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Answers to Exercises
CHAPTER 6 • CHAPTER
6
CHAPTER 6 • CHAPTER
LESSON 6.1
1.
3.
5.
6.
50°
30°
76 in.
possible answer:
2. 55°
4. 105°
11. Construct a line and label a point T on it. On
the line, mark off two points, L and M, each on
the same side of T at distances r and t from T,
respectively. Construct circle L with radius r.
Construct circle M with radius t.
r
L
M
T
t
Answers to Exercises
Target
7. Possible answer: The perpendicular to the
tangent passes through the center of the circle. Use
the T-square to find two diameters of the Frisbee.
The intersection of these two lines is the center.
. Construct a line through point T
8. Construct OT
.
perpendicular to OT
r
T
O
, OY
, and OZ
. Construct tangents
9. Construct OX
through points X, Y, and Z.
Z
O
t
Y
X
10. Construct tangent circles M and N. Construct
equilateral MNP. Construct circle P with radius s.
M
s
s
N
P
74
ANSWERS TO EXERCISES
12. Start with the construction from Exercise 11.
, mark off length TK so that TK s
On line LM
and K is on the opposite side of T from L and M.
Construct circle K with radius s.
r
L
M
T
t
K
s
13. Sample answer: If the three points do not lie
on the same semicircle, the tangents form a circumscribed triangle. If two of the three points are
on opposite sides of a diameter, the tangent lines to
those two points are parallel. If the points lie on the
same semicircle, they form a triangle outside the
circle, with one side touching (called an exscribed
triangle).
14. sample answer: internally tangent: wheels on a
roller-coaster car in a loop, one bubble inside
another; externally tangent: touching coins, a
snowman, a computer mouse ball and its roller
balls
15. Constructions will vary.
16. 360° 30° 90° 90° 150° and
150°
%
41.6
360°
17. 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°.
18a. rhombus
18b. rectangle
18c. kite
18d. parallelogram
19. 78°
20. 9.7 km
21. x 55° 55° 180° and 40° y y 180°, so x y 70°
11
22. 2
1
LESSON 6.2
5 cm
5 cm
15. M(4, 3), N(4, 3), O(4, 3)
16. The center of the circle is the circumcenter of
the triangle. Possible construction:
17. possible construction:
O
18. 13.8 cm
19. They can draw two chords and locate the
intersection of their perpendicular bisectors. The
radius is just over 5 km.
20. See flowchart below.
20. (Lesson 6.2)
1
AB CD
? Given
ᎏ
2
AO CO
All radii of a circle
are congruent
3
AOB COD
4
? ᎏ
?
ᎏ
? SSS Congruence
ᎏ
Conjecture
5
COD
?
AOB ᎏ
? CPCTC
ᎏ
BO DO
? All radii of a circle are congruent
ᎏ
ANSWERS TO EXERCISES
75
Answers to Exercises
1. 165°, definition of measure of an arc
2. 84°, Chord Arcs Conj.
3. 70°, Chord Central Angles Conj.
4. 8 cm, Chord Distance to Center Conj.
68°; mB 34° (Because OBC is
5. mAC
isosceles, mB mC, mB mC 68°,
and therefore mB 34°.)
6. w 115°, x 115°, y 65°; Chord Arcs
Conjecture
7. 20 cm, Perpendicular to a Chord Conj.
8. w 110°, x 48°, y 82°, z 120°; definition
of arc measure
9. x 96°, Chord Arcs Conjecture; y 96°,
Chord Central Angles Conjecture; z 42°,
Isosceles Triangle Conjecture and Triangle Sum
Conjecture.
10. x 66°, y 48°, z 66°; Corresponding
Angles Conjecture, Isosceles Triangle Conjecture,
Linear Pair Conjecture
11. The length of the chord is greater than the
length of the diameter.
12. The perpendicular bisector of the segment
does not pass through the center of the circle.
13. The longer chord is closer to the center; the
longest chord, which is the diameter, passes
through the center.
14. The central angle of the smaller circle is larger,
because the chord is closer to the center.
1
21. y 7 x; (0, 0) is a point on this line.
22a. true
C
A
X
B
Answers to Exercises
D
is the perpendicular
Possible explanation: If AB
, then every point on AB
is
bisector of CD
equidistant from endpoints C and D. Therefore
AD
and BC
BD
. Because CD
is not the
AC
, C is not equidistant
perpendicular bisector of AB
from A and B. Likewise, D is not equidistant from
and BC
are not congruent, and
A and B. So, AC
and BD
are not congruent. Thus ACBD
AD
has exactly two pairs of consecutive congruent
sides, so it is a kite.
22b. false, isosceles trapezoid
22c. false, rectangle
23. 140°, 160°, 60°; 180° minus the measure of
the intercepted arc
76
ANSWERS TO EXERCISES
24. Using the Tangent Conjecture, 2 and 4
are right angles, so m2 m4 180°.
According to the Quadrilateral Sum Conjecture,
m1 m2 m3 m4 360°, so by the
Subtraction Property of Equality, m1 m3 180°, or m1 180° m3. The
measure of a central angle equals the measure of
.
its arc, so by substitution, m1 180° mAB
25. the circumcenter of the triangle formed by
the three light switches
26. 7
27. Station Beta is closer.
Downed aircraft
N
4.6 mi
48
7.2 mi
72
38
Alpha
28. D
Beta
312
8.2 mi
LESSON 6.3
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°
23. (2, 1); Possible method: Plot the
three points. Construct the midpoint and the
perpendicular bisector of the segments connecting
two different pairs of points. The center is the point
of intersection of the two lines. To check, construct
the circle through the three given points.
24. See flowchart below.
25. Start with an equilateral triangle whose vertices
are the centers of the three congruent circles. Then
locate the incenter/circumcenter/orthocenter/
centroid (all the same point because the triangle is
equilateral) to find the center of the larger circle. To
find the radius, construct a segment from the
incenter of the triangle through the vertex of the
triangle to a point on the circle.
It works on acute and right triangles.
20. The camera can be placed anywhere on the
major arc (measuring 268°) of a circle such that the
row of students is a chord intersecting the circle to
form a minor arc measuring 92°. This illustrates
the conjecture that inscribed angles that intercept
the same arc are congruent (Inscribed Angles
Intercepting Arcs Conjecture).
21. two congruent externally tangent circles with
half the diameter of the original circle
26. mA 60°, mB 36°, mC 90°;
60° 36° 90° 180°
24. (Lesson 6.3)
1 OR CD
2
ORC and ORD
are right angles
? Definition of
ᎏ
perpendicular lines
3
6
OCD is isosceles
7 C ? D
ᎏ
? Given
ᎏ
5
OC OD
?
ᎏ
All radii of
a circle are
congruent
? Definition of
ᎏ
isosceles triangle
mORC ⫽ 90
mORD ⫽ 90
? Definition of
ᎏ
right angle
?
ᎏ
7. Isosceles Triangle
Conjecture
4
mORC ⫽ mORD
(ORC ORD)
Substitution
?
ᎏ
property
8 OCR ?
ᎏ
?
ᎏ
ODR
SAA
9 CR ? DR
ᎏ ?
ᎏ
CPCTC
10 OR bisects CD
? Definition of bisect
ᎏ
ANSWERS TO EXERCISES
77
Answers to Exercises
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.
17. BFE DFA (Vertical Angles Conjecture).
BGD FHD (all right angles congruent).
Therefore, B D (Third Angle Conjecture).
, mD 1 mEC
, AC
EC
mB 21 mAC
2
18. Possible answer: Place the corner so that it is
an inscribed angle. Trace the inscribed angle. Use
the side of the paper to construct the hypotenuse
of the right triangle (which is the diameter). Repeat
the process. The place where the two diameters
intersect is the center.
19. possible answer:
Answers to Exercises
LESSON 6.4
mABD by the
1. Proof: mACD 21mAD
Inscribed Angle Conjecture. ACD ABD.
2. Proof: By the Inscribed Angle Conjecture,
mACB 21mAD
B 21(180°) 90°. ACB is a
right angle.
3. Proof: By the Inscribed Angle Conjecture,
and mL 1mYCI
1(360° mC 21mYLI
2
2
1 mYLI ) 180° 2mYLI 180° mC. L
and C are supplementary. (A similar proof can be
used to show that I and Y are supplementary.)
2m2 4. Proof: 1 2 by AIA. mBC
2m1 mAD by the Inscribed Angle Conjecture.
AD
.
BC
5. 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.
and DB
. Diagonal RG
6. Construct diagonals RG
is the perpendicular bisector of BD by the Kite
Diagonal Bisector Conjecture. Because kite BRDG
and RG
are chords of
is inscribed in the circle, BD
the circle. By the Perpendicular Bisector of a Chord
Conjecture, the perpendicular bisector of BD
passes through the center of the circle. Because RG
is a chord of the circle that passes through the
is a
center, by the definition of diameter, RG
diameter.
, , and AR
. Because they
7. Draw in radii GR
ER , TR
TR
.
are radii of the same circle, GR
ER and AR
By the Parallel Lines Intercepted Arcs Conjecture,
ET
. Their central angles must also be
GA
congruent, so GRA ERT. Thus GRA ET
by CPCTC. GATE is
ERT by SAS, so GA
an isosceles trapezoid.
8. x 65°, y 40°, z 148°; half the measure of
the intercepted arc
9. Because the radii of a circle are congruent,
OB
, so OAB is isosceles. By the Isosceles
OA
Triangle Conjecture, 2 3, so m2 m3.
By the Triangle Sum Conjecture, m1 m2 m3 180°, so by substitution, m1 2(m2) 180°, or m1 180° 2(m2).
78
ANSWERS TO EXERCISES
, so
BC
By the Tangent Conjecture, OB
mOBC 90°. By Angle Addition, m2 m4 mOBC, so m2 m4 90°,
or m2 90° m4. Substitute 90° m4
for m2 in the earlier equation: m1 180° 2(90° m4). Simplifying gives m1 2(m4).
m1, substitution gives mAB
Because mAB
.
2(m4), or m4 21mAB
10a. S; An equilateral triangle is equiangular, but a
rhombus is not equiangular.
10b. A
10c. N; Only one diagonal is the perpendicular
bisector of the other.
10d. A
10e. S; An equilateral triangle has rotational
symmetry and three lines of symmetry; a
parallelogram has rotational symmetry but no line
of symmetry.
11. L
12. J
13. K
14. D
15. E
16. B
17. H
18. G
19. N
20. a b b a 180°, so a b 90°
2
21. 2
1
RS
. OP
OQ
OR
22. It is given that PQ
because all radii in a circle are congruent.
OS
Therefore, OPQ ORS by SSS and 2 1
by CPCTC. Now we move on to the smaller right
triangles inside OPQ and ORS. It is given that
PQ
and OV
RS
. Therefore, OTQ and
OT
OVS are right angles by the definition of
perpendicular lines and OTQ OVS because
all right angles are congruent. Thus, OTQ OV
by CPCTC.
OVS by SAA and OT
LESSON 6.5
1. d 5 cm
2. C 10 cm
12
3. r m
11
4. C 5.5 or 2
18. (Lesson 6.5)
1
2
? MA
MT
ᎏ
Given
? SA
ST
ᎏ
Given
3
MAST is a kite
? Definition of kite
ᎏ
is the perpendicular bisector of AT
MS
4 ?
ᎏ
? Kite Diagonal Bisector Conjecture
ᎏ
ANSWERS TO EXERCISES
79
Answers to Exercises
5. C 12 cm
6. d 46 m
7. C 15.7 cm
8. C 25.1 cm
9. r 7.0 m
10. C 84.8 in.
11. 565 ft
12. C 6 cm
13. 16 in.
14. Trees grow more in years with more rain;
244 yr
15. 1399 tiles
16. g 40°, n 30°, x 70°; y 142°; z 110°;
Conjecture: The measure of the angle formed by
two intersecting chords is one-half the sum of the
measures of the two intercepted arcs. In the
mLG
.
diagrams, mAEN 21mAN
and mLNG 1mLG
17. mAGN 21mAN
2
by the Inscribed Angle Conjecture. mAEN mAGN mLNG by the Triangle Exterior
mLG
Angle Conjecture. So, mAEN 21mAN
by substitution and the distributive property.
18. See flowchart below.
19. b 90°, c 42°, d 70°, e 48°, f 132°,
g 52°
5
150°
20. 360° or 12
21. The base angles of the isosceles triangle have a
measure of 39°. Because the corresponding angles
are congruent, m is parallel to n.
22. 10x 2y
LESSON 6.6
Answers to Exercises
1. 4398 km/h
2. 11 m/s
3. 37,000,000 revolutions
4. 637 revolutions
5. Mama; C 50 in.
6. 168 cm
7. d 7.6 ft. The table will fit, but the chairs may
be a little tight in a 12-by-14 ft room. 12 chairs 192 in., 12 spaces 96 in., C 288 in., d 91.7 in.
7.6 ft.
8. 0.35 ft/s
9. a 37.5°, s 17.5°, x 20°; y 80°; z 61°;
Conjecture: The measure of an angle formed by
two secants that intersect outside a circle is
one-half the difference of the larger arc measure
and the smaller arc measure.
80
ANSWERS TO EXERCISES
and mSAN 1mSN
by
10. mESA 21mEA
2
the Inscribed Angle Conjecture. mSAN mESA mECA by the Triangle Exterior Angle
1mEA
mECA by
Conjecture. So, 21mSN
2
substitution. With a little more algebra, mECA 1
mSN mEA .
2
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.
12. C
13. 38°
14. 48°
15. 30 cm
400 rev 2 . 2 6 ft 1 min
16. 1 min 1 rev 60 s 1089 ft/s
17. 12 cm third side 60 cm. This is based on
the Triangle Inequality Conjecture.
USING YOUR ALGEBRA SKILLS 6
50
40
Cost ($)
1
1. 2, 3
2. (3, 3)
2
3. 5, 3
4. (7, 2)
5. infinitely many solutions
6. no solution
7. 4. The lines intersect at the solution point, (7, 2).
The circumcenter is 272 , 178 . Only two
perpendicular bisectors are needed because
the third bisector intersects at the same point.
9a. Plan A: y 4x 20; Plan B: y 7x;
x 6 h 40 min, y $46.67
y
9b.
y
y 2x 16
5
30
Plan B
20 (0, 20)
10
2
5
( 7, 2)
x
10
4
6
Hours
x
The point of intersection shows when both plans
cost the same, the answer for 9a.
9c. Plan B; 721 hours, with Plan A
10. (3, 1), (0, 3), (3, 1)
y 12 x 32
–5
–10
y
y 13 x 2
x
6. The lines are parallel (the slopes are the same,
but the y-intercepts are different).
y
There is no solution.
y = –2x + 9
5
x
–5
–5
y = –2x – 1
–10
8. Possible solution using the equations
y 21x 1 and y 32x 134 :
1
y 2x 1
2
14 1
3x 3 2x 1
4x 28 3x 6
22 7x
22
x
7
1 22
y 2 7 1
11a.
y 43x 23
y x 12
11b. (6, 6)
11c. (6, 6); (6, 6)
11d. The diagonals intersect at their midpoints,
which supports the conjecture that the diagonals of
a parallelogram bisect each other.
2
13
12. y 3x 3
13. (4, 7)
3
15. 3, 2
69 6
14. 1,
4 7
16. The circumcenter is the midpoint of the
hypotenuse. For a right triangle, the perpendicular
bisectors of the legs are two of the midsegments
of the triangle. As with any pair of triangle
midsegments, they intersect at the midpoint of
the third side.
18
y 7
ANSWERS TO EXERCISES
81
Answers to Exercises
5. The lines are the same. There are infinitely many
solutions.
4
6 _23 , 46 _23
Plan A
Answers to Exercises
LESSON 6.7
4
1. 3 in.
2. 8 m
3. 14 cm
4. 9 m
5. 6 ft
6. 4 m
7. 27 in.
8. 100 cm
9. 217 m/min
10. 4200 mi
11. Desks are about 17 meters from the center.
About four desks will fit because an arc with
one-half the radius and the same central angle
will be one-half as long as the outer arc.
12. The measure of the central angle is 7.2°
because of the Corresponding Angles Conjecture.
Therefore, 500 376.20 C, so C 25,000 mi.
13. 18°/s. No, the angular velocity is measured in
degrees per second not in distance per second, so it
is the same at every point on the carousel.
14. Outer horse 2.5 m/s, inner horse 1.9 m/s.
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).
15. a 50°, b 75°, x 25°; y 45°; z 35°;
Conjecture: The measure of the angle formed by
an intersecting tangent and secant to a circle is
one-half the difference of the larger intercepted arc
measure and the smaller intercepted arc measure.
16. Let z represent the measure of the exterior
.
and tangent PB
angle of PBA formed by AB
1 By the Tangent Chord Conjecture, z 2mAB . By
.
the Inscribed Angle Conjecture, mBAP 21mBC
By the Triangle Exterior Angle Conjecture, z mBPA mBAP, or mBPA z mBAP.
82
ANSWERS TO EXERCISES
1mBC
.
By substitution, mBPA 21mAB
2
Using the distributive property, mBPA 1
(mAB mBC ).
2
17. 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°
18. possible answer:
19. 170°
8
289
20. y 1x
5 15
21. 45°
22. The sum of the lengths is 8 cm for Case 1,
Case 2, Case 3, and Case 10.
23. 6
24. Yes, as long as the three points are noncollinear; possible answer: connect the points with
segments, then find the point of concurrency of
the perpendicular bisectors (same as circumcenter
construction).
CHAPTER 6 REVIEW
23. Sample answer: Construct a right angle and
and RT
with any
label the vertex R. Mark off RE
lengths. From point E, swing an arc with radius RT.
From point T, swing an arc with radius RE. Label
the intersection of the arcs as C. Construct the
and RC
. Their intersection is the
diagonals ET
center of the circumscribed circle. The circle’s
radius is the distance from the center to a vertex. It
is not possible to construct an inscribed circle in a
rectangle unless it is a square.
E
C
R
T
24. Sample answer: Construct acute angle R. Mark
off equal lengths RM and RH. From points M and
H, swing arcs of lengths equal to RM. Label the
intersection of the arcs as O. Construct RHOM. The
intersection of the diagonals is the center of the inscribed circle. Construct a perpendicular to a side to
find the radius. It is not possible to construct a circumscribed circle unless the rhombus is a square.
H
O
R
M
4
32
25. 4x 3y 32, or y 3x 3
26. (3, 2)
27. d 0.318 m
28. Melanie: 151 m/min or 9 km/h; Melody:
94 m/min or 6 km/h.
2(6357)
2(6378)
29. 1.849 1.852 1.855 360 60
360 60
30.
Possible
location
7700 ft
5280 ft
5500 ft
D
T
Possible
location
ANSWERS TO EXERCISES
83
Answers to Exercises
1. Answers will vary.
2. Draw two nonparallel chords. The intersection
of their perpendicular bisectors is the center of the
circle.
Fold the paper so that two semicircles coincide.
Repeat with two different semicircles. The center is
the intersection of the two folds.
Place the outside or inside corner of the L in the
circle so that it is an inscribed right angle. Trace the
sides of the corner. Draw the hypotenuse of the right
triangle (which is the diameter of the circle). Repeat.
The center is the intersection of the two diameters.
3. The velocity vector is always perpendicular to
the radius at the point of tangency to the object’s
circular path.
4. 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.
5. 55°
6. 65°
7. 128°
8. 118°
9. 91°
10. 66° 11. 125.7 cm 12. 42.0 cm
13. 15 cm
14. 14 ft
15. 2 57° 2 35° 180°
16. 84° 56° 56° 158° 360°
1 (180° 108°) 36° 17. mEKL 21 mEL
2
YL
by Converse of the Parallel
mKLY. KE
Lines Conjecture.
360° 56° 152° 152° mMI
.
18. mJI
1 mMI
mMJI. By the
mJMI 21 mJI
2
Converse of the Isosceles Triangle Conjecture,
JIM is isosceles.
2mKEM 140°. mKI
19. mKIM
140° 70° 70° mMI . mIKM 1 1 mMI mKI mIMK. By the Converse of the
2
2
Isosceles Triangle Conjecture, KIM is isosceles.
20. 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.
21. Sample answer: Construct
perpendicular bisectors of two
sides of the triangle. The point
at which they intersect (the
circumcenter) is the center of
the circle. The distance from
the circumcenter to each vertex
is the radius.
22. Sample answer: Construct the incenter (from
the angle bisectors) of the triangle. From the incenter,
which is the center of the circle, construct a perpendicular to a side. The distance from the incenter to
the foot of the perpendicular is the radius.
200
31. ft 63.7 ft
32. 8 m 25.1 m
8
33. The circumference is 346
0 2(45) 12; the
diameter is 12 cm.
34. False. 20° 20° 140° 180°. An angle with
measure 140° is obtuse.
35. true
D
36. false
C
Answers to Exercises
A
53. False. The ratio of the circumference to the
diameter is .
24 cm
24 cm
54. false; 24 24 48 48 96
48 cm
55. true
56. This is a paradox.
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°
58. TAR YRA by SAS, TAE YR E
by SAA
59. FTO YTO by SAA, SAS, or SSS;
FLO YLO by SAA, SAS, or SSS;
FTL YTL by SSS, SAS, or ASA
60. PTR ART by SSS or SAS;
TPA RAP by SSS, SAS, SAA, or ASA;
TLP RLA by SAA or ASA
61. ASA
84°, length of AC
11.2 35.2 in.
62. mAC
63. x 63°, y 27°, w 126°
64. sample answer:
B
37. true 38. true 39. true 40. true
41. False. (7 2) 180° 900°. It could have seven
sides.
42. False. The sum of the measures of any triangle
is 180°.
43. 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 counterexamples.
44. False. The consecutive angles between the
bases are supplementary.
45. False. 48° 48° 132° 180°
46. False. Inscribed angles that intercept the same
arc are congruent.
47. False. The measure of an inscribed angle is
half the measure of the arc.
48. true
and BD
bisect each other, but AC
is
49. False. AC
.
not perpendicular to BD
D
C
A
B
48 cm
30
150
65. See table below.
66a. The circle with its contents has 3-fold rotational symmetry, the entire tile does not.
66b. No, it does not have reflectional symmetry.
67.
68. 9.375 cm
50. False. It could be isosceles.
51. False. 100° 100° 100° 60° 360°
CD
52. false; AB
C
A
69.
90
90
70.
D
B
65. (Chapter 6 Review)
84
n
1
2
3
4
5
6
f (n)
5
1
3
7
11
15
ANSWERS TO EXERCISES
...
n
...
20
. . . 9 4n . . .
71