, 1) * U)(3, 1 ' 4) = U)(3, '3)Graph image and preimage.

9. The image of (x, y) is (x + 3, y + 2).J('2, 2) * J)('2 + 3, 2 + 2) = J)(1, 4)K('1, 2) * K)('1 + 3, 2 + 2) = K)(2, 4)L('1, '2) * L)('1 + 3, '2 + 2) = L)(2, 0)M('3, '1) * M)('3 + 3, '1 + 2) = U)(0, 1)Graph image and preimage.

10. The second polygon has coordinates (1, 5 ' 4) = (1, 1), (2, 3 ' 4) = (2, '1), (1, 1 ' 4) = (1, '3), and (0, 3 ' 4) = (0, '1).The third polygon has coordinates(1, 1 ' 4) = (1, '3), (2, '1 ' 4) = (2, '5), (1, '3 ' 4) = (1, '7), and (0, '1 ' 4) = (0, '5).The fourth polygon has coordinates(1, '3 ' 4) = (1, '7), (2, '5 ' 4) = (2, '9), (1, '7 ' 4) = (1, '11), and (0, '5 ' 4) = (0, '9).



12. No; not all points have moved the same distance.





Step 2 Measure the length of the vector. Then, from each vertex mark off this distance in the same direction as the vector, on each of the . lines.



Step 2 Measure the length of the vector. Then, from each vertex mark off this distance in the same direction as the vector, on each of the . lines.


17. The image of (x, y) is (x ' 3, y).P('1, 2) * P)('1 ' 3, 2) = P)('4, 2)Q(1, '1) * Q)(1 ' 3, '1) = Q)('2, '1)R(3, 1) * R)(3 ' 3, 1) = R)(0, 1)S(2, 3) * S)(2 ' 3, 3) = S)('1, 3)Graph the image and the preimage.

18. The image of (x, y) is (x ' 3, y ' 3).A(1, 3) * A)(1 ' 3, 3 ' 3) = A)('2, 0)B('1, 2) * B)('1 ' 3, 2 ' 3) = B)('4, '1)C(2, 1) * C)(2 ' 3, 1 ' 3) = C)('1, '2)D(4, 2) * D)(4 ' 3, 2 ' 3) = D)(1, '1)Graph the image and the preimage.


19. The image of (x, y) is (x + 5, y ' 20).D(0, 15) * D)(0 + 5, 15 ' 20) = D)(5, '5)E('10, 5) * E)('10 + 5, 5 ' 20) = E)('5, '15)F(10, '5) * F)(10 + 5, '5 ' 20) = F)(15, '25)Graph the image and the preimage.

20a. The ladybug starts at ('2.5, '1.5).The second position is at ('2.5 + 1, '1.5 + 1) = ('1.5, '0.5).The third position is at ('1.5 + 2, '0.5 + 2)

= (0.5, 1.5).The final position is at (0.5 + 3, 1.5 + 3) = (3.5, 4.5).

b. 21, 13 + 22, 23 + 23, 33 = 26, 63

21.

22.

23a. Possible images are (3 ' 3, 2) = (0, 2), (3 ' 1, 2 ' 4) = (2, '2), (3 + 3, 2 ' 2) = (6, 0), and (3 + 2, 2 + 3) = (5, 5).The points in the fourth quadrant: (2, '2); P = 1 __

4

b. Points on an axis: (0, 2) and (6, 0). P = 2 __ 4

= 1 __ 2

c. Points at origin: none. P = 0 __ 4 = 0

24a. ball travels from (1, 2) to (3, 3)vector is 23 ' 1, 3 ' 23 = 22, 13

b. ball travels from (3, 3) to (7, 1)vector is 27 ' 3, 1 ' 33 = 24, '23

c. sum of vectors is 22 + 4, 1 + ('2)3 = 26, '13ball travels from (1, 2) to (7, 1)vector is 27 ' 1, 1 ' 23 = 26, '13

25.

26.

27. No; there are no fixed points because, by definition of a translation, every point must move by the same distance.

28. First use the adjustable .s to draw a line through the given pt. that is . to the given vector. Then use a ruler to measure a distance along this line that is equal to the magnitude of the vector. The only additional tool needed to do this construction is the ruler.

29. vector is 21 ' ('3), 2 ' 23 = 24, 03;('3, 2) * (1, 2)

30. vector is 2'3 ' 1, 2 ' 23 = 2'4, 03;(1, 2) * ('3, 2)

31. vector is 20 ' 3, '3 ' ('1)3 = 2'3, '23;(3, '1) * (0, '3)

32. vector is 21 ' ('4), 2 ' ('3)3 = 25, 53;('4, '3) * (1, 2)

33. vector is 20 ' 3), 0 ' ('1)3 = 2'3, 13;(3, '1) * (0, 0)

34. The overlap is a rectangle with

! = 8 ' 2 __ 3 (8) = 2 2 __

3 in. and

w = 3 ' 2 __ 3 (3) = 1 in.

A = !w = (2 2 __ 3 ) (1) = 2 2 __

3 in. 2

35. The distance between P and its image is equal to the magnitude of the translation vector 2a, b3. By Distance Formula,

the magnitude of this vector is + ,,,, a 2 + b 2 .



39. AP(1, 3) * P)(1 ' 3, 3 + 5) = P)('2, 8)

40. GA('6, '2) * B('4, '4) = B('6 + 2, '2 '2)(3, '1) * (3 + 2, '1 ' 2) = (5, '3)


41. CQ(3, '1) * P(1, 3) = P(3 ' 2, '1 + 4)vector is 2'2, 43


42. vector 2a, b3 satisfies b = 2a and + ,,,, a 2 + b 2 = + , 5 . a 2 + (2a ) 2 = 5 a 2 + 4 a 2 = 5 5 a 2 = 5 a 2 = 1 a = ±1b = 2(±1) = ±2M(1, 2) * M)(1 ' 1, 2 ' 2) = M)(0, 0)or * M)(2, 4) = M)(2, 4)

43a. the vector 444/ PQ

b. 444/ PQ =

/ u +

/ v +

/ w

= 22, 0, 03 + 22, 0, 03 + 22, 0, 03= 22, 2, 23

0 444/ PQ 1 = + ,,,,,, 2 2 + 2 2 + 2 2 = + ,, 12 - 3.46 cm

44. Draw $$$

AA) and $$$

BB) . By def. of translation, $$$

AA) # $$$

BB) and

$$$ AA) .

$$$ BB) . Opposite sides

$$$ AA) and

$$$ BB) are #

and .. So AA)B)B is a parallelogram. Opposite sides of a quadrilateral are #. So

$$ AB #

$$$ A)B) .

45. Use fact that the translation of a segment is # to its preimage and def. of # segs.

46. Draw $$

AC and $$$

A)C) . Use the fact that the translation of a segment is # to its preimage to prove %ABC # %A)B)C) by SSS. By CPCTC, "ABC # "A)B)C). So m"ABC = m"A)B)C) by def. of # !.

47. Use the fact that the translation of a segment is # to its preimage to prove %ABC # %A)B)C) by SSS.

48. Since C is between A and B, AC + BC = AB. Use the fact that the translation of a segment is # to its preimage to prove A)C) + B)C) = A)B). Then use def. of betweenness to prove that C) is between A) and B).

49. Since A, B, and C are collinear, one point is between other two. Case 1: If C is between A and B, AC + BC = AB. Use the fact that the translation of a segment is # to its preimage to prove A)C) + B)C) = A)B). Then C) is between A) and B). So A), B), and C) are collinear. Prove the other two cases similarly.


50. 5

6 7 '5x ' 2y = 17 (1)

6x ' 2y = '5 (2)

(2) ' (1):6x ' ('5x) = '5 ' 17 11x = '22 x = '2Substitute in (2):6('2) ' 2y = '5 '12 ' 2y = '5 '2y = 7 y = '3.5Solution is: ('2, '3.5) or ('2, ' 7 __

2 )

51. 5 6 7 2x ' 3y = '7 (1)

6x + 5y = 49 (2)

(2) ' 3(1):6x ' 3(2x) + 5y ' 3('3y) = 49 ' 3('7) 6x ' 6x + 5y + 9y = 49 + 21 14y = 70 y = 5Substitute in (1):2x ' 3(5) = '7 2x ' 15 = '7 2x = 8 x = 4Solution is: (4, 5)

52. 5

6 7 4x + 4y = '1 (1)

12x ' 8y = '8 (2)

2(1) + (2):2(4x) + 12x + 2(4y) ' 8y = 2('1) ' 8 8x + 12x + 8y ' 8y = '2 ' 8 20x = '10 x = '0.5Substitute in (1):4('0.5) + 4y = '1 '2 + 4y = '1 4y = 1 y = 0.25Solution is: ('0.5, 0.25) or (' 1 _ 2 , 1 _

4 )

53. Think: All ! in the diagram are right. !.x + 15y = 90 (1)3x + 9y = 90 (2)3(1) ' (2):3x ' 3x + 3(15y) ' 9y = 3(90) ' 90 36y = 180 y = 5Substitute in (1):x + 15(5) = 90 x + 75 = 90 x = 15

54. Think: 4x ° is a right ".4x = 90 x = 22.5Think: y ° and 2x ° ! are comp. y + 2x = 90y + 2(22.5) = 90 y + 45 = 90 y = 45

55. General point (x, y) * (x, 'y)M('2, 0) * M)('2, 0)N('3, 2) * N)('3, '2)P(0, 4) * P)(0, '4)

56. General point (x, y) * ('x, y)M('2, 0) * M)(2, 0)N('3, 2) * N)(3, 2)P(0, 4) * P)(0, 4)

57. General point (x, y) * (y, x)M('2, 0) * M)(0, '2)N('3, 2) * N)(2, '3)P(0, 4) * P)(4, 0)


CONNECTING GEOMETRY TO ALGEBRA: TRANSFORMATIONS OF FUNCTIONS, PAGE 838

TRY THIS, PAGE 838 1. function rule:

y = (x ' 4 ) 2 ' 1graph:

2. function rule:y = ' + , x graph:

3. function rule:y = 0x + 11 + 2graph:

12-3 ROTATIONS, PAGES 839–845

CHECK IT OUT! PAGES 839–841 1a. No; the figure appears to be translated, not turned.

b. Yes; the figure appears to be turned around a point.

2. Step 1 Draw a segment from each vertex to point Q.

Step 2 Construct an " # to "X onto each segment. Measure the distance from each vertex to point Q and mark off this distance on the corresponding ray to locate the image of each vertex.


3. The rotation of (x, y) is ('x, 'y).A(2, '1) * A)(2, '1)B(4, 1) * B)(4, 1)C(3, 3) * C)(3, 3)


4. Step 1 Find the " of rotation. 6 min is 6 ___ 30

= 1 __ 5 of a

complete rotation, or 1 __ 5 (360) = 72°.

Step 2 Draw a right % to represent the car’s location (x, y) after a rotation of 72° about the origin.Step 3 Use the cosine ratio to find the x-coordinatecos 72° = x ____

67.5

x = 67.5 cos 72° - 20.9Step 4 Use the sine ratio to find the y-coordinate

sin 72° = y ____

67.5

y = 67.5 sin 72° - 64.2The car’s location after 6 min is approximately (20.9, 64.2).

THINK AND DISCUSS, PAGE 841 1. The image is in the same position as the preimage.

2. $$

AP # $$$

A)P

3.



1. Yes; the figure appears to be turned around a point.

2. No; the figure appears to be flipped, not turned.

3. No; the figure appears to be translated, not turned.


5. Step 1 Draw a segment from each vertex to point P.

Step 2 Construct an " # to "A onto each segment. Measure distance from each vertex to point P and mark off this distance on the corresponding ray to locate image of each vertex.



Step 2 Construct an " # to "A onto each segment. Measure the distance from each vertex to point P and mark off this distance on the corresponding ray to locate image of each vertex.Step 3 Connect the images of the vertices.

7. The rotation of (x, y) is ('y, x).A(1, 0) * A)(0, 1)B(3, 2) * B)('2, 3)C(5, 0) * C)(0, 5)


8. The rotation of (x, y) is ('y, x).J(2, 1) * J)('1, 2)K(4, 3) * K)('3, 4)L(2, 4) * L)('4, 2)M('1, 2) * M)('2, '1)

9. The rotation of (x, y) is ('x, 'y).D(2, 3) * D)('2, '3)E('1, 2) * E)(1, '2)F(2, 1) * F)('2, '1)

10. The rotation of (x, y) is ('x, 'y).P('1, '1) * P)(1, 1)Q('4, '2) * Q)(4, 2)R(0, '2) * R)(0, 2)

11. Step 1 Draw a right % to represent the car’s location (x, y) after a rotation of 30° about the origin.Step 2 Use the cosine ratio to find the x-coordinatecos 30° = x ___

10

x = 10 cos 30° - 8.7Step 3 Use the sine ratio to find the y-coordinate

sin 30° = y ___

10

y = 10 sin 30° = 5The car’s final position is approximately (8.7, 5).


12. No; the figure appears to be flipped and translated, not turned.



15. No; the figure appears to be enlarged, not turned.


Step 2 Construct an " # to "A onto each segment. Measure the distance from each vertex to point P and mark off this distance on the corresponding ray to locate the image of each vertex.



Step 2 Construct an " # to "A onto each segment. Measure the distance from each vertex to point P and mark off this distance on the corresponding ray to locate the image of each vertex.Step 3 Connect the images of the vertices.

18. The rotation of (x, y) is ('y, x).E('1, 2) * E)('2, '1)F(3, 1) * F)('1, 3)G(2, 3) * G)('3, 2)

19. The rotation of (x, y) is ('y, x).A('1, 0) * A)(0, '1)B('1, '3) * B)(3, '1)C(1, '3) * C)(3, 1)D(1, 0) * D)(0, 1)


20. The rotation of (x, y) is ('x, 'y).P(0, 2) * P)(0, '2)Q(2, 0) * Q)('2, 0)R(3, '3) * R)('3, 3)

21. The rotation of (x, y) is ('x, 'y).L(2, 0) * L)('2, 0)M('1, '2) * M)(1, 2)N(2, '2) * N)('2, 2)

22. Step 1 Find the " of rotation. 6 min is 6 ___ 72

= 1 ___ 12

of a

complete rotation, or 1 ___ 12

(360) = 30°.

Step 1 Draw a right % to represent the table’s location (x, y) after a rotation of 30° about the origin.Step 2 Use the cosine ratio to find the x-coordinatecos 30° = x ___

50

x = 50 cos 30° - 43.3Step 3 Use the sine ratio to find the y-coordinate

sin 30° = y ___

50

y = 50 sin 30° = 25The table’s final position is approximately (43.3, 25).

23.

24.

25.

26a. tan " = 2 __ 3

" = tan '1 ( 2 __ 3 ) - 34°

b. The x-coordinate of the image is 0. The y-coordinate of the image is

OQ = + ,,,, 2 2 + 3 2 = + ,, 13 - 3.6The location of the image is approximately (0, 3.6).

27. T 28. M

29. $$

ST 30. $$

NP

31a. 4 ___ 20

(360) = 72°

b. A(7, 3) = A(4 + 3, 3 + 0)A)(4 + 3 cos 72°, 3 + 3 sin 72°) - A)(4.9, 5.9)

32.

33.

34.

35a. Possible answer: 90° b. x ___ 24

= 90 ____ 360

= 1 __ 4

x = 1 __ 4 (24) = 6 h

36a. Possible answer: 45°. Check students’ estimates.

b. Draw $$

AP and $$$

A)P and use the protractor to measure "A)PA.

c. Check students’ constructions; possible answer: 50°.

37. No; although all points are rotated around the center of rotation by the same ", points that are farther from the center of rotation move a greater distance than points that are closer to the center of rotation.

38. The rotation of (x, y) is ('y, x).A(2, '3) * A)(3, 2)B(3, 0) * B)(0, 3)C(0, 3) * C)('3, 0)D('3, 0) * D)(0, '3)E('2, '3) * E)(3, '2)


39. The rotation of (x, y) is ('x, 'y).A(2, '3) * A)('2, 3)B(3, 0) * B)('3, 0)C(0, 3) * C)(0, '3)D('3, 0) * D)(3, 0)E('2, '3) * E)(2, 3)

40. The image of ABCDE under a rotation of 180° is not the same as its reflection across the x –axis because the images of the specific points are in different locations. For example, A(2, '3) * A)('2, 3) is under a 180° rotation, but * A)(2, 3) is under a reflection in the x–axis.


TEST PREP, PAGE 845

42. C(x, y) * ('y, x)('2, 5) * ('5, '2)

43. H 1 _ 3 (360) = 120°

44. 180('3, 4) * (3, '4) = ('('3), '(4))(x, y) * ('x, 'y): rotation of 180°


45. Gear B has 8 teeth. So one complete counterclockwise rotation of gear B will move gear A by 8 teeth clockwise. Gear A has 18 teeth. So 8

teeth is 4 __ 9 of a complete rotation, or 4 __

9 (360) = 160°

clockwise.

46. Draw auxiliary segs. $$

AP , $$

BP , $$$

A)P , and $$$

B)P . By the def. of rotation, $$

AP # $$$

A)P and $$

BP # $$$

B)P . Also by the def. of rotation, "A)PA # "B)PB. By the Common " Thm., "B)PA) # "BPA. Thus %B)PA) # %BPA by SAS, and

$$ AB #

$$$ A)B) by CPCTC.

47. Use the fact that the rotation of a segment is # to its preimage and the def. of # segs.

48. Draw auxiliary segs. $$

AC and $$$

A)C) . Use the fact that the rotation of a segment is # to its preimage to prove %ABC # %A)B)C) by SSS. By CPCTC, "ABC # "A)B)C). So m"ABC = m"A)B)C) by the def. of # !.

49. Use the fact that the rotation of a segment is # to its preimage to prove %ABC # %A)B)C) by SSS.

50. Since C is between A and B, AC + BC = AB. Use the fact that the rotation of a segment is # to its preimage to prove A)C) + B)C) = A)B). Then use the def. of betweenness to prove C) is between A) and B).

51. Since A, B, and C are collinear, one point is between other two. Case 1: If C is between A and B, then AC + BC = AB. Use the fact that the rotation of a segment is # to its preimage to prove A)C) + B)C) = A)B). Then C) is between A) and B). So A), B), and C) are collinear. Prove the other two cases similarly.


52. 3 = x 2 ' 4x + 70 = x 2 ' 4x + 40 = (x ' 2 ) 2 x = 2

53. 3 = 2 x 2 ' 5x ' 9 0 = 2 x 2 ' 5x ' 12 0 = (2x + 3)(x ' 4)

x = ' 3 __ 2 or 4

54. 3 = x 2 ' 25 = x 2 x = ± + , 5

55. QRYX is an isosceles trapazoid.m"XYR = 180 ' m"QRY

= 180 ' m"XQR= 180 ' 86 = 94°

56. QR ' PS ________ XY ' PS

= PQ ___ PX

= 2

QR ' PS = 2(XY ' PS) QR ' PS = 2XY ' 2PS QR = 2XY ' PS

= 2(4.2) ' (4)= 8.4 ' 4 = 4.4

57. A(1, 3) * D(5, '6) = D(1 + 4, 3 ' 9)translation vector: 24, '93

58. D(5, '6) * B(5, 0) = B(5 + 0, '6 + 6)translation vector: 20, 63

59. C('3, '2) * (0, 0) = ('3, + 3, '2, + 2)translation vector: 23, 23

12-3 TECHNOLOGY LAB: EXPLORE TRANSFORMATIONS WITH MATRICES, PAGES 846–847


1. [A] · [B] = 8

9 : '1 '2 '5 0 4 3

;

< =

a reflection across the y-axis

2. [A] · [B] = 8

9 : 0 4 3 1 2 5

;

< =

a reflection across the line y = x



3. [A] + [B] = 8

9 : '1 2 1 4 5 8

;

< =

a translation 1 unit left and 4 units up

4. Let [A] = 8

9 : a a a b b b

;

< = . Add [A] + [B] and use the

solution matrix to graph the image of the %.


5. [A] · [B] = 8

9 : '1 '4 '1 '1 '1 '2

;

< =

a 180° rotation about the origin

6. [A] · [B] = 8

9 : 1 1 2 '1 '4 '1

;

< =

a 270° rotation about the origin

12-4 COMPOSITIONS OF TRANSFORMATIONS, PAGES 848–853

CHECK IT OUT! PAGES 849–850 1. Step 1 The reflection image

of (x, y) is (x, 'y).J(1, '2) * J)(1, 2)K(4, '2) * K)(4, 2)L(3, 0) * L)(3, 0)Step 2 The rotation image of (x, y) is ('x, 'y).J)(1, 2) * J>('1, 2)K)(4, 2) * K>('4, 2)L)(3, 0) * L>('3, 0)Step 3 Graph the preimage and the image.

2. By Thm. 12-4-2, the composition of the two reflections across . lines is equivalent to a translation ( to these lines, n and p. By Thm. 12-4-2, the distance of translation is 2(3) = 6 in.

3. Step 1 Draw $$$

MM) and locate its midpoint X.Step 2 Draw the ( bisectors of

$$ MX and

$$$ XM) .

THINK AND DISCUSS, PAGE 850 1. Theorem 12-4-1; a composition of two isometries is

an isometry.

2. / v . !; possible answer; Translate the preimage along the vector then reflect the image across the line.

3.



1. Draw a figure and translate it along a vector. Then reflect the image across a line.

2. Step 1 Translate %DEF along / u .

Step 2 Reflect the image across line !.

3. Step 1 Reflect rectangle PQRS across line m.Step 2 Translate the image along

/ v .


4. Step 1 The reflection image of (x, y) is('x, y).A(1, '1) * A)('1, '1)B(4, '1) * B)('4, '1)C(3, 2) * C)('3, 2)Step 2 The translation image of (x, y) is (x, y ' 2).A)('1, '1) * A>('1, '3)B)('4, '1) * B>('4, '3)C)('3, 2) * C>('3, 0)Step 3 Graph the preimage and the image.

5. By Thm. 12-4-2, the composition of the two reflections across the intersecting lines is equivalent to a rotation about the point of intersection. By Thm. 12-4-2, the " of rotation is 2(50) = 100°.

6. Step 1 Draw $$

FF) and locate its midpoint X.Step 2 Draw the ( bisectors of

$$ FX and

$$ XF) .

7. Step 1 Draw "APA). Draw its " bisector ??@ PX .

Step 2 Draw " bisectors of "APX and "A)PX.


8. Step 1 Translate %RST along / u .

Step 2 Translate the image along / v .

9. Step 1 Rotate %ABC 90° around point P.Step 2 Reflect the image across the line !.

10. Step 1 The reflection image of (x, y) is (y, x).G(1, '1) * G)('1, 1)H(3, 1) * H)(1, 3)J(3, '2) * J)('2, 3)Step 2 The reflection image of (x, y) is (x, 'y).G)('1, 1) * G>('1, '1)H)(1, 3) * H>(1, '3)J)('2, 3) * J>('2, '3)Step 3 Graph the preimage and the image.

11a. The move is a horizontal or vertical translation by 2 spaces followed by a vertical or horizontal translation by 1 space.

b.

c.

12. Step 1 Draw $$

BB) and locate its midpoint X.Step 2 Draw the ( bisectors of

$$ BX and

$$ XB) .

13. Step 1 Draw "JQJ). Draw its " bisector

??@ QX .

Step 2 Draw " bisectors of "JQX and "J)QX.

14. Solution A is incorrect because the endpts. are not written in the same order as they are written for the preimage.


15.

16. sometimes (if reflection lines intersect)

17. never (by def. of isometry)

18. always (Thm. 12-4-1)

19. always (Th. 12-4-3)

20. Yes; the order matters, as shown in the figures.

21. R('3, '2) * ('3, 2) * R)(2, 2)S('1, '2) * ('1, 2) * S)(4, 2)T('1, 0) * ('1, 0) * T)(2, 0)(x, y) * (x, 'y); (x, y) * (x + 5, y)The line of reflection is the x-axis, and the translation vector is 25, 03.

22a. 21, 33 + 23, 13

b. Possible answer: 23, '33 + 24, 43 + 2'3, 33

TEST PREP, PAGE 853

23. A(x, y) * ('x, y); (x, y) * ('y, x)A(2, 1) * ('2, 1) * ('1, '2)

24. GThe rotation maps %ABC into 3rd quadrant; reflection maps image into 4th quadrant.

25. CAdd the two translation vectors to get the vector for the new translation.


26. A(3, 1) = A('1 + 4, 2 + ('1)) * ('1 ' ('1), 2 + 4)) = (0, 6);(0, 6) = (0, 5 + 1) * A)(0, 5 ' 1) = A)(0, 4)

27. Possible answer: Reflect %MNP across a horizontal line !. %M)N)P) is a translation of this image. This means there are two . lines such that the composition of the reflections across these lines is equivalent to the translation. These lines can be found as shown in the figure.

28. First reflection: (x, y) * (y ' 1, x + 1),second reflection: (x, y) * (y ' 3, x + 3),composition: (x, y) * (y ' 1, x + 1) * ((x + 1) ' 3, (y ' 1) + 3) = (x ' 2, y + 2)translation vector: 2'2, 23


29. yes (no x-coordinate is repeated)

30. no ('3 would be mapped to '1 and to 1)

31. 5(EJ) = 4(8) 5EJ = 32 EJ = 6.4

32. 5(5 + CD) = 4(4 + 12) 25 + 5CD = 64 5CD = 39 CD = 7.8

33. FH 2 = 4(4 + 16) FH 2 = 64 FH = 8

34. rotation sends (x, y) * ('y, x)F(2, 3) * F)('3, 2)

35. rotation sends (x, y) * ('x, 'y)N('1, '3) * N)(1, 3)

36. rotation sends (x, y) * ('y, x)Q('2, 0) * Q)(0, '2)


1. translation vector: / v = 25 ' 1, 1 ' 33 = 24, '23

0 / v 1 = + ,,,, 4 2 + 2 2 = + ,, 20 = 2 + , 5 - 4.5 m

2. (2, 4); First reflect H across $$

DC . Image is H)(5, 7). Then draw

$$$ TH) and find this segment’s intersection

with $$

DC .

3. path is 21, 13 + 23, '33

distance is + ,,,, 1 2 + 1 2 + + ,,,, 3 2 + 3 2 = + , 2 + + , 8 = + , 2 + 3 + , 2 = 4 + , 2 - 5.7 m

4. The turntable rotates 2 ___ 16

(360) = 45° in 2 s.

Coordinates of pillar: (4, 2) = (3 + 1, 2) * (3 +

+ , 2 ___ 2 , 2 +

+ , 2 ___ 2 ) - (3.7, 2.7)



1. Yes; the image appears to be flipped across a horizontal line.


3. Step 1 Through each vertex draw a line ( to the the line of reflection.Step 2 Measure the distance from each vertex to the line of reflection. Locate the image of each vertex on the opposite side of the line of reflection and the same distance from it.Step 3 Connect the images of the vertices.

4. Step 1 Through each vertex draw a line ( to line of reflection.Step 2 Measure distance from each vertex to line of reflection. Locate image of each vertex on opposite side of line of reflection and same distance from it.Step 3 Connect images of vertices.

5. No; not all points have moved the same distance.

6. Yes; all points have moved the same distance in the same direction.

7. The image of (x, y) is (x ' 4, y ' 3).(1, 0) * (1 ' 4, 0 ' 3) = ('3, '3)(4, 0) * (4 ' 4, 0 ' 3) = (0, '3)(4, 2) * (4 ' 4, 2 ' 3) = (0, '1)(1, 2) * (1 ' 4, 2 ' 3) = ('3, '1)Graph the image.



10. The rotation of (x, y) is ('x, 'y).P(0, 2) * P)(0, '2)Q(2, 0) * Q)('2, 0)R(3, '3) * R)('3, 3)

11. The rotation of (x, y) is ('y, x).A('1, 0) * A)(0, '1)B('1, '3) * B)(3, '1)C(1, '3) * C)(3, 1)D(1, 0) * D)(0, 1)

12.

13. A(1, 0) * ('1, 0) * A)('1, 0)B(1, 3) * ('1, 3) * B)('1, '3)C(2, 3) * ('2, 3) * C)('1, '3)(x, y) * ('x, y) * ('x, 'y)a rotation by 180° about the origin

12-5 SYMMETRY, PAGES 856–862

CHECK IT OUT! PAGES 856–858 1a. yes; two lines of

symmetry

b. yes; one line of symmetry

c. yes; one line of symmetry

2a. yes; 120°; order: 3 b. yes; 180°; order: 2

c. no rotational symmetry

3a. line symmetry and rotational symmetry; " of rotational symmetry: 72°; order: 5


b. line symmetry and rotational symmetry; ! of rotational symmetry: 51.4°; order: 7

4a. both; plane symmetry and symmetry about an axis

b. neither

THINK AND DISCUSS, PAGE 858 1. First, fold the paper in half. Cut out a design that

goes up the fold, then unfold paper.

2. The ! of rotational symmetry is 360° ____ n .

3.



1. The line of symmetry is " bisector of base.

2. line symmetry

3. yes; two lines of symmetry

4. yes; one line of symmetry

5. no line symmetry

6. yes; 180°; order: 2 7. no rotational symmetry

8. yes; 120°; order: 3

9. The ! of rotational symmetry is 72°; order: 5.

10. plane symmetry

11. both; plane symmetry and symmetry about an axis



13. yes; one line of symmetry

14. yes; three lines of symmetry

15. no line of symmetry

16. yes; 60°; order: 6 17. yes; 72°; order: 5

18. yes; 72°, order: 5.

19. The ! of rotational symmetry is 90°; order: 4.




21. neither


23. isosceles

24. equilateral

25. scalene

26. (area for x < 0) = (area for x > 0) (area for x < 0) + (area for x > 0)

= (area under whole curve)2(area for x > 0) = 1 (area for x > 0) = 0.5

27. (area for x > 3) - 0

28. P = (area for '1 < x < 1)

_____________________ (area under whole curve)

= 2(area for 0 < x < 1)

_________________ 1

= 2(0.34) = 0.68

29. line symmetry

30. line symmetry and rotational symmetry; 90°; order: 4

31. line symmetry

32. line symmetry and rotational symmetry; 180°; order: 2

33. rotational symmetry of order 4

34. line symmetry; x = 0

35. line symmetry; x = 2

36. rotational symmetry; 180°; order: 2

37a. No; there is no line that divides the woodcut into two identical reflected halves.

b. yes; 180°; order: 2

c. Yes; if color is not taken into account, the " of rotational symmetry is 90°, and the order is 4.

38. rectangle

39. parallelogram

40. rhombus

41. square

42. kite

43. The " of rotational symmetry is 1 ___ 24

(360) = 15°.


44. All rectangles have two lines of symmetry, one through midpoints of each opposite pair of sides. The only other possible lines of symmetry are through opposite vertices, which means that adjacent sides are #. Therefore, the rectangle must be a square.

45. rotational symmetry with an " of 120°, order 3

46. The figures all have a rotational symmetry of order 2.

47.

48.

49.

50. If the " of rotational symmetry is x °, then the

order of rotational symmetry is 360 ____ x . If the order

of rotational symmetry is x, then the " of rotational

symmetry is ( 360 ____ x ) °.TEST PREP, PAGE 862

51. A 52. J

53. C 54. H


55. If the polygon has n sides, then the " of rotational

symmetry is 360 ____ n .

360 ____ n = 5

n = 360 ____ 5 = 72

56. If n is even, the lines of symmetry pass through each pair of opposite vertices and each pair of opposite midpoints of sides.

There are n __ 2 + n __

2 = n lines of symmetry.

If n is odd, the lines of symmetry pass through a vertex and the midpoint opposite it. So there are n lines of symmetry.

57. The line of symmetry is vertical and passes through the vertex of parabola, ('4, 0). So the line of symmetry is x = '4.

58. The line of symmetry is vertical and passes through vertex of the graph, (2, 0). So the line of symmetry is x = 2.

59. The line of symmetry is vertical and passes through the vertex of the parabola, (0, 5). So the line of symmetry is x = 0.

60. 13; a line through the centers of each pair of opposite faces (3 lines), a line through the midpoints of each pair of opposite edges (6 lines), and a line through each pair of opposite vertices (4 lines).

61. 8; a line through each vertex and the center of the opposite face (4 lines) and a line through the midpoints of each pair of opposite edges (4 lines).

62. 13; a line through the centers of each pair of opposite faces (4 lines), a line through the midpoints of each pair of opposite edges (6 lines), and a line through each pair of opposite vertices (3 lines).


63. E ______ 197.12

= 20 ___ 16

E = 20 ___ 16

(197.12) = $246.40

64. S = # r 2 + #r!61# = #(5 ) 2 + #(5)!61# = 25# + 5#!36# = 5#! 36 = 5! ! = 7.2 in.

65. Step 1 Find the base edge length. S = B + L65.25 = s 2 + 4520.25 = s 2 s = 4.5 cmStep 2 Find the slant height. L = 1 _

2 P!

45 = 1 _ 2 (18)! = 9!

! = 5 cm

66. Step 1 Find the base area.b = 1 _

3 P = 8 + , 3 m

h = (4 + , 3 ) + , 3 = 12 mB = 1 _

2 bh = 1 _

2 (8 + , 3 ) (12) = 48 + , 3 m 2

Step 2 Find the slant height. S = B + L120 + , 3 = 48 + , 3 + 1 _

2 P!

72 + , 3 = 1 _ 2 (24 + , 3 ) !

72 = 12! ! = 6 m

67. P('1, 4) * (4, '1) * P)(4 + 2, '1 ' 4) = P)(6, '5)

68. P('1, 4) * ('4, '1) * P)(4, '1)

69. P('1, 4) * ('1 + 1, 4) = (0, 4) * P)(0, '4)

12-6 TESSELLATIONS, PAGES 863–869

CHECK IT OUT! PAGES 863–865 1a. translation symmetry

b. translation symmetry and glide reflection symmetry


2. Step 1 Rotate the quadrilateral 180° about the midpoint of one side.

Step 2 Translate the resulting pair of quadrilaterals to make a row of quadrilaterals.

Step 3 Translate the row of quadrilaterals to make

a tessellation.

3a. Only hexagons are used. The tessellation is regular.

b. Different vertices have different arrangements of polygons. The tessellation is neither regular nor semiregular.

c. Two hexagons and two A meet at each vertex. The tessellation is semiregular.

4a. Yes; three hexagons meet at each vertex.120° + 120° + 120° = 360°

b. No; no combination of 90° and 120° ! is equal to 360°.

THINK AND DISCUSS, PAGE 866 1. In a pattern that has glide reflection symmetry, you

go from one figure to the next by a composition of a translation and a reflection.

2. It is not possible. No matter how circles are arranged to cover a plane, there will always be overlaps or gaps between them.

3.



1. Possible answer:

2. Possible answers: checkerboard, hexagonal floor tiles, honeycomb

3. translation symmetry, and glide reflection symmetry



6. Step 1 Rotate % 180° about the midpoint of one side.

Step 2 Translate the resulting pair of A to make a row of A.

Step 3 Translate the row of A to make a tessellation.

7. Step 1 Translate the rectangle to make a row of rectangles.

Step 2 Translate the row of rectangles to make a tessellation.


8. Step 1 Rotate the kite 180° about the midpoint of one side.

Step 2 Translate the resulting pair of kites to make a row of kites.

Step 3 Translate the row of kites to make a tessellation.

9. Only hexagons are used. The tessellation is regular.

10. Irregular quadrilaterals (rectangles) are used. The tessellation is neither regular nor semiregular.

11. Two hexagons and two A meet at each vertex. The tessellation is semiregular.

12. No; each " of the octagon measures 135°, and 135 is not a divisor of 360.

13. Yes; possible answer: two hexagons and two A meet at each vertex.120° + 120° + 60° + 60°= 360°

14. No; no combination of 90° and 72° ! is = to 360°.


15. translation symmetry

16. translation symmetry and glide reflection symmetry

17. translation symmetry

18. Step 1 Rotate quadrilateral 180° about the midpoint of one side.

Step 2 Translate the resulting pair of quadrilaterals

to make a row of quadrilaterals

Step 3 Translate the row of quadrilaterals to make a tessellation.

19. Step 1 Rotate the % 180° about the midpoint of one side.

Step 2 Translate the resulting pair of A to make a

row of A.






21. Different vertices have different arrangements of polygons. The tessellation is neither regular nor semiregular.

22. Two hexagons, two squares, and one % meet at each vertex. The tessellation is semiregular.

23. Irregular pentagons are used. The tessellation is neither regular nor semiregular.

24. No; each " of the heptagon measures

5 __ 7

(180) - 128.6°, which is not a divisor of 360.

25. No; no combination of 60° and 135° ! is = to 360°.

26. Yes; two hexagons, two squares, and one % meet at each vertex.2(120°) + 2(90°) + 60° = 360°

27. translation and glide reflection symmetry

28. translation

29. translation, glide reflection, rotation

30. translation, glide reflection, rotation

31. always 32. sometimes

33. always 34. never

35. never

36a. equilateral % 36b. parallelogram

37.

38.

39.

40.

41.

The tessellation has translation symmetry, reflection symmetry, and order 3 rotation symmetry.

42. No, because the interior " measures of a regular pentagon and a regular hexagon are 120° and 108°, and there is no possible arrangement of both

these " measures that adds to 360°.

43.

44.

45. Equilateral A, squares, and regular hexagons are the only regular polygons with interior " measures that divide evenly into 360°.

TEST PREP, PAGE 869

46. B 47. H

48. D2(90°) + 3(60°) = 360°Two squares and three A fit around vertex.



50. No; you cannot fit cylinders together in any way without creating gaps or overlaps.

51. Each prism is made up of quadrilaterals and since any quadrilateral can be used to tessellate the plane, the prism will tessellate three-dimensional space.

52. Each prism is made up of quadrilateral and since any quadrilateral can be used to tessellate the plane, the prism will tessellate three-dimensional space.


53. Let r be the tax rate.8.00(0.85)(1 + r) + 2.69 = 10.00 6.8(1 + r) = 7.71 1 + r = 1.075 r = 0.075 = 7.5%

54. Let m be the minutes before 7:00 AM

Louis’ jogging time is 5 mi ______ 6 mi/h

(60 min/h) = 50 min

Andrea’s jogging time is 3.9 mi _______ 6.5 mi/h

(60 min/h) = 36 min50 = 36 + mm = 14Louis should leave home 14 min before 7:00 AM, or at 6:46 AM

55. (x ' ('2) ) 2 + (y ' 3 ) 2 = ( + , 5 ) 2 (x + 2 ) 2 + (y ' 3 ) 2 = 5

56. x 2 + y 2 = r 2 3 2 + 4 2 = r 2 25 = r 2 x 2 + y 2 = 25

57. (x ' 5 ) 2 + (y + 3 ) 2 = r 2 (1 ' 5 ) 2 + ('1 + 3 ) 2 = r 2 20 = r 2 (x ' 5 ) 2 + (y + 3 ) 2 = 20


58. no rotational symmetry

59. " of rotational symmetry: 72°; order: 5

60. " of rotational symmetry: 180°; order: 2

12-6 GEOMETRY LAB: USE TRANSFORMATIONS TO EXTEND TESSELLATIONS, PAGES 870–871

ACTIVITY 1, TRY THIS, PAGE 870 1–3. Check students’ work.

ACTIVITY 2, TRY THIS, PAGE 871 4. Check students’ work.

5. Possible answer: This tessellation has rotational and translation symmetry, but the tessellation from Activity 1 has only translation symmetry.


12-7 DILATIONS, PAGES 872–879

CHECK IT OUT! PAGES 872–874 1a. No; the figures are not similar.

1b. Yes; the figures are similar and the image is not turned or flipped.

2. Step 1 Draw a ray through Q and each vertex.Step 2 On each ray, mark three times the distance from Q to each vertex.Step 3 Connect the vertices of the image.

3. Scale factor is 4. So a 1 in. by 1 in. square on the photo represents a 4 in. by 4 in. square on the painting.Find the dimensions of the painting.s = 4(10) = 40 in.Find the area of the painting.A = s 2 = (40 ) 2 = 1600 in. 2

4. The dilation of (x, y) is (' 1 _ 2 x, ' 1 _

2 y) .

R(0, 0) * R) (' 1 _ 2 (0), ' 1 _

2 (0)) = R)(0, 0)

S(4, 0) * S) (' 1 _ 2 (4), ' 1 _

2 (0)) = S)('2, 0)

T(2, '2) * T) (' 1 _ 2 (2), ' 1 _

2 ('2)) = T)('1, 1)

U('2, '2) * U) (' 1 _ 2 ('2), ' 1 _

2 ('2)) = U)(1, 1)

Graph the preimage and the image.

THINK AND DISCUSS, PAGE 874 1. Measure the length of one side of the image and the

length of the corresponding side of the preimage. Form the ratio of these two lengths to find the scale factor.

2. dilation by a scale factor of 'k

3.



1. The center is the origin; the scale factor is 3.

2. Yes; the figures are similar and the image is not turned or flipped.


4. No; the figures are not similar.


6. Step 1 Draw a ray through P and each vertex.Step 2 On each ray, mark two times the distance from P to each vertex.Step 3 Connect vertices of the image.

7. Step 1 Draw a ray through P and each vertex.Step 2 On each ray,

mark 1 __ 2 the distance

from P to each vertex.Step 3 Connect the vertices of the image.

8. The scale factor of the dilation from the room to

the blueprint is 1 ___ 50

.

Therefore, the scale factor of the dilation from the blueprint to the room is 50. So a 1 in. by 1 in. square on the blueprint represents a 50 in. by 50 in. square in the room.Find the dimensions of the room.! = 50(8) = 400 in.w = 50(6) = 300 in.Find perimeter of room.A = 2! + 2w

= 2(400) + 2(300) = 1400 in. = 116 ft 8 in.


9. The dilation of (x, y) is (2x, 2y).A(1, 0) * A)(2(1), 2(0)) = A)(2, 0)B(2, 2) * B)(2(2), 2(2)) = B)(4, 4)C(4, 0) * C)(2(4), 2(0)) = C)(8, 0)Graph the preimage and the image.

10. Dilation of (x, y) is ( 1 _ 2 x, 1 _

2 y) .

J('2, 2) * J) ( 1 _ 2 ('2), 1 _ 2 (2)) = J)('1, 1)

K(4, 2) * K) ( 1 _ 2 (4), 1 _ 2 (2)) = K)(2, 1)

L(4, '2) * L) ( 1 _ 2 (4), 1 _ 2 ('2)) = L)(2, '1)

M('2, '2) * M) ( 1 _ 2 ('2), 1 _ 2 ('2)) = M)('1, '1)



3 y) .

D('3, 3) * D) (' 1 _ 3 ('3), ' 1 _ 3 (3)) = D)(1, '1)

E(3, 6) * E) (' 1 _ 3 (3), ' 1 _ 3 (6)) = E)('1, '2)

F(3, 0) * F) (' 1 _ 3 (3), ' 1 _ 3 (0)) = F)('1, 0)


12. The dilation of (x, y) is ('2x, '2y).P('2, 0) * P)('2('2), '2(0)) = P)(4, 0)Q('1, 0) * Q)('2('1), '2(0)) = Q)(2, 0)R(0, '1) * R)('2(0), '2('1)) = R)(0, 2)S('3, '1) * S)('2('3), '2('1)) = S)(6, 2)Graph the preimage and the image.




15. No; the image is flipped.


17. Step 1 Draw a ray through P and each vertex.Step 2 On each ray, mark 3 times the distance from P to each vertex.Step 3 Connect the vertices of the image.

18. Step 1 Draw a ray through P and each vertex.Step 2 On each ray, mark one half the distance from P to each vertex.Step 3 Connect the vertices of the image.

19. Scale factor is 1.5, so a 1 cm by 1 cm square on photo represents a 1.5 cm by 1.5 cm square on enlargement.Find dimensions of enlargement.! = 1.5(6) = 9 cmw = 1.5(8) = 12 cm# of tiles = area of enlargement

= !w = (9)(12) = 108 tiles


3 y) .

M(0, 3) * M) (' 1 _ 3 (0), ' 1 _ 3 (3)) = M)(0, '1)

N(6, 0) * N) (' 1 _ 3 (6), ' 1 _ 3 (0)) = N)('2, 0)

P(0, '3) * P) (' 1 _ 3 (0), ' 1 _ 3 ('3)) = P)(0, 1)



21. The dilation of (x, y) is ('x, 'y).A('1, 3) * A)('('1), '(3)) = A)(1, '3)B(1, 1) * B)('(1), '(1)) = B)('1, '1)C('4, 1) * C)('('4), '(1)) = C)(4, '1)Graph the preimage and the image.

22. The dilation of (x, y) is ('2x, '2y).R(1, 0) * R)('2(1), '2(0)) = R)('2, 0)S(2, 0) * S)('2(2), '2(0)) = S)('4, 0)T(2, '2) * T)('2(2), '2('2)) = T)('4, 4)U('1, '2) * U)('2('1), '2('2)) = U)(2, 4)Graph the preimage and the image.

23. The dilation of (x, y) is (' 1 _ 2 x, ' 1 _ 2 y) .D(4, 0) * D) (' 1 _ 2 (4), ' 1 _ 2 (0)) = D)('2, 0)

E(2, '4) * E) (' 1 _ 2 (2), ' 1 _ 2 ('4)) = E)('1, 2)

F('2, '4) * F) (' 1 _ 2 ('2), ' 1 _ 2 ('4)) = F)(1, 2)

G('4, 0) * G) (' 1 _ 2 ('4), ' 1 _ 2 (0)) = G)(2, 0)

H('2, 4) * H) (' 1 _ 2 ('2), ' 1 _ 2 ('4)) = H)(1, '2)

J(2, 4) * J) (' 1 _ 2 (2), ' 1 _ 2 (4)) = J)('1, '2)


24. %FGH & %KLM 25. ABCDE & MNPQR

26. Find the the dimensions of the image.! = 4(5) = 20 cmw = 4(2) = 8 cmh = 4(3) = 12 cmFind the surface area and the volume.S = L + B

= (2! + 2w)h + 2!w= (2(20) + 2(8))(12) + 2(20)(8)= 56(12) + 40(8) = 992 cm 2

V = Bh= !wh= (20)(8)(12) = 1920 cm 3

27.

28.

29.

30a. Let the scale factor be k.poster length = k · original length 82.8 = k(27.6) k = 3

b. A = !w= 82.8(3 · 19.9)= 4943.16 cm 2

31. Solution B is incorrect. The area of the rectangle A)B)C)D) is (2.5 ) 2 · 6 = 37.5.

32a. 8 = k(6)

k = 1 1 __ 3

b. before dilation:A = #(3 ) 2 = 9# - 28.3 mm 2 after dilation:A = #(4 ) 2 = 16# - 50.3 mm 2

c. light increases by a factor m, where50.3 - (1 + m)(28.3)1.78 - 1 + m m = 0.78 = 78%

33a. Possible answer: 2.5. Check students’ estimates.

b. Measure $$

AB and $$$

A)B) , then calculate A)B) ____ AB

.

c. 16 ___ 6.5

- 2.5

34a. The dilation of (x, y) is (2x, 2y).A('1, 1) * A)('2, 2) * A>('2, '2)B(2, 1) * B)(4, 2) * B>(4, '2)C(2, 2) * C)(4, 4) * C>(4, '4)


b. The dilation of (x, y) is (2x, 2y).A('1, 1) * A)('1, '1) * A>('2, '2)B(2, 1) * B)(2, '1) * B>(4, '2)C(2, 2) * C)(2, '2) * C>(4, '4)

c. The images are the same. The order of transformations does not matter.

35. scale factor = '0.25 in. __________ 870,000 mi

= '0.25 in. ___________________________ (870,000 mi)(5280 ft/mi)(12 in./ft)

- '4.5 B 1 0 '12

36. A)(k('2), k(2)) = A)('4, 4) k = 2B(1, 3) * B)(2, 6)C(1, '1) * C)(2, '2)

37. C)(k(1), k('1)) = C)('2, 2) k = '2A('2, 2) * A)(4, '4)B(1, 3) * B)('2, '6)

38. B)(k(1), k(3)) = B) ('1, '3) k = '1A('2, 2) * A)(2, '2)C(1, '1) * C)('1, 1)

39. For k = 1, the image and the preimage are the same figure. For k = '1, the dilation is equivalent to a 180° rotation. A rotation is an isometry. So the image is # to the preimage.

40. A dilation is equivalent to a 180° rotation when the scale factor is '1. In this case, the image has the same size as the preimage. So the only effect of transformation is the rotation by 180°.

41. Yes; first dilation multiplies all the linear measures of the preimage by m. The second dilation multiplies all the linear measures of the image by n. The overall effect is to multiply all the linear measures by mn, which is equivalent to a single dilation with the scale factor mn.


TEST PREP, PAGE 878

46. BD = D(0, 2), so D)(0, 2k) = D)(0, '2); k = '1

47. Horiginal dimensions: ! = 4, w = 2dilated dimensions: ! = 2.5(4) = 10, w = 2.5(2) = 5P = 2! + 2w = 2(10) + 2(5) = 30

48. 4.2(k('2), k(3)) = ('8.4, 12.6) '2k = '8.4 k = 4.2

49. No; the dimensions of the enlargement are 1.5(6) = 9 cm by 1.5(8) = 12 cm; A = 9(12) = 108 cm 2


50a. A(0, 2) = A(2 + ('2), 2 + 0) * A)(2 + 2('2), 2 + 2(0)) = A)('2, 2)B(1, 2) = B(2 + ('1), 2 + 0) * B)(2 + 2('1), 2 + 2(0)) = B)(0, 2)C(1, 0) = C(2 + ('1), 2 + ('2)) * C)(2 + 2('1), 2 + 2('2)) = C)(0, '2)D(0, 0) = D(2 + ('2), 2 + ('2)) * D)(2 + 2('2), 2 + 2('2)) = D)('2, '2)

b. The transformation is the composition of the dilation centered at the origin with the scale factor two followed by the translation along the vector 2'2, '23.

c. The transformation is the composition of the dilation centered at the origin with the scale factor k followed

by the translation along the vector 2'a, 'b3.

51. The dilation sends (x, y) * (3x, 3y).For (x, y) on line !, y = 'x + 2(3y) = '(3x) + 6So the image point (3x, 3y) lies on the line y = 'x + 6; therefore this line is the image of ! after dilation.



52. The data fit the equation y = 1.6x ' 4. For $68 in tips,68 = 1.6x ' 472 = 1.6x x = 45Jerry would need to serve 45 customers.

53. P = JK + KL + LM + JM

= + ,,,, 3 2 + 4 2 + (7 ' 0) + + ,,,, 3 2 + 4 2 + (4 ' ('3)) = 5 + 7 + 5 + 7 = 24 unitsJKLM is a C with b = JM = 7 and h = 2 ' ('2) = 4.A = bh = 7(4) = 28 units 2

54. P = DE + EF + DF

= + ,,,, 4 2 + 2 2 + + ,,,, 2 2 + 6 2 + + ,,,, 2 2 + 4 2 = 2 + , 5 + 2 + ,, 10 + 2 + , 5 = (4 + , 5 + 2 + ,, 10 ) units $$

DE and $$

DF are (, so %DEF is a right % with

b = DE = 2 + , 5 and h = DF = 2 + , 5 .

A = 1 _ 2 bh = 1 _

2 (2 + , 5 ) (2 + , 5 ) = 10 units 2

55. Yes; for example, you can fit two squares and two right A around a vertex.

56. No; the internal ! are 7(180)

______ 9 = 140° and 60°, and

no combination of these ! is = to 360°.

USING TECHNOLOGY, PAGE 879

2. 8

9 : 6 10 2 8 4 2

;

< =

3. 8

9 : ' 1 _ 2 0

0 ' 1 _

2 ;

< =

4. [A] · [B] = 8

9 : '1.5 '2.5 '0.5 '2 '1 '0.5

;

< =


1. A: none; B: .; C: none; D: intersecting; E: none

2. A: no; B: no; C: yes, 60°, order: 6; D: yes, 120°, order: 3; E: no

3. A: yes, C; B: yes, parallelogram; C: no; D: yes, equilateral %; E: no

4. 21,098.88 = k 2 · (13.2)(11.1) = k 2 · 146.52 k 2 = 144 k = 12


1. yes

2. yes

3. no 4. yes; 180°; order: 2





8. Step 1 Rotate the trapezoid 180° about the midpoint of one side.

Step 2 Translate the resulting pair of trapezoids to make a row of trapezoids.

Step 3 Translate the row of trapezoids to make a tessellation.





10. Only the regular hexagons are used. The tessellation is regular.

11. Irregular quadrilaterals (rhombuses) are used. The tessellation is neither regular nor semiregular.

12. Two squares and three equilateral A meet at each vertex. The tessellation is semiregular.

13. No; each internal " of a regular octagon measures 135°, which is not a divisor of 360°.




17. The dilation sends (x, y) to (2x, 2y).A(0, 2) * A)(2(0), 2(2)) = A)(0, 4)B('1, 0) * B)(2('1), 2(0)) = B)('2, 0)C(0, '1) * C)(2(0), 2('1)) = C)(0, '2)D(1, 0) * D)(2(1), 2(0)) = D)(2, 0)

18. The dilation sends (x, y) to (' 1 _ 2 x, ' 1 _

2 y) .

P('4, '2) * P) (' 1 _ 2 ('4), ' 1 _

2 ('2)) = P)(2, 1)

Q(0, '2) * Q) (' 1 _ 2 (0), ' 1 _

2 ('2)) = Q)(0, 1)

R(0, 0) * R) (' 1 _ 2 (0), ' 1 _

2 (0)) = R)(0, 0)

S('4, 0) * S) (' 1 _ 2 ('4), ' 1 _

2 (0)) = S)(2, 0)


VOCABULARY, PAGE 884 1. regular tessellation 2. frieze pattern

3. isometry 4. composition of transformations

LESSON 12-1, PAGE 884 5. Yes; the image appears to be flipped.

6. No; the image appears to be shifted as well as flipped.

7. No; the figure appears to be turned.

8. Yes; the image appears to be flipped.

9. The image of (x, y) is (x, 'y).E('3, 2) * E)('3, '2)F(0, 2) * F)(0, '2)G('2, 5) * G)('2, '5)

10. The image of (x, y) is ('x, y).J(2, '1) * J)('2, '1)K(4, '2) * K)('4, '2)L(4, '3) * L)('4, '3)M(2, '3) * M)('2, '3)


11. The image of (x, y) is (y, x).P(2, '2) * P)('2, 2)Q(4, '2) * Q)('2, 4)R(3, '4) * R)('4, 3)

12. The image of (x, y) is (y, x).A(2, 2) * A)(2, 2)B('2, 2) * B)(2, '2)C('1, 4) * C)(4, '1)

LESSON 12-2, PAGE 885 13. No; the image is smaller than the preimage.

14. Yes; the image appears to be # to the preimage.

15. No; the image is flipped.

16. No; the image is smaller than the preimage.

17. R(1, '1) * R)(1 ' 5, '1 + 2) = R)('4, 1)S(1, '3) * S)(1 ' 5, '3 + 2) = S)('4, '1)T(4, '3) * T)(4 ' 5, '3 + 2) = T)('1, '1)U(4, '1) * U)(4 ' 5, '1 + 2) = U)('1, 1)

18. A('4, '1) * A)('4 + 6, '1) = A)(2, 1)B('3, 2) * B)('3 + 6, 2) = B)(3, 2)C('1, '2) * C)('1 + 6, '2) = C)(5, '2)

19. M(1, 4) * M)(1 ' 3, 4 ' 3) = M)('2, 1)N(4, 4) * N)(4 ' 3, 4 ' 3) = N)(1, 1)P(3, 1) * P)(3 ' 3, 1 ' 3) = P)(0, '2)

20. D(3, 1) * D)(3 ' 6, 1 + 2) = D)('3, 3)E(2, '2) * E)(2 ' 6, '2 + 2) = E)('4, 0)F(3, '4) * F)(3 ' 6, '4 + 2) = F)('3, '2)G(4, '2) * G)(4 ' 6, '2 + 2) = G)('2, 0)

LESSON 12-3, PAGE 885 21. Yes; the image appears to be turned.

22. Yes; the image appears to be turned.


24. No; the image appears to be flipped.

25. (x, y) * ('y, x)A(1, 3) * A)('3, 1)B(4, 1) * B)('1, 4)C(4, 4) * C)('4, 4)

26. (x, y) * ('x, 'y)A(1, 3) * A)('1, '3)B(4, 1) * B)('4, '1)C(4, 4) * C)('4, '4)


27. (x, y) * ('y, x)M(2, 2) * M)('2, 2)N(5, 2) * N)('2, 5)P(3, '2) * P)(2, 3)Q(0, '2) * Q)(2, 0)

28. (x, y) * ('x, 'y)G('2, 1) * G)(2, '1)H('3, '2) * H)(3, 2)J('1, '4) * J)(1, 4)

LESSON 12-4, PAGE 886 29. Step 1 Translate ABCD along

/ v .

Step 2 Reflect the image across line m.

30. Step 1 Reflect %JKL across line m.Step 2 Rotate the image around P.

LESSON 12-5, PAGE 886 31. yes

32. yes

33. yes; 120°; order: 3 34. no


LESSON 12-6, PAGE 887 37. Step 1 Rotate % 180° about the midpoint of one

side.













41. Irregular pentagons are used. The tessellation is neither regular nor semiregular.

42. Two regular hexagons and two equilateral A meet as each vertex. The tessellation is semiregular.

LESSON 12-7, PAGE 887 43. Yes; the figures appear to be similar, and the image

is not flipped or turned.

44. Yes; the figures appear to be similar, and the image is not flipped or turned.

45. The dilation sends (x, y) to (' 1 _ 2 x, ' 1 _ 2 y) .R(0, 0) * R) (' 1 _ 2 (0), ' 1 _ 2 (0)) = R)(0, 0)

S(4, 4) * S) (' 1 _ 2 (4), ' 1 _ 2 (4)) = S)('2, '2)

T(4, '4) * T) (' 1 _ 2 (4), ' 1 _ 2 ('4)) = T)('2, 2)

46. The dilation sends (x, y) to ('2x, '2y).D(0, 2) * D)('2(0), '2(2)) = D)(0, '4)E('2, 2) * E)('2('2), '2(2)) = E)(4, '4)F('2, 0) * F)('2('2), '2(0)) = F)(4, 0)


1. No; the image appears to be shifted, not flipped.

2. Yes; the figures are # and the image is not flipped or turned.


4. Yes; the figures are similar and the image is not flipped or turned.

5. (1, 3) * (1 ' 1, 3 ' 1) = (0, 2)(2, 2) * (2 ' 1, 2 ' 1) = (1, 1)(5, 5) * (5 ' 1, 5 ' 1) = (4, 4)(4, 6) * (4 ' 1, 6 ' 1) = (3, 5)

6. Yes; the figures appear # and the image appears to be turned.

7. Yes; the figures appear # and the image appears to be turned.

8. (x, y) * ('x, 'y)D(1, '1) * D)('1, 1)E(4, '1) * E)('4, 1)F(4, '3) * F)('4, 3)G(1, '3) * G)('1, 3)

9. Rectangle ABCD with vertices A(3, '1), B(3, '2), C(1, '2), and D(1, '1) is reflected across the y–axis, and then its image is reflected across the x–axis. Describe a single transformation that moves the rectangle from its starting position to its final position.

10. yes

11. yes; 180°; order: 2











15. One square and two regular octagons meet at each vertex. The tessellation is semiregular.

16. Yes; the figures are similar, and the image is not turned or flipped.


18. The dilation sends (x, y) to (' 1 _ 2 x, ' 1 _

2 y) .

A(2, '1) * A) (' 1 _ 2 (2), ' 1 _

2 ('1)) = A)('1, 0.5)

B(1, '4) * B) (' 1 _ 2 (1), ' 1 _

2 ('4)) = S)('0.5, 2)

C(4, '4) * C) (' 1 _ 2 (4), ' 1 _ 2 ('4)) = C)('2, 2)


1. Af('x) = ('x ) 4 ' 2 = x 4 ' 2 = f(x)

2. G(1, '3) * ('1, '7) = (1 ' 2, '3 ' 4)('4, 5) * ('4 ' 2, 5 ' 4) = ('6, 1)

3. D(x, y) * (x, 'y)('2, '5) * ('2, 5)

4. HA(1, 4) * ('1, 4) * ('1, 4 ' 6) = C('1, '2)B(4, 2) * ('4, 2) * ('4, 2 ' 6) = D('4, '4)

5. C







= 2.25(0.8) = 1.8 mi





c. T









n _ 28,800

= 7 _ 18,327

n = 7 ______ 18,327

(28,800) ! 11 mo





2 (4) = 2 in.; base is b = 6 in.;


6 (40 % 2 & ' 3 )





4 & '' 9.8 ! 12.52 cm.


438.4 ! 20t t ! 22 s



3. Let P and Q be pts. on ))

AD directly below B and C. Then AP = PB = 310 ft, PQ = BC = 60 ft, and QD satisfies Q D 2 + Q C 2 = C D 2 Q D 2 + 310 2 = 314.8 2 Q D 2 = 2999.04 QD = & '''' 2999.04 ! 54.8 ft AD = AP + PQ + QD ! 310 + 60 + 54.8 ! 425 ft

4. XY = 1 _ 2 (AD + BC)

! 1 _ 2 (424.8 + 60) ! 242 ft


THE JOHN HANCOCK CENTER, PAGE 582 1. By Alt. Int. * Thm., height and horiz. dist. x are opp.

and adj. sides for 10° #.tan 10° = 1000 _____ x

x = 1000 ______ tan 10°

! 5671 ft

2. tan 61° = h _____ 818.2

h = 818.2 tan 61° ! 1476 ft

3. tan 39° = (818.2 tan 61°)

____________ x

x = 818.2 tan 61° ___________ tan 39°

! 1823 ft

4. Shadow is longest when # of elevation is smallest, on Dec 15.

tan 25° = (818.2 tan 61°)

____________ x

x = 818.2 tan 61° ___________ tan 25°

! 3165 ft

ERNEST HEMINGWAY’S BIRTHPLACE , PAGE 583 1. perim. of dining room on plan ! 3.5 in.

3.5 ___________ actual length

! 1 ___ 16

actual length ! 16(3.5) ! 56 ft

2. area of parlor and living room on plan ! 1 1 __ 8 i n. 2

1.125 __________ actual area

! ( 1 ___ 16

) 2 = 1 ____ 256

actual area ! 256(1.125) = 288 f t 2

3. ! = w + 4 and 2! + 2w = 402(w + 4) + 2w = 40 4w + 8 = 40 4w = 32 w = 8 ! = 8 + 4 = 12plan dimensions are 12 ___

16 = 3 __

4 in. by 8 ___

16 = 1 __

2 in.


THE MELLON ARENA, PAGE 581 1. The area is a circle with a diameter of 400 ft.

area in square feet: A = " r 2 = "(200 ) 2 ! 126,000 ft 2 Area in acres: A = 126,000 ft 2 · 1 acre ________

43,560 ft 2 ! 3 acres

2. ! __ w = 40 ___ 17

! = 40 ___ 17

w

P = 2! + 2w

570 = 2 ( 40 ___ 17

w) + 2w

570(17) = 80w + 34w 9690 = 114w w = 85 ft! = 40 ___

17 (85) = 200 ft

The dimensions are 200 ft by 85 ft.

3. ! = w + 130 P = 2! + 2w740 = 2(w + 130) + 2w740 = 2w + 260 + 2w480 = 4w w = 120 ft ! = (120) + 130 = 250 ftThe dimensions are 250 ft by 120 ft.

4. Step 1 Find the probability of sitting under the fixed sections.area under fixed sections: A = 2 _

8 "(200 ) 2 = 10,000" ft 2

total area: A = "(200 ) 2 = 40,000" ft 2

P = 10,000"

_______ 40,000"

= 1 __ 4

Step 1 Find the probability of sitting under the open sky.area under open sky: A = 40,000" % 10,000" = 30,000" ft 2

P = 30,000"

_______ 40,000"

= 3 __ 4

THE U.S. MINT, PAGE 582 1. Assume the quarters are stamped out in a

rectangular grid pattern, with each quarter occupying a 1-in. square. Then the number of quarters that can be stamped out of each strip equals the area of the strip in square inches.# quarters = A = !w = (13 in.) (1500 ft · 12 in. _____

1ft )

! 234,0002(234,000) < 700,000 < 3(234,000)So, for 700,000 quarters, 3 strips are needed.

2. V penny = " r 2 h = "(9.525 ) 2 (1.55) ! 442 mm 3

% (copper) = V copper

______ V penny

· 100% ! 11 ____ 442

· 100% ! 2.5%


3. Assume that all the metal is used up in making the nickels. V nickel = "(10.605 ) 2 (1.95) $ 689 mm 3 V cube = (1000 ) 3 = 1 ( 1 0 9 mm 3

# nickels $ V cube

_____ V nickel

$ 1 ( 1 0 9 _______ 689

$ 1.45 ( 10 6 , or 1.45 million

4. The plastic forms the surface area of a cylinder with a height of 50 dimes.S = 2"rh + 2" r 2

= 2"(8.955) (50(1.35)) + 2"(8.955 ) 2 $ 4302 mm 2


SANDY HOOK LIGHTHOUSE, PAGE 894

1. A = " r 2 ( m ____ 360

) = "(19 ) 2 ( 60 ____ 360

) = 361 ____ 6 " $ 189 mi 2

2. Let the distance from the top of the tower to the horizon be x mi.

4000 2 + x 2 = (4000 + 85 _____ 5280

) 2

16,000,000 + x 2 = 16,000,128.79 x 2 = 128.79 x $ 11.3 mi

3. 8.6(8.6) = 2(d ' 2) 73.96 = 2d ' 4 77.96 = 2d d $ 39 in. or 3 ft 6 in.The order of the lens is third.

MOVEABLE BRIDGES, PAGE 895 1. rotation; 14.1 ft

The rotation forms a right triangle, whose legs are each 10 ft. Thus, by Pythagorean Theorem: c 2 =1 0 2 + 1 0 2 c = " ##### 100 + 100 = " ## 200 $ 14.1

2. speed = 138 ft ' 35 ft ___________ 2 min

= 51.5 ft/min

3. time = 10 ft _________ 51.5 ft/min

$ 0.194 min $ 12 s

4. time = 135 ft ' 49 ft ___________ 51.5 ft/min

$ 1.670 min $ 100 s

5. height = 151 sin 20° $ 51.6 ft


Holt Geometry. Complete Solutions Manual (2007)

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Transcript of Holt Geometry. Complete Solutions Manual (2007)