geometry rbc a answers - · pdf file(thm. 10.6) 3. up uq ut us ≅≅ ≅ 3 ... 2.25...

Answers _ Ch10

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 sidelengths do not satisfy the Pythagorean Theorem(Thm. 9.1).

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

7. no; ABC is not a right triangle because the sidelengths 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 nopoints of intersection and no segment joining thecenters of the circles.

A

B

C

Answers 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 segmentsfrom a common point to a circle are congruent,

and .KI KL KM KJ= = By the AdditionProperty of Equality, .KI KM KL KJ+ = + TheSegment Addition Postulate (Post. 1.2) shows thatIM KI KM= + and .JL KL KJ= + So, by theTransitive Property of Equality, IM JL= and so

IM JL≅ by the definition of congruent segments.

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

10.3 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 theircorresponding 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

STATEMENTS REASONS

1. PQ is the diameter

of .U PT QS≅

1. Given

2. PT QS≅ 2. CongruentCorrespondingChords Theorem(Thm. 10.6)

3. UP UQ UT

US

≅ ≅

≅

3. Definition ofradius of circle

4. PUT QUS≅ 4. SSS CongruenceTheorem (Thm. 5.8)

P QA

O

80°

1.5

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 arecongruent by the Inscribed Angles of a CircleTheorem (Thm. 10.11).

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

7. 39, 29x y= =

8. Opposite angles should be supplementary, notcongruent; 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 anglesand 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 theInscribed Angles of a Circle Theorem(Thm. 10.11).

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

with hypotenuse .AB So, AB is a diameter of thecircle 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°

STATEMENTS REASONS

1. P 1. Given

2. AED BEC∠ ≅ ∠ 2. Vertical AnglesCongruenceTheorem (Thm. 2.6)

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

4. AED BES 4. AAA Similarity

Theorem (Thm. 8.3)

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 theangle labeled .x° The measure of the angle shouldbe one-half the sum of the measures of the arcsintercepted 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 themeasures of the arcs intercepted by the angle and itsvertical 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°

Answers 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 arecongruent. 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. Congruentsegments have equal lengths, so Q is equidistantfrom points R, U, S and T that lie on the circle. So,Q is the center of the circle.

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 theVertical 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 and12

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• = •

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 toobtain 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

y

4

6

2

42−2

x2 + (y − 3)2 = 4

x

y

8

8

x2 + y2 = 100

Answers 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 distancefrom 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 onthe 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+ + − + + =

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. 6c. 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.

geometry rbc a answers - · pdf file(thm. 10.6) 3. up uq ut us ≅≅ ≅ 3 ... 2.25...

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