1. Chapter 6 Geometry
Homework Answers

2. Sec 6.1 x = 165°, definition of measure of an arc.
The central angle of the smaller circle is larger, because the chord is closer to the center.
z = 84°, Chord Arcs Conj.
w = 70° Chord Central Angles Conj.
M (-4, 3), N (-4, -3), O (4, -3)
y = 96°, Chord Arcs Conj.
≈ 13.8 cm
1) Given; 3) All radii of a circle are congruent; 4) ΔBOA ê ΔCOD, SSS Congruence Conjecture; 5) ‹COD, CPCTC
8 cm, Chord Distance to Center Conj.
20 cm, Perpendicular to a Chord Conj.
AC = 68°; ‹B = 34° (Since Δ OBC is isosceles, ‹B = ‹C, ‹B + ‹C = 68°, and therefore ‹B = 34°.)
They can draw two chords and locate the intersection of their perpendicular bisectors. The radius is just over 5 km.
The length of the chord is greater than the length of the diameter.
y = ⅟₇ x; (0,0) is a point on this line.
The perpendicular bisector of the segment does not pass through the center of the circle.
a) rhombus; b) rectangle; c) kite; d) parallelogram
The longer chord is closer to the center; the longest chord, which is the diameter, passes through the center.

3. Sec 6.2
50°
1) Given; ) Definition of perpendicular lines; 3) Definition of right angle; ) Substitution property; ) All radii or a circle are congruent; 6) Definition of isosceles triangle; ) D, Isosceles Triangle Conjecture; ) ΔODR, SAA; ) DR, CPCTC; ) Definition of bisect
55°
30°
105° 76
Answers will vary.
y = -⁸⁄₁₅ x + ²⁸⁹⁄₁₅
45°
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°.
a) False; b) False, isosceles trapezoid; c) False, rectangle

4. Sec 6.3
65°
180°
30°
75°
70°
The two inscribed angles intercept the same arc, so they should be congruent.
50°
140°, 42°
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°
90°, 100°
148°
44°
142°
120°, 60°
140°, 111°
71°, 41°

5. Sec 6.4
See proof.
True. Opposite angles of a parallelogram are congruent. If it is inscribed in a circle, the opposite angles are also supplementary. So they are right angles, and the parallelogram is indeed equiangular, or a rectangle.
Proof: m ‹ ACD = ½mAD = m ‹ ABD by Inscribed Angle Conjecture. Therefore, ‹ ACD ê ‹ ABD.
Proof: By the Inscribed Angle Conjecture, m ‹ ACB = ½mADB = ½ (180°) = 90°. Therefore, ‹ ACB is a right angle.
True. mGA = mET by Parallel Lines Intercepted Arcs Conjecture. Therefore, the chords that intercept these arcs are congruent (Converse of the Chord Arcs Conjecture), that is GA ê ET. Therefore GATE is isosceles.
Proof: By the Inscribed Angle Conjecture, m ‹ C = ½mYLI and m ‹ L = ½mYCI = ½(360° - mYLI) = ° - m ‹ C. Therefore ‹ L and ‹ C are supplementary. (A similar proof can be used to show that ‹ I and ‹ Y are supplementary.)
a. S b. A c. N d. A e. S
The base angles of the isosceles triangle have a measure of 39°. Because the corresponding angles are congruent, m is parallel to n.
Proof: ‹ 1 ê ‹ 2 by AIA. mBC = 2m ‹ 2 = 2m ‹ 1 = mAD by the Inscribed Angle Conjecture. Therefore BC ê AD.

6. Sec 6.5
d = 5 cm
1399 tiles
C = 10π cm
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).
r = ¹²⁄π m
C = 5.5π
C = 12π cm
d = 46 m
C ê 15.7 cm
19. b = 90°, c = 42°, d = 70°, e = 48°, f = 132°, g = 52°
C ê 25.1 cm
r ê 7.0 m
C ê 84.8 in
565 ft
C = 6π cm
16 in.
244 yr

7. Sec 6.6 ê 4398 km/hr C ê 11 m/sec 38° 37,000,000 revolutions 48°
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.
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.

8. Sec 6.7
⁴π⁄3 in.
Outer horse ê 2.5 m/sec, inner horse ê 1.9 m/sec. One horse has traveled farther in the same amount of time (tangential velocity), but both horses have rotated the same number of times (angular velocity).
8π m
14π cm
9 m
6π ft
4π m
a = 70°, b = 110°, c = 110°, d = 70°, e = 20°, f = 20°, g = 90°, h = 70°, k = 20°, m = 20°, n = 20°, p = 140°, r = 80°, s = 100°, t = 80°, u = 120°
27 in.
100 cm
217 m/min
ê 4200 miles
18°/sec. No, the angular velocity is measured in degrees per second, so it is the same at every point on the carousel.

9. Ch 6 Review
The velocity vector is always perpendicular to the radius at the point of tangency to the object’s circular path.
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.
Sample answer: An arc measure is between 0° and 360°. An arc length is proportional to arc measure and depends on the radius of the circle.
mJI = 360°-56°-152° = 152° = mML. m‹JMI = ½ mJI = ½ mMI = m‹MJI. ΔJIM is isosceles.
b = 55°
mKIM = 2m‹KEM = 140°. mKI = 140°-70° = 70° = mMI. m‹IKM = ½ mMI = ½ mKI = m‹IMK. ΔKIM is isosceles.
a = 65°
c = 128°
e = 118°
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!
d = 91°
f = 66°
125.7 cm
42.0 cm
15π cm
14π ft
2 · 57° +2 · 35° ≠ 180°

10. Ch 6 Review (cont)
False. 20° + 20° + 140° = 180°. An angle with measure 140° is obtuse.
False. The consecutive angles between the bases are supplementary.
True
False. 48° + 48° + 132° ≠ 180°
False. See diagram.
False. Inscribed angles that intercept the same arc are congruent.
False. The measure of an inscribed angle is half the measure of the arc.
False. AC and BD bisect each other but AC is not perpendicular to BD. (see diagram)
False. (7 – 2) · 180° = 900°. It could have seven sides.
False. It could be isosceles.
False. The sum of the measures of any triangle is 180°.
False. 100°+100°+100°+60° = 360°
False. The sum of the measures of one set of exterior angles for any polygon is 360°. The sum of the measures of the interior angles of a triangle is 180° and of a quadrilateral is 360°. Neither is greater than 360°, so these are two counter examples.
False. AB is not congruent to CD. (see diagram)
False. The ratio of the circumference to the diameter is π.

11. 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.