Big Ideas Math Chapter 8 Answer Key PDF

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Answer key for chapter 8 of big ideas math geometry...

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Chapter 8 Chapter 8 Maintaining Mathematical Proficiency (p. 415)

⋅ ⋅

5 5 7 35 1. — = — = — 3 3 7 21 35 5 — and — form a proportion. 21 3

P = 2 + 5 + 6 + 3 = 16 centimeters

⋅ ⋅ ⋅ = — ⋅ 3 ⋅ (8) = 3 ⋅ (4) = 12 square centimeters k A = 3 ⋅ 12 = 9 ⋅ 12 = 108 square centimeters

kP = 3 16 = 48 centimeters A = 1—2 3 (2 + 6) 1 2

9÷3 3 24 ÷ 3 8 24 ÷ 8 3 —=— 64 ÷ 8 8 9 24 — and — form a proportion. 24 64

2. — = —

2

2

c. k = 4

P = 2 + 5 + 6 + 3 = 16 centimeters

8÷8 1 56 ÷ 8 7 3 6÷2 —=— 28 ÷ 2 14 6 8 — and — do not form a proportion. 56 28

3. — = —

18 ÷ 2 9 2 4÷2 27 ÷ 9 3 — =—=3 1 9÷9 27 18 — and — do not form a proportion. 9 4

4. — = —

15 ÷ 3 5 5. — = — 21 ÷ 3 7 55 ÷ 11 5 —= — 77 ÷ 11 7 55 15 — and — form a 77 21 proportion.

b. k = 3

26 ÷ 2 13 6. — = — 4 8÷2 39 ÷ 3 13 —=— 4 12 ÷ 3 39 26 — and — form a 12 8 proportion.

⋅ ⋅ ⋅ = — ⋅ 3 ⋅ (8) = 3 ⋅ (4) = 12 square centimeters k A = 4 ⋅ 12 = 16 ⋅ 12 = 192 square centimeters

kP = 4 16 = 64 centimeters A = 1—2 3 (2 + 6) 1 2

2

2

2. a. k = 2

P = 2 + 5 + 2 + 5 = 14 feet

⋅ ⋅

kP = 2 14 = 28 feet A = 2 4 = 8 square feet

⋅

⋅

k2A = 22 8 = 4 8 = 32 square feet b. k = 3

P = 2 + 5 + 2 + 5 = 14 feet

⋅ ⋅

kP = 3 14 = 42 feet A = 2 4 = 8 square feet

⋅

⋅

k2A = 32 8 = 9 8 = 72 square feet c. k = —1 2

6 3 CP′ CP 14 7 The scale factor is 3 k = —. 7

CP′ 24 8 3 CP 9 The scale factor is 8 k = —. 3

7. — =— = —

8. — = — = —

M′K′ 28 MK 14 The scale factor is k = 2.

9. — = — = 2

10. yes; All of the ratios are equivalent by the Transitive Property

P = 2 + 5 + 2 + 5 = 14 feet

⋅ ⋅

kP = 1—2 14 = 7 feet A = 2 4 = 8 square feet

( ) ⋅ 8 = — ⋅ 8 = 2 square feet

k2A = —21

2

⋅

height ⋅ length ⋅ width++22 ⋅length ⋅ ⋅ width Volume of a rectangular prism = length ⋅ width ⋅ height

3. Surface area = 2 height

a. k = 2

⋅ ⋅

of Equality.

1 4

⋅ ⋅

⋅ ⋅

S=2 5 3+2 5 4+2 3 4

Chapter 8 Mathematical Practices (p. 416) 1. a. k = 2

P = 2 + 5 + 6 + 3 = 16 centimeters

⋅

kP = 2 16 = 32 centimeters

⋅ ⋅ = — ⋅ 3 ⋅ (8) = 3 ⋅ (4)

A = 1—2 3 (2 + 6)

⋅ ⋅ ⋅ k V = 2 ⋅ 60 = 8 ⋅ 60 = 480 cubic inches k2S = 4 94 = 376 square inches

V = 3 4 5 = 60 cubic inches 3

3

b. k = 3

⋅ ⋅

⋅ ⋅

⋅ ⋅

S=2 5 3+2 5 4+2 3 4

1 2

= 30 + 40 + 24 = 94 square inches

⋅

k2S = 9 94 = 846 square inches

= 12 square centimeters

⋅

= 30 + 40 + 24 = 94 square inches

⋅

k2A = 22 12 = 4 12 = 48 square centimeters

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⋅ ⋅ ⋅

V = 3 4 5 = 60 cubic inches

⋅

k3V = 33 60 = 27 60 = 1620 cubic inches

Geometry Worked-Out Solutions

271

Chapter 8 c. k = 4

4. The perimeter of Gazebo B is 15 + 9 + 12 + 15 + 18 =

⋅ ⋅

⋅ ⋅

⋅ ⋅

S=2 5 3+2 5 4+2 3 4 = 30 + 40 + 24 = 94 square inches

⋅ ⋅ ⋅ k V = 4 ⋅ 60 = 64 ⋅ 60 = 3840 cubic inches k2S = 16 94 = 1504 square inches

V = 3 4 5 = 60 cubic inches 3

3

8.1 Explorations (p. 417) 1. Check students’ work. a. The corresponding angles are congruent.

69 meters. Let y be the perimeter of Gazebo A. 10 y —= — 69 15 2 y —= — 69 3 2 y = — 69 = 2 23 = 46 3 So, the perimeter of Gazebo A is 46 meters.

⋅

⋅

(NPJK ) 84 = 7 (21 ) A

Area of GHJK Area of LMNP

5. —— = —

2

b. The ratios are equal to the scale factor.

49 84 441 A 441 84 = A 49 37,044 = 49A 756 = A —=—

2. Check students’ work.

⋅

a. The perimeter of △A′B′C′ is equal to k times the

perimeter of △ABC. b. The area of △A′B′C′ is equal to k2 times the area of c. Yes, the results are similar. 3. Polygons that are similar have corresponding angles that

are congruent and corresponding side lengths that are proportional. 4. k = 3

⋅ ⋅

area of △R′S′T′ = k2 area of △RST = 32 1 =9

PQ 9 3 6 2 JK QR = 6— = 3— — 4 2 KL 12 3 PR = — =— — 2 8 JL

1. — = — = —

hexagons are equal. Because the sides of the red hexagon are all equal and the sides of the blue hexagon are equal, both hexagons are regular. So, corresponding angles are congruent and corresponding side lengths are proportional.

2. The question that is different is “What is the ratio of their areas?”

5 4 = __ DF = ___ 3 = 1__, DE FE = ___ 1 1, ___ ___ ___ = ___ = __

3 The scale factor is —. The congruent angles are ∠ J ≅ ∠ P, 2 ∠ K ≅ ∠ Q, and ∠ L ≅ ∠ R. Because the ratios above are equal, PQ QR PR — = — = —. JK KL JL

272

7. yes; Based on the markings, all of the interior angles of both

Vocabulary and Core Concept Check 1. For two figures to be similar, the corresponding angles must be congruent and the corresponding side lengths must be proportional.

8.1 Monitoring Progress (pp. 418–422)

8=x

interior angles are equal. The blue hexagon, based on the markings, has varying side lengths. So, it is not regular and corresponding side lengths are not proportional.

8.1 Exercises (pp. 423–426)

The area of △R′S′T′ is 9 square inches.

BC RS x — 4 x 2=— 4

⋅

6. no; The red hexagon is a regular, all the sides and all of the

△ABC.

AB QR 12 —= 6

—

—

c. Yes, the results are similar, if not the same.

2. — = —

2

KM JM 3. — = — FH EH KM 48 —= — 40 35 KM 6 —= — 5 35 6 KM = — 35 5 KM = 42

Geometry Worked-Out Solutions

⋅

12 4 AB 20 4 CB 16 4 AC 1 The scale factor is — and the ratio of corresponding side 4 1 lengths and perimeters is— . The ratio of their areas is 4 2 1 1 k2 = — = —. 4 16

()

Monitoring Progress and Modeling with Mathematics 900 = _________ 9 = ____ 900 ÷ 225 = 4__ LM = ____ 3. ___ AB

6.75

675

675 ÷ 225

3

8 = 4__ MN = __ ____

BC 6 3 ÷ 15 NL ___ 6 4 ___ = 60 = 60 — = __ —= 4.5 45 45 ÷ 15 3 CA 4 . The congruent angles are ∠A ≅ ∠L, The scale factor is __ 3 ∠B ≅ ∠M, and ∠C ≅ ∠N. Because the ratios above are CA BC = ___ AB = ____ equal, ___ . LM MN NL

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 1 4 =— RS = ___ 4. ___

FG 12 3 1 ____ SP = __2 = __ GD 6 3 PQ __ ___ = 3 = __1 DE 9 3 QR __ ___ =1 3 EF 1 . The congruent angles are ∠D ≅ ∠P, The scale factor is__ 3 ∠E ≅ ∠Q, ∠F ≅ ∠R, and ∠G ≅ ∠S. Because the ratios EF = FG ___ = ___ ___ = GD ____. above are equal, DE PQ QR RS SP

5.

LK = KJ ___ ___

6.

QR

RP x 18= — ___ 12 20 x 18= ___ ___ 20 12 x __3 = ___ 20 2 x 3 = 20 ___ 20 __ 20 2 30 = x

⋅

⋅

⋅

DE FD ___ = ___ GJ JH x 16 = — ___ 12 18 x 16 = — ___ 12 18 4 x —= — 3 18 4 x 18 — = 18 — 3 18 24 = x

⋅

MN NP JK HJ 9 = x__ __ 8 6 15 = x— ___ 10 8 3 = x— __ 2 8 x 3 __ =8 — 8 2 8

KL = ___ LM 7. ___

8. — = —

QR x 13 = ___ ___ 26 22 x 6 = ___ ___ 12 22 x __1 = — 2 22

PQ

⋅

⋅

11 = x 12 = x

AB = 24 ___ = 4 __ AB + BC + CD + DA = ___ 12. ___________________ RS + ST + TU + UR

18

3

2 smaller 48 x 3 larger 3 48 = x 2 144 = 2x

13. — = — = —

⋅

⋅

72 = x The perimeter of the larger polygon is 72 centimeters. smaller 66 3 14. — = — = __ larger

4 x 4 66 = x 3 264 = 3x

⋅

⋅

88 = x The perimeter of the larger polygon is 88 feet. smaller x __ 15. — = ____ = 1 larger

120 6 6 x = 120 6 6x = 120

⋅

⋅

x = 20 The perimeter of the smaller polygon is 20 yards. smaller larger

x 85 5 5 x = 85 2 5x = 170

2 16. — = ___ = __

⋅

⋅

x = 34 The perimeter of the smaller polygon is 34 meters. 17. Perimeter of NCAA court = 2

⋅

⋅

94 + 2 50 = 188 + 100 = 288 feet

x = ___ 27 9. ___

18 16 x = __3 ___ 2 16 __ x = 3 16 2

10 288 = ___ ____

9 P 9 288 = 10 P

⋅

⋅

⋅ ⋅

2592 = 10 P 259.2 = P

x = 24 The blue segment is an altitude and x = 24. y y − 1 ___ _____ = 10. 16 18 18(y − 1) = 16y

The perimeter of the new court in the school is 259.2 feet. 18. Perimeter of backyard = 2

⋅ 45 + 2 ⋅ 20 = 90 + 40

= 130 feet 45 130 = ___ ____

18y − 18 = 16y

P

−18 = −2y

18

5 130 = __ ____

9=y The blue segment is a median and y = 9. BC = ___ 8 = __2 AB + BC + CD + DA = ___ 11. ___________________ RS + ST + TU + UR

RS

4. The ratio of the perimeters is__ 3

ST

The ratio of the perimeters is2__. 3

12

3

2 P 130 = __ 5 P 2 ____ 2 P 2 130 = 5 P

⋅ ⋅

⋅ ⋅ 260 = 5 ⋅ P 52 = P

The perimeter of the new court in the school is 52 feet. Copyright © Big Ideas Learning, LLC All rights reserved.

Geometry Worked-Out Solutions

273

Chapter 8 19.

() ()

3 2 27 = __ ___

A 6 1 27 = __ ___ 2 A 27 = 1__ ___ A 4 4 27 = A

2

⋅

108 = A The area of the larger quadrilateral is 108 square feet. 20.

( ) ()

4 2 10 = ___ ___

12 A 1 2 ___ 10 = __ A 3 10 = __1 ___ 9 A 9 10 = A

⋅

90 = A The area of the larger triangle is 90 square centimeters.

( ) ()

4 A = ___ 21. ____

2

100 20 1 2 A = __ ____ 100 5 A = ___ 1 ____ 100 25

⋅

1 A = 100 ___ 25 A=4 The area of the smaller quadrilateral is 4 square inches.

( 12 ) (14)

A = ___ 3 22. ___ 96 A = ___ 96 A = ___ 96

2

__ 2

___ 1

16

⋅

1 A = 96 ___ 16 A=6 The area of the smaller triangle is 6 square centimeters. 23. Because the fi rst ratio has a side length of B over a side

length of A, the second ratio should have the perimeter of B over the perimeter of A. 5 = ___

x ___

10 28 10x = 140 x = 14 24. The square of the ratio of their corresponding side lengths

should be set equal to the ratio of their areas.

(18 ) (13)

2

24 = ___ x 24 __ 2 = ___ x 1__ = 24 ___ 9 x x = 216

6 ___

274

Geometry Worked-Out Solutions

25. No.

225 ÷ 75 = 3__ 225 = ________ 22.5 = ____ ____

300 300 ÷ 75 4 30 30 ÷ 10 3 30 = _______ = __ ___ 40 40 ÷ 10 4 Even though both polygons have the same apparent shape and their side lengths are proportional, their corresponding angles are not congruent. 26. yes; corresponding angles are congruent and corresponding

side lengths are proportional. BC = ___ AC and CA ___ = BC ___. 27. A, D; Because △ABC ∼ △DEF ,___ EF

DF

FD

EF

8 EF = — = 2__ 28. ___ JK 20 5 2. The scale factor of JKLM to EFGH is__ 5 JK = ___ 20 = 5__ 29. ___ 8 EF 2 5. The scale factor of EFGH to JKLM is__ 2 FG EF = ___ 30. ___ JK

KL

11 8 = ___ ___ x 20 11 2__ = ___ x 5 2x = 55 55 = 27.5 x = ___ 2 EH EF = ___ ___ JK

JM

y 8 = ___ ___

30 20 y 2— — = 5 30 60 = 5y 12 = y

m∠E = m∠J = z = 65° So, x = 27.5, y = 12, and z = 65. 31. Perimeter of JKLM = 20 + 30 + 7.5 + 27.5 = 85 units

3 8 =— ___

20 LM 8LM = 60 LM = 7.5 Perimeter of EFGH = 12 + 3 + 11 + 8 = 34 units JK + KL + LM + MJ = JK ___ = 20 ___ = 5 __ 32. ____________________ EF + FG + GH + HG

EF

8

2

5. The ratio of the perimeters is__ 2

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Chapter 8 1

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

33. Area of EFGH = —2 11

Area of

⋅ (8 + 3) ⋅

38. (y − 73)° = 360° − (116° + 61° + 90°)

1 = —2 11 11 1 = —2 121 = 60.5 square 1 JKLM = —2 x (ML + 20) 1 = —2 27.5 (7.5 + 20) 1 275 275 — = —2 — 10 10 1 55 55 — = —2 — 2 2 3025 =— = 378.125 square 8

⋅

⋅ ⋅

y − 73 = 360 − 267 y − 73 = 93

units

y = 166 x = __ 6 __

⋅

5

4

⋅

x = __46 5 units

( ) ( ) ()

2 20 2 = __ 5 2 = 25 = ___ — Area of EFGH EF 8 2 4 The ratio of the areas of JKLM to EFGH is 25 : 4.

JK Area of JKLM = ___ 34. _____________

35. B, D;

___ = 7.5 30 = 15 x = ___

4 2 So, x = 7.5 and y = 166. Perimeter of A Side length of A Perimeter of B Side length of B 72 = 18 ____ ___ 120 x

39. —— = ——

__ ___ 3 = 18

A 12

B

18 or

x

B 18

6

___ __6 = 12

18 x 18 6 = 12 x

⋅

⋅

108 = 12x

or

⋅ 18

6x = 216

9=x x = 36 So, the length of the other side of rectangle B could be 9 or 36. 36. no; If the table and the tennis court are similar, then the side 9 5 . By cross products, lengths would be proportional and78 — =— 36 5 9 . ≠— 9 36 = 324 does not equal 5 78 = 390, therefore— 78 36

⋅

⋅

So, corresponding side lengths are not proportional.

37.

___ 27 = 24 ___

y

⋅

Perimeter of A Perimeter of B 24 = ___ 36 2= __ 3

Side length of A Side length of B x ___ 12 x ___ 12

40. —— = ——

⋅

2=x 12 __ 3 8=x The corresponding side length of Figure A is 8 inches. Area of A Area of B 48 = ___ 75

(

)

Side length of A Side length of B __ 6 2 x ___ 16 = 36 ___ 25 x2 16x2 = 36 25

41. — = ——

18

___ 27 = 24 ___

y

3x = 90

x = 30 The corresponding side length of Figure B is 30 meters.

x

12 6 —= — 18 x 6x = 12

5 x 3x = 18 5

18

27 = 4__ ___

y 3 27 3 = 4y

⋅

()

2

⋅ ⋅

81 = 4y

36 25 x2 = — 16

20.25 = y

√ 16⋅

—

39 __4 = _____

3 x−6 4(x − 6) = 3 39

⋅

4x − 24 = 117 4x = 141

— 36 25 √ x2 = —

⋅

6 5 30 15 = 7.5 = ___ x = — = ___ 4 2 4 The corresponding side length of Figure B is 7.5 feet.

x = 35.25 So, x = 35.25 and y = 20.25.

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Geometry Worked-Out Solutions

275

Chapter 8

)

(

Area of A Side length of A 2 42. — = —— Area of B 18 = ___ 98 9 = ___ 49 — 9 = ___ 49

52.

Side length of B x 2 ___ 14 x 2 ___ 14 — x 2 ___ 14 x __3 = ___ 7 14 14 __73 = x 6=x The corresponding side length of Figure A is 6 feet.

√

( ) ( ) √( )

x

N

M

PQ + QR + RS + SP

53.

P x

N

47. Two regular polygons are sometimes similar.

49. yes; If the length and width of the fi rst rectangle is

50. yes; The light, object, and image form similar triangles.

BD

DA = 9.3 × 107 AB = 4.325 × 105 EC 2.4 × 105 = ___________ _________ 4.325 × 105

(4.3251× 10 )( 9.32.4 ×× 1010 ) = EC _________5 7

(4.325 × 2.4)(10 × 10 ) _____________________ = EC 5

9.3 × 107

(4.325 × 2.4)(1010) 9.3 × 10

—— = EC 7

( 4.3259.3× 2.4) (10 ) = EC (10 ) = EC ( 10.38 9.3 ) __________

3

_____

3

( 1.116) (103 ) = EC 1116 = EC The approximate length of EC is 1116 miles.

SP

Q

L ky

M

S

R

( ) ( ) ( ) ( )

54. no; Three angles determine a unique triangle in a sphere. 55.

DE = 2.4 × 105

RS

()

proportional to the length of the width of the second rectangle and all four angles are 90°, by definition, then the two rectangles are similar.

5

kx

QR

Let KLMN and PQRS be similar rectangles as shown. The 1. The KL = __ x = __ ratio of corresponding side lengths is___ PQ kx k area of KLMN is xy and the area of PQRS is (kx)(ky) = k2xy. xy __ 1 2 So, the ratio of the areas is____ = 12 = __ . Because k k2xy k the ratio of corresponding side lengths is1__, any pair of k corresponding side lengths can be substituted for1__ . k KL 2 LM 2 MN 2 = ___ NK 2. of KLMN = ___ _____________ ___ ____ So, Area = = Area of PQRS PQ QR RS SP

48. A right triangle and an equilateral triangle are never similar.

___________5

PQ

y

46. Two squares are always similar.

9.3 × 107

R

KL + LM + MN + NK = ___ KL = LM MN = ___ NK. ____________________ ___ = ____

45. Two rhombuses are sometimes similar.

AB

S

Let KLMN and PQRS be similar rectangles as shown. KL = __ x = 1__. The ratio of corresponding side lengths is___ kx k PQ The perimeter of KLMN is 2x + 2y and the perimeter of PQRS is 2kx + 2ky. So, the ratio of the perimeters is 2x + 2y 2x + 2y 1, _________ = _________ = 1__. Because both ratios equal __ k 2kx + 2ky k(2x + 2y) k the ratios are equal. So,

K

CD EC = ___ DE = ___ 51. Given △DEC ∼ △DAB, then ___ .

Q ky

43. Two isosceles triangles are sometimes similar.

DA

kx

L y

⋅

44. Two isosceles trapezoids are sometimes similar.

P K

x−1 1 1 x x2 − x = 1

—= —

x2 − x − 1 = 0

—

± √b2 − 4ac . _______________ Use the Quadratic Formula: x =−b 2a —— −(−1) ± √(−1)2 − 4 1 (−1) x = ——— 2 1 —

⋅

⋅ ⋅

—

1 ±√ 5 1 ± √1 + 4 x=—= — 2 2 — 1 + √5 x = — ≈ 1.618 2 — 1 − √5 x = — ≈ −0.618 2 Because x ≈ −0.618 is negative, only consider the positive solution x ≈ 1.618. x−1 1

1 1.618 − 1 0.618 1 1 x 1.618 1 1 1.618 = 1.618 0.618 = 0.999924, which is approximately 1.

—= —= —= —= — =—

⋅

—

1 + √5 x−1 1 So, x =...