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Chapter 6 Geometry
Homework Answers
Sec 6.1
1. x = 165°, definition of measure of
an arc.
2. z = 84°, Chord Arcs Conj.
3. w = 70° Chord Central Angles Conj.
4. y = 96°, Chord Arcs Conj.
5. 8 cm, Chord Distance to Center Conj.
6. 20 cm, Perpendicular to a Chord Conj.
7. AC = 68°; ‹B = 34° (Since Δ OBC is
isosceles, ‹B = ‹C, ‹B + ‹C = 68°, and
therefore ‹B = 34°.)
8. The length of the chord is greater
than the length of the diameter.
9. The perpendicular bisector of the
segment does not pass through the
center of the circle.
10. The longer chord is closer to the
center; the longest chord, which is the
diameter, passes through the center.
11. The central angle of the smaller circle is
larger, because the chord is closer to
the center.
12. M (-4, 3), N (-4, -3), O (4, -3)
15. ≈ 13.8 cm
16. 1) Given; 3) All radii of a circle are
congruent; 4) ΔBOA ê ΔCOD, SSS
Congruence Conjecture; 5) ‹COD,
CPCTC
17. They can draw two chords and locate
the intersection of their perpendicular
bisectors. The radius is just over 5 km.
18. y = ⅟₇ x; (0,0) is a point on this line.
19. a) rhombus; b) rectangle; c) kite;
d) parallelogram
Sec 6.2
1. 50°
2. 55°
3. 30°
4. 105°
5. 76
6. Answers will vary.
17. y = -⁸⁄₁₅ x + ²⁸⁹⁄₁₅
18. 45°
19. Angles A and B must be right angles,
but this would make the sum of the
angle measures in the quadrilateral
shown greater than 360°.
21. 1) Given;
2) Definition of perpendicular lines;
3) Definition of right angle;
4) Substitution property;
5) All radii or a circle are congruent;
6) Definition of isosceles triangle;
7) D, Isosceles Triangle Conjecture;
8) ΔODR, SAA;
9) DR, CPCTC;
10) Definition of bisect
22. a) False; b) False, isosceles trapezoid;
c) False, rectangle
Sec 6.3
1. 65°
2. 30°
3. 70°
4. 50°
5. 140°, 42°
6. 90°, 100°
7. 50°
8. 148°
9. 44°
10. 142°
11. 120°, 60°
12. 140°, 111°
13. 71°, 41°
14. 180°
15. 75°
16. The two inscribed angles intercept the
same arc, so they should be congruent.
22. a = 108°; b = 72°; c = 36°; d = 108°;
e = 108°; f = 72°; g = 108°; h = 90°;
l = 36°; m = 18°; n = 54°; p = 36°
Sec 6.4
1. See proof.
6. True. Opposite angles of a parallelogram
are congruent. If it is inscribed in a circle,
2. Proof: m ‹ ACD = ½mAD = m ‹ ABD by
the opposite angles are also
Inscribed Angle Conjecture. Therefore, ‹
supplementary. So they are right angles,
ACD ê ‹ ABD.
and the parallelogram is indeed
3. Proof: By the Inscribed Angle Conjecture,
equiangular, or a rectangle.
m ‹ ACB = ½mADB = ½ (180°) = 90°.
7. True. mGA = mET by Parallel Lines
Therefore, ‹ ACB is a right angle.
Intercepted Arcs Conjecture. Therefore,
4. Proof: By the Inscribed Angle Conjecture,
the chords that intercept these arcs are
m ‹ C = ½mYLI and
m ‹ L = ½mYCI =
congruent (Converse of the Chord Arcs
½(360° - mYLI) = 180° - m ‹ C. Therefore
Conjecture), that is GA ê ET. Therefore
‹ L and ‹ C are supplementary. (A similar
GATE is isosceles.
proof can be used to show that ‹ I and ‹ Y
8. a. S b. A
c. N
d. A
e. S
are supplementary.)
9. The base angles of the isosceles triangle
have a measure of 39°. Because the
5. Proof: ‹ 1 ê ‹ 2 by AIA. mBC = 2m ‹ 2 = 2m
corresponding angles are congruent, m is
‹ 1 = mAD by the Inscribed Angle
parallel to n.
Conjecture. Therefore BC ê AD.
Sec 6.5
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
d = 5 cm
C = 10π cm
r = ¹²⁄π m
C = 5.5π
C = 12π cm
d = 46 m
C ê 15.7 cm
C ê 25.1 cm
r ê 7.0 m
C ê 84.8 in
565 ft
C = 6π cm
16 in.
244 yr
15. 1399 tiles
16. Conjecture: The measure of the angle
formed by two intersecting chords is
equal to one-half the sum (average) of
the measures of the two intercepted
arcs. In diagrams, m ‹ NEA = ½ (mAN +
mGL).
19. b = 90°, c = 42°, d = 70°, e = 48°,
f = 132°, g = 52°
Sec 6.6
12. C
ê 4398 km/hr
13. 38°
ê 11 m/sec
37,000,000 revolutions
14. 48°
15. 30 cm
ê 637 revolutions
Mama; C ê 50 in.
ê 168 cm
m ‹ ECA = ½ (mSN – mEA)
Conjecture: The measure of an angle
formed by two intersecting secants
through a circle is equal to one-half the
difference of the larger arc measure
and the smaller arc measure.
11. Both triangles are isosceles so the base
angles in each triangle are congruent.
But one of each base angle is part of a
vertical pair. So, a = b by the Vertical
Angles Conjecture and transitivity.
1.
2.
3.
4.
5.
6.
9.
Sec 6.7
1. ⁴π⁄3 in.
14. Outer horse ê 2.5 m/sec, inner horse ê
1.9 m/sec. One horse has traveled
2. 8π m
farther in the same amount of time
3. 14π cm
(tangential velocity), but both horses
4. 9 m
have rotated the same number of
5. 6π ft
times (angular velocity).
6. 4π m
15. a = 70°, b = 110°, c = 110°,
7. 27 in.
d = 70°, e = 20°, f = 20°, g = 90°, h = 70°,
k = 20°, m = 20°, n = 20°, p = 140°,
8. 100 cm
r = 80°, s = 100°, t = 80°, u = 120°
9. 217 m/min
10. ê 4200 miles
13. 18°/sec. No, the angular velocity is
measured in degrees per second, so it
is the same at every point on the
carousel.
Ch 6 Review
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
The velocity vector is always
16.
perpendicular to the radius at the point of 17.
tangency to the object’s circular path.
Sample answer: An arc measure is
between 0° and 360°. An arc length is
18.
proportional to arc measure and depends
on the radius of the circle.
b = 55°
19.
a = 65°
c = 128°
20.
e = 118°
d = 91°
f = 66°
125.7 cm
42.0 cm
15π cm
14π ft
2 · 57° +2 · 35° ≠ 180°
84° + 56° + 56° + 158° ≠ 360°
m‹EKL = ½ mEL= ½ (180°-108°) = 36° =
m‹KLY. KE is parallel to YL by Converse of
the Parallel Lines Conjecture.
mJI = 360°-56°-152° = 152° = mML. m‹JMI
= ½ mJI = ½ mMI = m‹MJI. ΔJIM is
isosceles.
mKIM = 2m‹KEM = 140°. mKI = 140°-70° =
70° = mMI. m‹IKM = ½ mMI = ½ mKI =
m‹IMK. ΔKIM is isosceles.
Ertha can trace the incomplete circle on
paper. She can lay the corner of the pad
on the circle to trace an inscribed right
angle. Then Ertha should mark the
endpoints of the intercepted arc and use
the pad to construct the hypotenuse of
the right triangle, which is the diameter of
the circle. Math is magical!
Ch 6 Review (cont)
34. False. 20° + 20° + 140° = 180°. An angle
44.
with measure 140° is obtuse.
35. True
45.
36. False. See diagram.
46.
37. True
47.
38. True
39. True
48.
40. True
49.
41. False. (7 – 2) · 180° = 900°. It could have
seven sides.
50.
42. False. The sum of the measures of any
51.
triangle is 180°.
43. False. The sum of the measures of one set of 52.
exterior angles for any polygon is 360°. The
sum of the measures of the interior angles 53.
of a triangle is 180° and of a quadrilateral is
360°. Neither is greater than 360°, so these
are two counter examples.
False. The consecutive angles between the
bases are supplementary.
False. 48° + 48° + 132° ≠ 180°
False. Inscribed angles that intercept the
same arc are congruent.
False. The measure of an inscribed angle is
half the measure of the arc.
True
False. AC and BD bisect each other but AC is
not perpendicular to BD. (see diagram)
False. It could be isosceles.
False. 100°+100°+100°+60° = 360°
False. AB is not congruent to CD. (see
diagram)
False. The ratio of the circumference to the
diameter is π.
Ch 6 Review (cont)
54.False. 24+24+48+48 ≠ 96. (see diagram)
55.True
56.This is a paradox. Isn’t Math fun?
57.a= 58°, b= 61°, c= 58°, d= 122°, e= 58°, f= 64°,
g= 116°, h= 52°, i= 64°, k= 64°, l= 105°, m=
105°, n= 105°, p= 75°, q= 116°, r= 90°, s= 58°,
t= 122°, u= 105°, v= 75°, w= 61°, x= 29°, y=
151°
62. mAC = 84°, length of AC = 11.2π ≈ 35.2 in.