Download Chapter Resources Volume 2: Chapters 7-12

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Date ————————————
Practice A
LESSON
10.4
For use with pages 699–707
1. m∠A
2. m∠A
1588
LESSON 10.4
Find the indicated measure.
3. m∠B
B
A
C
B
C
C
A
C
A
24� B
C
4. mBC
C
5. mBC
6. mBC
A
A
B
768
C
B
A
7. m∠C
C
8. m∠A
C
9. mBC
A
C
328
B
A
508
Copyright © Holt McDougal. All rights reserved.
B
C
C
408
A
648
B
B
C
Find the indicated measure in (M.
10. m∠PNO
11. m∠QNP
12. mPQ
13. mQO
14. m∠NMO
15.
16. m∠QMP
17.
C
N
1188
C
C
mNOP
mC
OQN
M
Q
P
O
688
Name two pairs of congruent angles.
18.
X
19.
E
Y
F
W
D
C
Z
Geometry
Chapter 10 Resource Book
263
Name ———————————————————————
Practice A
LESSON
LESSON 10.4
10.4
For use with pages 699–707
Date ————————————
continued
Decide whether a circle can be circumscribed about the quadrilateral.
20.
21.
92�
22.
130�
120�
70�
Find the values of the variables.
23.
B
A
y8
x8
24.
C
628
628
D
O
25.
N
M
x8
y8
E
D
y8
x8
1048
988
C
688
F
P
27.
28.
J
E
x8
R
y8
U
x8
y8 K
M
T
F
y8
S
408
758
1088
1148
528
x8
808
L
H
1048
Find m∠ A and m∠ C.
29.
B
A
408
C
30.
31.
B
318
458
B
C
(2x 1 8)8
A
328
D
738
D
C
A
(3x 2 24)8
D
32. Construct the tangents to (Q that pass through a
point C not on the circle. Label the points on the
circle the tangents pass through R and S. Justify
why aQRC and aQSC are right angles and
}
}
why CR and CS are tangent to (Q.
264
Geometry
Chapter 10 Resource Book
1
G
#
Copyright © Holt McDougal. All rights reserved.
1708
26.
Lesson 10.3, continued
22. Sample answer: Because three points
determine a plane, let P be the plane that contains
X, Y, and Z. Let line j be the perpendicular
}
bisector of XY in P, and let line k be the
}
perpendicular bisector of YZ in P. Because X, Y,
}
}
and Z are noncollinear points, XY and YZ are not
collinear or parallel. Then j and k are not parallel
either, which means that j and k intersect in some
point C. By the Perpendicular Bisector Thm.,
CX 5 CY, and CY 5 CZ. Therefore, the circle
}
centered at C with radius CY contains X, Y, and Z.
23. Sample answer: If one chord is a perpendicular
bisector of another chord, then the first chord is a
diameter of the circle, so you can construct
}
}
CK of DG and @##$
CL of EH
perpendicular bisectors @##$
that contain diameters of the circle, where C is the
}
center of the (, K is the midpoint of DG,
}
and L is the midpoint of EH. Because > chords
} }
are equidistant from the center, CK > CL.
} }
CJ > CJ by Reflexive Prop. so nKJC > nLJC
by HL. Therefore, ∠DJI > ∠HJI because
corresponding parts of > ns are >. 24. no; The
top board bisects the two boards it rests on whose
inside edges form > 10-foot chords. However,
the top board is not perpendicular to the other
boards, so it does not lie on a diameter of the pool.
Therefore, the center of the top board is not the
center of the pool.
Review for Mastery
1. 1008 2. 958 3. 708
}
}
4. Step 1: Draw segments AB and AC as shown.
B
A
C
}
Step 2: Draw perpendicular bisectors of AB and
}
AC. By Theorem 10.4, these are diameters of the
circle containing A, B, and C.
B
}
1. 10 units 2. 3Ï 5 units
3. no; no; Sample answer:
P
808 A
1.5
O
4. AC 5 AD 5 AE 5 EF. n ACD > n EAF
} }
(SAS congruence postulate).Therefore, CD > EF.
5. 184 6. 30
Lesson 10.4
Practice Level A
1. 798 2. 128 3. 908 4. 808 5. 1528 6. 1808
7. 408 8. 168 9. 528 10. 348 11. 318
12. 628 13. 1308 14. 1128 15. 1808 16. 628
17. 2488 18. ∠ C > ∠ D, ∠ E > ∠ F
19. ∠ W > ∠ Z, ∠ X > ∠Y 20. no; The 708 and
1308 angles are not supplementary. 21. yes
22. yes 23. x 5 118, y 5 118 24. x 5 112,
y 5 76 25. x 5 90, y 5 82 26. x 5 105, y 5 75
27. x 5 66, y 5 66 28. x 5 91.5, y 5 106
29. m∠ A 5 408, m∠ C 5 328 30. m∠ A 5 738,
m∠ C 5 318 31. m∠ A 5 458, m∠ C 5 728
; ∠QRC and ∠QSC
are right angles
because they
#
are inscribed in
1
}
semicircles. CR is
perpendicular
3
}
}
to radius QR at its outer endpoint, and CS is
}
perpendicular to radius QS at its outer endpoint.
}
}
So, by Theorem 10.1, CR and CS are tangent to
(Q.
32.
2
1. B 2. 588 3. 1408 4. 468 5. 638 6. 288
C
Step 3: Find the point where these bisectors
intersect. This is the center of the circle containing
points A, B, and C, so it is equidistant from each
city. This is were the cellular tower should be
located.
B
tower
C
A44
Challenge Practice
Practice Level B
A
A
5. 808 6. 10
Geometry
Chapter 10 Resource Book
7. 1238 8. 908 9. 428 10. 588 11. 488
12. 588 13. 428 14. 968 15. 1808
16. x 5 14, y 5 38 17. x 5 58, y 5 29
18. x 5 72, y 5 90 19. x 5 39, y 5 29
20. x 5 16, y 5 14 21. x 5 6, y 5 36.5
22. D 23. ∠ AED > ∠ BEC; Theorem 10.8;
Angle-Angle Similarity Postulate
Copyright © Holt McDougal. All rights reserved.
ANSWERS
17. D 18. A 19. E 20. B 21. F