Copy

right

© H

olt M

cDou

gal

337Geometry

Solutions Key

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.

Florida Spiral Review

40. C

Lesson 10.5

10.5 Guided Practice (pp. 708–710)

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. 711–714)

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

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

Chapter 10, continued

LAHGE11FLSOL_c10.indd 337LAHGE11FLSOL_c10.indd 337 2/4/09 3:29:51 PM2/4/09 3:29:51 PM

Copy

right

© H

olt M

cDou

gal

338GeometrySolutions Key

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 GE 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)

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.

Chapter 10, continued

LAHGE11FLSOL_c10.indd 338LAHGE11FLSOL_c10.indd 338 2/4/09 3:29:54 PM2/4/09 3:29:54 PM

Copy

right

© H

olt M

cDou

gal

339Geometry

Solutions Key

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.

Chapter 10, continued

LAHGE11FLSOL_c10.indd 339LAHGE11FLSOL_c10.indd 339 2/4/09 3:29:56 PM2/4/09 3:29:56 PM

Copy

right

© H

olt M

cDou

gal

340GeometrySolutions Key

Prove: m∠CAB 5 1 }

2 m C APB

A

P

C

B

Q

Case 2

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.

Chapter 10, continued

LAHGE11FLSOL_c10.indd 340LAHGE11FLSOL_c10.indd 340 2/4/09 3:29:59 PM2/4/09 3:29:59 PM

Copy

right

© H

olt M

cDou

gal

341Geometry

Solutions Key

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.

Florida Spiral Review

32. D; Find the slope:

m 5 rise

} run 5 3 2 2

} 2 2 0

5 1 }

2

y-intercept of the top line is 2.

y-intercept of the bottom line is 22.

The equations of the two lines are

y 5 1 }

2 x 1 2 and y 5

1 }

2 x 2 2.

33. Each side is one unit greater than the next smaller side.

The shortest that the shortest side can be is 1. Then

x 1 4 5 1

x 5 23

Then the other two sides would be (23) 1 5 5 2 and (23) 1 6 5 3. But since the sum of the two shorter sides is equal to the longer side, x cannot equal 23.

The next shortest integer length that the shortest sides can be is 2. Then

x 1 4 5 2

x 5 22

Then the other two sides would be (22) 1 5 5 3 and (22) 1 6 5 4. Since this makes a triangle and its the shortest integer length, x 5 22.

Ready To Go On? Quiz for 10.4–10.5 (p. 714)

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.

Chapter 10, continued

LAHGE11FLSOL_c10.indd 341LAHGE11FLSOL_c10.indd 341 2/4/09 3:30:02 PM2/4/09 3:30:02 PM

Copy

right

© H

olt M

cDou

gal

342GeometrySolutions Key

Problem Solving Connections (p. 715)

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 from 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 15 5 248

The angle between any two spokes is 248.

b. There are four sections between cars A and E.

248 p 4 5 968

The arc measure is 968.

5. Check students’ drawings.

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

Exploring Geometry Activity 10.6 (p. 716)

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

10.6 Guided Practice (pp. 718–720)

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

x 5 8 (x 5 218 is extraneous.)

Chapter 10, continued

LAHGE11FLSOL_c10.indd 342LAHGE11FLSOL_c10.indd 342 2/4/09 3:30:04 PM2/4/09 3:30:04 PM