ISBN #978-1-929099-16-0 · 30. The side opposite the longest angle is not the longest side....

C

Edition

Core ommon

eometryorkbook WG

Topical Review Book CompanyISBN #978-1-929099-16-0

Bar Code Here

StandardsFounded in 1936 C

Edition

Core ommon

eometryorkbook WG

Topical Review Book CompanyISBN #978-1-929099-16-0

Bar Code Here

StandardsFounded in 1936

Answer Key

TEST 1Part I

1. 2 5. 4 9. 2 13. 1 17. 2 21. 32. 4 6. 3 10. 3 14. 2 18. 4 22. 3 3. 1 7. 2 11. 2 15. 3 19. 1 23. 2 4. 3 8. 4 12. 4 16. 4 20. 3 24. 4For parts II, III, and IV, answers may vary, partial credit should be given for answers that include, but not limited to, the following; - correct answer, but no work shown - incorrect answer, but rest of work is appropriate - appropriate work is shown, but one computational or rounding error is made

Part II

25.

26. Yes, by SSS

27. a) b) A ′ (–5, 4) B ′ (–10, 4) C ′ (–11, 0) D′ (–6, 0)

28. The container can hold approximately 4189 cubic centimeters of ice cream.29. S.A. = 480 square units

y

x

76543210

-1-2-3

B

C

A

D

•

•

•

•

• •

••B ′ A ′

C ′ D ′-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4

P

Q

P ′

Q ′

30.

31. 125°

32. Statement Reason 1. LMNO is a parallelogram 1. Given MO is a diagonal of LMNO 2. LM || NO and LO || MN 2. Definition of parallelogram 3. ∠OML ≅ ∠MON 3. Alternate interior angles are congruent ∠LOM ≅ ∠NMO 4. MO ≅ MO 4. Reflexive property 5. DMOL ≅ DMON 5. ASA

6. LM ≅ NO and MN ≅ LO 6. Corresponding parts of congruent triangles are congruent

Part III33. Possible response is T(2, –1) and Ry-axis

34. a)

b) Yes, DABC and DDEF would be congruent by SAS.

y

x

543210

-1-2-3

D

F EB C

A

Figure 1 Figure 2l

((

L″L

MNM″

A B

N″

35. a) Perimeter using the distance formula is approximately 24 units or 240 feet. b) Cost: $2388.00 Note: Point D on the Property Diagram, in the book, should be at (7, 7) and not at (8, 7).

36.

Part IV

37. Statement Reason 1. ABCD is a parallelogram. 1. Given EF bisect AB at E. EF bisects CD at F

2. AB ≅ CD & AB||CD 2. Opposite sides of a parallelogram are ≅ and ||.

3. EB ≅ EA & FD ≅ FC 3. Definition of bisector

4. EB ≅ FD 4. If equal quantities are divided equally, the results are equal.

5. ∠GBE ≅ ∠GDF and 5. When parallel lines are cut by a transversal, ∠BEG ≅ ∠DFG alternate interior angles are congruent.

6. DBEG ≅ DDFG 6. ASA

7. EG ≅ FG 7. Corresponding Parts of Congruent Triangles are Congruent

•

TEST 2Part I

1. 1 5. 4 9. 3 13. 3 17. 3 21. 2 2. 4 6. 3 10. 3 14. 3 18. 3 22. 3 3. 3 7. 3 11. 2 15. 4 19. 4 23. 1 4. 2 8. 1 12. 1 16. 1 20. 2 24. 2For parts II, III, and IV, answers may vary, partial credit should be given for answers that include, but not limited to, the following; - correct answer, but no work shown - incorrect answer, but rest of work is appropriate - appropriate work is shown, but one computational or rounding error is made

Part II

25. Yes, for example, AC = BC =3 5 and appropriate justification.

26.

27. 8.4 square inches28.

29. Statement Reason

1. Isosceles DABC, bisector AD 1. Given 2. AB ≅ AC 2. Definition of isosceles triangle 3. AD ≅ AD 3. Reflexive property 4. BD ≅ DC 4. Definition of perpendicular as bisector 5. DADB ≅ DADC 5. SSS

6. < B ≅ ≅ C 6. Corresponding parts of congruent triangles are congruent

30. x = 5

)

T

B

A

G

•

•

•

•

31.

32. The volume of the cylinder is three times the volume of the cone, or the volume of the cone is 1/3 the volume of the cylinder.

Part III

33.

a) Use slope to prove that both pair of opposite sides of the quadrilateral are parallel.

m

m

BC AD

BC

AD

=−

=−

4545

m

m

AB DC

AB

DC

=

=

5454

ABCD is a parallelogram because it is a quadrilateral with two pair of parallel sides.

b) From the slope values found in part a, AB ⊥ BC because their slopes are negative reciprocals of each other. Since perpendicular lines form right angles, ABCD contains a right angle and is a rectangle.

34. a) Use the construction of a perpendicular bisector of each side to find the midpoint. Connect each vertex to the midpoint of the opposite side to construct the the medians.

b) 1 point of intersection

y

x

B(-2, 3)

A(-6, -2)

D(-1, -6)

C(3, -1)•

•

•

•

P

Q

R

•

•

•

•

•

•

•

TBA

C

•

•

• •

||||

35. The diagonals of the quadrilaterals are congruent because they are both diameters of the circle. They are perpendicular by construction. A quadrilateral with congruent perpendicular diagonals is a square.

A = 12 bh A = 12 (12.5)2

A = 78.125 sq inches A = 4(78.125) A = 312.5 sq inches

36. a) The constant of dilation (scale factor) is 2. b) The constant of dilation gives the similarity ratio between the two triangles.

Part IV37. a) translation (x – 3, y + 2) b) Reflection in the y-axis c) Rotation of 90 degrees about the origin d) Yes, each of the transformations is a rigid motion, so the two triangles would be congruent. e) A dilation of 2 would result in a triangle that would not be congruent to DABC because the size of the triangle would change.

A•

B•C•

y

x-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11

6543210

-1-2-3-4-5-6

A′•

B′•C′•

A′′•

B′′•

C′′•

B′′′•

C′′′•

A′′′•

TEST 3Part I


Part II25. S. A. = 80 square units26. No, for example –– cos B is always equal to sin A cos B = BCAB and sin A = BCAB .

27. V = 46,16628.

Example –– Plan: Show that AD ≅ BC;AB ≅ DC, and that AD ⊥ DC.

d

d

AD

BC

= − − −( ) + − − −( ) = + − = =

= − + − − −( )

1 1 4 1 0 3 9 3

5 5 1 4

2 2 2 2

2

( ) ( ) ( )

( ) ( ) 22 2 2

2 2 2 2

0 3 9 3

1 5 1 1 6 0 36 6

5

= + = =

∴

= − − + − − −( ) = + = =

= − −

AD BC

d

d

AB

DC

( ) ( )

( 11 4 4 6 0 36 6

1 11 4

03

2 2 2 2) ( )

( )( )

( ) + − − −( ) = + = =

∴

= − − −− − −

=

=

AB DC

m

m

AD

DC55 14 4

60

− −− − −

=( )( )

: .Undefined

\ AD ⊥ DC because they have slopes that are negative reciprocals of each other. Conclusion: ABCD is a rectangle because it is a quadrilateral in which both pair of opposite sides are congruent and a pair of adjacent sides that is perpendicular.

≅

≅

yx0

-1-2-3-4-5

-3 -2 -1 1 2 3 4 5 6 7 8 9 B

C

A

D

29. Yes, by ASA30.

31.

32. a) 112 security guards b) 110 people

Part III 33. d C D' ' = 2a

34. Reflection: (x, y) (x, –y) (Reflection over the x-axis) Followed by Translation: (x, y) (x + 5, y + 1) Triangle 2 labels: A′(0, 6), B′(3, 4) and C′ is at (–2, 2) D1 ≅ D2 because reflections and translations are rigid motion and preserve angle measure and distance.

S

O•

•

U

T

•

•

y

x

76543210

-1-2-3-4-5-6-7

-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

1

A

C

2

B

B′

A′

C′

35. AB = AD

AB = AD AD + DC =AC AC AB AC BC

(AB)2 = (AC)(AD) (BC)2 = (AC)(DC)

(AB)2 + (BC)2 = (AC)(AD) + (AC)(DC) (AB)2 + (BC)2 = AC(AD + DC) (AB)2 + (BC)2 = AC(AC) (AB)2 + (BC)2 = (AC)2

36. a) and b)

c) ry-axis

Part IV

37. a) cone b) 44 gallons more

y

xY

Z

X

•

•

••

Z′′

•

••

Y′′

X′′

TEST 4Part I


Part II25. S1 = 67.08 and S2 = 33.5426. x = 3 and y = 227.

28.

29. T3,–5 then a Rot180º

30. The side opposite the longest angle is not the longest side. Therefore the triangle drawn is not plausible.31. Rot360° and reflection across line of symmetry

32. a) possible answers may include a square, triangle, rectangle, pentagon, and hexagon b) possible answers include a circle or octagon

• •

Part III 33. 7.6 units

34. a) AB = 41 b)

c) (x – 3)2 + y2 = 41

35. Area = 12 square units36. a)

b) y = 3

2x + 1 or equivalent response.

Part IV37. a) and b)

c) A′ equation is (x – 4)2 + (y + 7)2 = 4 A′′ equation is (x – 4)2 + (y + 7)2 = 16

y

x

•

-2 -1 1 2 3 4 5 6 7 8 9 10

•

10

-1-2-3-4-5-6-7-8-9

-10-11-12

y

x

76543210

-1-2-3-4-5-6-7

-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

••

•A

B

y

x

76543210

-1-2-3-4-5-6-7

-4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10

B

CA •

•

•

A

A′

•

•

A′′

TEST 5Part I


Part II25. 120π26. The third side has to be longer than 4 cm and shorter than 24 cm.27. Distance from (0, 0) to (1, 3 ) = 2 Distance from (0, 0) to (0, 2) = 2 Therefore, the point (1, 3 ) lies on the circle centered at the origin and containing the point (0, 2).28. 972πm3


1. AB || CD,EF intersects AB and 1. Given CD at P and T respectively. 2. ∠x ≅ ∠CTP 2. When 2 lines are cut by a transversal corresponding angles are congruent. 3. ∠CTP + ∠y are supplementary. 3. Two angles that form a straight line are supplementary. (Or linear pairs of angles are supplementary. 4. ∠x + ∠y are supplementary. 4. Substitution

30. a) b) ABCD rx-axis A′B′C′D′

A′B′C′D′ t0, 3 A′′B′′C′′D′′

y

x -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 B

CA •••

•D

•••

•

9876543210

-1-2-3-4-5-6

A′B′

C′

D′A′ ′

D′ ′

C′ ′

B′ ′

••

• (7, 8)

(7, 5)(7, 5)

(7, 2)

(4, 5)

(5, 4)

(4, 2)(5, 1)

y

x--8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

AB

C

876543210

-1-2-3-4-5-6-7

C ′

B ′

A ′ A ′ ′

B ′ ′

C ′ ′

•

31. Yes, by SAS. DABC ≅ DCDA

32. 3 + 13 + 2 2 + 2 5 33.

34. 133 inches35. a)

b) Rot180°

36.


1. AB || ED and C is the 1. Given midpoint of AD.

2. ∠BAC ≅ ∠EDC 2. If 2 parallel lines are cut by a transversal, the alternate interior angles are congruent.

3. AC ≅ DC 3. Definition of midpoint 4. ∠ACB ≅ ∠DCE 4. Vertical angles are congruent. 5. DABC ≅ DDEC 5. ASA ≅ ASA

6. DE ≅ AB 6. Corresponding parts of congruent triangles are congruent.

TEST 6Part I


Part II25. Statement Reason 1. LO ≅ NQ & LP ⊥ NP 1. Given and PL ⊥ NP 2. ∠P ≅ Right angle 2. Perpendicular lines form right angles. 3. ∠P ≅ ∠P 3.Reflexiveproperty 4. DQPN & DOPL are 4. A triangle containing a right angle right triangles is a right triangle. 5. DQPN ≅ DOPL 5. HL ≅ HL QP ≅ OP 6. Corresponding parts of congruent 6. triangles are congruent

26. Statement Reason 1. m∠RSU + m∠RST = 180° 1. Linear angle pairs are supplementary 2. m∠RST + m∠TSV = 180° 2. Linear angle pairs are supplementary 3. m∠RSU + m∠RST = m∠RST + m∠TSV 3. Substitution 4. m∠RSU ≅ m∠TSV 4. Subtraction

27.

28. (4.8, –1)

29.

30.

a) d

d

AD

AB

= − + + + = + = =

= − − + + = + = =

( ) ( )

( ) ( )

1 4 1 2 9 9 18 3 2

1 3 1 3 16 16 32 4 2

2 2

2 2

PP l w

P

A lw

A

A

= +

= +( )=

=

=( )( )=

2 2

2 3 2 4 2 14 2

3 2 4 2

24

b)

31. a) Similar b) Example: A dilation with a constant of dilation of ½ would create the image. In a dilation the angles are preserved, but distance is changed by the factor of the constant of dilation. If 2 angles on a triangle to 2 corresponding angles in another, the triangles are similar.

32. 3 13 + 5

•

•

•

A

BA′•

••

•

••B′

y

x

A(–1, 1)

D(–4, 2) B(3, –3)

C(0, –6)

Part III 33. Jessica’s container; .3125 g/cubic inch vs .2381 g/cubic inch

34. a)

b) S ′(7, –3) T ′(13, –3) U ′(11, –6) V ′(5, –6) c) S ′(7, 3) T ′(13, 3) U ′(11, 6) V ′(5, 6)

35. Midpoint of WY = midpoint of XZ = (1, – 52

). Since the diagonals of WXYZ share the same midpoint, they bisect each other. Therefore WXYZ is a parallelogram. WY = 13 XZ = 101 WY ≠ XZ (should be not congruent). Since the diagonals of WXYZ are not congruent, it is not a rectangle.

36.

y

x

76543210

-1-2-3-4-5-6-7

S T

V

••

• •U

-4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13 14

S′• T′•

V′• •

U′

• •

•S′′

•

V′′ U′′

T′′

B

C

DE

B′

D′

E′

C′••

• •

•

••

•• center

Part IV37. Statement Reason 1. m∠1 ≅ m∠2 1. Given AE bisects BD 2. BC ≅ DC 2. Definition of bisector 3. ∠DEC and ∠2 are supplementary. 3. They are linear pairs. ∠BAC and ∠1 are supplementary. 4. If two angles are each supplementary 4. ∠DEC ≅ ∠BAC to the same or equal angles, they are congruent to each other.

5. ∠BCA ≅ ∠DCE 5. Vertical angles are congruent

6. DACB ≅ DECD 6. AAS ≅ AAS 7. AB ≅ DE 7. Corresponding parts of congruent triangles are congruent. (CPCTC)

ISBN #978-1-929099-16-0 · 30. The side opposite the longest angle is not the longest side....

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