Geometry RBC A answers - Mr. Reevesmrdreeves.weebly.com/.../unit_10_practice_resources_answers.pdf · −8 4 8 (5, 0) A B C. Answers ... Answers Geometry Copyright © Big Ideas Learning,

Answers

Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A100

Chapter10 10.1 Start Thinking

1.

Sample answer: no; It does not pass through the center.

2.

Sample answer: radius; It connects the center of the circle with a point on the circle.

3. Sample answer: yes; no

10.1 Warm Up

1. 5r = 2. 9 2r = 3. 6r =

10.1 Cumulative Review Warm Up

1. yes; obtuse 2. yes; obtuse

3. yes; right 4. yes; acute

5. yes; obtuse 6. no

10.1 Practice A

1. A 2. ,AB AD 3. ,BD CH

4. CH

5. EG

6. no; ABC is not a right triangle because the side lengths do not satisfy the Pythagorean Theorem (Thm. 9.1).

7. yes; ABC is a right triangle because the side lengths satisfy the Pythagorean Theorem (Thm. 9.1).

8. 8r = 9. 20r =

10. 5 11. 3

and 42

−

12. Sample answer:

13. a. 40 ft; By the External Tangent Congruence Theorem (Thm. 10.2), the sidewalks are the same length.

b. 60 ft

10.1 Practice B

1. ,CE EF 2. ,CF BD 3. CF

4. BD

5. ,AG H

6. yes; ABC is a right triangle because the side lengths satisfy the Pythagorean Theorem (Thm. 9.1).

7. no; ABC is not a right triangle because the side lengths do not satisfy the Pythagorean Theorem (Thm. 9.1).

8. 12r = 9. 1.4r =

10. 5 11. 23

and 7−

12. when the two circles are concentric; There are no points of intersection and no segment joining the centers of the circles.

x

y

4

2

−4

−2

42−4

(−4, −3)

(3, 4)

x

y

4

2

−4

−2

42−4 −2

(3, −4)

x

y8

4

−8

84−8

(5, 0)

A

B

C

Answers

Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers

A101

13. a. about 19.2 ft

b. and ,AE BC DE CD= = so BD AD= by

the SSS Similarity Theorem (Thm. 8.4).

10.1 Enrichment and Extension

1. 9 2. 412 20.3≈

3. 19.6

4. It is given that andIM JL are tangent segments.

They intersect at point K. Because tangent segments from a common point to a circle are congruent,

and .KI KL KM KJ= = By the Addition Property of Equality, .KI KM KL KJ+ = + The Segment Addition Postulate (Post. 1.2) shows that IM KI KM= + and .JL KL KJ= + So, by the Transitive Property of Equality, IM JL= and so

IM JL≅ by the definition of congruent segments.

10.1 Puzzle Time

HE WAS SERVING PI

10.2 Start Thinking

60 min or 1 h

1. 180° 2. 270° 3. 60°

4. 54° 5. 288° 6. 312°

10.2 Warm Up

1. 29% 2. 119° 3. 35°


1. 13

7 2. 12 3. 62 4.

8

3

10.2 Practice A

1. minor arc; 55° 2. major arc; 245°

3. semicircle; 180° 4. minor arc; 120°

5. a. 32°

b. 208°

c. 105°

d. 260°

6. yes; They are arcs of congruent circles and .m EF mGH=

7. no; They are arcs of the same circle, but 120mSTV = ° and 150 .mUVT = °

8. a. 45°

b. 14.4°

10.2 Practice B

1. semicircle; 180° 2. minor arc; 74°

3. major arc; 286° 4. minor arc; 42°

5. yes; They are arcs of the same circle and .m AC m BD=

6. no; NM and OP have the same angle measure, but they are arcs of circles that are not congruent.

7. yes; They are arcs of the same circle and 42 .m AB mCD= = °

8. 22.5°

9. a. 135°

b. 225°

10. a. 170°

b. 34 sec


1. 18.6 in. 2. about 19.1 cm

3. a. 6 times

b. 60.4%

4.

a. 72°

b. about 5.9 in.

c. about 29.4 in.

d. about 259.4 in.

e. 21 180 180cos sin

4A nd

n n

° ° =

5. 73 6. 12 7. 126

10.2 Puzzle Time

THE CRAB APPLE

Answers


10.3 Start Thinking

1. sometimes true; If a chord passes through the center of the circle, then it is a diameter.

2. always true; By definition, a chord is a segment whose endpoints are on a circle and a diameter always satisfies this definition.

3. sometimes true; Because a radius is half the measure of the diameter, it is possible to draw infinitely many chords within the circle that have a measure equal to the radius. However, there are also infinitely many chords that do not have the same measure as the radius. For example, all the diameters do not have the same measure.

4. never true; It is possible for a chord to have the same measure as a diameter, but it will never be longer. A diameter is the longest possible chord in a circle.

10.3 Warm Up

1. 6 2 2. 3 3 3. 11


1. Given B is the midpoint of EC and ,DA you can

conclude that EB BC≅ and .AB BD≅ Because andEBA CBD∠ ∠ are vertical angles, you can

conclude that they are also congruent. Then by the SAS Congruence Theorem (Thm. 5.5), you can conclude that .AEB DCB≅

2. You are given BDE BED∠ ≅ ∠ and .A C∠ ≅ ∠

Then if you conclude DE DE≅ by the Reflexive Property of Segment Congruence (Thm. 2.1), you have AED CDE≅ by the AAS Congruence

Theorem (Thm. 5.11).

10.3 Practice A

1. 115° 2. 160° 3. 11

4. 65° 5. 4

6. a. yes; AB is a perpendicular bisector of .MN

b. no; AB is not perpendicular to .MN

7. 18 8. 6

9. 6 10 19 units≈ 10. D

10.3 Practice B

1. In a circle, if two chords are congruent, then their corresponding minor arcs are congruent.

2. 10 3. 110m AD mBE= = °

4. 100° 5. 7 6. 11 7. 3

8. yes; AB is a perpendicular bisector of .QR

9. about 12.8 units 10. about 30.4 units

11. Sample answer:

12. Sample answer: You could also use the SAS

Congruence Theorem (Thm. 5.5). ,PT QS≅ so

m PUT m QUS∠ ≅ ∠ by the Congruent Central

Angles Theorem (Thm. 10.4).


1. 60° 2. 19.2° 3. 53.1°

4. 90° 5. 103.5° 6. 180°

7. no; no; Sample answer:

8. 30 units

10.3 Puzzle Time

BECAUSE IT WANTED THE SCHOOL TO HAVE A LITTLE SPIRIT

STATEMENTS REASONS

1. PQ is the diameter

of .U

PT QS≅

1. Given

2. PT QS≅ 2. Congruent Corresponding Chords Theorem (Thm. 10.6)

3. UP UQ UT

US

≅ ≅

≅

3. Definition of radius of circle

4. PUT QUS≅ 4. SSS Congruence Theorem (Thm. 5.8)

P QA

O

80°

1.5

Answers


A103

10.4 Start Thinking

; ;m BMC mBC m A m B∠ = ∠ = ∠ Sample answer:

Because MB and MA are radii of the same circle, we can conclude that they are congruent. With this information, we can conclude that A B∠ ≅ ∠ by the

Base Angles Theorem (Thm. 5.6); m BMC m A m B∠ = ∠ + ∠ by the Exterior Angle Theorem (5.2). Because ,A B∠ ≅ ∠ by substitution

m BMC m A m A∠ = ∠ + ∠ or 2 .m BMC m A∠ = ∠Because , then 2m BMC mBC m BC m A∠ = ∠ = ∠ or

1.

2m BC m A∠ = ∠

10.4 Warm Up

1. 100 , 132m C m D∠ = ° ∠ = °

2. 97 , 50 , 33m X m Y m Z∠ = ° ∠ = ° ∠ = °

3. 115 , 115 , 65 ,

65

m P m Q m R

m S

∠ = ° ∠ = ° ∠ = °∠ = °


1. 18 square units

2. about 39.3 square units

3. about 15.9 square units

10.4 Practice A

1. 20° 2. 144° 3. 58°

4. B; Sample answer: andRQS RPS∠ ∠ are inscribed

angles that intercept the same arc, so the angles are congruent by the Inscribed Angles of a Circle Theorem (Thm. 10.11).

5. 110, 67x y= = 6. 99, 90x y= =

7. 39, 29x y= =

8. Opposite angles should be supplementary, not congruent; 95m B∠ = °

9. a. 62.3°

b. 83.1°

c. acute, scalene; Sample answer: Because 34.6 , 62.3 ,m A m B∠ = ° ∠ = ° and

83.1 ,m C∠ = ° ABC has three acute angles and no congruent sides.

10.4 Practice B

1. 90° 2. 42° 3. 58° 4. 48°

5. 58° 6. 42° 7. 96° 8. 180°

9. 14, 38x y= = 10. 72, 90x y= =

11. 16, 14x y= =

12. Sample answer:

13. yes; Sample answer: andADB BCA∠ ∠ intercept

the same arc, so the angles are congruent by the Inscribed Angles of a Circle Theorem (Thm. 10.11).

14. yes; Sample answer: 60m CAB∠ = ° by the Measure of an Inscribed Angle Theorem (Thm. 10.10) and 90m ACB∠ = ° by the Triangle Sum Theorem (Thm. 5.1). ABC is a right triangle

with hypotenuse .AB So, AB is a diameter of the circle by the Inscribed Right Triangle Theorem (Thm. 10.12).


1. 1 4 45 , 2 20 , 3 70m m m m∠ = ∠ = ° ∠ = ° ∠ = °

2. 27.70°

3. 1 60 , 2 60 , 3 120 ,

4 30

m m m

m

∠ = ° ∠ = ° ∠ = °∠ = °

4. 1 40 , 2 25 , 3 40m m m∠ = ° ∠ = ° ∠ = °

5. 24° 6. 48°

7. 45°, 135°, 75°, 105°

10.4 Puzzle Time

IT USED ITS HEAD

STATEMENTS REASONS

1. P 1. Given

2. AED BEC∠ ≅ ∠ 2. Vertical Angles Congruence Theorem (Thm. 2.6)

3. CAD DBC∠ ≅ ∠ 3. Inscribed Angles of a Circle Theorem (Thm. 10.11).

4. AED BES 4. AAA Similarity

Theorem (Thm. 8.3)

Answers


10.5 Start Thinking

Sample answer:

Two chords intersect at the center of the circle.

The circle is divided into four arcs, and opposite arcs are congruent.

Two chords intersect within the circle, but not at the center.

The circle is divided into four arcs. In the diagram of ,C none of the arcs have the same measure. In the

diagram of ,M m AD mBE= and the chords are

congruent.

Two chords intersect at a point on the circle.

The circle is divided into three arcs. Of the three arcs, none may be congruent, two may be congruent, or all three may be congruent.

Two chords do not intersect.

The circle is divided into four arcs. Of the four arcs, you may have none that are congruent, or two, three, or all four congruent.

10.5 Warm Up

1. 120° 2. 74° 3. 84°


1. 1 2. 2 3. 4

10.5 Practice A

1. 202° 2. 102° 3. 56° 4. 133

5. 42 6. 35 7. 26

8. Sample answer: This finds the supplement of the angle labeled .x° The measure of the angle should be one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle;

( )166 , so 666 6 .

2m x m x° + °∠ = ∠ = °

9. 21°

10.5 Practice B

1. 60° 2. 30° 3. 60°

4. 60° 5. 30° 6. 60°

7. D; The measure of 4∠ is one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. So,

( )14 75 125 100 90 .

2m∠ = ° + ° = ° ≠ °

8. 50 9. 7 10. 70

11. a. 120°

b. 100°

c. 140°

12. about 6.8°


1. a. 164°

b. 196°

c. 48°

d. 32°

e. 64°

f. 80°

BA

C

ED

B

A

C

E

F

D

A

D

M

B

E

C

C M Q

A

Y

ZW

X

B

X

YZ

W

CY

Z W

XD

Y

Z

W

X

FY

Z

W X

Answers


A105

2. a. 60°

b. 60°

c. 2.25

d. 1.125

e. ( )1.125 3 1.95≈

f. ( )2.25 2.25 3 6.1+ ≈

3. Sample answer: Draw chords and .RU ST It is

given that .RU ST≅ Because congruent arcs have

congruent chords, .RU ST≅ It is given that .RS TU≅ , , ,RUS URT TSU∠ ∠ ∠ and STR∠ are

all inscribed angles that intercept either or .RS TU

So, all four angles have the same measure and are congruent. By the SAS Congruence Theorem (Thm. 5.5), QRU and QST are congruent

triangles. Also, the base angles are all the same,

so they are isosceles triangles. So , , ,RQ UQ SQ

and TQ are congruent because corresponding parts

of congruent triangles are congruent. Congruent segments have equal lengths, so Q is equidistant from points R, U, S and T that lie on the circle. So, Q is the center of the circle.

10.5 Puzzle Time

A PARALLEL

10.6 Start Thinking

1. Sample answer: 1 17.7, 11.6PX PY= =




Each pair of segments has approximately the same period.

10.6 Warm Up

1. 16x = 2. 1x = 3. 3, 1

2x =

4. 2, 5x = − 5. 6 2 11x = ±

6. 7, 1x = −


1. 6 2. 13 5 53

2

− + 3. 3 10

10.6 Practice A

1. 15 2. 2 3. 12 4. 5

5. 6 6. 7 7. 15 8. 12

9. 4 10. 4 11. 4 12. 7

13. about 14.2 ft

10.6 Practice B

1. 10 2. 8 3. 4 4. 4

5. 8 6. 15 7. 9 8. 5

9. 30 10. about 20.1 in.

11. about 139.8 in.


1. 16.5, 16.8AC BD= =

2. 40.5

3. a. 60°

b. Sample answer: ACB FCE∠ ≅ ∠ by the Vertical Angles Congruence Theorem (Thm. 2.6). Because 60m CAB∠ = ° and

60 ,m EFD∠ = ° then .CAB EFD∠ ≅ ∠ Using the AA Similarity Theorem (Thm. 8.3),

.ABC FEC

c. Sample answer: 10 10

;3 6 2

y x xy

+ += =

d. Sample answer: ( )2 16y x x= +

e. 2, 6x y= =

f. 2 30; Sample answer: Because

ABC FEC and 12

2,6

CF

AC= = then

2.

1

CE

CB= Let 2CE x= and .CB x= So,

22 60x = by the Segments of Chords Theorem

(Thm. 10.18), which implies 30x = and

2 30.CE =

4. 2OT OP OQ= • and 2OT OR OS= • by the

Segments of Secants and Tangents Theorem (Thm. 10.20). So, .OP OQ OR OS• = •

Answers


10.6 Puzzle Time

BECAUSE THE ELLIPSES ARE TOO ECCENTRIC FOR THE CIRCLES

10.7 Start Thinking 2 2 4;x y+ =

10.7 Warm Up

1. ( )4, midpoint 0, 8PQ = =

2. 753, midpoint , 6

2PQ

= = −

3. ( )10 2, midpoint 5, 4PQ = = −

4. 117 3, midpoint 1,

2 4PQ

= = − −


1. 63° 2. 42° 3. 138°

4. 117° 5. 180°

10.7 Practice A

1. 2 2 49x y+ =

2. ( ) ( )2 25 1 25x y− + − =

3. 2 2 64x y+ =

4. ( )22 5 4x y+ + =

5. 2 2 25x y+ =

6. ( ) ( )2 23 2 841x y− + + =

7. B 8. A 9. C

10. center: ( )0, 3 , radius: 2

11. Sample answer: The distance from point ( )3, 3− to

the origin is 3 2, but the radius of the circle is 4, so the point does not lie on the circle.

12. a. from left to right, top row:

( ) ( )2 228 44 169,x y− + − =

( ) ( )2 257 44 169,x y− + − =

( ) ( )2 286 44 169;x y− + − =

from left to right, bottom row:

( ) ( )2 242.5 31 169,x y− + − =

( ) ( )2 271.5 31 169x y− + − =

b. Sample answer: Subtract 3 from the radius to obtain 100 on the right side of each equation.

10.7 Practice B

1. 2 2 9x y+ =

2. ( ) ( )2 23 2 4x y− + − =

3. ( ) ( )2 24 7 16x y− + + =

4. ( )2 23 25x y+ + =

5. 2 2 1x y+ =

6. ( ) ( )2 24 1 25x y− + + =

7. ( ) ( )2 22 4 169x y− + − =

8. center: ( )0, 0 , radius: 10

x

y4

2

−4

−2

42−2−4

x

y

4

6

2

42−2

x2 + (y − 3)2 = 4

x

y

8

8

x2 + y2 = 100

Answers


A107

9. center: ( )2, 9 , radius: 2

10. center: ( )0, 2 ,− radius: 6

11. center: ( )1, 0 , radius: 2

12. Sample answer: The statement is true. The distance from point ( )3, 4− to the origin is 5, and the radius

of the circle is 5, so the point lies on the circle.

13. Sample answer: The statement is false. The distance

from point ( )2, 3 to the origin is 7, but the

radius of the circle is 3, so the point does not lie on the circle.

14. a.

b. ( )2, 4−

c. no; The point ( )4, 5− is about 10.8 miles away

from the epicenter.


1. Sample answer: ( ) ( )2 23 3 9x y+ + + =

2. ( ) ( ) ( )2 23, 2 , 3 2 26x y− − + + =

3. ( )22 9.5 56.25x y+ + =

4. ( ) ( )2 212 19 56.25x y− + − =

5. a. 14 and 10h h= − =

b. 4

c. ( ) ( )2 22 4 16x y+ + + = and

( ) ( )2 22 4 484x y+ + + =

6. a. ( ) ( )2 225 4 121x y z+ + + − =

b. ( ) ( ) ( )2 2 210 6 2 169x y z− + + + − =

c. ( ) ( ) ( )2 2 21 2 4 59x y z+ + − + + =

10.7 Puzzle Time

COINCIDE

Cumulative Review

1. 2 10 21x x− + 2. 2 4 3j j+ +

3. 2 4 96c c+ − 4. 2 12 20m m− +

5. 2 21 110y y+ + 6. 2 11 30s s+ +

7. 25 17 12q q− − 8. 212 40 7p p− −

9. 22 17 84f f− − + 10. 254 36 6b b− +

11. 215 46 24g g− + − 12. 221 35 56k k+ −

13. 8 and 9x x= − = 14. 10 and 12x x= =

15. 6 and 12x x= = 16. 7 and 6x x= − =

17. 1 and 7x x= = 18. 4 and 8x x= − =

19. 10 and 11x x= − = 20. 4 and 9x x= =

21. 9 and 1x x= − = − 22. 1 and 8x x= − =

23. 4 and 2x x= − = 24. 8 and 5x x= − = −

25. 6 and 1x x= − = − 26. 2 and 8x x= − =

x

y

8

12

4

84−4

(x − 2)2 + (y − 9)2 = 4

x

y

−4

84−4−8

x2 + (y + 2)2 = 36

x

y

2

−2

42−2

(x − 1)2 + y2 = 4

x

y8

−4

8−4

(x − 2)2 + (y − 1)2 = 25

(x + 2)2 + (y + 2)2 = 36

(x + 6)2 + (y − 4)2 = 16

A

B

C

Answers


27. 8 and 7x x= − = 28. 4 and 7x x= − =

29. 43x = 30. 102x = 31. 112x =

32. 18x = 33. 26x = 34. 84x =

35. 133x = 36. 26x = 37. 19x =

38. 1275x = 39. 34 40. 25

41. 82 42. 34 43. 8

44. 12 45. 9 46. 6

47. 18

48. a. 2 11x +

b. 6

c. 11

d. 17

49. ( )3, 3− − 50. 112,

2 −

51. 11,

2 − −

52. 15 7,

2 2 −

53. 13 7,

2 2 − −

54. 11, 6

2 −

55. 5 9,

2 2 −

56. 5 1,

2 2 − −

57. 745

58. 2 29 59. 233 60. 2 145

61. 365 62. 109 63. 569

64. 8

65. Sample answer: , , andABC ABD CBD∠ ∠ ∠

66. Sample answer: , , andGFJ GFH JFH∠ ∠ ∠

67. 107° 68. 113° 69. 2 130

70. 218 71. 3 65 72. 466

73. 8 2 74. 2 102

75. a. 18.4 in.

b. 44.4 in.

76. a. 7.6 in.

b. 17.6 in.

Chapter 11 11.1 Start Thinking

10 31.4 cmπ ≈

1. 5 15.7 cmπ ≈ 2. 57.9 cm

2

π ≈

3. 25

13.1 cm6

π ≈

11.1 Warm Up

1. 120 , 8π° 2. 90 ,15π° 3. 135 ,16π°


1. 130° 2. 54

3. 164 2 41=

11.1 Practice A

1. 21 m 2. about 169.6 ft

3. about 47.1 in. 4. 12.4 cm

5. Divide the circumference of the tree by π to find the diameter of the tree. Because the diameter is 50 15.9 inches,π÷ ≈ which is less than 18 inches, the tree is not suitable for tapping.

6. about 6.28 cm 7. about 47.1 in.

8. about 7.33 ft 9. about 36.57 mm

10. about 86.85 in. 11. 3

π

12. 225° 13. 1257 ft

11.1 Practice B

1. 36

mπ

2. 10.8 ftπ

3. about 44.0 cm 4. 160°

5. 200° 6. about 19.54 m

7. about 24.43 m 8. 280°

9. about 34.21 m 10. about 114°

11. about 58.03 ft 12. about 20.53 cm

13. 7

12

π 14. 150°

15. a. 35.6 in. b. about 71 teeth

Geometry RBC A answers - Mr. Reevesmrdreeves.weebly.com/.../unit_10_practice_resources_answers.pdf · −8 4 8 (5, 0) A B C. Answers ... Answers Geometry Copyright © Big Ideas Learning,

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