Name ———————————————————————
Date ————————————
Practice A
LESSON
10.6
For use with pages 716–723
Fill in the blanks. Then find the value of x.
1. x p _____ 5 5 p _____
3
2. 6 p _____ 5 3 p _____
3. x p _____ 5 8 p _____
5
9
6
x
x
x
6
3
8
4
4. 4 p _____ 5 5 p _____
5. 3 p _____ 5 4 p _____
4
6. 3 p _____ 5 5 p _____
5
13
4
x
5
10
3
4
8
7. x 2 5 4 p _____
8. x 2 5 2 p _____
9. x 2 5 6 p _____
x
4
12
x
2
x
4
In Exercises 10–24, find the value of x.
10.
11.
5
6
12.
2
5
8
10
6
x
x
13.
x
3
14.
3
15.
x
4
x
x
10
8
5
7
LESSON 10.6
6
16
Copyright © Holt McDougal. All rights reserved.
3
x
x
4
6
8
Geometry
Chapter 10 Resource Book
283
Name ———————————————————————
Practice A
LESSON
10.6
For use with pages 716–723
16.
Date ————————————
continued
17.
18.
8
12
24
x
x
16
3
x
20.
7
6
21.
3
x
2x
3x
4
22.
20
23.
6
9
24.
3x
13
9
x
x
5x
LESSON 10.6
25. Doorway An arch over a doorway is an arc that
is 160 centimeters wide and 50 centimeters high.
You are curious about the size of the entire circle.
By drawing and labeling a circle passing through
the arc as shown in the diagram, you can use the
following steps to find the radius of the circle.
a. Find AB.
b. Find AC and AD.
c. Use AB, AC, and AD to find EA.
d. Find EB.
e. Find the length of the radius, EO.
Geometry
Chapter 10 Resource Book
8
x�2
3
284
31
x
18
160 cm
B
C
50 cm
A
D
O
E
Copyright © Holt McDougal. All rights reserved.
19.
Lesson 10.5, continued
Lesson 10.6
1. 57.58 2. 528 3. 408 4. 288 5. 638 6. 498
ANSWERS
7. 1808 8. 568 9. 568 10. 628 11. 37 12. 11
13. 8 14. 7 or 8 15. 208 16. 1008
17. Sample answer: Because it is given that m is
C
tangent to both circles at T, mTU 5 2m∠ KTU
5 mTV . Then mTWU 5 3608 2 2m∠ KTU
C
C
C. Because m∠ TJU 5 12 1 mC
5 mTXV
TWU 2 mC
TU 2
1
C 2 mTV
C 2, the
and m∠ TKV 5 2 1 mTXV
}
}
Substitution and Transitive Properties of Equality
can be used to show m∠ TJU 5 m∠ TKV.
Therefore, j i k by the Corresponding Angles
Converse.
18. Sample answer: Because
1
m∠ D 5 }2 (mAG 2 mCE ), you can use the given
C C
C 5 2mC
information mAG
CE to substitute:
1
C 2 mCE
C) 5 12 mCE
C. Because in
m∠ D 5 2 (2mCE
1
C 2 mBF
C), you
the small circle m∠ D 5 2 (mBHF
1
C 2 mBF
C) 5
can write m∠ D 5 2 (3608 2 mBF
C 5 1808 2 mBF
C. Therefore, 12 mCE
C
1808 2 BF
C 5 3608.
which gives mC
CE 1 2mBF
}
}
}
}
}
Review for Mastery
1. 2268 2. 828 3. 1808 4. 1098 5. 408
Problem Solving Workshop:
Mixed Problem Solving
1. a. 40 ft b. 58 ft 2. a. Check sketches; 368,
368, 728, 728 b. 368, 368 3. Answers will vary.
4. a. 468 b. 468, 1348, 1348 c. 628 5. 5
6. a. 308 b. 1508 c. 4 in., 2 in.; Because the side
length of the square is 6 in. and 1 in. on each side
of the square is the cardboard, then the circle has a
diameter of 4 in., and the radius is 2 in.
Challenge Practice
1. 42.58
C C
1
C)
BC 1 mAD
m∠ CPB 5 2 (mC
Thus, m∠ APD 5 m∠ CPB. Therefore,
C. 2Ï5 128
AD 5 mCB
mC
1
2. m∠ APD 5 } (mAD 1 mBC ) and
2
}
3.
A46
Geometry
Chapter 10 Resource Book
}
4.
Practice Level A
1. 3, 9; 15 2. 4, x; 8 3. 4, 6; 12 4. x, 18; 22.5
5. x, 12; 16 6. x, 15; 25 7. 16; 8 8. 18; 6
}
35
9. 10; 2Ï 15 10. 12 11. 4 12. 12.5 13. }
3
14. 12 15. 7.4 16. 9 17. 9 18. 5.7 19. 7.7
20. 3 21. 28.1 22. 2 23. 1.9 24. 5.6
25. a. 50 cm b. 80 cm, 80 cm c. 128 cm
d. 178 cm e. 89 cm
Practice Level B
1. 15 2. 2 3. 4 4. AB 5 16, DE 5 17
5. AB 5 21, DE 5 23 6. AB 5 32, DE 5 24
7. 5 8. 6 9. 7 10. RT 5 20, TV 5 16
11. RT 5 35, TV 5 45 12. RT 5 36, TV 5 27
13. 15 14. 12 15. 4 16. 18 17. 30 18. 50
19. 4 20. 8 21. 10 22. 8 23. 4 24. 6
25. 15 26. 8 27. 4 28. 20.125 in. 29. 140 in.
30. 14.2 ft
Practice Level C
1. 3.7 2. 2.3 3. 7.4 4. 2.5 5. 1 6. 3.9
7. 14.3 8. 5 9. 10 10. 3 11. 6 12. 5
13. Sample answer: When you use the theorem
to solve for x and y you get x 5 26 and y 5 39.
These segments are not possible in the given
diagram, so Thm. 10.14 cannot be applied.
14. AP 5 3, PQ 5 5, QB 5 7, PD 5 18, EQ 5 4
OQ
OP
15. 908 16. } 5 } 17. Sample answer: By
OQ
OR
the Segments of Chords Thm., OP(OR) 5 OS(OQ).
But OS 5 OQ, so OP(OR) 5 (OQ)2, and by the
OP
OQ
Division Prop. of Equality, }
5}
.
OQ
OR
18. Sample answer: By the Segments of Secants
and Tangents Thm., OP(OQ) 5 (OT )2 and
OR(OS) 5 (OT)2. Therefore, OP(OQ) 5 OR(OS)
by the Transitive Prop. of Equality.
19. a.
b. 39.9 ft
23 ft
r
10.5 ft
C
Copyright © Holt McDougal. All rights reserved.
Practice Level C