Download Answers for the lesson “Use Inscribed Angles and Polygons”

LESSON
10.4
Answers for the lesson “Use Inscribed Angles
and Polygons”
Skill Practice
18. A
1. inscribed
19. 908
2. The diagonals of a rectangle
20. Yes; opposite angles are 908 and
create two right triangles.
Theorem 10.9 tells you the
hypotenuse of each of these
triangles is a diameter of
the circle.
3. 428
4. 858
5. 108
6. 1348
7. 1208
8. 1008
9. The measure of the arcs add up to
3708; change the measure of
ŽQ to 408 or change the measure
of C
QS to 908.
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10. Ž ADB, Ž ACB, and
ŽCAD, Ž DBC
thus are supplementary.
21. Yes; opposite angles are 908 and
thus are supplementary.
22. No; opposite angles are not
always supplementary.
23. No; opposite angles are
not supplementary.
24. No; opposite angles are
not supplementary.
25. Yes; opposite angles
are supplementary.
12
26. }
5
11. Ž JMK, ŽJLK and
ŽLKM, ŽLJM
12. ŽWXZ, ŽWYZ and
Problem Solving
B
27.
Ž XWY, Ž XZY
13. x 5 100, y 5 85
14. k 5 60, m 5 120
20,000 km
A
C
15. a 5 20, b 5 22
16. B
17. a. 368; 1808
100,000 km
220,000 km
b. about 25.78; 1808
c. 208; 1808
Geometry
Answer Transparencies for Checking Homework
307
28. Place the tool so the outer corner
touches the circumference and
the two outer edges intersect the
circumference at two points. Then
connect the two points to form a
diameter.
29. Double the length of the radius.
30. mC
GDE , Measure of an
Inscribed Angle, 2mŽ F,
mŽD 1 mŽF 5 1808 and thus
are supplementary, ŽE and ŽG
are supplementary.
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31. Given: ŽB is inscribed in (Q.
Let mŽB 5 x8. Point Q lies on
}
BC. Since all radii of a circle are
} > BQ
}. Using the
congruent, AQ
Base Angles Theorem, ŽB > ŽA,
which implies mŽ A 5 x8.
Using the Exterior Angles
Theorem, mŽ AQC 5 2x8, which
implies mC
AC 5 2x8. Solving
for x, you get }2mC
AC 5 x8.
1
Substituting you
get }2mC
AC 5 mŽB.
1
32. Given Ž ABC is inscribed in
(Q. Point Q is in the interior of
Ž ABC.
1 C
mAC .
Prove mŽ ABC 5 }
2
Construct the diameter }
BD of (Q
and show mŽ ABD 5 }
mC
AD
2
1
and mŽ DBC 5 }
mC
DC .
2
1
Use the Arc Addition
Postulate and the Angle
Addition Postulate to show
2mŽ ABC 5 mC
AD 1 mC
DC .
33. Given ŽABC is inscribed in (Q.
Point Q is in the exterior
of ŽABC.
Prove mŽABC 5 }
mC
AC .
2
1
BD of (Q
Construct the diameter }
AD
and show mŽABD 5 }2mC
1
CD . Use the
and mŽCBD 5 }2mC
1
Arc Addition Postulate and the
Angle Addition Postulate to
show mŽABD 2 mŽCBD 5
m ŽABC. Then use substitution to
AC .
show 2mŽABC 5 mC
Geometry
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308
34.
A
D
O
C
B
Given (O with inscribed ŽC
and ŽD both intercepting C
AB .
Prove ŽC > ŽD. Using the
Measure of an Inscribed Angle
Theorem, mŽC 5 }2 mC
AB and
1
AB . Using the
mŽD 5 }2 mC
1
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
Transitive Property of Equality,
mŽC 5 mŽD which implies
ŽC > ŽD.
36. In the figure n ABC is a right
triangle with ŽABC being the
right angle. By Theorem 10.1,
‹]›
since AB
is the perpendicular to
radius }
BC, it is tangent to (C at
point B.
GJ
HJ
37. } 5 }; in a right triangle,
FJ
GJ
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. 6 in., 2 in., 2Ï 3 in.; 4Ï 3 in.
39. yes
35. Case 1: Given (D with
inscribed n ABC where }
AC is a
diameter of (D. Prove n ABC is
a right triangle. Let E be a point
on C
AC . Show that mC
AEC 5 1808
and then that mŽB 5 908.
Case 2: Given (D with inscribed
n ABC with ŽB a right angle.
Prove }
AC is a diameter of (D.
Using the Measure of an
Inscribed Angle Theorem,
show that mC
AC 5 1808.
Geometry
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309