Chapter 10 - Houston Independent School · PDF file2. Two angles whose sides form two pairs...

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321Geometry

Worked-Out Solution Key

Prerequisite Skills (p. 648)

1. Two similar triangles have congruent corresponding angles and proportional corresponding sides.

2. Two angles whose sides form two pairs of opposite rays are called vertical angles.

3. The interior of an angle is all of the points between the sides of the angle.

4. c2 ? a2 1 b2 5. c2 ? a2 1 b2

0.92 ? 0.62 1 0.82 172 ? 112 1 122

0.81 ? 0.36 1 0.64 289 ? 121 1 144

0.81 < 1 289 > 265

The triangle is acute. The triangle is obtuse.

6. c2 ? a2 1 b2 7. 6x 2 8 5 5x

2.52 ? 1.52 1 22 x 5 8

6.25 ? 2.25 1 4

6.25 5 6.25

The triangle is right.

8. (8x 2 2) 1 (2x 1 2) 5 180

10x 5 180

x 5 18

9. 7x 5 5x 1 40

2x 5 40

x 5 20

Lesson 10.1

Investigating Geometry Activity 10.1 (p. 650)

1. Answers will vary.

2. Tangent segments from a common external point are congruent.

3. MQ 5 MP 5 5.5

MN 5 LM 5 7

LQ 5 LM 1 MQ 5 7 1 5.5 5 12.5

PN 5 PM 1 MN 5 5.5 1 7 5 12.5

4. Because AC 5 EC and BC 5 DC, it follows that AB 5 ED which makes

} AB > } ED .

10.1 Guided Practice (pp. 651–654)

1. }

AG is a chord because its endpoints are on the circle. }

CB is a radius because C is the center and B is a point on the circle.

2. A tangent is @##$ DE and a tangent segment is } DB .

3. The radius of (C is 3 units.

The diameter of (C is 6 units.

The radius of (D is 2 units.

The diameter of (D is 4 units.

4. 2 common tangents 5. 1 common tangent

6. 0 common tangents

7. CE2 0 CD2 1 DE2

(3 1 2)2 0 32 1 42

25 0 9 1 16

25 5 25

Yes, } DE is tangent to (C.

8. QT 2 5 QS2 1 ST 2

(r 1 18)2 5 r2 1 242

r2 1 36r 1 324 5 r2 1 576

36r 5 252

r 5 7

9. x2 5 9

x 5 63

10.1 Exercises (pp. 655–658)

Skill Practice

1. The points A and B are on (C. If C is a point on }

AB , then

} AB is a diameter.

2. When referring to a segment, “a radius” and “a diameter” is used. When referring to a length, “the radius” and “the diameter” is used.

3. G; B is a point of tangency.

4. H; @##$ BH is a common tangent.

5. C; }

AB is a chord. 6. E; @##$ AB is a secant.

7. F; @##$ AE is a tangent. 8. A; G is the center.

9. B; }

CD is a radius. 10. D; } BD is a diameter.

11. The error is that }

AB is not a secant, but rather a chord. The length of chord } AB is 6 units.

12. The radius of (C is 9 units.

The diameter of (C is 18 units.

13. The radius of (D is 6 units.

The diameter of (D is 12 units.

14.

3

x

y

23

C D

Chapter 10

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322GeometryWorked-Out Solution Key



17. 1 common tangent

18. Use the Converse of the Pythagorean Theorem. Because 32 1 42 5 52, nABC is a right triangle and

} AB ⊥

} AC .

By Theorem 10.1, }

AB is tangent to (C.

19. Use the Converse of the Pythagorean Theorem. Because 92 1 152 Þ 182,

} AB is not perpendicular to

} BC and

} AB is

not tangent to (C.

20. The diameter of (C is 20. Using the Converse of the Pythagorean Theorem, 202 1 482 5 522 and the triangle is a right triangle. This implies that

} CB ⊥

} BA . Using

Theorem 10.1, }

AB is tangent to (C.

21. (r 1 16)2 5 r2 1 242

r2 1 32r 1 256 5 r2 1 576

32r 5 320

r 5 10

22. (r 1 6)2 5 r2 1 92

r2 1 12r 1 36 5 r2 1 81

12r 5 45

r 5 3.75

23. (r 1 7)2 5 r2 1 142

r2 1 14r 1 49 5 r2 1 196

14r 5 147

r 5 10.5

24. 3x 1 10 5 7x 2 6 25. 2x2 1 5 5 13

16 5 4x 2x2 5 8

4 5 x x2 5 4

x 5 62

26. 3x2 1 4x 2 4 5 4x 2 1

3x2 5 3

x2 5 1

x 5 61

27. The common tangents are external because they do not intersect the segment that joins the centers of the two circles.

28. The common tangents are internal because they intersect the segment that joins the centers of the circles.

29. C; Using Theorem 10.1, }

RS ⊥ }

QR . Draw a congruent, parallel segment } XP as shown.

RS

Q P3x

5

Using Pythagorean Theorem, (XQ)2 1 (XP)2 5 (QP)2, 22 1 (XP)2 5 82. XP 5 2 Ï

}

15 , so RS 5 2 Ï}

15 .

30. Using Theorem 10.2, }

PA > } PB and } PB > }

PC . Using the Transitive Property of Segment Congruence, }

PA > } PB > }

PC .

31. Sample Answer: Two lines tangent to the same circle will not intersect when the lines are tangent at opposite endpoints of the same diameter. Using Theorem 10.1, the two lines are perpendicular to the same line, so they are parallel.

32. C is in the interior of ∠ ABD and AC 5 DC. By Theorem 5.6, ###$ BC bisects ∠ ABD.

33. For any point outside of a circle, there is never only one or more than two tangents to the circle that passes through the point. There will always be two tangents.

34.

Statements Reasons

1. AB 5 AC 5 12, BC 5 8 1. Given

2. radiusr 5 PD 5 PE 5 PF

2. Def. of radius

3. AB 5 AD 1 BD, AC 5 AF 1 CF,BC 5 BE 1 CE

3. Postulate 2, Segment Addition

4. AB 5 AC 4. Given

5. AD 1 BD 5 AF 1 CF 5. Substitution for AB and AC

6. BE 5 BD, CE 5 CF,AD 5 AF

6. Theorem 10.2

7. AD 1 BE 5 AD 1 CE 7. Substitution for BD, AF, and CF

8. BE 5 CE 8. Subtract AD

9. E and P are on the angle bisector of ∠A.

9. Theorem 5.6

10. AE 5 AP 1 PE 10. Postulate 2, Segment Addition

11. BC 5 8 11. Given

12. BE 1 CE 5 8 12. Substitution for BC

13. CE 1 CE 5 8 13. Substitution for BE

14. CE 5 4 14. Divide by 2.

15. } PE ⊥ }

CE 15. Theorem 10.1

Chapter 10, continued

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323Geometry


Statements Reasons

16. AC 5 AF 1 CF 16. Statement 3

17. 12 5 AF 1 4 17. Substitution for AC and CF

18. AF 5 8 18. Subtract 4

19. (CE)2 1 (AE)2 5 (AC)2 19. Pythagorean Theorem

20. 42 1 (AE)2 5 122 20. Substitution for CE and AC

21. AE 5 8 Ï}

2 21. Solve for AE.

22. } PF ⊥ }

AF 22. Theorem 10.1

23. (PF)2 1 (AF)2 5 (AP)2 23. Pythagorean Theorem

24. (PF)2 1 (AF)2

5 (AE 2 PE)224. Substitution for AP

25. r2 1 82 5 (8 Ï}

2 2 r)2 25. Substitution for PF, AF, AE, and PE

26. r 5 2 Ï}

2 26. Solve for radius, r.

Problem Solving

35. The wheel has radial spokes because each spoke has endpoints that are the center and a point on the circle.

36. The wheel has tangential spokes because each spoke intersects the circle in exactly one point.

37. BE2 5 EC2 1 CB2

(11,000 1 3959)2 5 39592 1 CB2

14,9592 5 39592 1 CB2

208,098,000 5 CB2

14,426 ø CB

BA 5 BC ø 14,426 miles

38. m∠ ARC 5 m∠ BSC 5 908, so ∠ ARC > ∠ BSC. Also, ∠ RCA > ∠ SCB because vertical angles are congruent. Therefore, nARC , nBSC by the AA Similarity

Postulate. So, AC

} BC

5 RC

} SC

because the ratio of corresponding

sides is the same.

39. a. Because R is exterior to (Q and P is on (Q, QR > QP.

b. Because }

QR is perpendicular to line m it must be the shortest distance from Q to line m. Thus, QR < QP.

c. It was assumed }

QP was not perpendicular to line m but }

QR was perpendicular to line m. Since R is outside of (Q you know that QR > QP but part (b) tells you that QR < QP which is a contradiction. Therefore, line m is perpendicular to

} QP .

40. Assume line m is not tangent to (Q. So, there is another point X on line m that is also on (Q. X is on (Q, so QX 5 QP. But the perpendicular segment from Q to line m is the shortest such segment, so QX > QP. QX cannot be both equal to and greater than QP. The assumption that point X exists must be false. Therefore, line m is tangent to (Q.

41.

Statements Reasons

1. } SR and }

ST are tangent to (P.

1. Given

2. } SR ⊥ } RP , }

ST ⊥ } TP 2. Tangent and radius are perpendicular.

3. RP 5 TP 3. Def. of circle

4. } RP > } TP 4. Def. of congruence

5. } PS > }

PS 5. Refl exive Property

6. nPRS > nPTS 6. HL Congruence Theorem

7. } SR > }

ST 7. Corresponding parts of congruent triangles are congruent.

42. a. The slope of the line perpendicular to line l through C

is 2 3 }

4 , so the slope of line l is

4 }

3 .

b. y 5 mx 1 b

3 5 4 }

3 (24) 1 b

3 5 2 16

} 3 1 b

25

} 3 5 b

The equation for l is y 5 4 }

3 x 1

25 }

3 .

c. 32 1 (24)2 5 r2

25 5 r2

5 5 r

d. The y-intercept is 25

} 3 and the radius is 5. So, the

distance from the y-intercept to (C is 25

} 3 2 5 5

10 }

3 .

Mixed Review

43. m∠ABD 1 m∠DBC 5 m∠ABC

258 1 m∠DBC 5 708

m∠DBC 5 458

44. x8 1 508 5 1808 45. x 5 102

x 5 130 x8 1 3y8 5 1808

y 5 x 5 130 102 1 3y 5 180

3y 5 78

y 5 26

46. (2x 1 3)8 1 1378 5 1808

2x 1 140 5 180

2x 5 40

x 5 20

(4y 2 7)8 5 1378

44 5 144

y 5 36


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47. x 1 8 > 13 8 1 13 > x

x > 5 21 > x

So, the third side must be greater than 5 and less than 21.

x 1 4 > 11 4 1 11 > x

x > 7 15 > x

So, the third side must be greater than 7 and less than 15.

Lesson 10.2


1. C TQ is a minor arc.

m C TQ 5 1208

2. C QRT is a major arc.

m C QRT 5 m C QR 1 m C RT 5 608 1 1808 5 2408

3. C TQR is a semicircle.

m C TQR 5 1808

4. C QS is a minor arc.

m C QS 5 m C QR 1 m C RS 5 608 1 1008 5 1608

5. C TS is a minor arc.

m C TS 5 808

6. C RST is a semicircle.

m C RST 5 1808

7. C AB > C CD because they are in congruent circles and m C AB 5 m C CD .

8. C MN and C PQ have the same measure but they are not congruent because they are arcs of circles that are not congruent.

10.2 Exercises (pp. 661–663)

Skill Practice

1. If ∠ ACB and ∠ DCE are congruent central angles of (C, then C AB and C DE are congruent.

2. You need to know that the radii of two circles are the same in order to show that the two circles are congruent.

3. C BC is a minor arc.

m C BC 5 708

4. C DC is a minor arc.

m C DC 5 1808 2 m C ED 2 m C BC

5 1808 2 458 2 708 5 658

5. C DB is a minor arc.

m C DB 5 m C DC 1 m C CB 5 658 1 708 5 1358

6. C AE is a minor arc.

m C AE 5 m C BC 5 708

7. C AD is a minor arc.

m C AD 5 m C AE 1 m C ED 5 708 1 458 5 1158

8. C ABC is a semicircle.

m C ABC 5 1808

9. C ACD is a major arc.

m C ACD 5 m C ABC 1 m C CD 5 1808 1 658 5 2458

10. C EAC is a major arc.

m C EAC 5 m C EB 1 m C BC 5 1808 1 708 5 2508

11. C; C QRS is a semicircle because you know that }

QS is a diameter.

12. m C CD 5 1808 2 708 2 408 5 708

C AB > C CD because they are in the same circle and m C AB 5 m C CD .

13. C LP and C MN have the same measure but they are not congruent because they are arcs of circles that are not congruent.

14. C VW > C XY because they are in congruent circles and m C VW 5 m C XY .

15. The statement is incorrect because you can tell that the circles are congruent. The circles have the same radius, }

CD .

16. C B

A D 20°

P

m C ACD 5 3608 2 m C AD 5 3608 2 208 5 3408

m C AC 5 1808 2 m C AD 5 1808 2 208 5 1608

17. A; nAPB is a right triangle.

AP2 1 PB2 5 AB2

32 1 32 5 AB2

18 5 AB2

3 Ï}

2 5 AB

18.

C

120°

100°E

F

G

m C GE 5 3608 2 1008 2 1208 5 1408

If m C GH 5 1508, point H must be 108 beyond point E, placing it on C EF . Or, point H is 308 beyond point F, placing it on C EF .

19.

R

A

BC

DE

60° 25°

70°

20°

m C AE 5 m C AB 1 m C BC 1 m C CD 1 m C DE

5 608 1 258 1 708 1 208 5 1758

m C AE 5 3608 2 (m C AB 1 m C BC 1 m C CD 1 m C DE ) 5 3608 2 1758 5 1858

20. nPAQ is a right triangle with m∠APQ 5 308 and m∠AQP 5 608. nPBQ is a right triangle with m∠BPQ 5 308 and m∠BQP 5 608. So, m∠AQB 5 m∠AQP 1 m∠BQP 5 608 1 608 5 1208. Therefore, m C AUB 5 1208.


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325Geometry


21. a. tan X 5 3 }

4

m∠ X ø 36.98

m C BD ø 36.98

b. tan Y 5 4 }

3

m∠Y ø 53.18

m C AD ø 53.18

c. m C AB 5 m C AD 2 m C BD

5 53.18 2 36.98 5 16.28

Problem Solving

22. The measure of the arc is 608.

23. 3608 4 20 5 188

The measure of each arc in the outermost circle of the dartboard is 188.

24. a. 3608 2 908 5 2708

The measure of the arc surveyed by the camera is 2708.

b. 2708 4 108 per minute 5 27 minutesIt takes the camera 27 minutes to survey the area once.

c. The camera must go 2708 2 858 5 1858 counterclockwise and another 1858 clockwise to return to the same position. So, 1858 1 1858 5 3708. 3708 4 108 per minute 5 37 minutes. It will take the camera 37 minutes.

d. In 15 minutes, the camera can go 108(15) 5 1508. The camera will go 508 counterclockwise and then 1008 clockwise. So, the camera will be 1008 from wall A after 15 minutes.

25. a. 3608 } 60 min

5 68 per minute

After 20 minutes, the minute hand will move 20(68) 5 1208.

3608

} 12 h

5 308 per hour

The hour hand moves 308 per hour.

308 }

60 min (20 min) 5 108 in 20 minutes

At 1:20, the hour hand is 308 1 108 5 408 from the 12. The minute hand is 1208 from 12. The minor arc between the hour and minute hand is 1208 2 408 5 808.

b. You want the arc between the hour and minute hand to be 1808. Use guess, check, and revise. At 1:40, the minute hand is 40(68) 5 2408 from 12. The hour hand

moves 308 }

60 min (40 min) 5 208 in 40 minutes. The hour

hand is 308 1 208 5 508 from 12. The arc is 2408 2 508 5 1908. At 1:38, the minute hand is 38(68) 5 2288 from 12. The hour hand moves

308 }

60 min (38 min) 5 198 in 38 minutes. The hour hand is

308 1 198 5 498 from 12. The arc is 2288 2 498 5 1798.

Mixed Review

26. y 5 5x 1 2

y 5 5(1 2 x) 5 5 2 5x

The slopes of the lines are different, so they are not parallel.

27. 2y 1 2x 5 5

2y 5 22x 1 5

y 5 2x 1 5 }

2

y 5 4 2 x

The slopes are the same, so the lines are parallel.

28.

X Y

Z

X9

Y9Z9

P

145°

29. (x 1 2)(x 1 3) 5 x2 1 2x 1 3x 1 6

5 x2 1 5x 1 6

30. (2y 2 5)( y 1 7) 5 2y2 2 5y 1 14y 2 35

5 2y2 1 9y 2 35

31. (x 1 6)(x 2 6) 5 x2 1 6x 2 6x 2 36 5 x2 2 36

32. (z 2 3)2 5 (z 2 3)(z 2 3)

5 z2 2 3z 2 3z 1 9 5 z2 2 6z 1 9

33. (3x 1 7)(5x 1 4) 5 15x2 1 12x 1 35x 1 28

5 15x2 1 47x 1 28

34. (z 2 1)(z 2 4) 5 z2 2 4z 2 z 1 4 5 z2 2 5z 1 4

Lesson 10.3


1. m C BC 5 1108

2. 2m C AB 1 m C AC 5 3608

2m C AB 1 1508 5 3608

2m C AB 5 2108

m C AB 5 1058

3. 9x8 5 (80 2 x)8

10x 5 80

x 5 8

m C CD 5 9x8 5 9(88) 5 728

4. 9x8 5 (80 2 x)8

10x 5 80

x 5 8

m C DE 5 (80 2 x)8 5 (80 2 8)8 5 728

5. m C CE 5 m C CD 1 m C DE 5 728 1 728 5 1448

6. QR 5 ST 5 32

7. QU 5 1 }

2 QR 5

1 }

2 (32) 5 16


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8. QC2 5 QU2 1 UC2

QC2 5 162 1 122

QC2 5 400

QC 5 20

10.3 Exercises (pp. 667–670)

Skill Practice

1. Sample answer: Point Y bisects C XZ if C XY > C YZ .

2. If two chords of a circle are perpendicular and congruent, one of them does not have to be a diameter. A square inscribed in a circle is an example of two chords that are congruent and perpendicular.

3. m C AB 5 m C ED 5 758

4. 2m C AB 1 m C AD 5 3608

2m C AB 1 1288 5 3608

2m C AB 5 2328

m C AB 5 1168

5. EG 5 EJ 5 8

6. By Theorem 10.5, } BD bisects }

AC .

4x 5 3x 1 7

x 5 7

7. By Theorem 10.5, } LN bisects } PM .

5x 2 6 5 2x 1 9

3x 5 15

x 5 5

8. Because }

SQ and }

TQ are radii, they have the same measure.

6x 1 9 5 8x 2 13

22 5 2x

11 5 x

9. }

AB and }

CD are equidistant from the center, so by Theorem 10.6, they are congruent.

5x 2 7 5 18

5x 5 25

x 5 5

10. }

AD and }

BC are equidistant from the center, so by Theorem 10.6, they are congruent. By Theorem 10.5, AD 5 2(3x 1 2) 5 6x 1 4.

6x 1 4 5 22

6x 5 18

x 5 3

11. } EF and }

HG are congruent, so by Theorem 10.6, they are equidistant from Q.

4x 1 1 5 x 1 8

3x 5 7

x 5 7 }

3

12. Because }

AB is a perpendicular bisector of }

CD , it is a diameter by Theorem 10.4.

13. Because } JH is a diameter and it is perpendicular to }

FG , it bisects

} FG and C FG by Theorem 10.5.

14. Because } NP and } LM are equidistant from the center, } NP > } LM by Theorem 10.6.

15. D; Choice A is true because they are radii of the circle. By Theorem 10.5 choice C is true. Choice B is true because the two triangles are congruent by the SAS Congruence Postulate.

16. From the diagram, CD 5 12 and EF 5 14. However, by Theorem 10.6, CD 5 EF.

17. You do not know that }

AC ⊥ } BD , therefore you cannot show that C BC > C CD .

18. }

AB is a perpendicular bisector of }

CD , so by Theorem 10.4, it is a diameter.

19. The two triangles are congruent using the SAS Congruence Postulate. So,

} AB is the perpendicular

bisector of }

CD . By Theorem 10.4, }

AB is a diameter.

20. By the Pythagorean Theorem, ED 5 4. CE Þ ED, so }

AB is not a perpendicular bisector of

} CD . Therefore,

} AB is

not a diameter.

21. Using the facts that nAPB is equilateral which makes it equiangular and that m C AC 5 308, you can conclude that m∠ APD 5 m∠ BPD 5 308. You now know that m C BC 5 308 which makes

} AC >

} BC . nAPD > nBPD

using the SAS Congruence Postulate because } BP > }

AP and } PD > } PD . Using corresponding parts of congruent triangles are congruent,

} AD > } BD . Along with

}

DC > }

DC , you have nADC > nBDC using the SSS Congruence Postulate.

22. a. The converse of Theorem 10.5 is: If a diameter of a circle bisects a chord and its arc, then the diameter is perpendicular to the chord. The converse is different from Theorem 10.4 in that in the hypothesis, it is known that the fi rst chord is a diameter.

b. P

R

CQS

T

Statements Reasons

1. } PT > } TR , C PS > C SR 1. Given

2. } CT > }

CT 2. Refl exive Property

3. } PC > }

CR 3. All radii in a circle are congruent.

4. nPCT > nRCT 4. SSS Congruence Postulate

5. ∠CPT > ∠CTR 5. Congruent parts of congruent ns are congruent.

6. m∠CTP 1 m∠CTR5 1808

6. Linear Pair Postulate

7. m∠CTP 5 908 7. Substitution Property

8. } CT ⊥ } PR 8. Def of ⊥


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327Geometry


c. Using the converse of Theorem 10.5, you know that }

PT > } RT and }

QT ⊥ } PR . By the Refl exive Property, }

QT > }

QT . Using the SAS Congruence Postulate, nPQT > nRQT.

} PQ >

} RQ because corresponding

parts of congruent triangles are congruent. So, PQ 5 RQ.

23. From the diagram, m C AB 5 x8. By Theorem 10.3, m C BC 5 m C CA . Let y8 5 m C BC 5 m C CA . Then

m C AB 1 m C BC 1 m C CA 5 3608

x8 1 y8 1 y8 5 3608

x 5 360 2 2y

x is an even number because (360 2 2y) is even.

24.

AB

T

R

P

In nPAT, PA 5 10 and AT 5 8.

m∠APT 5 sin21 1 8 } 10

2 5 53.138.

In nPBR, PB 5 10 and BR 5 6.

m∠BPR 5 sin21 1 6 } 10

2 5 36.868.

m C AB 5 m C AR 2 m C BR 5 53.138 2 36.868 5 16.268

Problem Solving

25. In order for C AB > C BC , }

AB should be congruent to }

BC .

26. To fi nd the center of the cross section, you can (1) construct the perpendicular bisector of the control panel, extend this segment to the other control panel, and then fi nd the midpoint of the segment. Or you can (2) draw two diagonals from the top of one control panel to the bottom of the other. The center will be at their intersection.

27.

Statements Reasons

1. } AB > }

CD 1. Given

2. } PA , } PB , }

PC , and } PD are radii of (P.

2. Given

3. } PA > } PB > }

PC > } PD 3. All radii of same circle are congruent.

4. nPCD > nPAB 4. SSS Congruence Postulate

5. ∠CPD > ∠APB 5. Corr. parts of > ns are >.

6. m∠CPD 5 m∠ APB 6. Def. of >

7. ∠CPD and ∠ APBare central ∠s.

7. Given

8. m C CD 5 m C AB 8. Def. of arc measure

9. C CD > C AB 9. Def. of >

28. Because C AB > C CD , ∠APB > ∠CPD by the defi nition of congruent arcs.

} PA , } PB ,

} PC , and } PD are all radii of (P,

so }

PA > } PB > }

PC > } PD . Then nAPB > nCPD by the SAS Congruence Postulate, so corresponding sides

} AB

and }

CD are congruent.

29. a.

D

CB

A The longer chord is closer to

the center.

b. The length of a chord in a circle increases as the distance from the center of the circle to the chord decreases.

c.

xyb

d Let c be the radius.

Let x < y.

By the Pythagorean Theorem,

c2 5 b2 1 x2 and c2 5 d2 1 y2

So, b2 1 x2 5 d2 1 y2. Since x < y, b > d. Therefore, the length of a chord in a circle increases as the distance from the center decreases.

30. a. r2 5 1602 1 (r 2 80)2

r2 5 25,600 1 r2 2 160r 1 6400

160r 5 32,000

r 5 200

The radius of the circle is 200 feet.

b. S 5 3.86 Ï}

fr 5 3.86 Ï}

(0.7)(200) ø 45.67

The car’s speed is about 45.7 miles per hour.

31. Given }

QS is the perpendicular bisector of } RT in (L. Suppose L is not on

} QS . Since } LT and } LR are radii of the

circle, they are congruent. With } PL > } PL by the Refl exive Property, you now have nRLP > nTLP using the SSS Congruence Postulate. Corresponding angles ∠RPL and ∠TPL are now congruent and they form a linear pair. This makes them right angles and leads to

} QL being

perpendicular to } RT . Using the Perpendicular Postulate, L must be on

} QS and thus

} QS must be a diameter.

32. Draw radii } LD and } LF . So, } LD > } LF , }

LC > }

LC (Refl exive Property) and since

} EG ⊥ } DF ,

nLDC > nLFC by the HL Congruence Theorem. Then, corresponding sides

} DC and

} FC are congruent,

as are corresponding angles ∠DLC and ∠FLC. By the defi nition of congruent arcs, C DG > C FG .

33. Case 1: Given: } EF ⊥ }

AB , }

EG ⊥ }

DC , }

EG ⊥ } EF

Prove: }

AB > }

DC

Draw radii } EB and }

EC . } EB > }

EC and } EF > }

EG . Also, since } EF ⊥

} AB and

} EG ⊥

} DC , nEFB and

nEGC are right triangles and are congruent by the HL Congruence Theorem. Corresponding sides } BF and

} CG

are congruent, so } BF 5 }

CG and, by the Multiplication Property of Equality, 2BF 5 2CG. By Theorem 10.5, } EF bisects

} AB and

} EG bisects

} CD , so AB 5 2BF and

CD 5 2CG. Then by the Substitution Property, AB 5 CD or

} AB >

} CD .


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Case 2: Given: } EF ⊥ }

AB , }

EG ⊥ }

DC , }

AB > }

DC

Prove: } EF > }

EG

Draw radii } EB and }

EC , } EB > } EC . By Theorem 10.5, } EF bisects

} AB and

} EG bisects

} CD , so AB 5 2BF and

CD 5 2CG. We know AB 5 CD, so by the Substitution Property, 2BF 5 2CG. By the Division Property of Equality, BF 5 CG, or } BF >

} CG . nEFB and nEGC are

right triangles and are congruent by the HL Congruence Theorem. It follows that corresponding sides } EF and

} EG

are congruent.

34.

T

Q

P

A B

The point where the tire touches the ground is a point of tangency. By Theorem 10.1, line g is perpendicular to radius of the circle, } TP . Because

} AB i g, } PT ⊥

} AB . By

Theorem 10.5, }

PQ bisects }

AB and C AB .

Mixed Review

35. 1008 1 1408 1 (x 1 20)8 1 (2x 1 10)8 5 3608

3x 1 270 5 360

3x 5 90

x 5 30

36.J K

M L

1

x

y

21

JKLM is a rectangle.

37. JK

ML

2

x

y

21

JKLM is a rhombus.

Quiz 10.1–10.3 (p. 670)

1. CA2 0 CB2 1 BA2

152 0 92 1 122

225 0 81 1 144

225 5 225

}

AB is tangent to (C at B because m∠ ABC 5 908 and radius

} CB is perpendicular to

} AB .

2. CB2 5 CA2 1 AB2

142 0 52 1 122

196 0 25 1 144

196 Þ 169

}

AB is not tangent to (C at A because m∠ BAC Þ 908 and radius

} CA is not perpendicular to

} AB .

3. m C EFG 5 m C EF 1 m C FG

1958 5 808 1 m C FG

1158 5 m C FG

m C EG 5 3608 2 m C EFG 5 3608 2 1958 5 1658

4.

CA

D

B

m C ABD 5 m C AB 1 m C BD

m C ABD 5 m C AB 1 m C AB

m C ABD 5 2m C AB

1948 5 2m C AB

978 5 m C AB

Lesson 10.4


1. Answers will vary. 2. Answers will vary.

3. The measure of an inscribed angle is one half the measure of the corresponding central angle.


1. m∠HGF 5 1 }

2 m C HF 5

1 }

2 (908) 5 458

2. m C TV 5 2m∠TUV 5 2(388) 5 768

3. m∠ ZXW 5 m∠ ZYW 5 728

4. To frame the front and left side of the statue in your picture, make the diameter of your circle the diagonal of the rectangular base. This diagonal connects the upper left corner to the bottom right corner.

5. m∠ B 1 m∠ D 5 1808

x8 1 828 5 1808

x 5 98

m∠ C 1 m∠ A 5 1808

688 1 y8 5 1808

y 5 112

6. m∠S 1 m∠U 5 1808

c8 1 (2c 2 6)8 5 1808

3c 5 186

c 5 62

m∠T 1 m∠V 5 1808

10x8 1 8x8 5 1808

18x 5 180

x 5 10


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329Geometry


10.4 Exercises (pp. 676–679)

Skill Practice

1. If a circle is circumscribed about a polygon, then the polygon is inscribed in the circle.

2. The diagonals of a rectangle create two right triangles. Theorem 10.9 tells you the hypotenuse of each of these triangles is a diameter of the circle.

3. m∠A 5 1 }

2 m C BC 5

1 }

2 (848) 5 428

4. m C FD 5 3608 2 1208 2 708 5 1708

m∠G 5 1 }

2 m C FD 5

1 }

2 (1708) 5 858

5. m C LM 5 1808 2 1608 5 208

m∠N 5 1 }

2 m C LM 5

1 }

2 (208) 5 108

6. m C RS 5 2m∠ Q 5 2(678) 5 1348

7. m C TV 5 2m∠U 5 2(308) 5 608

m C UV 5 1808 2 m C TV 5 1808 2 608 5 1208

8. m C YW 5 2m∠ X 5 2(758) 5 1508

m C WX 5 3608 2 m C XY 2 m C YW

5 3608 2 1108 2 1508 5 1008

9. From the diagram, the measure of C RS is 908. So, the measures of the arcs add up to 3708. You can either change the measure of ∠Q to 408 or change the measure of C QS to 908.

10. ∠ ADB > ∠ ACB because they intercept the same arc, C AB .∠ CBD > ∠ DAC because they intercept the same arc, C CD .

11. ∠ JMK > ∠ KLJ because they intercept the same arc, C JK . ∠ MKL > ∠ LJM because they intercept the same arc, C LM .

12. ∠WXZ > ∠ZYW because they intercept the same arc, C ZW . ∠XWY > ∠YZX because they intercept the same arc, C XY .

13. m∠ R 1 m∠T 5 1808

x8 1 808 5 1808

x 5 100

m∠ S 1 m∠Q 5 1808

y8 1 958 5 1808

y 5 85

14. m∠D 1 m∠F 5 1808

608 1 2k8 5 1808

2k 5 120

k 5 60

m∠E 1 m∠G 5 1808

m8 1 608 5 1808

m 5 120

15. m∠J 5 1 }

2 m C KLM 5

1 }

2 (1308 1 1108) 5 1208

m∠ J 1 m∠L 5 1808

1208 1 3a8 5 1808

3a 5 60

a 5 20

m∠K 5 1 }

2 m C JML 5

1 }

2 (548 1 1308) 5 928

m∠K 1 m∠M 5 1808

928 1 4b8 5 1808

4b 5 88

b 5 22

16. B; m C AC 5 608

m∠ B 5 1 }

2 m C AC 5

1 }

2 (608) 5 308

17. a. There are 5 congruent arcs. 3608 4 5 5 728. The

measure of each inscribed angle is 1 }

2 the measure of

the arc. 1 }

2 (728) 5 368. There are 5 inscribed angles.

Sum 5 5(368) 5 1808.

b. There are 7 congruent arcs. 3608 4 7 ø 51.48. The


2 the measure of

the arc. 1 }

2 (51.48) ø 25.78. There are 7 inscribed angles.

Sum ø 7(25.78) ø 1808.

c. There are 9 congruent arcs. 3608 4 9 5 408. The


2 the measure of

the arc. 1 }

2 (408) 5 208. There are 9 inscribed angles.

Sum 5 9(208) 5 1808.

18. A; 2m∠EFG 5 m C EG

2(8x 1 10)8 5 (12x 1 40)8

16x 1 20 5 12x 1 40

4x 5 20

x 5 5

19. In a parallelogram, the opposite angles are congruent. In an inscribed parallelogram, opposite angles are supplementary. A rectangle is a parallelogram with congruent supplementary opposite angles. The m∠ R is 908.

20. A square can always be inscribed in a circle because its opposite angles are 908 and thus are supplementary.

21. A rectangle can always be inscribed in a circle because its opposite angles are 908 and thus are supplementary.

22. A parallelogram cannot always be inscribed in a circle because its opposite angles are not always supplementary.

23. A kite cannot always be inscribed in a circle because its opposite angles are not always supplementary.

24. A rhombus cannot always be inscribed in a circle because its opposite angles are not always supplementary.

25. An isosceles trapezoid can always be inscribed in a circle because its opposite angles are supplementary.


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26.3

4

5

JK

A B

C

} JK is a diameter. The altitude from C to }

AB is also a

diameter. nABC is a right triangle, so sin A 5 4 } 5 . Using

the smaller triangle, sin A 5 alt

} 3 . So,

4 } 5 5

alt }

3 → alt 5

12 } 5 .

Therefore, the diameter } JK is also 12

} 5 .

Problem Solving

27.

CA

B

100,000 km

20,000 km

AC 5 100,000 1 20,000 1 100,000 5 220,000

Moon A is 220,000 km from moon C.

28. Place the carpenter’s square so the endpoints of the square and the vertex of the square are on the circumference of the circle, then connect the endpoints.

29. The hypotenuse of the right triangle inscribed in the circle is the diameter of the circle. So, double the length of the radius to fi nd the length of the hypotenuse.

30. By the Arc Addition Postulate, m C EFG 1 m C GDE 5 3608 and m C FGD 1 m C DEF 5 3608. Using the Measure of an Inscribed Angle Theorem, m C EDG 5 2m∠F, m C EFG 5 2m∠D, m C DEG 5 2m∠G, and m C FGD 5 2m∠E. By the Substitution Property, 2m∠D 1 2m∠F 5 3608, so m∠D 1 m∠ F 5 1808. Similarly, m∠E 1 m∠G 5 1808.

31. Let m∠B 5 x8. Because }

QA and }

QB are both radii of (Q,

} QA > } QB and nAQB is isosceles. Because ∠ A and

∠B are base angles of an isosceles triangle, ∠ A > ∠B. So, by substitution, m∠A 5 x8. By the Exterior Angles Theorem, m∠AQC 5 m∠A 1 m∠B 5 2x8. So, by the defi nition of the measure of a minor arc, m C AC 5 2x8.

Divide each side by 2 to show that x8 5 1 }

2 m C AC . Then,

by substitution, m∠B 5 1 }

2 m C AC .

32. Given: ∠ ABC is inscribed in (Q. Point Q is in the interior of ∠ ABC.

Prove: m∠ ABC 5 1 }

2 m C AC

Plan for Proof: Construct the diameter } BD of (Q and

show m∠ ABD 5 1 }

2 m C AD and m∠DBC 5

1 }

2 C DC . Use

the Arc Addition Postulate and the Angle Addition Postulate to show 2m∠ABC 5 m C AD 1 m C DC .

33. Given: ∠ABC is inscribed in (Q. Point Q is in the exterior of ∠ABC.

Prove: m∠ ABC 5 1 }

2 m C AC

Plan for Proof: Construct the diameter } BD of (Q and

show m∠ ABD 5 1 }

2 m C AD and m∠DBC 5

1 }

2 m C DC . Use

the Arc Addition Postulate and the Angle Addition Postulate to show 2m∠ ABC 5 m C AD 2 m C DC .

34. Given: ∠ ACB and ∠ ADB are A

C

D

B

O

inscribed angles.

Prove: ∠ ADB > ∠ ACB

Paragraph Proof:

By Theorem 10.7,

m∠ADB 5 1 }

2 m C AB and m∠ ACB 5

1 }

2 m C AB . By

substitution, m∠ ADB 5 m∠ ACB. By the defi nition of congruence, ∠ ADB > ∠ ACB.

35. Case 1: Given: (T with inscribed

TA C

B

E

nABC. }

AC is a diameter of (T.

Prove: nABC is a right triangle.

Plan for Proof:

Use the Arc Addition Postulate to show

that m C AEC 5 m C ABC and thus m C ABC 5 1808. Then use the Measure of an Inscribed Angle Theorem to show m∠B 5 908, so that ∠B is a right angle and nABC is a right triangle.

Case 2: Given: (T with inscribed nABC. ∠B is a right angle.

Prove: }

AC is a diameter of (T.

Plan for Proof: Use the Measure of an Inscribed Angle Theorem to show the inscribed right angle intercepts an arc with measure 2(908) 5 1808. Since

} AC intercepts an

arc that is half of the measure of the circle, it must be a diameter.

36. In the fi gure, nABC is a right triangle with ∠ABC being the right angle. Using Theorem 10.1, since @##$ AB is perpendicular to radius

} BC , it is tangent to (C at point B.

37. HJ

} GJ

5 GJ

} FJ ; in a right triangle, the altitude from the right

angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of the lengths of these two segments.

38. FJ 5 6 in., JH 5 2 in.

HJ

} GJ

5 GJ

} FJ

2 }

x 5

x }

6

x2 5 12

x 5 2 Ï}

3

GJ 5 2 Ï}

3 in.

GK 5 2GJ 5 2(2 Ï}

3 ) 5 4 Ï}

3 in.


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331Geometry


39. A

B

C PD

1.5

1.5

2

2

Show that m∠ APB is less than or equal to 458. The

diagram shown represents 1 }

11 of the fuel booster design.

∠ACB is equal to 1 }

11 of 3608 or about 32.78.

} CP bisects

∠ACB and ∠APB, so m∠ACD ø 16.358. }

CP is also the perpendicular bisector of

} AB , and contains point D.

nACD and nAPD are right triangles with ∠D 5 908 in both triangles.

sin 16.358 5 AD

} 2 → AD ø 0.563

cos ∠PAD 5 AD

} 1.5

5 0.563

} 1.5

ø 0.375

m∠PAD 5 cos21 0.375 ø 688

Since nAPB is isosceles, the base angles are equal.

m∠PAB 5 m∠PBA 5 688

m∠APB 5 1808 2 m∠PAB 2 m∠PBA

5 1808 2 688 2 688 5 448

So m∠APB is less than 458.

Mixed Review

40. x2 5 552 1 602 41. x2 5 382 1 822

x2 5 3025 1 3600 x2 5 1444 1 6724

x2 5 6625 x2 5 8168

x ø 81.4 x ø 90.4

42. x2 5 262 1 162

x2 5 676 1 256

x2 5 932

x ø 30.5

43.

C 9

A9

B9

1

y

21 x

AB

C

44. E

F

H9

E9

F9

G9

1

y

21 x

G

H

45.

R9

S9

Q9 1

y

22 x

Q

R

S

46.

x

y2

2

AA9

47.

x

y4

22

A

A9 y 5 4x

y 5 2 x 1 114

Lesson 10.5


1. m∠1 5 1 }

2 (2108) 5 1058

2. m C RST 5 2m∠T 5 2(988) 5 1968

3. m C XY 5 2m∠X 5 2(808) 5 1608

4. 1808 2 1028 5 788

788 5 1 }

2 (958 1 y8)

156 5 95 1 y

61 5 y

5. m∠FJG 5 1 }

2 (m C FG 2 m C KH )

308 5 1 }

2 (a8 2 448)

60 5 a 2 44

104 5 a

6. sin TQS 5 3 } 5

m∠TQS ø 36.878

m∠TQR 5 2(m∠TQS) ø 2(36.878) ø 73.748

m∠TQR 5 1 }

2 (m C TUR 2 m C TR )

73.748 5 1 }

2 (x8 2 (360 2 x8))

147.48 5 x 2 360 1 x

507.48 5 2x

253.74 ø x

10.5 Exercises (pp. 683–686)

Skill Practice

1. The points A, B, C, and D are on a circle and @##$ AB

intersects @##$ CD at P. If m∠ APC 5 1 }

2 (m C BD 2 m C AC )

then P is outside the circle.

2. If m C AB 5 08, then the two chords intersect on the circle.

By Theorem 10.12, m∠1 5 1 }

2 (m C DC 1 m C AB )

5 1 }

2 (m C DC 1 08) 5

1 }

2 m C DC .

This is consistent with the measure of an Inscribed Angle Theorem (Lesson 10.4).

3. m∠A 5 1 }

2 m C AB 4. m∠D 5

1 }

2 m C DEF

658 5 1 }

2 m C AB 1178 5

1 }

2 m C DEF

1308 5 m C AB 2348 5 m C DEF

5. m∠1 5 1 }

2 (2608) 5 1308

6. D; Because }

AB is not a diameter, m C AB Þ 1808.

m∠ A 5 1 }

2 m C AB Þ

1 } 2 (1808) Þ 908


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7. x8 5 1 }

2 (1458 1 858) 5

1 }

2 (2308) 5 1158

8. 1808 2 122.58 5 57.58

57.58 5 1 }

2 (x8 1 458)

115 5 x 1 45

70 5 x

9. (180 2 x)8 5 1 }

2 (308 1 (2x 2 30)8)

2(180 2 x) 5 2x

360 2 2x 5 2x

360 5 4x

90 5 x

10. 3608 2 2478 5 1138

x8 5 1 }

2 (2478 2 1138)

x 5 1 }

2 (134)

x 5 67

11. 298 5 1 }

2 (1148 2 x8)

58 5 114 2 x

x 5 56

12. 348 5 1 }

2 [(3x 2 2)8 2 (x 1 6)8]

68 5 (3x 2 2) 2 (x 1 6)

68 5 2x 2 8

76 5 2x

38 5 x

13. D; m∠4 5 1 }

2 (808 1 1208) 5

1 }

2 (2008) 5 1008

14. The error is that the given measurements imply two different measures for C BE . Using Theorem 10.12,

508 5 1 }

2 (m C BE 1 608)

1008 5 m C BE 1 608

408 5 m C BE

Using Theorem 10.13,

158 5 1 }

2 (608 2 m C BE )

308 5 608 2 m C BE

m C BE 5 308

15. If ###$ PL is perpendicular to } KJ at K, then m∠LPJ 5 908, otherwise it would measure less than 908. So, m∠LPJ ≤ 908.

16. a. 1 }

2 (1108) 5 558

558 5 1 }

2 (x8 2 408)

1108 5 x8 2 408

150 5 x

b. 1 }

2 c8 5

1 }

2 (b8 2 a8)

c 5 b 2 a

17. m∠Q 5 1 }

2 (m C EGF 2 m C EF )

608 5 1 }

2 [(360 2 x)8 2 x8]

120 5 360 2 2x

2x 5 240

x 5 120

So, m C EF 5 1208.

m∠R 5 1 }

2 (m C GEF 2 m C FG )

808 5 1 }

2 [(360 2 x)8 2 x8]

160 5 360 2 2x

2x 5 200

x 5 100

So, m C FG 5 1008.

m C GE 5 3608 2 m C EF 2 m C FG

5 3608 2 1208 2 1008 5 1408

So, m C GF 5 1408.

18. m∠B 5 1 }

2 (m C AD 2 m C AC )

408 5 1 }

2 (7x8 2 3x8)

80 5 4x

20 5 x

m C AD 5 7x8 5 7(20)8 5 1408

m C AC 5 3x8 5 3(20)8 5 608

m C CD 5 3608 2 m C AD 2 m C AC

5 3608 2 1408 2 608 5 1608

19. a.

C

A

B

tC

A

B

t

b. For the diagram on the left,

m∠BAC 5 1 }

2 m C AB

2m∠BAC 5 m C AB

For the diagram on the right,

m∠BAC 5 1 }

2 (3608 2 m C BA )

2m∠BAC 5 3608 2 m C BA

m C BA 5 3608 2 2m∠BAC

m C BA 5 2(1808 2 m∠BAC)


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333Geometry


c. 2m∠BAC 5 2(1808 2 m∠BAC)

m∠BAC 5 1808 2 m∠BAC

2m∠BAC 5 1808

m∠BAC 5 908

So, these equations give the same value for m C AB when }

AB is perpendicular to t at point A.

20. Let x8 5 m C XW 5 m C ZY .

Then m C WZ 5 (200 2 x)8 and m C XY 5 (3608 2 (200 1 x)8) 5 (160 2 x)8

m∠P 5 1 }

2 (m C WZ 2 m C XY )

5 1 }

2 [(200 2 x)8 2 (160 2 x)8]

5 1 }

2 (40) 5 208

21. m∠CHD 5 1808 2 1158 5 658

m∠CHD 5 1 }

2 (m C CD 1 m C EA )

658 5 1 }

2 (858 1 m C EA )

1308 5 858 1 m C EA

458 5 m C EA

m C AF 5 m C EA 2 m C EF 5 458 2 208 5 258

m∠J 5 1 }

2 (m C AB 2 m C AF )

308 5 1 }

2 (m C AB 2 258)

608 5 m C AB 2 258

858 5 m C AB

m∠FGH 5 1 }

2 (m C AB 1 m C FD )

908 5 1 }

2 (858 1 (208 1 m C ED ))

1808 5 1058 1 m C ED

758 5 m C ED

Problem Solving

22. m∠A 5 808

m∠B 5 1 }

2 (808 2 308) 5

1 }

2 (508) 5 258

m∠B 5 1 }

2 (808) 5 408

23. x8 5 1 }

2 (1808 2 808) 5

1 }

2 (1008) 5 508

24. m∠ B 5 1 } 2 (808 2 x8)

308 5 1 } 2 (808 2 x8)

608 5 808 2 x8

x8 5 208

Camera B will have a 308 view of the stage when the arc measuring 308 is reduced to an arc measuring 208. You should move the camera closer to the stage.

25.

A

E

B D

C

4000 mi

4001.2 mi

Not drawn to scale

sin BCA 5 4000

} 4001.2

→ m∠BCA ø 88.68

m∠ BCD ø 2(88.68) ø 177.28

Let m C BD 5 x8.

m∠ BCD 5 1 }

2 (m C DEB 2 m C BD )

177.28 ø 1 } 2 [(3608 2 x8) 2 x8]

x ø 2.8

The measure of the arc from which you can see is about 2.88.

26.

14° x° (180 2 x)°

148 5 1 }

2 [(180 2 x)8 2 x8]

28 5 180 2 2x

2x 5 152

x 5 76

(180 2 x)8 5 (180 2 76)8 5 1048

The measures of the arcs between the ground and the cart are 768 and 1048.

27. Case 1: Given: tangent @##$ AC intersects chord }

AB at point A on (Q.

} AB contains the center of (Q.

Prove: m∠CAB 5 1 }

2 m C AB

AC

B

Q

Case 1

Paragraph proof: By Theorem 10.1, @##$ CA ⊥ }

AB . By the defi nition of perpendicular, m∠ CAB 5 908. Because

}

AB is a diameter, m C AB 5 1808. So, m∠ CAB 5 1 }

2 m C AB .

Case 2: Given: tangent @##$ AC intersects chord }

AB at point A on (Q. The center of the circle, Q, is in the interior of ∠CAB.

Prove: m∠CAB 5 1 }

2 m C APB

A

P

C

B

Q

Case 2


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Plan for proof: Draw diameter }

AP . Use the Angle Addition Postulate and Theorem 10.7 to show that

m∠CAB 5 908 1 1 }

2 m C PB . Use the Arc Addition Postulate

to show that m C APB 5 1808 1 m C PB .

Case 3: Given: Tangent @##$ AC intersects chord }

AB at point A on (Q. The center of the circle, Q, is in the exterior of ∠CAB.

Prove: m∠ CAB 5 1 }

2 m C AB

A

P

C

B

Q

Case 3

Plan for Proof: Draw diameter }

AP . Use the Angle Addition Postulate and Theorem 10.7 to show that

m∠ CAB 5 908 2 1 }

2 m C PB . Use the Arc Addition Postulate

to show that m C AB 5 1808 2 m C PB .

28. Given: Chords }

AC and } BD

A

B

C

D

1intersect.

Prove: m∠ 1 5 1 }

2 (m C DC 1 m C AB )

Statements Reasons

1. Chords }

AC and } BD intersect.

1. Given

2. Draw }

BC . 2. Given

3. m∠ 1 5 m∠ DBC 1 m∠ ACB

3. Exterior Angle Theorem

4. m∠ DBC 5 1 }

2 m C DC 4. Theorem 10.7

5. m∠ ACB 5 1 }

2 m C AB 5. Theorem 10.7

6. m∠ 1 5 1 }

2 m C DC 1

1 }

2 m C AB 6. Substitution Property

7. m∠ 1 5 1 }

2 (m C DC 1 m C AB ) 7. Distributive Property

29.

C

1

2

A

B

Case 1

Case 1: Draw }

BC . Use the Exterior Angle Theorem to show that m∠ 2 5 m∠ 1 1 m∠ ABC, so that m∠ 1 5 m∠ 2 2 m∠ ABC. Then use Theorem 10.11

to show that m∠ 2 5 1 }

2 m C BC and the Measure of an

Inscribed Angle Theorem to show that m∠ABC 5 1 }

2 m C AC .

Then, m∠ 1 5 1 }

2 (m C BC 2 m C AC ).

R

2

4

3

PQ

Case 2

Case 2: Draw } PR . Use the Exterior Angle Theorem to show that m∠ 3 5 m∠ 2 1 m∠ 4, so that m∠ 2 5 m∠ 3 2 m∠ 4. Then use Theorem 10.11 to show that

m∠ 3 5 1 }

2 m C PQR and m∠ 4 5

1 }

2 m C PR . Then,

m∠ 2 5 1 }

2 (m C PQR 2 m C PR ).

3 4

W

Z

X

Y

Case 3

Case 3: Draw } XZ . Use the Exterior Angle Theorem to show that m∠ 4 5 m∠ 3 1 m∠ WXZ, so that m∠ 3 5 m∠ 4 2 m∠ WXZ. Then use the Measure of an

Inscribed Angle Theorem to show that m∠ 4 5 1 }

2 m C XY and

m∠ WXZ 5 1 }

2 m C WZ . Then, m∠ 3 5

1 }

2 (m C XY 2 m C WZ ).

30.

P

Q

R

T

Given: }

PQ and } PR are tangents to a circle.

Prove: }

QR is not a diameter.

Paragraph Proof: Assume }

QR is a diameter. Then

m C QTR 5 m C QR 5 1808. By Theorem 10.13,

m∠ P 5 1 }

2 (m C QTR 2 m C QR ) 5

1 }

2 (1808 2 1808) 5 08

The m∠ P cannot equal 08, so }

QR cannot be a diameter.

31.

113 cm

D

EB C

15 cm

15 cm

DE 5 113, EC 5 15

DE2 5 DC2 1 EC2

1132 5 DC2 1 152

12,544 5 DC2

112 5 DC

m∠ DEC 5 sin21 1 112 }

113 2 ø 82.48

m C BC 5 m∠ BED 1 m∠ DEC

5 908 1 82.48 5 172.48

172.48

} 3608

ø 48%

About 48% of the circumference of the bottom pulley is not touching the rope.


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335Geometry


Mixed Review

32. Because CP9

} CP

5 12

} 16

5 3 }

4 , the scale factor is K 5

3 }

4 . This is

a reduction.

33. Because CP9

} CP

5 15

} 9 5

5 }

3 , the scale factor is K 5

5 }

3 . This is

an enlargement.

34. x2 1 7x 1 6 5 0

x 5 27 6 Ï

}}

72 2 4(1)(6) }}

2(1)

5 27 6 Ï

}

25 } 2

5 27 6 5

} 2 → x 5 21 or x 5 26

35. x2 2 x 2 12 5 0

x 5 2(21) 6 Ï

}}

(21)2 2 4(1)(212) }}}

2(1)

5 1 6 Ï

}

49 }

2 5

1 6 7 }

2 → x 5 4 or x 5 23

36. x2 1 16 5 8x

x2 2 8x 1 16 5 0

x 5 2(28) 6 Ï

}}

(28)2 2 4(1)(16) }}}

2(1)

5 8 6 Ï

}

0 }

2 5 4

37. x2 1 6x 5 10

x2 1 6x 2 10 5 0

x 5 26 6 Ï

}}

62 2 4(1)(210) }}

2(1)

5 26 6 Ï

}

76 }

2 → x ø 1.36 or x ø 27.36

38. 5x 1 9 5 2x2

22x2 1 5x 1 9 5 0

x 5 25 6 Ï

}}

52 2 4(22)(9) }}

2(22)

5 25 6 Ï

}

97 }

24 → x ø 21.21 or x ø 3.71

39. 4x2 1 3x 2 11 5 0

x 5 23 6 Ï

}}

32 2 4(4)(211) }}

2(4)

5 23 6 Ï

}

185 } 8 → x ø 1.33 or x ø 22.08

Quiz 10.4–10.5 (p. 686)

1. m∠ D 1 m∠ B 5 1808

x8 1 858 5 1808

x 5 95

m∠ C 1 m∠ A 5 1808

y8 1 758 5 1808

y 5 105

m C ABC 5 2m∠ D 5 2(958) 5 1908

So, z 5 190.

2. m∠ E 1 m∠ G 5 1808

x8 1 1128 5 1808

x 5 68

m∠ F 1 m∠ H 5 1808

2m∠ F 5 180

m∠ F 5 908

m C GHE 5 2m∠ F 5 2(908) 5 1808

So, z 5 180.

3. m∠ K 1 m∠ M 5 1808

7x8 1 1318 5 1808

7x8 5 49

x 5 7

m∠ L 1 m∠ J 5 1808

(11x 1 y)8 1 998 5 1808

(11(7) 1 y) 1 99 5 180

77 1 y 5 81

y 5 4

m C JKL 5 2m∠ M 5 2(1318) 5 2628

So, z 5 262.

4. x8 5 1 }

2 (1078 1 838) 5

1 }

2 (1908) 5 958

5. x8 5 1 }

2 (748 2 228) 5

1 }

2 (528) 5 268

6. 618 5 1 }

2 (x8 2 878)

122 5 x 2 87

209 5 x

7.

A

E

B D

C

4000 mi

4001.37 mi

Not drawn to scale

sin BCA 5 4000

} 4001.37

→ m∠ BCA ø 88.58

m∠ BCD ø 2(88.58) ø 1778

Let m C BD 5 x8.

m∠ BCD 5 1 }

2 (m C DEB 2 m C BD )

1778 ø 1 }

2 [(3608 2 x8) 2 x8]

354 ø 360 2 2x

26 ø 2x

x ø 3 The measure of the arc from which you can see is

about 38.


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Mixed Review of Problem Solving (p. 687)

1. a. r2 1 32 5 (r 1 2)2

r2 1 9 5 r2 1 4r 1 4

0 5 4r 2 5

5 5 4r

1.25 5 r

The radius of the discus circle is 1.25 meters.

b. 1.25 1 2 5 3.25

The offi cial is 3.25 meters form the center of the discus circle.

2. m C XYZ 5 3608 2 m C XQZ

5 3608 2 1998 5 1618

m C YZ 5 1 }

2 m C XYZ 5

1 }

2 (1618) 5 80.58

3. a. 3608 4 3 5 1208

b. x 5 361 2 2(131) 5 361 2 262 5 99

The distance between the lowest point reached by the blades and the ground is 99 feet.

c.

30° 30°

131131

xy

cos 308 5 x }

131

x ø 113.45

y 5 2x ø 2(113.45) ø 226.9

The distance from the tip of one blade to the tip of another is about 226.9 feet.

4. a. 3608 4 40 5 98

b. 728 4 98 5 8; 8 angles and 9 total spokes. There are 7 spokes between the two spokes.

c. The wheel itself is 150 2 10 5 140 feet tall. The diameter is the total height, or 140 feet. The radius is half the height, or 70 feet.


6. a. x8 5 1 }

2 (m C NM 1 m C LK )

5 1 }

2 (358 1 938) 5 648

b. m∠ LDK 5 m∠ MDN 5 648

m∠ LDM 5 1808 2 m∠ MDN 5 1808 2 648 5 1168

m∠ KDN 5 m∠ LDM 5 1168

7. x8 5 1 }

2 (m C BC 2 m C AD )

2x8 5 m C BC 2 m C AD

m C BC 5 2x8 1 m C AD

y8 5 1 }

2 (m C BC 1 m C AD )

2y8 5 m C BC 1 m C AD

m C BC 5 2y8 2 m C AD

2x8 1 m C AD 5 2y8 2 m C AD

2m C AD 5 2y8 2 2x8

m C AD 5 y8 2 x8

Lesson 10.6


1. The products AE p CE and BE p DE are the same.

2. The products AE p CE and BE p DE are the same.

3. AE p CE 5 BE p DE

4. PT p QT 5 RT p ST

9 p 5 5 15 p ST

45 5 15ST

3 5 ST


1. 6 p (6 1 9) 5 5 p (5 1 x) 2. 3 p x 5 6 p 4

90 5 25 1 5x 3x 5 24

65 5 5x x 5 8

13 5 x

3. 3 p [3 1 (x 1 2)] 5 (x 1 1) p [(x 1 1) 1 (x 2 1)]

3(x 1 5) 5 (x 1 1)(2x)

3x 1 15 5 2x2 1 2x

2x2 2 x 2 15 5 0

(2x 1 5)(x 2 3) 5 0

x 5 3 1 x 5 2 5 } 2 is extraneous. 2

4. x2 5 1(3 1 1)

x2 5 4

x 5 2 (x 5 22 is extraneous.)

5. 72 5 5 p (5 1 x)

49 5 25 1 5x

24 5 5x

24

} 5 5 x

6. 122 5 x p (x 1 10)

144 5 x2 1 10x

x2 1 10x 2 144 5 0

(x 2 8)(x 1 18) 5 0



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337Geometry


7. Use Theorem 10.16.

152 5 x p (x 1 14)

225 5 x2 1 14x

x2 1 14x 2 225 5 0

x 5 214 6 Ï

}}

142 2 4(1)(2225) }}}

2(1)

x 5 214 6 2 Ï

}

274 }}

2

x 5 27 6 Ï}

274

x 5 27 1 Ï}

274

(x 5 27 2 Ï}

274 is extraneous.)8. Use Theorem 10.14.

x p 18 5 9 p 16

18x 5 144

x 5 8

9. Use Theorem 10.15.

18 p (18 1 22) 5 x p (x 1 29)

720 5 x2 1 29x

x2 1 29x 2 720 5 0

(x 2 16)(x 1 45) 5 0


10. EC cannot be equal to EA because that would make }

EC tangent to the circle. EC must be less than EA.

11. It is appropriate to use the approximation symbol in the solution to Example 4 because it is stated in the problem that each moon has a nearly circular orbit.

10.6 Exercises (pp. 692–695)

Skill Practice

1. The part of the secant segment that is outside the circle is called an external segment.

2. A tangent segment intersects the circle in only one point while the secant segment intersects the circle in two points.

3. 12 p x 5 10 p 6

12x 5 60

x 5 5

4. 9 p (x 2 3) 5 10 p 18

9x 2 27 5 180

9x 5 207

x 5 23

5. x p (x 1 8) 5 6 p 8

x2 1 8x 5 48

x2 1 8x 2 48 5 0

(x 2 4)(x 1 12) 5 0


6. 8 p (8 1 x) 5 6 p (6 1 10)

64 1 8x 5 96

8x 5 32

x 5 4

7. x p (x 1 4) 5 5 p (5 1 7)

x2 1 4x 5 60

x2 1 4x 2 60 5 0

(x 2 6)(x 1 10) 5 0


8. (x 2 2) p [(x 2 2) 1 (x 1 4)] 5 4 p (4 1 5)

(x 2 2)(2x 1 2) 5 36

2x2 2 2x 2 4 5 36

2x2 2 2x 2 40 5 0

x2 2 x 2 20 5 0

(x 2 5)(x 1 4) 5 0


9. x2 5 9 p (9 1 7)

x2 5 144


10. 242 5 12 p (12 1 x)

576 5 144 1 12x

432 5 12x

36 5 x

11. (x 1 4)2 5 x p (x 1 12)

x2 1 8x 1 16 5 x2 1 12x

16 5 4x

4 5 x

12. The error is that the wrong segment lengths are being multiplied. Use Theorem 10.15.

FD p FC 5 FA p FB

4 p (4 1 CD) 5 3 p (3 1 5)

16 1 4CD 5 24

4CD 5 8

CD 5 2

13. 15 p (x 1 3) 5 2x p 12 14. x p 45 5 27.50

15x 1 45 5 24x 45x 5 1350

45 5 9x x 5 30

5 5 x

15. ( Ï}

3 )2 5 x p (2 1 x)

3 5 2x 1 x2

x2 1 2x 2 3 5 0

(x 2 1)(x 1 3) 5 0


16. D; x p x 5 2 p (2x 1 6)

x2 5 4x 1 12

x2 2 4x 2 12 5 0

(x 2 6)(x 1 2) 5 0


17. MN2 5 PN p (PN 1 PQ)

122 5 6 p (6 1 PQ)

144 5 36 1 6PQ

108 5 6PQ

18 5 PQ


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18. RQ2 5 RS p (RS 1 SP)

RQ2 5 14 p (14 1 12)

RQ2 5 364

PQ2 1 RQ2 5 RP2

PQ2 1 364 5 262

PQ2 5 312

PQ ø 17.7

19. AB2 5 BC p (BC 1 CD 1 DE)

122 5 8 p (8 1 CD 1 6)

144 5 112 1 8CD

4 5 CD

P

E

C

D

r

446

CD p DE 5 (r 1 PD) p (r 2 PD)

4 p 6 5 (r 1 4) p (r 2 4)

24 5 r2 2 16

40 5 r2

2 Ï}

10 5 r

Problem Solving

20. x p x 5 62 p (250 2 62)

x2 5 11,656

x ø 107.96

The distance from the end of the passage to either side of the mound is about 108 feet.

21. Given: }

AB and }

CD are chords

A

C

DB

E

that intersect at E.

Prove: EA p EB 5 EC p ED

Statements Reasons

1. } AB and }

CD are chords that intersect at E.

1. Given

2. Draw } BD and }

AC . 2. Two points determine a line.

3. ∠C > ∠B 3. ∠ C and ∠ B intercept the same arc.

4. ∠ A > ∠ D 4. ∠ A and ∠ D intercept the same arc.

5. nAEC , nDEB 5. AA Similarity Postulate

6. EA } ED 5

EC } EB 6. Corresponding sides are

proportional.

7. EA p EB 5 EC p ED 7. Cross Product Property

22. AD p (AD 1 DE) 5 AB p (AB 1 BC)

(AD)2 1 (AD)(DE) 5 (AB)2 1 (AB)(BC)

(AD)2 1 (AD)(DE) 2 (AB)2 5 (AB)(BC)

(AD)2 1 (AD)(DE) 2 (AB)2

}}} AB

5 BC

23. Given: }

EA is a tangent A

DE CT

segment and } ED is a secant segment passing through the center.

Prove: (EA)2 5 EC p ED

Proof: Draw radius }

AT . By Theorem 10.1, }

EA ⊥ }

AT . By the Pythagorean Theorem, EA2 1 AT 2 5 ET 2. By the Segment Addition Postulate, ET 5 EC 1 CT. So, EA2 1 AT 2 5 (EC 1 CT)2 by substitution. Simplify the equation.

EA2 1 AT 2 5 (EC 1 CT)2

EA2 1 AT 2 5 EC2 1 (EC )(CT ) 1 CT2

AT and CT are both radii, so they are equal. Substitute CT for AT.

EA2 1 CT2 5 EC2 1 2(EC)(CT) 1 CT 2

EA2 5 EC(EC 1 2CT)

EA2 5 EC(EC 1 CD)

EA2 5 EC p ED

24. 4 p CN 5 6 p 8

CN 5 12

The length CN is 12 centimeters. Sparkles travel from

C to D in 6 cm

} 2 cm/sec

5 3 seconds. So, sparkles must travel

from C to N in 3 seconds. The sparkles must travel at a

rate of 12 cm

} 3 sec

5 4 cm/sec.

25. Given: } EB and } ED are secant segments.

Prove: EA p EB 5 EC p ED

A

D

EC

B

Paragraph proof: Draw }

AD and }

BC . ∠ B and ∠ D intercept the same arc, so ∠ B > ∠ D. ∠ E > ∠ E by the Refl exive Property of Congruence, so nBCE , nDAE by the AA Similarity Theorem. Then, since lengths of corresponding sides of similar triangles are proportional,

EA

} EC

5 ED

} EB . By the Cross Product Property,

EA p EB 5 EC p ED.


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339Geometry


26. Given: }

EA is a tangent segment. } ED is a secant segment.

Prove: EA2 5 EC p ED

A

D

EC

Paragraph proof: Draw }

AC and }

AD . m∠ D 5 1 }

2 m C AC and

m∠ EAC 5 1 }

2 m C AC . So, ∠ D > ∠ EAC. ∠ E > ∠ E by the

Refl exive Property of Congruence, so nDAE , nACE by the AA Similarity Theorem. Then, since lengths of corresponding sides of similar triangles are proportional,

EA

} EC

5 ED

} EA

. By the Cross Product Property,

(EA)2 5 EC p ED.

27. a. m∠ CAB 5 1 }

2 m C BD 5

1 }

2 (1208) 5 608

b. m∠ CAB 5 m∠ EFD 5 608, so ∠ CAB > ∠ EFD. Also, ∠ ACB > ∠ FCE by the Vertical Angles Theorem. So, nABC , nFEC by the AA Similarity Theorem.

c. EF

} BA

5 FC

} AC

y }

3 5

x 1 10 }

6

y 5 3 1 x 1 10 }

6 2

y 5 x 1 10

} 2

d. Use Theorem 10.16.

EF2 5 FD p FA

y2 5 x p (x 1 10 1 6)

y2 5 x(x 1 16)

e. y2 5 x(x 1 16)

1 x 1 10 }

2 2 2 5 x(x 1 16)

x2

} 4 1 5x 1 25 5 x2 1 16x

x2 1 20x 1 100 5 4x2 1 64x

3x2 1 44x 2 100 5 0

(3x 1 50)(x 2 2) 5 0

x 2 2 5 0 or 3x 1 50 5 0

x 5 2 x 52 50

} 3

Length cannot be negative, so x 5 2.

y 5 x 1 10

} 2 5

2 1 10 }

2 5 6

f. EF

} AB

5 6 }

3 5

2 }

1 , so the ratio of nFEC to nABC is 2 to 1.

So, CF

} CB

5 2 }

1 . Let CE 5 2x and CB 5 x.

By Theorem 10.14,

CE p CB 5 AC p CD

2x p x 5 6 p 10

2x2 5 60

x2 5 30

x 5 6 Ï}

30

Length cannot be negative, so x 5 Ï}

30 .

CE 5 2x 5 2 Ï}

30

28. N

O

P

2

5 P9

The distance from the origin to P9 is 5 (Pythagorean Theorem).

} ON is the diameter, so its length is 2.

(NP9)2 5 (ON)2 1 (OP9)2 5 22 1 52 5 29

NP9 5 Ï}

29 . By Theorem 10.16,

(OP9)2 5 PP9 p NP9

52 5 PP9 p Ï}

29

25 }

Ï}

29 5 PP9

25 Ï

}

29 }

29 5 PP9

The distance d is 25 Ï

}

29 }

29 units.

Mixed Review

29. Ï}}

(210)2 2 82 5 Ï}

100 2 64 5 Ï}

36 5 6

30. Ï}}

25 1 (24) 1 (6 2 1)2 5 Ï}

29 1 52

5 Ï}

29 1 25

5 Ï}

16 5 4

31. Ï}}}

[22 2 (26)]2 1 (3 2 6) 2 5 Ï}

42 1 (23)2

5 Ï}

16 1 9 5 Ï}

25 5 5

32.

50°

40°8P

R

Q

cos 408 5 8 }

QR → QR 5

8 }

cos 408 ø 10.4

tan 508 5 8 } PR → PR 5

8 }

tan 508 ø 6.7

33. C

E F

5

8

FC2 5 EC2 1 EF2 5 52 1 82 5 89

FC 5 Ï}

89

34. m C AB 5 m∠ AFB 5 1358

35. m C CD 5 m∠ CFD 5 608

36. m C BCA 5 3608 2 m C AB 5 3608 2 1358 5 2258

37. m C CBD 5 3608 2 m C CD 5 3608 2 608 5 3008

38. C CDA is a semicircle, so m C CDA 5 1808.

39. C BAE is a semicircle, so m C BAE 5 1808.

40. }

AB does not bisect chord }

RS , so }

AB is not a diameter.


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41. 102 5 82 1 x2

100 5 64 1 x2

36 5 x2

6 5 x

}

AB bisects }

CD and is perpendicular to }

CD , so }

AB is a diameter.

42. }

AB does not bisect chord }

CD , so }


Problem Solving Workshop 10.6 (p. 696)

1. RS

} RQ

5 RQ

} RT

4 }

RQ 5

RQ }

4 1 9

4 }

RQ 5

RQ }

13

RQ2 5 52

RQ 5 Ï}

52

RQ 5 2 Ï}

13

2. a.

A

C

DB

E

m∠ BED 5 1 }

2 m C BCD

m∠ AEB 5 1808 2 m∠ BED 5 1808 2 1 }

2 m C BCD

m∠ ACD 5 1 }

2 m C BED

5 1 }

2 (3608 2 m C BCD ) 5 1808 2

1 }

2 m C BCD

So, m∠ AEB 5 m∠ ACD or ∠ AEB > ∠ ACD. Also ∠ A > ∠ A by the Refl exive Property of Congruence. So, nAEB , nACD by the AA Similarity Theorem.

b. AC

} AE

5 AD

} AB

15 1 5

} 12

5 12 1 DE

} 15

20

} 12

5 12 1 DE

} 15

300 5 144 1 12DE

156 5 12DE

13 5 DE

3. 72 5 5(5 1 x)

49 5 25 1 5x

24 5 5x

24

} 5 5 x

4. x(x 1 w) 5 y( y 1 z)

x 1 w 5 y(y 1 z)

} x

w 5 y(y 1 z)

} x 2 x

10.6 Extension (pp. 697–698)

1.

P1 in.

2. l

k

m

1in.

1in.

3.

C1 in.

4. l

k

m

1in.

1in.

5. The locus of points consists of two points on l each 3 centimeters away from P.

P l

3 cm 3 cm

6. The locus of points consists of four points on a circle with center at Q and a radius of 5 centimeters. The four points are the intersections of the circle and two lines parallel to and 3 centimeters away from m.

Q

m3 cm

5 cm

3 cm

7. The locus of points consists of a semi-circle centered at R with radius 10 centimeters. The diameter bordering the semi-circle is 10 centimeters from k and parallel to k.

R10 cm

10 cm

k

8. The portions of line l and m that are no more than 8 centimeters from point P and the portion of the circle, including its interior, with center P and radius 8 centimeters that is between lines l and m.

8 cm 8 cm

5 cm

5 cm

m

P

l


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341Geometry


9.

6 ft

3 ft

9 ft

dog

2 ft

4 ft5 ft

2 ft doghouse

Lesson 10.7


1. x2 1 y2 5 r2

x2 1 y2 5 2.52

x2 1 y2 5 6.25

2. (x 2 h)2 1 ( y 2 k)2 5 r2

(x 2 (22))2 1 ( y 2 5)2 5 72

(x 1 2)2 1 ( y 2 5)2 5 49

3. r 5 Ï}}

(3 2 1)2 1 (4 2 4)2

5 Ï}

22 1 02 5 2

(x 2 h)2 1 ( y 2 k)2 5 r2

(x 2 1)2 1 ( y 2 4)2 5 22

(x 2 1)2 1 ( y 2 4)2 5 4

4. r 5 Ï}}

(21 2 2)2 1 (2 2 6)2

5 Ï}}

(23)2 1 (24)2 5 5

(x 2 h)2 1 ( y 2 k)2 5 r2

(x 2 2)2 1 (y 2 6)2 5 52

(x 2 2)2 1 ( y 2 6)2 5 25

5. (x 2 4)2 1 ( y 1 3)2 5 16

(x 2 4)2 1 ( y 2 (23))2 5 42

The center is (4, 23) and the radius is 4.

1

x

y

21

(4, 23)

6. (x 1 8)2 1 ( y 1 5)2 5 121

(x 2 (28))2 1 ( y 2 (25))2 5 112

The center is at (28, 25) and the radius is 11.

(28, 25)

6

x

y

23

7. Three seismographs are needed to locate an earthquake’s epicenter because two circles intersect in two points. You would not know which point is the epicenter. You need the third circle to determine which point is the epicenter.

10.7 Exercises (pp. 702–705)

Skill Practice

1. The standard equation of a circle can be written for any circle with known center and radius.

2. The location of the center and one point on a circle is enough information to draw the rest of the circle because the distance from the center to the known point is the radius of the circle.

3. The radius is 2.

x2 1 y2 5 4

4. The center is (2, 3). The radius is 2.

(x 2 2)2 1 (y 2 3)2 5 4

5. The radius is 20.

x2 1 y2 5 202

x2 1 y2 5 400


(x 2 5)2 1 ( y 2 0)2 5 102

(x 2 5)2 1 y2 5 100


(x 2 50)2 1 ( y 2 50)2 5 102

(x 2 50)2 1 ( y 2 50)2 5 100

8. The center is (23,23). The radius is 9.

(x 2 (23))2 1 ( y 2 (23))2 5 92

(x 1 3)2 1 ( y 1 3)2 5 81

9. x2 1 y2 5 72

x2 1 y2 5 49

10. (x 2 (24))2 1 ( y 2 1)2 5 12

(x 1 4)2 1 ( y 2 1)2 5 1

11. (x 2 7)2 1 ( y 2 (26))2 5 82

(x 2 7) 1 ( y 1 6)2 5 64

12. (x 2 4)2 1 ( y 2 1)2 5 52

(x 2 4) 1 ( y 2 1)2 5 25

13. (x 2 3)2 1 ( y 2 (25))2 5 72

(x 2 3)2 1 ( y 1 5)2 5 49

14. (x 2 (23))2 1 ( y 2 4)2 5 52

(x 1 3)2 1 ( y 2 4)2 5 25

15. If (h, k) is the center of a circle with radius r, the equation of the circle should be

(x 2 h)2 1 (y 2 k)2 5 r2 not (x 1 h)2 1 (y 1 k)2 5 r2.

(x 2 (23))2 1 ( y 2 (25))2 5 32

(x 1 3)2 1 (y 1 5)2 5 9

16. C; r2 5 16 so r 5 4.

d 5 2r 5 2(4) 5 8


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17. r 5 Ï}}

(0 2 0)2 1 (0 2 6)2 5 Ï}

(26)2 5 6

x2 1 y2 5 62

x2 1 y2 5 36

18. r 5 Ï}}

(1 2 4)2 1 (2 2 2)2 5 Ï}

(23)2 5 3

(x 2 1)2 1 (y 2 2)2 5 32

(x 2 1)2 1 (y 2 2)2 5 9

19. r 5 Ï}}

(23 2 1)2 1 (5 2 8)2

5 Ï}}

(24)2 1 (23)2 5 5

(x 2(23))2 1 (y 2 5)2 5 52

(x 1 3)2 1 (y 2 5)2 5 25

20. x2 1 y2 5 49 21. (x 2 3)2 1 y2 5 16

The center is (0, 0). The center is (3, 0).The radius is 7. The radius is 4.

(0, 0)2

x

y

22

1

x

y

1

(3, 0)

22. x2 1 ( y 1 2)2 5 36

The center is (0, 22). The radius is 6.

(0, 22)

2

x

y

22

23. (x 2 4)2 1 ( y 2 1)2 5 1


1

x

y

1

(4, 1)

24. (x 1 5)2 1 ( y 2 3)2 5 9


(25, 3)

1

x

y

21

25. (x 1 2)2 1 ( y 1 6)2 5 25


(22, 26)

2

x

y

28

26. D; (x 1 2)2 1 ( y 2 4)2 5 25

(0 1 2)2 1 (5 2 4)2 0 25

22 1 12 0 25

5 Þ 25

27. x2 1 y2 2 6y 1 9 5 4

x2 1 ( y2 2 6y 1 9) 5 4

x2 1 (y 2 3)2 5 4

The equation is a circle.

28. x2 2 8x 1 16 1 y2 1 2y 1 4 5 25

(x2 2 8x 1 16) 1 (y2 1 2y 1 1) 5 25 2 3

(x 2 4)2 1 ( y 1 1)2 5 22


29. x2 1 y2 1 4y 1 3 5 16

x2 1 (y2 1 4y 1 3 1 1) 5 16 1 1

x2 1 (y2 1 4y 1 4) 5 17

x2 1 (y 1 2)2 5 17


30. x2 2 2x 1 5 1 y2 5 81

(x2 2 2x 1 5 2 4) 1 y2 5 81 2 4

(x2 2 2x 1 1) 1 y2 5 77

(x 2 1)2 1 y2 5 77


31. (x 2 4)2 1 ( y 2 3)2 5 9

The center is (4, 3).

y 5 23x 1 1

3 0 23(4) 1 6

3 Þ 26

The line does not contain the center of the circle. Solve for y in both equations and set them equal to fi nd points of intersection.

(x 2 4)2 1 (y 2 3)2 5 9

(y 2 3)2 5 9 2 (x 2 4)2

y 2 3 5 Ï}}

9 2 (x 2 4)2

y 5 3 1 Ï}}

9 2 (x 2 4)2

23x 1 6 5 3 1 Ï}}

9 2 (x 2 4)2

23x 1 3 5 Ï}}

9 2 (x 2 4)2

(23x 1 3)2 5 9 2 (x 2 4)2

9x2 2 18x 1 9 5 9 2 x2 1 8x 2 16

10x2 2 26x 1 16 5 0

5x2 2 13x 1 8 5 0


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343Geometry


Use the quadratic equation to solve for x.

x 5 13 6 Ï

}}

(213)2 2 4(5)(8) }}

2(5)

x 5 13 6 3

} 10 → x 5 1 or x 5 1.6

Because x has two solutions, the line intersects the circle at two points. The line is a secant.

32. (x 1 2)2 1 (y 2 2)2 5 16


y 5 2x 2 4

2 0 2(22) 2 4

2 Þ 28

The line does not contain the center of the circle. Solve for y in both equations and set them equal to fi nd points of intersection.

(x 1 2)2 1 ( y 2 2)2 5 16

( y 2 2)2 5 16 2 (x 1 2)2

y 2 2 5 Ï}}

16 2 (x 1 2)2

y 5 2 1 Ï}}

16 2 (x 1 2)2

2x 2 4 5 2 1 Ï}}

16 2 (x 1 2)2

2x 2 6 5 Ï}}

16 2 (x 1 2)2

(2x 2 6)2 5 16 2 (x 1 2)2

4x2 2 24x 1 36 5 16 2 x2 2 4x 2 4

5x2 2 20x 1 24 5 0

Use the quadratic equation to solve for x.

x 5 20 6 Ï

}}

(220)2 2 4(5)(24) }}

2(5)

x 5 20 6 Ï

}

280 } 10

The square root of a negative number does not exist. There are no solutions, so the line does not intersect the circle. The line is not a secant or a tangent.

33. (x 2 5)2 1 (y 1 1)2 5 4


y 5 1 } 5 x 2 3

21 0 1 } 5 (5) 2 3

21 Þ 22

The line does not contain the center of the circle.

At x 5 5,

y 5 1 } 5 x 2 3 5

1 } 5 (5) 2 3 5 22.

The point (5, 22) is in the interior of the circle, so the line is a secant.

34. (x 1 3)2 1 ( y 2 6)2 5 25


y 5 2 4 } 3 x 1 2

6 0 2 4 } 3 (23) 1 2

6 5 6

The line contains the center of the circle, so it is a secant that contains a diameter.

35. Let radius (O be k, then center (C: (15k, 0), point: (63, 16).

r 5 Ï}}}

(15k 2 63)2 1 (0 2 16)2

r 5 Ï}}

225k2 2 1890k 1 4225

4k 5 Ï}}

225k2 2 1890k 1 4225

16k2 5 225k2 2 1890k 1 4225

0 5 209k2 2 1890k 1 4225

k 5 Ï}}

18902 2 4(209)(4225) }}

2(209)

k 5 1890 6 200

} 418 → k 5 5 or k 5 4.04

k 5 5 because k is an integer.

(x 2 15)2 1 y2 5 100

Problem Solving

36. a.

x

y

2

2

Zone 1

Zone 2

Zone 3

CityCenter

b. (3, 4) is in Zone 2.

(6, 5) is in Zone 3.




37. r 5 1 } 2 d 5

1 } 2 (4.8) 5 2.4

x2 1 y2 5 2.42

x2 1 y2 5 5.76

r 5 1 } 2 d 5

1 } 2 (0.6) 5 0.3

x2 1 y2 5 0.32

x2 1 y2 5 0.09

38. center: (23, 0)

radius: AD 5 1

(x 2 (23))2 1 ( y 2 0)2 5 12

(x 1 3)2 1 y2 5 1


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39. center: (3, 0)

radius: BD 5 7

(x 2 3)2 1 (y 2 0)2 5 72

(x 2 3)2 1 y2 5 49


41. The height (or width) always remains the same as the fi gure is rolled on its edge.

42. a.

4

x

y

24 X

YZ

Yes, cell towers X and Y cover part of the same area, so this area may receive calls from Tower X or Tower Y.

b. Your home is within range of Tower Y, but your school is not in the range of any tower. So, you can use your phone at home.

c. City B has better cell phone coverage because it is located entirely within the range of Tower Z. City A is only partially covered by Tower X and Tower Y.

43. a. The center C can be found at the intersection of the lines perpendicular to the tangents.

y 5 2 4 } 3 x 1 b

5 5 2 4 }

3 (4) 1 b

31

} 3 5 b

y 5 2 4 }

3 x 1

31 }

3

y 5 4 }

3 x 1 b

13 5 2 4 } 3 (4) 1 b

23

} 3 5 b

y 5 4 }

3 x 1

23 }

3

2 4 } 3 x 1

31 }

3 5

4 }

3 x 1

23 }

3

24x 1 31 5 4x 1 23

8 5 8x

1 5 x

y 5 4 }

3 x 1

23 }

3 5

4 }

3 (1) 1

23 }

3 5 9

The center C is at (1, 9).

r 5 Ï}}

(1 2 4)2 1 (9 2 5)2 5 Ï}

(23)2 1 42 5 5

b. center: (1, 9)

radius: 5

(x 2 1)2 1 ( y 2 9)2 5 25

3

x

y

23

(1, 9)

44. Given: A circle passing through the points (21, 0) and (1, 0).

Prove: The equation of the circle is x2 2 2yk 1 y2 5 1 with center at (0, k).

Construct the perpendicular bisector of chord with the endpoints (21, 0) and (1, 0). Using Theorem 10.4, this new chord is a diameter of the circle. Since the new chord is a segment of the y-axis, the center of the circle is located at some point (0, k) which makes the equation of the circle x2 1 (y 2 k)2 5 r2 or x2 1 y2 2 2yk 5 r2 2 k2. Now consider the right triangle whose vertices are (0, 0), (0, k), and (1, 0) with the distance from (0, k) to (1, 0) being r, the radius of the circle. Using the Pythagorean Theorem, you get k2 1 12 5 r2 or r2 2 k2 5 1. Substituting you get x2 2 2yk 1 y2 5 1.

45. a. If r 5 2, there are two possible points of intersection. Use Theorem 10.1. If the center of the circle is (8, 8), then the point of intersection for m and n is (10, 6). If the center of the circle is (10, 6), then the point of intersection is (8, 8). If r 5 10, these are also two possible points of intersection. (C has two possible centers. To fi nd these centers, set the equation of the circle with radius 10 and center (8, 6) equal to the equation of the circle with radius 10 and center (10, 8).

rm 5 10 center (8, 6) (x 2 8)2 1 (y 2 6)2 5 100

→ y 5 6 1 Ï}}

100 2 (x 2 8)2

rn 5 10 center (10, 8) (x 2 10)2 1 (y 2 8)2 5 100

→ y 5 8 1 Ï}}

100 2 (x 2 10)2

6 1 Ï}}

100 2 (x 2 8)2 5 8 1 Ï}}

100 2 (x 2 10)2

100 2 (x 2 8)2 5 (2 1 Ï}}

100 2 (x 2 10)2 )2

100 2 (x 2 8)2 5 4 1 4 Ï}}

100 2 (x 2 10)2 1 100 2 (x 2 10)2

(x 2 10)2 2 (x 2 8)2 2 4

}} 4 5 Ï

}}

100 2 (x 2 10)2

24x 1 32

} 4 5 Ï

}}

100 2 (x 2 10)2

(2x 1 8)2 5 100 2 (x 2 10)2

x2 2 18x 1 32 5 0

(x 2 2)(x 2 16) 5 0

x 5 2 or x 5 16


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345Geometry


Possible centers for (C are (2, 14) and (16, 0). If the

center of (C is (16, 0), then line m is y 5 4 }

3 x 2

14 }

3 and

line n is y 5 3 }

4 2 x 1

1 }

2 .

4 }

3 x 2

14 }

3 5

3 }

4 x 1

1 }

2

16x 2 56 5 9x 1 6

x 5 8 6 } 7 → y 5 7

1 } 7

The point of intersection is 1 8 6 } 7 , 7

1 } 7 2 .

If the center of (C is (2, 14), then line m is y 5 3 }

4 x

and line n is y 5 4 }

3 x 2

16 }

3 .

3 }

4 x 5

4 }

3 x 2

16 }

3

9x 5 16x 2 64

x 5 9 1 } 7 → y 5 6

6 } 7

The point of intersection is 1 9 1 } 7 , 6

6 } 7 2 .

The points of intersection and the center of (C are collinear.

b. The locus of intersection points of m and n for all possible values of r for (C is the intersection of all circles centered (8, 6) and (10, 8).

(x 2 8)2 1 ( y 2 6)2 5 (x 2 10)2 1 ( y 2 8)2

x2 2 16x 1 64 1 y2 2 12y 1 36

5 x2 2 20x 1 100 1 y2 2 16y 1 64

4x 1 4y 2 64 5 0

x 1 y 2 16 5 0

x

y

2

2

C

m

n(8, 6)

(10, 8)r 5 2

r 5 10

Mixed Review

46. P 5 2l 1 2w 5 2(22) 1 2(9) 5 44 1 18 5 62 in.

47. P 5 4s 5 4(18) 5 72 ft

48. x2 1 402 5 572

x2 5 1649

x ø 40.6

P ø 40.6 1 40 1 57 ø 137.6 m

49. C 5 2πr 5 2(3.14)(7) 5 43.96 cm

50. C 5 πd 5 (3.14)(160) 5 502.4 in.

51. C 5 πd 5 (3.14)(48) 5 150.72 yd

52. 152 1 r2 5 (9 1 r)2

225 1 r2 5 81 1 18r 1 r2

144 5 18r

8 5 r

53. r2 1 212 5 (r 1 15)2

r2 1 441 5 r2 1 30r 1 225

216 5 30r

7.2 5 r

54. r2 1 282 5 (r 1 20)2

r2 1 784 5 r2 1 40r 1 400

384 5 40r

9.6 5 4

Quiz 10.6–10.7 (p. 705)

1. 6 p x 5 8 p 9 2. 7 p (7 1 5) 5 6 p (6 1 x)

6x 5 72 7(12) 5 36 1 6x

x 5 12 48 5 6x

8 5 x

3. 162 5 12(x 1 12) 4. center: (1, 4)

256 5 12x 1 144 radius: 6

112 5 12x (x 2 1)2 1 (y 2 4)2 5 62

28

} 3

5 x (x 2 1)2 1 (y 2 4)2 5 36

5. center: (5, 27)

point: (5, 23)

r 5 Ï}}}

(5 2 5)2 1 (27 2 (23))2 5 Ï}

(24)2 5 4

(x 2 5)2 1 (y 2 (27))2 5 42

(x 2 5)2 1 (y 1 7)2 5 16

6.

4

x

y

24

center of tire: (24, 3)

radius of tire: 12.1

(x 2 (24))2 1 (y 2 3)2 5 12.12

(x 1 4)2 1 ( y 2 3)2 5 146.21

center of rim: (24, 3)

radius of rim: 7

(x 2 (24))2 1 ( y 2 3)2 5 72

(x 1 4)2 1 ( y 2 3)2 5 49

Mixed Review of Problem Solving (p. 706)

1. a. center: (20, 30)

radius: 20

(x 2 20)2 1 ( y 2 30)2 ≤ 202

(x 2 20)2 1 ( y 2 30)2 ≤ 400


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b. At E(25, 25):

(x 2 20)2 1 ( y 2 30)2 ≤ 400

(25 2 20)2 1 (25 2 30)2 ≤? 400 52 1 (25)2 ≤? 400 50 ≤ 400 You can receive the signal at E.

At F(10, 10):

(x 2 20)2 1 ( y 2 30)2 ≤ 400

(10 2 20)2 1 (10 2 30)2 ≤? 400

(210)2 1 (220)2 ≤? 400

500 > 400

You cannot receive the signal at F.

At G(20, 16):

(x 2 20)2 1 (y 2 30)2 ≤ 400

(20 2 20)2 1 (16 2 30)2 ≤? 400

(214)2 ≤? 400

196 ≤ 400

You can receive the signal at G.

At H(35, 30):

(x 2 20)2 1 ( y 2 30)2 ≤ 400

(35 2 20)2 1 (30 2 30)2 ≤? 400

152 ≤? 400

225 ≤ 400

You can receive the signal at H.

2. a.

1

x

y

21

X

YZ

The areas covered by the cell towers overlap, so there are areas that can transmit calls using more than one tower.

b. Your house is located within range of tower X. You can use your cell phone at your house. You cannot use your cell phone at your friend’s house.

c. City B has better coverage from the cell phone towers.

3. 40 p 43 0 41 p 42

1720 0 1722

The track is not circular because the product of the lengths of one chord is not equal to the product of the lengths of the other chord.

4. a. 172 5 6(6 1 2r)

289 5 36 1 12r

253 5 12r

21.1 ø r The radius of the tank is about 21.1 feet.

b. x2 5 4(4 1 12)

x2 5 4(16)

x2 5 64

x 5 8

You are 8 feet from a point of tangency.

5. a.

2

x

y

26

AC

B

b. The circles intersect at point (2, 3), so the epicenter is at (2, 3).

c. center: (2, 3)

radius: 12

(x 2 2)2 1 ( y 2 3)2 5 122

(x 2 2)2 1 (y 2 3)2 5 144

At (14, 16):

(x 2 2)2 1 (y 2 3)2 5 144

(14 2 2)2 1 (16 2 3)2 0 144

122 1 132 0 144

313 > 144

The point (14, 16) lies outside the range of the earthquake, so you could not feel the earthquake.

6. a. y(y 1 2x) 5 152

y2 1 2xy 5 225

y2 1 2xy 2 225 5 0

y 5 22x 6 Ï

}}

(2x)2 2 4(1)(2225) }}}

2(1)

y 5 22x 6 Ï

}

4x2 1 900 }}

2

y 5 2x 1 Ï}

x2 1 225

b. 8 p x 5 x p x

8 5 x

c. y 5 28 6 Ï}

82 1 225

y 5 9

Chapter 10 Review (pp. 708–711)

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

2. A

C

O B

An inscribed angle is an angle whose vertex is on the circle and whose sides contain chords of the circle. The arc that lies in the interior of an inscribed angle and has endpoints on the angle is called the intercepted arc of the angle.


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347Geometry


3. The measure of the central angle and the corresponding minor arc are the same. The measure of the major arc is 3608 minus the measure of the minor arc.

4. B; } KL is a tangent segment.

5. C; } LN is a secant segment.

6. A; } LM is an external segment.

7. XY 5 XZ

9a2 2 30 5 3a

3(3a2 2 a 2 10) 5 0

(3a 1 5)(a 2 2) 5 0

3a 1 5 5 0 or a 2 2 5 0

a 5 2 5 } 3 a 5 2

Length cannot be negative, so a 5 2.

8. XY 5 XZ

2c2 1 9c 1 6 5 9c 1 14

2c2 2 8 5 0

2(c2 2 4) 5 0

(c 2 2)(c 1 2) 5 0

c 2 2 5 0 or c 1 2 5 0

c 5 2 c 5 22

Length cannot be negative, so c 5 2.

9. WZ 2 1 XZ 2 5 WX 2

r2 1 92 5 (r 1 3)2

r2 1 81 5 r2 1 6r 1 9

72 5 6r

12 5 r

10. m C KL 5 m∠ KPL 5 1008

11. m C LM 5 1808 2 m C MN 5 1808 2 1208 5 608

12. m C KM 5 m C KL 1 m C LM 5 1008 1 608 5 1608

13. m C KN 5 1808 2 m C KL 5 1808 2 1008 5 808

14. m C ED 5 m∠ ECD 5 618

m C AB 5 m C ED 5 618

15. m C AB 5 m C AD 5 658 16. m C AB 5 m C ED 5 918

17. m∠ YXZ 5 1 }

2 m C YZ 18. m∠ BAC 5

1 }

2 m C BC

c8 5 1 }

2 (568) 408 5

1 }

2 x8

c 5 28 80 5 x

19. m∠ E 1 m∠ G 5 1808

q8 1 808 5 1808

q 5 100

m∠ F 1 m∠ D 5 1808

1008 1 4r8 5 1808

4r 5 80

4 5 20

20. x8 5 1 }

2 [2508 2 (3608 2 2508)]

x 5 1 }

2 (250 2 110)

x 5 1 }

2 (140)

x 5 70

21. 408 5 1 }

2 (968 2 x8) 22. x8 5

1 }

2 (1528 1 608)

80 5 96 2 x x 5 1 }

2 (212)

x 5 16 x 5 106

23. 202 5 12 p (12 1 2r)

400 5 144 1 24r

256 5 24r

10 2 }

3 5 r

The radius of the rink is 10 2 }

3 feet.

24. center: (4, 21)

radius: 3

(x 2 4)2 1 (y 2 (21))2 5 32

(x 2 4)2 1 (y 1 1)2 5 9

25. center: (8, 6)

radius: 6

(x 2 8)2 1 (y 2 6)2 5 62

(x 2 8)2 1 (y 2 6)2 5 36

26. center: (0, 0)

radius: 4

x2 1 y2 5 42

x2 1 y2 5 16

27. x2 1 y2 5 92

x2 1 y2 5 81

28. (x 2 (25))2 1 (y 2 2)2 5 1.32

(x 1 5) 1 (y 2 2)2 5 1.69

29. (x 2 6)2 1 (y 2 21)2 5 42

(x 2 6)2 1 (y 2 21)2 5 16

30. (x 2 (23))2 1 (y 2 2)2 5 162

(x 1 3)2 1 (y 2 2)2 5 256

31. (x 2 10)2 1 (y 2 7)2 5 3.52

(x 2 10)2 1 (y 2 7)2 5 12.25

32. x2 1 y2 5 5.22

x2 1 y2 5 27.04

Chapter 10 Test (p. 712)

1. AB 5 AD

5x 2 4 5 3x 1 6

2x 5 10

x 5 5


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2. AD2 1 DC2 5 AC2

122 1 r2 5 (r 1 6)2

144 1 r2 5 r2 1 12r 1 36

108 5 12r

9 5 r

3. AB 5 AD

2x2 1 8x 2 17 5 8x 1 15

2x2 5 32

x2 5 16

x 5 4

4. m C CD 5 m∠ CED 5 608 5 m C AB

C AB > C CD

5. m C FG 5 m∠ FHG 5 1368

m C JL 5 3608 2 2248 5 1368

C FG > C JL

6. C MN is not congruent to C QR because they are not part of two congruent circles.

7. }

AB is perpendicular to and bisects }

CD , so }

AB is a diameter.

8. }

AB is perpendicular to and bisects }

CD , so }

AB is a diameter.

9. 202 1 ZY2 5 252

ZY2 5 225

ZY 5 15

}

AB is perpendicular to but does not bisect } XY , so }


10. m∠ ABC 5 1 }

2 m C AC 5

1 }

2 (1068) 5 538

11. m C FD 5 2m∠ FED 5 2(828) 5 1648

12. m C JG 5 2m∠ JHG 5 2(438) 5 868

m C GHJ 5 3608 2 m C JG 5 3608 2 868 5 2748

13. m∠ 1 5 1 }

2 (2388) 5 1198

14. m∠ 2 5 1 }

2 (528 1 1128) 5

1 }

2 (1648) 5 828

15. m∠ B 5 1 }

2 (m C AD 2 m C AC )

428 5 1 }

2 (1688 2 m C AC )

848 5 1688 2 m C AC

m C AC 5 848

16. 8 p x 5 14 p 4

8x 5 56

x 5 7

17. 122 5 9(9 1 x)

144 5 81 1 9x

63 5 9x

7 5 x

18. x(x 1 28) 5 20(20 1 32)

x2 1 28x 5 1040

x2 1 28x 2 1040 5 0

x 5 228 6 Ï

}}

282 2 4(1)(21040) }}}

2(1)

x 5 228 6 Ï

}

4944 }}

2

x ø 21.2

19. (x 1 2)2 1 (y 2 5)2 5 169

(x 2 (22))2 1 (y 2 5)2 5 132

The center is (22, 5) and the radius is 13.

Chapter 10 Algebra Review (p. 713)

1. 6x2 1 18x4 5 6x2 (1 1 3x2) 2. 16a2 2 24b 5 8(2a2 2 3b) 3. 9r2 2 15rs 5 3r(3r 2 5s)

4. 14x5 1 27x3 5 x3(14x2 1 27) 5. 8t4 1 6t2 2 10t 5 2t(4t3 1 3t 2 5) 6. 9z3 1 3z 1 21z2 5 3z(3z2 1 1 1 7z) 7. 5y6 2 4y5 1 2y3 5 y3(5y3 2 4y2 1 2) 8. 30v7 2 25v5 2 10v4 5 5v4(6v3 2 5v 2 2) 9. 6x3y 1 15x2y3 5 3x2y(2x 1 5y2) 10. x2 1 6x 1 8 5 (x 1 4)(x 1 2)

11. y2 2 y 2 6 5 (y 2 3)(y 1 2)

12. a2 2 64 5 (a 2 8)(a 1 8)

13. z2 2 8z 1 16 5 (z 2 4)(z 2 4) 5 (z 2 4)2

14. 3s2 1 2s 2 1 5 (3s 2 1)(s 1 1)

15. 5b2 2 16b 1 3 5 (5b 2 1)(b 2 3)

16. 4x4 2 49 5 (2x2 2 7)(2x2 1 7) 17. 25r2 2 81 5 (5r 2 9)(5r 1 9)

18. 4x2 1 12x 1 9 5 (2x 1 3)(2x 1 3) 5 (2x 1 3)2

19. x2 1 10x 1 21 5 (x 1 3)(x 1 7)

20. z2 2 121 5 (z 2 11)(z 1 11)

21. y2 1 y 2 6 5 ( y 1 3)(y 2 2)

22. z2 1 12z 1 36 5 (z 1 6)(z 1 6) 5 (z 1 6)2

23. x2 2 49 5 (x 1 7)(x 2 7)

24. 2x2 2 12x 2 14 5 2(x2 2 6x 2 7) 5 2(x 2 7)(x 1 1)

Standardized Test Preparation (p. 715)

1. You can eliminate choice A (208) because m C NQ must be greater than m∠ P, which is 208.

2. You can eliminate choice A (217) because if x 5 217, then m∠ E 5 (x 1 8)8 5 (217 1 8)8 5 298. m∠ E cannot be negative.

Standardized Test Practice (pp. 716–717)

1. C; You do not know if C MP > C NQ .


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349Geometry


2. B; }

SQ is a diameter that is perpendicular to } PR , so }

SQ also bisects } PR .

PV 5 1 }

2 PR

5x 2 2 5 1 }

2 (4x 1 14)

5x 2 2 5 2x 1 7

3x 5 9

x 5 3

3. B; (x 1 2)2 1 (y 2 4)2 5 9

(x 2 (22))2 1 (y 2 4)2 5 9

The center is (22, 4)

4. D; m∠ P 5 1808 2 1058 2 278 5 488

m∠ P 5 1 }

2 m C QR

488 5 1 }

2 m C QR

968 5 m C QR

5. B; A central angle in a regular hexagon measures 3608 1 6 5 608.

m C KL 5 608

6. B; nTSV is a right triangle.

ST 2 1 SV 2 5 TV2

122 1 182 5 TV2

468 5 TV2

22 in. ø TV

7. D; m∠ TRS 5 1 }

2 m C RT

628 5 1 }

2 m C RT

1248 5 m C RT

m C RPT 5 3608 2 m C RT 5 3608 2 1248 5 2368

8. C; 3 common tangents

9. B; m C EF 5 (146 2 x)8

m C HG 5 (172 2 x)8

m C EF 1 m C FG 1 m C GH 1 m C HE 5 3608

(146 2 x)8 1 x0 1 (172 2 x)8 1 3x8 5 3608

2x 1 318 5 360

2x 5 42

x 5 21

10. KL 5 ML

1 }

2 x 1 5 5 x 2 1

6 5 1 }

2 x

12 5 x

11. m C AB 5 1808 2 m C AD 5 1808 2 1118 5 698

m∠ AHB 5 m C AB 5 698

12. 202 5 2x(2x 1 6x)

400 5 2x(8x)

400 5 16x2

25 5 x2

5 5 x

13. ∠ P and ∠ T intercept the same arc, C QS , so they are congruent. Also, ∠ R > ∠ R by the Refl exive Property of Congruence. So, nPSR , nTQR by the AA Similarity Theorem.

14. The measure of an inscribed angle is 1 }

2 the measure of

its intercepted arc. So, x 5 1 }

2 y or y 5 2x.

The slope of the line is 2.

45

x

y

45

15. By Theorem 10.3, C AC > C EG .

By Theorem 10.5, C EF > C FG > C CB > C BA .

By Theorem 10.6, C AC > C EG .

All radii of the same circle are congruent, so } JB > } JF .

16. a–b.

C

A

B

D


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C

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B

D

O

The point of intersection is the center of the plate.

c. The archeologist could measure the radius found in part (b) and double it. The diameter is always twice the radius.

17. a. Point P(3, 28), center C(22, 4)

r 5 Ï}}}

(22 2 3)2 1 (4 2(28))2

5 Ï}}

(25)2 1 122

5 Ï}

169

5 13

(x 2 (22))2 1 (y 2 4)2 5 132

(x 1 2)2 1 (y 2 4)2 5 169

b. m 5 y1 2 y2

} x1 2 x2

5 4 2 (28)

} 22 2 3

5 12

} 25 5 2

12 } 5

y 5 mx 1 b

28 5 2 12

} 5 (3) 1 b

2 40

} 5 5 2 36

} 5 1 b

2 4 } 5 5 b

y 5 2 12

} 5 x 2 4 } 5

c. m 5 5 }

12

y 5 mx 1 b

28 5 5 }

12 (3) 1 b

2 32

} 4 5 5 }

4 1 b

2 37

} 4 5 b

y 5 5 }

12 x 2

37 }

4


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Chapter 10 - Houston Independent School · PDF file2. Two angles whose sides form two pairs...

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