CHAPTER 9 - · PDF filex 8 3 2 192 x2 256 2x 16 21 x 441 x2 841 20 2x 29 13 x 13 x 9 x 2 23...

CHAPTER 9

182 GeometryChapter 9 Worked-out Solution Key

Copyright © McDougal Littell Inc. All rights reserved.

Lesson 9.1

Think and Discuss (p. 525)

1. Since angle B and angle D are both right angles, theskyscrapers must be parallel.

2. According to Theorem 4.8, if the hypotenuse and leg ofone right triangle are congruent to the hypotenuse andcorresponding leg of another right triangle, the twotriangles are congruent.

Skill Review (p. 526)

1.

is a right triangle.

2. 3.

4.

Developing Concepts Activity (p. 527)

3. All of the triangles are similar.

9.1 Guided Pratice (p. 531)

1. geometric mean 2. 3. MK

4. JM 5. JK 6. LJ 7. KM 8. KM 9. LK

10.

DF � 48.3

53.4FD

�FD43.6

DC � 53.4

9772

�72DC

�KML; �JMK

92

� x

9 � 2x

3x � 9 � 5x

3�x � 3� � 5x

x � 3

5�

x3

y

x1

1

leg, altitude

leg, altitude

altitude

hypotenuse

K

J

L

�JKL

m�L � 180� � �30� � 60�� 90�

9.1 Practice and Applications (pp. 531–534)

11.

12.

13. 14. 15.

16.

17.

18.

19.

20.

21.

22.

23.

x � 4

x2 � 16

x4

�4x

�CBA ~ �DBC ~ �DCA

x � 50

32x � 1600

x

40�

4032

�RSQ ~ �TRQ ~ �TSR

x � 68.27

15x � 1024

x

32�

3215

�LKJ ~ �MLJ ~ �MKL

x � 16

25x � 400

x

20�

2025

�GEF ~ �HGF ~ �HEG

x � 9

144 � 16x

1216

�x

12

�CBA ~ �DBC ~ �DCA

�GFE ~ �HFG ~ �HGE; EH

�SQR ~ �TQS ~ �TSR; RQ

�ZYX ~ �WYZ ~ �WZX; ZW

x � �15 x � 6 x � 33.3

15 � x2 36 � x2 12x � 400

5x

�x3

4x

�x9

x

20�

2012

�SRQ ~ �TSQ ~ �TRS

S

Q

T

S

R T

Q

S R

5.

Two angles of are con-gruent to two angles of therefore the two triangles aresimilar.

�ABC,�JKL

30°

60°

A

B

C

Geometry 183Chapter 9 Worked-out Solution Key


Chapter 9 continued

24.

25. 26. 27.

28.

29.

30.

Solution:

31. About 76 cm; and are congruent righttriangles by the SSS Congruence Postulate, so is aperpendicular bisector of . By Geometric MeanTheorem 9.3, the altitude from D to hypotenuse divides into segments of lengths 23.7 cm and 61.1 cm. By Geometric Mean Theorem 9.2, the length of the altitude to the hypotenuse of each right triangle isabout 38 cm long, so the crossbar should be about

or 76 cm long.2 � 38,BD

ACAC

BDAC

�ADC�ABC

x � 3

x � 3 x � �21

x � 3 � 0 x � 21 � 0

�x � 21��x � 3� � 0

x2 � 18x � 63 � 0

x2 � 18x � 81 � 144

x � 9

18�

8x � 9

z � 53.3 y � 40 x � 42.67

24z � 1280 y2 � 1600 24x � 1024

32z

�2440

66.67

y�

y24

32x

�2432

e � 6.78

14e � 94.88

121

4

14�

e�60

b � �60

b2 � 60

142 � b2 � 162

m � 1047

7m � 74 x � 15 x � 3�3

7m � 49 � 25 16x � 240 x2 � 27

57

�m � 7

5 x

20�

1216

x3

�9x

x � �126

x2 � 126

x7

�18x

�HGE ~ �FHE ~ �FGH 32.

33.

34. Given is a right triangle and altitude is drawnto hypotenuse ; and are right trianglesby the definition of right triangles; because all right angles are congruent; byreflexive property for angles; therefore by the AA Similarity Postulate; becauseall right angles are congruent; by the reflexiveproperty for angles; therefore by the AASimilarity Postulate; by theAngle Addition Postulate; because the two acute angles in a right triangle arecomplementary; by the Addition Property;

because all right angles are congruent;so by the AA Similarity Postulate.

35. From Ex. 34, . Corresponding side

lengths are in proportion, so

36. From Ex. 34, and .Corresponding side lengths are in proportion, so

and

37. Values of the ratios will vary, but will not be equal. Thetheorem says they are equal.

38. They become equal.

39. The ratios are equal when the triangle is a right trianglebut are not equal when the triangle is not a right triangle.

ABAC

�ACAD

.ABBC

�BCBD

�ABC ~ �ACD�ABC ~ �CBD

BDCD

�CDAD

.

�CBD ~ �ACD

�DCA ~ �DBC�CDA � �CDB

�ACD � �B

m�DCB � m�B � 90�m�ACD � m�DCB � 90�

�ACB ~ �ADC�A � �A

�CDA � �ACB�ACB ~ �CDB�B � �B

�CDB � �ACB�DCA�DBCAB

CD�ABC

� 1.5 m2

Area of �CAB �12�2��1.5�

� 0.96 m2

Area of �DAC �12�1.6��1.2�

� 0.54 m2

Area of �DCB �12�1.2��0.9�

DB � 0.9 m AD � 1.6 m CD � 1.2 m

DB1.5

�1.52.5

AC2

�2

2.5 CD2

�1.52.5

�DCB ~ �DAC ~ �CAB

h � 64.4 ft

5.5h � 30.25 � 324

h � 5.5

18�

185.5

xywy

�wyzy

18 ft

5 ft12

W Y

h

X

ZNot drawn to scale

xy � h � 5.5

d � 3.75

14d � 52.52

14

6.78�

�60d

c � 1214

16c � 196

1416

�c

14

Chapter 9 continued



40. Using the right triangle, calculate the values of

These proportions should be true:

and Now change the value of and

recalculate The values of the ratios

will vary but

41. D

42.

43. Method 1

Measure the distance from the ground to the person’s eyelevel and the distance from the person to the building.

Use the proportion and solve for One

advantage of this method is you only need two measure-ments. One disadvantage is you need a friend to help.

Method 2

Measure the length of the building’s shadow, the heightof the pole and the length of the pole’s shadow. Use the

proportion and solve for RS. One advantage

is it can be done by one person. One disadvantage is itmust be done when the building and pole cast a shadow.

9.1 Mixed Review (p. 534)

44. 45. 46.

47. If the measure of one of the angles of a triangle is greaterthan then the triangle is obtuse; true.

48. If the corresponding angles of two triangles are congru-ent, then the two triangles are congruent; false.

49. 50.

51.

� 62.5 m2

A �12�12 � 13��5�

� 31.5 cm2 � 36 in.2 A � �4.5� A �

12�6��12�

90�,

d � ±9 x � ±8

d2 � 81 x2 � 64 n � ±13

d2 � 18 � 99 14 � x2 � 78 n2 � 169

MPQS

�NPRS

BC.BCAC

�ACDC

DC � 6

DC �14424

DC12

�1224

ACAD

.ABAC

�CBDB

,ABCB

�

ACAD

.ABAC

,CBDB

,ABCB

,

�CABAC

�ACAD

.

ABCB

�CBDB

ACAD

.ABAC

,

CBDB

,ABCB

,Lesson 9.2

9.2 Guided Practice (p. 538)

1. In a right triangle, the square of the length of thehypotenuse is equal to the sum of the squares of thelengths of the legs.

2. A, C

3.

no

5.

no


7.

yes

9.

yes

11.

no

13.

no

15.

no

14�2 � x

392 � x2

196 � 196 � x2

142 � 142 � x2

x � 8�3

x2 � 192

64 � x2 � 256

82 � x2 � 162

x � 4�2

x2 � 32

49 � x2 � 81

72 � x2 � 92

x � 80

x2 � 6400

1521 � x2 � 7921

392 � x2 � 892

97 � x

9409 � x2

4225 � 5184 � x2

652 � 722 � x2

x � 4�3

x2 � 48

42 � x2 � 82

�5 � x

5 � x2

22 � 12 � x2 4.

yes

6.

8.

no

10.

yes

12.

no

14.

yes

16.

x � 8�3

x2 � 192

64 � x2 � 256

82 � x2 � 162

x � 21

x2 � 441

400 � x2 � 841

202 � x2 � 292

�13 � x

13 � x2

4 � 9 � x2

22 � 32 � x2

41 � x

1681 � x2

81 � 1600 � x2

92 � 402 � x2

x � 3�5

x2 � 45

36 � x2 � 81

62 � x2 � 92

d � �11 ft

d2 � 11

52 � d2 � 62

x � 6

x2 � 36

x2 � 82 � 102

C AD � 18

AD � 24 � 6



Chapter 9 continued

17.

19.

21.

23.

25.

27.

28.

� 21 m2

A �12�10.5��4�

base � a � b � 3 � 7.5 � 10.5 m

b � 7.5 m

b2 � 72.25 � 16 � 56.25

42 � b2 � 8.52

a � 3 m

a2 � 25 � 16 � 9

a2 � 42 � 52

� 25.2 cm2

A �12�7��7.2�

b � 7.2

b2 � 51.75

12.25 � b2 � 64

3.52 � b2 � 82

� 35.7 cm2

A �12�9��3�7 �

b � 3�7

b2 � 63

81 � b2 � 144

92 � b2 � 122

s � 12

s2 � 144

1225 � s2 � 1369

352 � s2 � 372

s � 24

s2 � 576

324 � s2 � 900

182 � s2 � 302

20 � t

400 � t2 144 � 256 � t2 122 � 162 � t2

2�13 � x

52 � x2

16 � 36 � x2

42 � 62 � x2

12 � 8 � 4

b � 8

b2 � 64

36 � b2 � 100

62 � b2 � 102 29.

30.

32. The minimum distance of the base of the ladder from the wallis or 2.5 feet. The ladder, ifplaced 2.5 feet from the wall,will reach feet up the wall.

33.

34.

35.

� 48 in.

� 12 � 12 � 24

r � 3 in.�4� � 6 in.�2� � 12 in.�2� h � 84.9 in.

h2 � 7201

36002 � h2 � 36012

300 ft 1 in. � 3601 in.

300 ft � 3600 in.

39 in. � 39 in. � 16 in. � 94 in.

39 in. � c

1521 � c2

1296 � 225 � c2

362 � 152 � c2

2 ft 6 in. � 30 inches

3 ft � 36 inches

�100 � 6.25 � 9.7

104

10 ft

ladder wall

2.5 ft

9.7 ft

� 100 m2

A �12�20��10�

d2 � 5 � 5 � 10

d1 � 12 � 8 � 20

b � 5

b2 � 25

144 � b2 � 169

122 � b2 � 132

� 104 cm2

A �12�8��10 � 16�

b � 8

b2 � 100 � 36 � 64

62 � b2 � 102

10 cm

10 cm

10 cm

6 cm

b

31.

Distance from pitcher’splate to home is 50 feet.The distance from secondbase to home is about 91.9feet so the distance fromsecond to the pitcher’splate is orabout 41.9 feet.

91.9 � 50

91.9 ft � c

8450 � c2

652 � 652 � c2

18.

20.

22.

24.

26.

� 32.7 m2

A �12�3�19 ��5�

b � 3�19

b2 � 171

25 � b2 � 196

52 � b2 � 142

r � 468

r2 � 219,024

354,025 � r2 � 573,049

5952 � r2 � 7572

r � 99

r2 � 9801

400 � r2 � 10,201

202 � r2 � 1012

15 � t

225 � t2 81 � 144 � t2 92 � 122 � t2

�58 � x

58 � x2

9 � 49 � x2

32 � 72 � x2

11 � 4 � 7

b � 4

b2 � 16

9 � b2 � 25

32 � b2 � 52

Chapter 9 continued



36.

Method 2 uses less ribbon.

37. The area of the large square is Also, the area ofthe large square is the sum of the areas of the four con-gruent right triangles plus the area of the small square, or

Thus,and so Subtracting fromeach side gives

38. The area of the trapezoid is Also, the area ofthe trapezoid is equal to the sum of the areas of the twocongruent right triangles plus the area of the isosceles triangle or

Thus

and so

Subtracting from eachside gives

39. a.

yes

c.

The length of the diagonal of the base is The length of the diagonal of the box is

40. The length of one side of the rhombus is

or Multiplying the length of one side by 4gives the perimeter of the rhombus, which is

or

41.


42. 43. 44.

45. 46.

47. 48. 4��9 �2 � 36��5�49 �2� �1225

�4�13 �2� 208�2�2 �2

� 8

��14 �2� 14��9 �2

� 9��6 �2� 6

a � 32, b � 0.75�32� � 24

32 � x

40 � 1.25x

40 � �1.5625x2

40 � �x2 � 0.5625x2

80 � 2�x2 � �0.75x�2

P � 2�a2 � b2; a � x, b � 0.75x

2�a2 � b2.4�12�a2 � b2 �

12�a2 � b2.

��12a�2

� �12b�2

��l 2 � w 2�2 � h2 � �l 2 � w 2 � h2.

�l2 � w2.

d � �l2 � w2 � h2

� 9.8 ft

BD � �80 � 16

� 4�5

AB � �82 � 42

a2 � b2 � c2.2aba2 � 2ab � b2 � 2ab � c2.

12�a � b�2 � a � b �

12c2,

�12 � a � b� � �1

2 � a � b� � �12 � c2�.

12�a � b�2.

a2 � b2 � c2.2aba2 � 2ab � b2 � 2ab � c2.

�a � b�2 � 4�12 � a � b� � c2,4�1

2 � a � b� � c2.

�a � b�2.

36.1 in. � r

1300 � r2

400 � 900 � r2

202 � 302 � r249. 50. no 51. no 52. yes 53. no

54. yes

55. Slope of slope of

Both pairs of opposite sides areparallel, so PQRS is a parallelogram by the definition ofa parallelogram.

56. slope of slope ofBoth pairs of opposites sides are

parallel, so PQRS is a parallelogram by the definition of aparallelogram.

Lesson 9.3

Activity 9.3 Investigating Sides and Angles of Triangles

Construct

Constructions may vary.

Investigate

Values in tables may vary.

Conjecture

4. when

when

when


1. If the square of the length of the longest side of a triangleis equal to the sum of the squares of the lengths of theother two sides, then the triangle is a right triangle.

2. acute:

right:

obtuse:

3. C 4. D 5. D 6. A

7. No; the sum of while Since the two numbers are not equal, the triangles formedby the crossbars and the sides are not right triangles sothe crossbars are not perpendicular.


8. 9.

right right

10. 11.

not right right

26 � 26529 < 542.89

��26�2 ? 12 � 52232 ? 20.82 � 10.52

7921 � 79219409 � 9409

892 ? 802 � 392972 ? 652 � 722

452 � 2025.222 � 382 � 1928,

c > 30c2 > 242 � 182

c � 30c2 � 242 � 182

c < 30c2 < 242 � 182

m�C < 90��AC�2 � �BC�2 > �AB�2

m�C > 90��AC�2 � �BC�2 < �AB�2

m�C � 90��AC�2 � �BC�2 � �AB�2

QR �38 � slope of PS.

PQ � �3 � slope of RS;

QR �54 � slope of PS.

PQ � �112 � slope of RS;

��7�3 �2� 147

b.

No. The longest space inthe room is the diagonal ofthe room which is onlyapproximately 17.5 ft.

� 17.5 ft

BD � �208 � 100

� 4�13

AB � �144 � 64



Chapter 9 continued

12. 13.

not right not right

14. 15.

right right

16. 17. not a triangle

obtuse

18. 19.

acute right

20. 21.

right acute

22. 23.

acute right

24. 25.

acute obtuse

26. The quadrilateral has two pairs of congruent oppositesides. The triangle formed by the diagonal is a righttriangle because Therefore, the quadrilateral has four right angles. Thequadrilateral is a rectangle.

27. Square; the diagonals bisect each other, so the quadri-lateral is a parallelogram; the diagonals are congruent,so the parallelogram is a rectangle. so the diagonals intersect at right angles to form perpen-dicular lines; thus, the parallelogram is also a rhombus. A quadrilateral that is both a rectangle and a rhombusmust be a square.

28. Rhombus; the diagonals bisect each other so the quadri-lateral is a parallelogram. so the diagonalsintersect at right angles to form perpendicular lines so theparallelogram is a rhombus.

32 � 42 � 52,

12 � 12 � ��2 �2,

142 � 82 � �2�65 �2; 260 � 260.

30 < 30.252501 > 2500

5 � 25 ? 30.25100 � 2401 ? 2500

��5 � � 52 ? 5.52102 � 492 ? 502

21,025 � 21,02541 > 25

289 � 20,736 ? 21,02516 � 25 ? 25

172 � 1442 ? 145242 � 52 ? 52

221 > 1961156 � 1156

100 � 121 ? 196256 � 900 ? 1156

102 � 112 ? 142162 � 302 ? 342

49 � 4983 > 81

13 � 36 ? 4916 � 67 ? 81

��13�2 � 62 ? 7242 � ��67 �2 ? 92

389 < 676

100 � 289 ? 676

102 � 172 ? 262

1225 � 122510,201 � 10,201

441 � 784 ? 1225400 � 9801 ? 10,201

212 � 282 ? 352202 � 992 ? 1012

560 < 56927 < 29

�4�35 �2 ? 202 � 132�3�3 �2 ? 22 � 5229. slope of

slope of

Since , so is a right

angle. Therefore, is a right triangle by the defini-tion of a right triangle.

30.

therefore is a right triangle.

31. Computing slopes is easier because it involves two calcu-lations, not three. Computing slopes also does not involvesquare roots.

32.

The triangle is an right triangle.

33.

The triangle is an acute triangle.

34. Since is obtuse and isobtuse. and are a linear pair and are thereforesupplementary. By the definition of supplementaryangles, Since is obtuse,

Therefore, is an acuteangle by definition of an acute angle.

�1m �1 < 90�.m�ABC > 90�.�ABCm�ABC � m�1 � 180�.

�ABC�1�ABC�ABCx2 � 32 < 42,

27 > 26

��10 �2 � ��17 �2 ? ��26�2

� �17

RQ � ��0 � 4�2 � ��1 � 1�2

� �10

PR � ��1 � 0�2 � �2 � 1�2

� �26

PQ � ��1 � 4�2 � �2 � 1�2

x2

1

R(0, �1)

P (�1, 2)Q(4, 1)

y

80 � 45 � 125

�4�5 �2 � �3�5 �2 ? �5�5�2

� 5�5

RQ � ��6 � 5�2 � ��2 � 0�2

� 3�5

PR � ��3 � 6�2 � �4 � 2�2

� 4�5

PQ � ��3 � 5�2 � �4 � 0�2y

x6

2

R(�6, �2)

P (�3, 4)

Q(5, 0)

�ABC25 � 25 � 50;

52 � 52 � �5�2 �2

� 5

distance from A to C � ��4 � 0�2 � �6 � 3�2

� 5�2

distance from B to A � ��3 � 4�2 � �7 � 6�2

� 5

distance from B to C � ��3 � 0�2 � �7 � 3�2

�ABC

�ABCAC � BC�34��

43� � �1,

BC � �7 � 33 � 0

� �43

;

AC �6 � 34 � 0

�34

;

Chapter 9 continued



35. Since is obtuse and is obtuse. By the Triangle Sum Theorem,

is obtuse, soIt follows that Vertical angles

are congruent, so By substitution,By the definition of an acute angle, is

acute.

36. If a, b, and c are a Pythagorean triple, then Let k represent a positive integer. Multiplying a, b, and cby k gives the equation Dividingboth sides of the equation by yields Bythe Converse of the Pythagorean Theorem, is aright triangle and a, b, and c represent the side lengths of

37. A, C, D 38. rectangle

39.

40.

Cincinnati is not directly north of Tallahassee. It is north-east of Tallahassee.

41. Reasons

1. Pythagorean Theorem

2. Given

3. Substitution property of equality

5. Converse of the Hinge Theorem

6. Given, def. of right angle, def. of acute angle, andsubstitution property of equality

7. Def. of acute triangle ( is the largest angle of.)

42. Given: In

Prove: is an obtuse triangle.

Plan for Proof: Draw right triangle PQR with side lengthsa, b, and x. Compare lengths c and x.

b

a

x

P R

Q

b

a

c

C B

A

�ABC

�ABC, c2 > a2 � b2

�ABC�C

509,796 < 521,210

7142 ? 5992 � 4032

343,768,681 � 343,768,681

18,5412 ? 13,5002 � 12,7092

44,209,201 � 44,209,20128,561 � 28,561

66492 ? 48002 � 460121692 ? 1192 � 1202

�ABC.

�ABCa2 � b2 � c2.k2

k2a2 � k2b2 � k2c2.

a2 � b2 � c2.

�1m�1 < 90�.m�ABC � m�1.

m�ABC < 90�.m�C > 90�.�Cm�A � m�ABC � m�C � 180�.

�C�ABC��10 �2 � 22 < 42, Statements Reasons

1. 1. Pythagorean Theorem

2. 2. Given

3. 3. Substitution prop. ofequality

4. 4. A property of square roots

5. 5. Converse of Hinge Thm.

6. is an obtuse angle 6. Given, def. of rt. angle, def.of obtuse angle, sub. prop.of equality

7. is an obtuse 7. Def. of obtuse triangletriangle ( is the largest of

).

43.

Statements Reasons

1. 1. Pythagorean Theorem

2. 2. Given

3. 3. Substitution prop. ofequality

4. 4. A property of square roots

5. 5. Converse of Hinge Thm.

6. is an right angle 6. Given, def. of rt. angle, def.of obtuse angle, sub. prop.of equality

7. is a right 7. Def. of right triangletriangle ( is the largest ).

44.

is obtuse is acute

is obtuse is acute

A

45.

B

46.

Statements Reasons

1. is an altitude 1. Given

2. 2. Def. of an altitude

3. 3. Given

4. is a right triangle 4. Theorem 9.2�MQN

rt

�ts

m�NPM � m�NPQ � 90�

NP

m�D < 90�, m�E � m�F > 90�

m�A > 90�, m�B � m�C < 90�

�D�A

�DEF�ABC

8324 > 82816925 < 7225

6724 � 1600 ? 82815629 � 1296 ? 7225

822 � 402 ? 912772 � 362 ? 852

��N�LMN

�N

m �N � m �R

c � x

c2 � x2

c2 � a2 � b2

x2 � a2 � b2

�ABC��C

�ABC

�C

m �C > m �P

c > x

c2 > x2

c2 > a2 � b2

x2 � a2 � b2



Chapter 9 continued

9.3 Mixed Review

47.

48.

49.

50.

51. 52.

53.

54.

55. an enlargement with center C and scale factor

56. reduction with center C and scale factor

57.

Quiz 1 (p. 549)

1. 2. BD

3. 4.

5. 6.

7.

8.

No; the square of the longest side is larger than the sumof the squares of the smaller sides.

Lesson 9.4

Activity 9.4 Developing Concepts (p. 550)

Exploring the Concept

1. Triangles may vary.

2. side length 3 cm:

side length 4 cm:

side length 5 cm:

c � 5�2 � 7.1 cm

52 � 52 � c2

c � 4�2 � 5.7 cm

42 � 42 � c2

c � 3�2 � 4.2 cm

32 � 32 � c2

47,961 > 47,824

2192 ? 1682 � 1402

x � 12�2 � 17.0

x2 � 62 � 182

x � 6�5 � 13.4 x � 2�10 � 6.3

x2 � 122 � 182 32 � x2 � 72

x � 12 AC � 25

180 � 15x 9AC � 225

9

15�

x20

915

�15AC

�CDB ~�BDA ~ �CBA

x � 9

y � 11 4x � 36

2y � y � 11 5x � x � 36

53

74

8�24

�8

2�6�

4�66

�2�6

3

12�18

�12

3�2�

4�22

� 2�2

4�5

�4�5

53

�11�

3�1111

�15 � �6 � �90 � 3�10

�14 � �6 � �84 � 2�21

�6 � �8 � �48 � 4�3

�22 � �2 � �44 � 2�11

Conjecture

3. The length of the hypotenuse is the product of the lengthof one side and

Exploring the Concept

4. Triangles may vary.

6. triangle with side length 4 cm: side lengths: 2 cm, 4 cm,

triangle with side length 6 cm: side lengths: 3 cm, 6 cm,

triangle with side length 8 cm: side lengths: 4 cm, 8 cm,

Conjecture

7.

ratio of hypotenuse: longer leg:shorter leg:


1. Right triangles with angle measures and

2. According to the AA Similarity Postulate, since twoangles of one triangle are congruent to two angles of theother triangle, the two triangles are similar.

3. true 4. false 5. false 6. true 7. true

8. true 9. 10.

11.


12. 13.

14.

15. 16.

17. 18.

19. 20.

h �16�3

3

f �8�3

3

n � 6 f � �3 � 8

m � 12; p � 6�3q � 16�2; r � 16

c � 4�2

d � 4�2

c � 5; d � 5�3d � �2 � 8

e � 2�2

a � 12�3; b � 24x � 5; y � 5�2

9�2

2� k � h

9�2

� k

9 � �2k

h � k

a � 2; b � 2�3x � 4�2

30��60��90�.45��45��90�

2:�3:1

longer legshorter leg

: 2�3

2� �3;

3�33

� �3; 4�3

4� �3

hypotenuseshorter leg

: 42

� 2; 63

� 2; 84

� 2

4�3 cm

3�3 cm

2�3 cm

�2.

Chapter 9 continued



21.

22.

23.

24.

25.

26.

27.

28. 29.

30.

31.

I used the Pythagorean theorem in each right triangle,working from left to right.

32. Going from left to right: triangle 1

33. Going from left to right: triangle 3

34.

35. Let Then By the PythagoreanTheorem,

by a property of square roots.Thus the hypotenuse is times as long as a leg.

36. Construct congruent to because they are corresponding parts

of congruent triangles. Therefore and

is equiangular so it is also equi-lateral. Since it is equilateral, If and

, then The sidelengths are in the following ratio: hypotenuse:longer leg:shorter leg: : Therefore, in a triangle, the hypotenuse is twice as long as the shorter legand the longer leg is times as long as the shorter leg.

37. C 38. A; 6 � 6�3 � 12 � 28.4 cm

�3

30��60��90��3 � a:a.2a

AC � ��2a�2 � a2 � �3 � a.AB � 2aBC � aAB � 2a.

�BAD30� � 30� � 60�.m�BAD � m�BAC � m�CAD �m�CAD � 30�.

m�D � 60��BAC � �CAD,

�B � �D,�ABC.�ADC

�2DE � �2x2 � �2 � x

2x2 � �DE�2;x2 � x2 � �DE�2;EF � x.DF � x.

h � �n � 1

w � �7u � �5s � �3

v � �6t � 2r � �2

y � 2�3 � 3.5 cm; x � 2 � 2 � 4 cm

x � �1.4��2 � � 2.0 cmx � �3 cm � 1.7 cm

A � 6�12�4��2�3�� 41.6 ft2

A � 5�2�3 � � 17.3 m2

A �12�6��6�3 � � 31.2 ft2

A �12�4�3 ��8� � 27.7 ft2

x � 18.4 in.

x2 � 338

2x2 � 67626 in.

x

x

x2 � x2 � 262

12.7 in. � x

162 � x2

81 � 81 � x2

92 � 92 � x2

s � 9 in.

9 in.

9 in.

9 in. x 9 in.

4s � 36

x � 4.3 cm

x2 � 18.75

6.25 � x2 � 25

2.52 � x2 � 52

5 cm 5 cmx

5 cm2.5 cm

39. Stage 1:

Stage 3:

40. The pattern of the lengths is where the number

of the stage.

41. Substitute 8 for n into the formula and

simplify.


42. third leg; third leg ;

43. 44. 45.

46. 47. AA Similarity Postulate

48. SAS Similarity Theorem 49. SSS Similarity Theorem

Math & History

1. area of triangles:

area of square:

2.

Lesson 9.5


1.

2. The sine function is a ratio of the sides. Since the anglemeasures are the same, the sides of will be in thesame ratio as the sides of and

3. 4. 5. 6. 7. 8.34

45

35

43

35

45

sin D � sin A.�DEF�ABC

tan A �BCAC

cos A �ACAB

sin A �BCAB

a2 � b2 � c2

2ab � b2 � 2ab � a2 � c2

�b � a�2 � b2 � 2ab � a2

4�12

ab� � 2ab

B��0, �10�

A��4, �5�P��8, 3�Q��1, 2�5 < third leg < 23� 9 > 1414 � 9 >

1�2n

1�28 �

116

n �1

�2n,

x �1

2�2

x2 �18

2x2 �14

x2 � x2 � �12�

2

x �1�2

x2 �12

2x2 � 1

x2 � x2 � 12 Stage 2:

Stage 4:

x �14

x2 �1

16

2x2 �18

x2 � x2 � � 12�2�

2

x �12

x2 �14

2x2 �12

x2 � x2 � � 1�2�

2



Chapter 9 continued

9.


10.

11.

12.

13.

14.

15.

16. 0.7431 17. 0.9744 18. 6.3138 19. 0.4540

20. 0.3420 21. 0.0349 22. 0.9781 23. 0.8090

24. 0.4245 25. 0.4540 26. 0.8290 27. 2.2460

28.

x � 10.0 y � 8.0

sin 37� �6x

tan 37� �6y

tan K �35

� 0.6cos K �5

�34� 0.8575

sin K �3

�34� 0.5145tan J �

53

� 1.6667

cos J �3

�34� 0.5145sin J �

5�34

� 0.8575

tan H �12

� 0.5cos H �2�5

� 0.8944

sin H �1�5

� 0.4472tan G �21

� 2

cos G �1�5

� 0.4472sin G �2�5

� 0.8944

tan F �247

� 3.4286cos F �7

25� 0.28

sin F �2425

� 0.96tan D �7

24� 0.2917

cos D �2425

� 0.96sin D �7

25� 0.28

tan y �23

� 0.6667cos y �3

�13� 0.8321

sin y �2

�13� 0.5547tan x �

32

� 1.5

cos x �2

�13� 0.5547sin �

3�13

� 0.8321

tan A �86

� 1.3333cos A �6

10� 0.6

sin A �8

10� 0.8tan B �

68

� 0.75

cos B �8

10� 0.8sin B �

610

� 0.6

tan S �2845

� 0.6222cos S �4553

� 0.8491

sin S �2853

� 0.5283tan R �4528

� 1.6071

cos R �2853

� 0.5283sin R �4553

� 0.8491

d � 16.6 ft

sin 25� �7d

29.

30.

31.

32.

33.

34.

35.

36.

37. 38.

39. vertical drop,

40. 41.

42.

43.

44. The tangent of one acute angle is the reciprocal of thetangent of the other acute angle. The sine of one acuteangle is the same as the cosine of the other acute angleand the cosine of one acute angle is the same as the sineof the other acute angle.

tan B �ba

cos B �ac

sin B �bc

tan A �ab

cos A �bc

sin A �ac

x � 2.9 ft

tan 20� �x8

x � 16.4 in.

x �30

sin 45�� 26 d � 714.1 m

sin 45� �30

26 � x tan 55� �

d500

� 1409.3 ft

sin 20� �482d

x � 5500 � 5018 � 482 ft

d � 36.0 m h � 13.4 m

tan 42� �d

40 tan 13� �

h58.2

A �12

�11��6 � 3� � 34.9 m2

A �12

�12��6.9� � 41.6 m2

A �12

�2�2 ��2�2 � � 4 cm2

y � 14.9 x � 16.0

tan 22� �6y

sin 22� �6x

w � 3.3 v � 9.6

tan 70� �9w

sin 70� �9v

u � 3.4 t � 7.3

cos 65� �u8

sin 65� �t8

s � 2.9 r � 4.9

tan 36� �s4

cos 36� �4r

s � 31.3 t � 13.3

cos 23� �s

34 sin 23� �

t34

Chapter 9 continued



45. Procedures may vary. One method is to reason that sincethe tangent ratio is equal to the ratio of the lengths of thelegs, the tangent is equal to 1 when the legs are equal inlength, that is, when the triangle is a triangle, when and when

since increasing the measure of in-creases the length of the opposite leg and decreasing themeasure of decreases the length of the opposite leg.

46. is not a right triangle, so you cannot use thetrigonometric ratios.

47. Reasons

1. Given

2. Phythagorean Theorem

3. Division property of equality

5. Substitution property of equality

48.

49.

50.

51.

�0.2250�2 � �0.9744�2 � 1

cos 13� � 0.9744sin 13� � 0.2250

��32 �

2

� �12�

2

�34

�14

� 1

cos 60� �12

sin 60� ��32

��22 �

2

� ��22 �

2

�24

�24

� 1

cos 45� ��22

sin 45� ��22

�12�

2

� ��32 �

2

�14

�34

� 1

cos 30� ��32

sin 30� �12

BC � 11.0 x � 9

sin 55� �9

BC sin 30� �

x18

x

C

B

A30�

18

55�

�ABC

�A

�Am�A < 45�,tan A < 1m�A > 45�,tan A > 1

45��45��90�

52. Statements Reasons

1. is a right 1. Giventriangle with sidelengths a, b, and c.

2. 2. def. of tangent

3. 3. def. of sine and cosine

4. 4. prop. of rational numbers

5. 5. transitive property

53. ; D 54. C

55.


56. enlargement scale

factor

57.

58. 59.

yes no

x � �69 x � 168

x2 � 1725 x2 � 28,224

x2 � 2500 � 4225 x2 � 9025 � 37,249

x2 � 502 � 652 x2 � 952 � 1932

NP � 7.73 QP � 3.27

NP2 � 59.7429 49 � 15QP

18.27NP

�NP3.27

715

�QP7

�MNP ~ �NQP ~ �MQN

P�R� � 8

Q�R� � 10

�63

� 2

5

10

R

R�

Q

Q�

P

P�

4

8

6

3

h � 46 ft

79.62 � 33.26 � h

x � 6 � h

y � 33.26 x � 79.62

tan 29� �y

60 tan 53� �

x60

sin 25� �8

CD

tan A �sin Acos A

sin Acos A

�

acbc

�ab

cos A �bc; sin A �

ac

tan A �ab

�ABC



Chapter 9 continued

60.

no

Quiz 2 (p. 566)

1.

2.

3.

4.

5.

6.

7.

d � 4887.3 ft

tan 11� �950d

y � 22.1 x � 9.3

cos 25� �20y

tan 25� �x

20

y � 15.9 x � 8.5

sin 62� �y

18 cos 62� �

x18

y � 11.9 x � 15.6

tan 40� �10y

sin 40� �10x

� 3.9 in.2

A �12

�3��1.5�3�h � 1.5�3 in.

3 in. 3 in.

3 in.1.5 in.

h

� 5.7 in.

x � 4�2 in.4 in.

4 in.

4 in. x 4 in.

� 3.5 m

x � 2�3 m

4 m 4 m

4 m2 m

x

82.1 � x

6740.41 � x2

1840.41 � 4900 � x2

42.92 � 702 � x2 Lesson 9.6

Activity 9.6 Developing Concepts (p. 567)

1.

2.

3.

4. The measured value is The approximated values areless.


1. To solve a right triangle is to find the measures of allangles and all sides of the triangle.

2. true 3. false 4. 5.

6. 7.

8.

9.

10.


11. 12.

13.

14. 15. 16.

17. 18. 19.

20. 21. m�A � 6.3�m�A � 65.6�

m�A � 50.2�m�A � 81.4�m�A � 20.5�

m�A � 30�m�A � 45�m�A � 26.6�

m�S � 41.1�

sin S �4873

73 � QS

m�Q � 48.9� 5329 � QS2

sin Q �5573

482 � 552 � QS2

x � 2 y � 3.5

cos 60� �x4

sin 60� �y4

d � 60

m�E � 56.6� m�D � 33.4�d2 � 3600

sin E �91

109 sin D �

60109

912 � d2 � 1092

65 � c

m�B � 30.5� m�A � 59.5� 4225 � c2

sin B �3365

sin A �5665

332 � 562 � c2

m�A � 84.3�m�A � 64.2�

m�A � 79.5�m�A � 35.0�

3.1�40�.

tan�1 0.75 � 36.9�cos�1 0.8 � 36.9�sin�1 0.6 � 36.9�

tan A �34

� 0.75cosA �45

� 0.8sin A �35

� 0.6

C

B

A

5 cm

4 cm

3 cm

Chapter 9 continued



22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

m�B � 90� � 56�

a � 7.4 c � 8.9

tan 56� �a5

cos 56� �5c

m�Y � 38�

m�Y � 90� � 52�

z � 13.8 x � 10.9

cos 52� �8.5z

tan 52� �x

8.5

m�T � 70�

m�T � 90� � 20�

t � 11.3 s � 4.1

cos 20� �t

12 sin 20� �

s12

m�P � 64�

m�P � 90� � 26�

p � 4.0 q � 2.0

cos 26� �p

4.5 sin 26� �

q4.5

TS � 11.0

m�R � 61.3� m�S � 28.7� TS2 � 120.25

cos R �6

12.5 sin S �

612.5

62 � TS2 � 12.52

NQ � 13.0

m�P � 72.9� m�N � 17.1� NQ2 � 168.96

cos P �4

13.6 sin N �

413.6

42 � NQ2 � 13.62

ML � 4.5

m�K � 29.6� m�L � 60.4� ML2 � 20.64

cos K �8

9.2 sin L �

89.2

82 � ML2 � 9.22

6.3 � GH

m�H � 18.4� m�G � 71.6� 40 � GH

sin H �2

�40 sin G �

6�40

22 � 62 � GH2

9.9 � DE

m�D � 45� m�E � 45� 98 � DE2

sin D �7

�98 sin E �

7�98

72 � 72 � DE2

29 � AB

m�A � 46.4� m�B � 43.6� 841 � AB2

sin A �2129

sin B �2029

202 � 212 � AB2

32.

33.

34. 35.

36. 37.

38.

40.

41. Answers may vary.

42. 43.

44. Sample answer: riser length: 6 in.; tread length: 12 in.

45. As the ratio approaches 1, the stairway becomes steeper.Steeper stairways are more dangerous than those thataren’t as steep.

x � 26.6�

tan x �6

12

x � 42.5� x � 32.5�

tan x �8.25

9 tan x �

711

8 in.

54.6 in.

55.2 in.8.33°

x � 15.4�

tan x �4855

17,625

77.8 in. � AB

m�A � 27.6� 6057 � AB2

sin A �36

77.8 692 � 362 � AB2

m�B � 62.4� � 1.9167

tan�1B �6936

tan B �6936

m�L � 56�

m�L � 90� � 34�

� � 5.9 m � 7.2

tan 34� �4�

sin 34� �4m

m�F � 39�

m�F � 90� � 51�

e � 4.8 d � 3.7

cos 51� �3e

tan 51� �d3

m�B � 34�

39.

y � 4.1�

sin y �17

240

x � 239.4 in. or 19.9 ft

57,311 � x2

2402 � 172 � x2



Chapter 9 continued

46. Draw an altitude from C to . Label it h.

and by def. of sine of an angle. Since the

If then by aConverse of the Hinge Theorem. Therefore,


47. 48. 49.

50. 51. 52.

53.

55.

57.

59. not a triangle

60. 61.

acute right

62. 63.

obtuse right

64. not a triangle

Lesson 9.7


1. The magnitude of a vector is the distance from its initialpoint to its terminal point. The direction of a vector is theangle it makes with a horizontal line.

2.

3. is parallel to is parallel to

4.

5.

6.

7. �MN\

� � ��3 � 1�2 � �4 � 1�2 � �29 � 5.4�2, �5��PQ

\

� � ��5 � 1�2 � �4 � 2�2 � 2�5 � 4.5��4, �2��AB

\

� � ��4 � 0�2 � �5 � 0�2 � �41 � 6.4�4, 5�

�2, 2�

PQ\

AB\

MN\

;VU\

AB\

: ��2, �2�; PQ\

: �3, 3�; MN\

: �0, �3�; UV\

: �0, 2�

12,769 � 12,76969,169 > 68,900

1132 ? 1122 � 1522632 ? 2502 � 802

72.25 � 72.2551,984 < 52,000

8.52 ? 7.72 � 3.622282 ? 2202 � 602

m � 14

m2

�71

g � 12.6

10g � 126

3

10�

g42

x � 25

6x � 150

x

30�

56

�3, 1��1, �2��1, 0�

��1, �3��2, 2��3, 2�

asin A

�b

sin B.

a � b�A �B,sin A � sin B, �A �B.

sin A �h12c

sin B �h12c

AB 8.

9.


10.

11.

12.

13.

14.

15.

16.

� 7.2

� �52

�PQ\

� � ��4 � 2�2 � ��3 � 7�2

�6, �4�y

x1�1

Q

P

� 10.8

� �116

�PQ\

� � ��3 � 7�2 � �2 � 6�2

�10, 4�y

x2

2

Q

P

� 5.8

� �34

�PQ\

� � ��5 � 2�2 � �1 � 6�2

��3, 5�y

x1

1

Q

P

� 7.3

� �53

�PQ\

� � ��2 � 0�2 � �7 � 0�2

�2, 7�y

x1

1

Q

P

�EF\

� � ��0 � 4�2 � ��1 � 5�2 � 4�2 � 5.7��4, �4��JK

\

� � ��2 � 1�2 � ��2 � 4�2 � 3�5 � 6.7��3, 6��RS

\

� � ��1 � 5�2 � �1 � 2�2 � �17 � 4.1�4, 1�

�4, 5� � ��4, �2� � �0, 3�

x � 51.3� north of east

tan x �54

54.

56.

58.

t � 22

88 � 4t

8t

�4

11

k � 216

7k � 1512

7

18�

84k

y � 112

7y � 784

7

16�

49y

Chapter 9 continued



17.

18.

19.

20.

21.

22.

x � 51.3� south of east

tan x �5040

� 64 miles per hour

�AB\

� � ��40 � 0�2 � ��50 � 0�2

�0, 0� and �40, �50�


tan x �1060


�ST\

� � ��60 � 0�2 � �30 � 20�2

�0, 20� and �60, 30�

� 3

� �9

�PQ\

� � ��0 � 3�2 � �5 � 5�2

�3, 0�y

x1

1

QP

� 4.1

� �17

�PQ\

� � ��6 � 5�2 � �0 � 4�2

�1, �4�y

x�1

�1

Q

P

� 8.2

� �68

�PQ\

� � ��6 � 2�2 � �3 � 1�2

��8, �2�y

x1

1Q

P

� 7.2

� �52

�PQ\

� � ��5 � 1�2 � �0 � 4�2

��6, �4�y

x1

1

Q

P

23.

24.

25. 26.

27. 28.

29. yes; no

30. Round 2; the vectors have the same magnitude. In Round1, team A won; since has a greater magnitude than

represents a greater force applied.

31.

32.

33.

34.

u\

� v\

� �3, �3�

v\

: �1, �6�

u\

: �2, 3�y

x1

1

u

u � v

v

u\

� v\

� �5, 2�

v\

: �3, 6�

u\

: �2, �4�y

x1

1

u

u � v

v

u\

� v\

� ��1, 5�

v\

: �5, 3�

u\

: ��6, 2�y

x�1

5

v

u

u � v

u\

� v\

� �6, 5�

v\

: �2, 4�

u\

: �4, 1�y

x1

1

v

u

u � v

CA\

CB\

,CA

\

GH\

and JK\

EF\

and CD\

EF\

and CD\

EF\

, CD\

, and AB\

x � 38.7� south of west

tan x �4050


�OP\

� � ��0 � 50�2 � �0 � 40�2

�0, 0� and ��50, �40�

x � 45� north of west

tan x �4040


�LM\

� � ��10 � 50�2 � �10 � 50�2

��10, 10� and ��50, 50�



Chapter 9 continued

35. 36. 37.

38. 39. 40.

41.

42.

43.

the speed at which the skydiver is falling, taking intoaccount the breeze.

44.

45. The new velocity is

46.

The component form must give the same magnitude as

�AB\

� � �10.JK\

.

AB\

� �3, 1��JK

\

� � �10

��30, �120�.

10

10EW

DOWN

UP

x � 71.6�

tan x �12040

�s\

� � ��20 � 60�2 � �140 � 20�2 � 126.5 mih

10

10EW

DOWN

UP

v

us

v\

: �40, 0�

u\

: �0, �120�

�0, 0��4, �4��2, �3�

�10, 10��8, 7��4, 11� 47. When the magnitude of is k times the magnitudeof and the directions are the same. When themagnitude of is times the magnitude of and the

direction of is opposite the direction of . Justificationsmay vary.

48. a.

b.

c. Answers may vary.

49.

50.

51.

Total distance

52. One represents the components of a vector and one isactual distance in feet.

53. Since and are right angles and all right anglesare congruent, . Since is equilateral,

so andby the Alternate Interior Angles

Theorem. An equilateral triangle is also equiangular, soBy the definition of con-

gruent angles and the substitution property of equality,. by the AAS

Congruence Theorem. Corresponding parts of congruenttriangles are congruent, so By the definiton ofmidpoint, B is the midpoint of

54. 55. 56.

57.

58.

59.

60. �7 � x�2 � 49 � 14x � x2

�x � 11�2 � x2 � 22x � 121

�x � 7�2 � x2 � 14x � 49

�x � 1�2 � x2 � 2x � 1

y � 60�y � 30�y � 90�

x � 30�x � 120�x � 45�

DE.DB EB.

�ADB �CEB�DBA �EBC

m�BAC � m�BCA � 60�.

�EBC �BCA�DBA �BACDE � AC, AB BC.

�ABC�D �E�E�D

� 59.1 � 50.9 � 62.6 � 172.6 ft

�CA\

� � ��18�2 � ��60�2 � 62.6 ft

�BC\

� � ��36�2 � �36�2 � 50.9 ft

�AB\

� � �542 � 242 � 59.1 ft

��18, �60� � �18, 60� � �0, 0�

CA\

: ��18, �60�

AB\

� BC\

� �18, 60�

BC\

: ��36, 36�

AB\

: �54, 24�


tan x �2

10

�s\

� � ��10 � 0�2 � �2 � 0�2 � 10.2 mih

2

2EW

S

N

v

u

s

u\

v\

u\

�k�v\

k < 0,u\

v\

k > 0,

Chapter 9 continued



Quiz 3 (p. 580)

1.

2.

3.

4.

5.

6.

7.

8.

� 7.8

�PQ\

� � ��2 � 4�2 � �2 � 3�2

PQ\

: �6, �5�y

x2

2

Q

P

� 5.1

�PQ\

� � ��3 � 2�2 � �4 � 3�2

PQ\

: ��5, �1�y

x1

1

Q

P

m�L � 90� � 14� � 76�

� � 12.0

�2 � 144.76 m�K � 14�

32 � �2 � 12.42 sin K �3

12.4

m�F � 90� � 52.1� � 37.9�

f � 4.7

f 2 � 21.76 m�G � 52.1�

62 � f 2 � 7.62 sin G �6

7.6

m�Q � 90� � 75� � 15�

q � 2.1 p � 7.7

cos 75� �q8

sin 75� �p8

m�N � 90� � 40� � 50�

q � 20.9 m � 13.4

cos 40� �16q

tan 40� �m16

m�y � 45�

y � 12 z � 17.0

tan 45� �12y

sin 45� �12z

m�A � 90� � 25� � 65�

a � 41.7 b � 19.4

cos 25� �a

46 sin 25� �

b46

9.

10.

11.

north of east

12. 13. 14.

15. 16. 17.

Chapter 9 Review (pp. 582–584)

9.1 Similar Right Triangles

1.

2.

3.

9.2 The Pythagorean Theorem

4. 5.

yes no

s � 4�5 � 8.9 20 � t

s2 � 80 400 � t2 82 � s2 � 122 122 � 162 � t2

z � 23.8 x � 48

z2 � 567 y � 21 27x � 1296

z

27�

21z

48 � 27 � y 3627

�x

36

y � 12 x � 15

y2 � 144 x2 � 225

y

16�

9y

25x

�x9

y � 3�5 x � 4

y2 � 45 36 � 9x

5y

�y9

6x

�96

�0, 3��6, 13��2, 1�

��2, �8��2, 4��4, 2�

x � 69.4�

tan x �83y

x

2

2

T

S

� 13.0

�PQ\

� � ��2 � 5�2 � �6 � 5�2

PQ\

: ��7, �11�y

x2

4

P

Q

� 5.8

�PQ\

� � ��3 � 0�2 � �4 � 1�2

PQ\

: �3, 5�y

x1

1

Q

P



Chapter 9 continued

6. 7.

yes no

9.3 The Converse of the Pythagorean Theorem

8. 9.

obtuse right

10. not a triangle

11.

acute

9.4 Special Right Triangles

12.

13.

14. ;

15.

9.5 Trigonometric Ratios

16.

17.

tan N �1235

� 0.3429cos N �3527

� 0.9459

sin N �1237

� 0.3243tan P �3512

� 2.9167

cos P �3537

� 0.3243sin P �3537

� 0.9459

tan L �6011

� 5.4545cos L �1161

� 0.1803

sin L �6061

� 0.9836tan J �1160

� 0.1833

cos J �6061

� 0.9836sin J �1161

� 0.1803

A �12

�18��9�3 � � 140.3 cm2

altitude � 9�3 cm

longer leg � 6�3 in.shorter leg �12

�12� � 6 in.

� 18 in.2 � 12�2 in.

A � �3�2 �2 P � 4�3�2 �

leg �6�2

� 3�2

hypotenuse � �2�3�2 � � 6

81 < 89

92 ? 32 � �4�5 �2

1681 � 1681100 > 85

412 ? 402 � 92102 ? 62 � 72

t � 2�13 � 7.2 r � 30

52 � t2 r2 � 900

42 � 62 � t2 r2 � 162 � 342

18.

9.6 Solving Right Triangles

19.

20.

21.

9.7 Vectors

22.

23.

24.

�PQ\

� � �9 � 4 � �13 � 3.6

PQ\

: �3, 2�y

x1

2 Q

P

�PQ\

� � �144 � 25 � 13

PQ\

: �12, �5�y

x�2

2

Q

P

�PQ\

� � �1 � 16 � �17 � 4.1

PQ\

: ��1, �4�y

x2

1

Q

P

17 � s

� 61.9� 289 � s2 � 28.1�

sin T �1517

82 � 152 � s2 tan R �8

15

� 40� f � 12.9 d � 15.3

m�F � 90� � 50� cos 50� �f

20 sin 50� �

d20

� 8.9

x � 4�5

Z � 41.8� X � 48.2� x2 � 80

sin Z �8

12 cos X �

812

82 � x2 � 122

tan A �7

4�2� 0.3094cos A �

4�29

� 0.6285

sin A �79

� 0.7778tan B �4�2

7� 0.8081

cos B �79

� 0.7778sin B �4�2

9� 0.6285

Chapter 9 continued



25.

Chapter 9 Chapter Test (p. 585)

1. E 2. A 3. C 4. D 5. B

6.

7. is a kite. The diagonals are perpendicular and thequadrilateral has two pairs of consecutive congruentsides, but opposite sides are not congruent.

8. 9.

acute

10.

side length

11.

12.

13.

� 4.5

m�R � 41.8� m�P � 48.2� QR � 2�5

sin R �46

cos P �46

42 � QR2 � 62

m�F � 90� � 25� � 65�

DE � 25.7 DF � 28.4

tan 25� �12DE

sin 25� �12DF

m�K � 90� � 30� � 60�

JL � 7.8 KL � 4.5

cos 30� �JL9

sin 30� �KL9

� 6 in.

� 31.2 in.2

�12

�6��6�3 �

A �12

d1d2

6 in.

6 in.6 in.

30�

3 3 in. 3 in.

60�

6 in.

40 < 54

�2�10 �2 ? ��29 �2 � 52

b � 112PR � �4 � 36 � 2�10

b2 � 12,544QR � �9 � 16 � 5

152 � b2 � 1132PQ � �25 � 4 � �29

WXYZ

DBA; DAC


tan x �9

14

�u\

� v\

� � �196 � 81 � �277 � 16.6

u\

� v\

� �14, 9� 14.

15.

16.

17. 18. 19.

Chapter 9 Standardized Test (pp. 586–587)

1.

C

2. C

3. D

4. B 5. D

6.

D

7.

E

8. 9.

`

A B

A � 38.0� x � 53.1�

sin A �8

13 tan x �

129

y � 20.5 x � 18.8

cos 67� �8y

tan 67� �x8

x � �128 � 8�2 in.

2x2 � 256

P � 4�8�2 � � 32�2 in. x2 � x2 � 162

P � 2��125 � � 2�14� � 50.4 in.

A � 11�14� � 154 in.2x � �11 x � 9

�x � 11��x � 9� � 0

x2 � 2x � 99 � 0

�x � 1�2 � 100

x � 1

20�

5x � 1

�2, 3��4, 1��2, �8�

DE � 32.7 BC � 22.9

tan 35� �22.9DE

sin 35� �BC40

m�BCA � 90� � 35� � 55�

AB � 13.1 CD � 6.4

sin 50� �10AB

sin 40� �CD10

x � 36.9� south of east

tan x �34

�LM\

� � �16 � 9 � 5

LM\

: �4, �3�y

x1

1

L

M



Chapter 9 continued

10.

11. D

12.

13.

14.

15.

16. ; ;

17.

18.

19. a. b.

c. d.

e.

20. As the sun rises, the value of b decreases.

21.

22. The value of the expression increases as the sunapproaches the horizon.

Chapter 9 Cumulative Practice (pp. 588–589)

1. No; if two planes intersect, then their intersection is aline. The three points must be collinear, so they cannot bethe vertices of a triangle.

2. always 3. never 4. always

x � 79.2�

tan x � 5.25

b � 1.8 cm

tan 70� �5b

b � 2.9 cm b � 4.2 cm

tan 60� �5b

tan 50� �5b

b � 6.0 cm b � 8.7 cm

tan 40� �5b

tan 30� �5b

A �12

�30��32.3 � 15� � 709.5 units2

AE � 37.3 FE � 18.6

cos 30� �32.3AE

tan 30� �FE

32.3

m�BGF � 135�m�FEA � 60�m�ABC � 75�

AF � 10�3 � 15 � 32.3DC � 30

GC � �450 � 21.2FG � 15

BC � 30�2 � 42.4BD � �10�3 ��3 � � 30

A � �25 � 12� � 12

�7��24� � 216 units2

y � 73.7� x � 16.3�

sin y �2425

sin x �7

25

p � 7 � 24 � 25 � 56

x � 7

242 � x2 � 252

�AB\

� � ��8 � 1�2 � �3 � 9�2 � 15

y � 7 x � 8

y � 4 � 11 �2 � x � 6 5. is the median from point B,and it is given that Thus, bythe SSS Congruence Postulate. Also,since corresponding parts of congruent triangles arecongruent. By the definition of an angle bisector,bisects

6. Given quadrilateral where Since there are in a

quadrilateral, . and are oppositeangles, and and are opposite angles. and and are consecutive angles.

therefore, they are supplementary.Since opposite angles are congruent and consecutiveangles are supplementary, quadrilateral is aparallelogram.

7. Yes; clockwise and counterclockwise rotational symmetryof

8.

9.

10.

11.

12.

scalene right triangle

13. A��1, �2�, B��3, �5�, C��5, 6�

AC � �36 � 64 � 10

BC � �4 � 121 � �125AB � �16 � 9 � 5

y �34

x �144

y � 2 �34

x �64

y � 2 �34

�x � 2�

midpoint: �5 � 12

, 6 � 2

2 � ��2, 2�

slope of perpendicular bisector �34

m �6 � 2

�5 � 1� �

86

� �43

z2 � 55�; x2 � 90�; y� � 65�

y � 113 x � 24

y � 67 � 180 8x � 192

y � 3�24� � 5 � 180 8x � 12 � 180

y � 3x � 5 � 180 3x � 5 � 5x � 7 � 180

y � 16�

x � 54 4y � 64�

26 � x � 10 � 90 26� � 4y � 90�

120�.

ABCD

m�A � m�B � 180�,�B�A�B � �D.

�A � �C�D�B�C�Am�D � 143�

360�m�C � 37�.m�B � 143�,m�A � 37�,ABCD

�ABC.BD

�ABD � �CBD�ABD � �CBDAB � CB.

BD � BD,AD � CD,BD

Chapter 9 continued



14.

15.

16. 17. 18.

19. No; in ABCD, the ratio of the length to width is 8:6 or4:3. In APQD, the ratio of length to width is 4:6, or 2:3.Since these ratios are not equal, the rectangles are notsimilar.

20. In triangle ABC, is themidsegment of By the Midsegment Theorem,

Since the lines con-taining these segments areparallel, and

since corresponding angles are congru-ent. Since two angles of are congruent to twoangles of BAC, the two triangles are similar by the AASimilarity Postulate.

21. Yes; the ratios all equal so the triangles aresimilar by the SSS Similarity Theorem.

22. segment 1 4.2 cm

segment 2 4.8 cm

23. The image with scale factor has endpoints

and its slope is The image with scale

factor has endpoints and its slope is

The two image segments are parallel.136

.

�6, 4.5�;�3, �2�12

133

2�

136

.�4, 3�;

2, �43

13

x � 4.2

15x � 63

� 8x � 63 � 7x

� x

9 � x�

78

23,6

9, 812, and 12

18

�BDE�BED � �BCA,

�BDE � �BAC

DE � AC.

AB and BC.DE

A

B

D E

C

y � 1012

2y � 21 x � 337

x � 445

9y � 7y � 21 7x � 24 24 � 5x

79

�y

y � 3 37

�x8

12x

�52

A��3, 6�, B��7, 9�, C��9, �2�

C��6, �5�

B��5, �3�

A��2, 1�y

x1

1

A

B

A�

B�

C�

C 24.

25. 26.

27. 28.

acute

29.

30.

31.

south of east

32. Construct circumscribed about circle X. Bisecteach angle of the triangle. The point at which the bisec-tors intersect is the center of the circle. Draw a segmentfrom the center to the circle. This is the radius.

33. 34.

35.

d � 189.4 mi

d2 � 35,856.25 mi

d2 � 127.52 � 1402

� � 11.25 in.

36 � 13.5 � 2�

w � 6.75 in.

16w � 108 x � 20 gallons

1636

�3w

376x � 7520

p � 3 � 3 � 5 � 5 � 16 37616

�470

x

�ABC

x � 7.1�

tan x �2

16

�u\

� v\

� � �4 � 256 � 16.1

u\

� v\

� �2, 16�y

x2

2

v

u

u � v

m�S � 90� � 57� � 33�

TR � 10.9 in. ST � 16.8 in.

cos 57� �TR20

sin 57� �ST20

tan x �8

15� 0.5333

cos x �1517

� 0.8824sin x �8

17� 0.4706

�2�6

3

361 < 369

8�24�3

�2�2�3

��3�3

192 ? 152 � 122

26 � XY 4 � XY

676 � XY2 12 � 3XY

102 � 242 � XY2 2�3XY

�3

2�3

4 � ZX ZP � 2�3

16 � ZX2 ZP2 � 12

�2�3 �2 � 22 � ZX2 ZP6

�2

ZP



Chapter 9 continued

Project: Investigating Fractals (pp. 590–591)

Investigation

1. stage 1: ; stage 2:

2. stage 3: ; stage 4:

3. The length continually gets shorter until it almostdisappears.

4. Yes

Stages of a Koch Snowflake

5. Sketches may vary.

6. Tables may vary.

7. P � n�43�n

181

127

19

13

CHAPTER 9 - · PDF filex 8 3 2 192 x2 256 2x 16 21 x 441 x2 841 20 2x 29 13 x 13 x 9 x 2 23...

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