Chapter 10
Chapter 10 Maintaining Mathematical Proficiency (p. 527)
1. (x + 7)(x + 4) = x
⋅x+x⋅4+x⋅7+7⋅4
w2 − 8w = 9
9.
w2 − 8w + (−4)2 = 9 + (−4)2
= x2 + 4x + 7x + 28
=
x2
(w − 4)2 = 25
+ 11x + 28
The product is x2 + 11x + 28.
2. (a + 1 )(a − 5) = a
=
w=4±5
⋅a−a⋅5+1⋅a−1⋅5
− 5a + a − 5
a2
The solutions are w = 4 + 5 = 9 and w = 4 − 5 = −1.
10. p 2 + 10p − 4 = 0
= a2 − 4a − 5
The product is
a2
p2 + 10p = 4
− 4a − 5.
3. (q − 9)(3q − 4) = q
p 2 + 10p + 52 = 4 + 52
⋅ 3q − q ⋅ 4 − 9 ⋅ 3q + 9 ⋅ 4
(p + 5)2 = 29
k2 − 4k − 7 = 0
11.
k2 − 4k = 7
= 10v2 − 33v − 7
k2
The product is 10v2 − 33v − 7.
− 4k + (−2)2 = 7 + (−2)2
(k − 2)2 = 11
5. (4h + 3)(2 + h) = (4h + 3)(h + 2)
= 4h
—
k = 2 ± √11
—
= 4h2 + 8h + 3h + 6
The solutions
are k = 2 + √ 11 ≈ 5.32 and
—
k = 2 − √ 11 ≈ −1.32.
= 4h2 + 11h + 6
6. (8 − 6b)(5 − 3b) = 8
−z2 + 2z = 1
12.
⋅ 5 − 8 ⋅ 3b − 6b ⋅ 5 − 6b ⋅ (−3b)
= 40 − 24b − 30b + 18b2
z2 − 2z = −1
z2
− 2z + (−1)2 = −1 + (−1)2
(z − 1)2 = 0
= 40 − 54b + 18b2
z−1=0
The product is 18b2 − 54b + 40.
z=1
x2 − 2x = 5
7.
The solution is z = 1.
x2 − 2x + (−1)2 = 5 + (−1)2
(x − 1)2 = 6
13. Sample answer: (2n + 1)(2n + 3); 2n + 1 is positive and
—
x − 1 = ±√ 6
odd when n is a nonnegative integer. The next positive, odd
integer is 2n + 3.
—
x = 1 ± √6
—
The solutions
are x ≈ 1 + √ 6 = 3.45 and
—
x = 1 − √6 ≈ −1.45.
8.
r2
Chapter 10 Mathematical Practices (p. 528)
1. Sketch circles A and B with radius 3 units and circle C so
that it passes through the centers of circles A and B. C must
be 3 units from A and B, so C must lie on an intersection of
circles A and B.
+ 10r = −7
r 2 + 10r + 52 = −7 + 52
(r + 5)2 = 18
—
k − 2 = ±√11
⋅ h + 4h ⋅ 2 + 3 ⋅ h + 3 ⋅ 2
The product is 4h2 + 11h + 6.
—
The solutions—
are p = −5 + √ 29 ≈ 0.39 and
p = −5 − √ 29 ≈ −10.39.
The product is 3q2 − 31q + 36.
= 10v2 + 2v − 35v − 7
—
p = −5 ± √29
= 3q2 − 31q + 36
⋅ 5v + 2v ⋅ 1 − 7 ⋅ 5v − 7 ⋅ 1
—
p + 5 = ±√29
= 3q2 − 4q − 27q + 36
4. (2v − 7)(5v + 1) = 2v
—
w − 4 = ±√25
—
r + 5 = ±√ 18
—
r = −5 ± 3√ 2
A
B
—
The solutions —are r = −5 + 3√ 2 ≈ − 0.76 and
r = −5 − 3√2 ≈ −9.24.
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C
Geometry
Worked-Out Solutions
341
Chapter 10
2. The distance from the center of circle D to any point on
4. The circles have 1 common external tangent.
circle D is greater than 6. Circles A, B, and C each have
diameters of 6 units, and because circle D is tangent to the
circles on the outside and whether circle A, circle B, or circle
C intersect or contain point D, the radius of circle D must be
greater than 6.
5. There are no common tangents.
A
B
D
C
6. Use the Converse of the Pythagorean Theorem (Thm. 9.2).
10.1 Explorations (p. 529)
1. Chord: a segment whose endpoints lie on the circle
52
32 + 42
25
9 + 16
Secant: a line that intersects a circle at two points
25 = 25
Tangent: a line in the plane that intersects the circle at
exactly one point
△CDE is a right triangle with the right angle at ∠D.
Radius: a segment whose endpoints are the center and any
point on a circle
Diameter: a chord that contains the center of the circle
— is tangent to ⊙C at point D.
Therefore, DE
(r + 18)2 = 242 + r2
7.
r2
+ 36r + 324 = 576 + r2
36r + 324 = 576
2. a. Check students’ work.
36r = 252
b. Check students’ work.
c. The distance between the pencil points are the radius and
half the diameter.
3. A chord is a segment with endpoints that lie on a circle.
A diameter is a chord that passes through the center of a
circle. A radius is a segment with one endpoint on the center
of a circle and one endpoint on the circle. A secant line is a
line that passes through two points on a circle. A tangent line
is a line that passes through only one point on a circle.
4. Diameters are a subset of chords. A diameter is a chord that
passes through the center of a circle.
5. To draw a circle with a diameter of 8 inches, use two pencils
tied together with a string that is 4 inches long.
10.1 Monitoring Progress (pp. 530 –533)
—
—
1. AG is a chord and CB is a radius.
— —
2. ⃖⃗
DE is a tangent line and DE or DB is a tangent segment.
3. The circles have 4 common tangents, 2 external tangents,
and 2 internal tangents.
r=7
The radius of ⊙Q is 7.
8. By the External Tangent Congruence Theorem (Thm. 10.2),
x2 = 9. By taking the square root of each side, x = ±3.
10.1 Exercises (pp. 534 –536)
Vocabulary and Core Concept Check
1. Chords and secants both intersect the circle in two points. A
chord is a segment and a secant is a line.
2. When the context is measure, it refers to length.
3. Coplanar circles that have a common center are called
concentric circles.
4. Tangent is the segment that does not belong, because a
tangent segment is on the outside of the circle. The other
three, chord, radius, and diameter, are segments that are on
the interior of the circle.
Monitoring Progress and Modeling with Mathematics
5. The name of the circle is C.
—
—
6. Two radii are CA and CD .
—
—
7. Two chords are BH and AD .
—
8. AD is a diameter.
342
Geometry
Worked-Out Solutions
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Chapter 10
⃖⃗ is a secant.
9. KG
20. Use the Converse of the Pythagorean Theorem (Thm. 9.2).
10. A tangent line is ⃖⃗
EG, and a point of tangency is F.
182
92 + 152
324
81 + 225
324 ≠ 306
11. There are 4 common tangents.
— is not a tangent
△ ABC is not a right triangle. Therefore, AB
segment.
21. Use the Converse of the Pythagorean Theorem (Thm. 9.2).
602
402 + 482
3600
1600 + 2304
3600 ≠ 3904
— is not a tangent
△DAB is not a right triangle. Therefore, AB
segment.
12. There are no common tangents.
22. Use the Converse of the Pythagorean Theorem (Thm. 9.2).
202 ____ 122 + 162
400 ____ 144 + 256
400 = 400
△ ABC is a right triangle, with the right angle at ∠A.
— is tangent to ⊙C at point A.
Therefore, AB
13. There are 2 common tangents.
(r + 16)2 = 242 + r 2
23.
r2
+ 32r + 256 = 576 + r 2
32r + 256 = 576
32r = 320
r = 10
14. There is 1 common tangent.
The radius of ⊙C is 10.
(r + 6)2 = 92 + r 2
24.
r 2 + 12r + 36 = 81 + r 2
12r + 36 = 81
12r = 45
r = 3.75
15. The common tangent is an external tangent because it does not
The radius of ⊙C is 3.75.
intersect the segment that joins the centers of the two circles.
16. The common tangent is an internal tangent because it
r2
intersects the segment that joins the centers of the two circles.
14r = 147
r = 10.5
18. The common tangent is an external tangent because it does not
The radius of ⊙C is 10.5.
intersect the segment that joins the centers of the two circles.
+
52
32
42
25
9 + 16
25 = 25
△CAB is a right triangle with the right angle at ∠A.
— is tangent to ⊙C at point A.
Therefore, AB
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+ 14r + 49 = 196 + r 2
14r + 49 = 196
17. The common tangent is an internal tangent because it
19. Use the Converse of the Pythagorean Theorem (Thm. 9.2).
(r + 7)2 = 142 + r 2
25.
intersects the segment that joins the centers of the two circles.
26.
(r + 18)2 = 302 + r 2
r 2 + 36r + 324 = 900 + r 2
36r + 324 = 900
36r = 576
r = 16
The radius of ⊙C is 16.
Geometry
Worked-Out Solutions
343
Chapter 10
27. Sample answer:
C
28. Sample answer:
C
A
2 in.
B
A
C
4.5 cm
29. 2x + 7 = 5x − 8
30. 3x + 10 = 7x − 6
7 = 3x − 8
10 = 4x − 6
15 = 3x
16 = 4x
5=x
4=x
31. 2x 2 + 4 = 22
32. 3x 2 + 2x − 7 = 2x + 5
2x 2 = 18
3x 2 − 7 = 5
=9
3x 2 = 12
x = ±3
x2 = 4
x2
38. a.
x = ±2
—
33. ∠Z is a right angle, not ∠YXZ. XY is not tangent to ⊙Z.
15
34. The diameter is 15 and the radius is —
2 = 7.5.
35. There are two possible points of tangency from a point
outside the circle, one from a point on the circle, and none
from a point inside the circle.
A
D
D
A
b. ∠ACB is a right angle because it was given that the radii
— and CB
— are radii of the
were perpendicular. Because CA
—
—
same circle, CA ≅ CB . The tangent lines are perpendicular
to the radii by the Tangent Line to Circle Theorem
(Thm. 10.1). So, ∠A and ∠B are right angles. Because
—
ACBD is a quadrilateral, ∠D is also a right angle. AD
—
≅ DB by the External Tangent Congruence Theorem
(Thm. 10.2). Therefore, quadrilateral ADBC is a square.
39. By the Tangent Line to a Circle Theorem (Thm. 10.1),
∠PEB and ∠PMB are right angles. By the External Tangent
Congruence Theorem (Thm. 10.2), BE = BM and because
all radii in a circle are equal, PE = PM. By the SAS
Congruence Theorem (Thm. 5.5), △PEB ≅ △PMB. Because
corresponding angles of congruent triangles are congruent,
∠PBE ≅ ∠PBM. Hence, the path of the bike trail will bisect
the angle formed by the nature trails.
40. To find the distance between the centers of the gears
represented by points L and P, divide the region into a
rectangle and a triangle.
M
1.8 in.
L
17.6 in.
1.8 in.
2.5 in.
B
C
N
17.6 in.
4.3 in.
P
36. When the tangent lines are parallel.
Sample answer: The tangents in the figure are perpendicular
to the same diameter and, therefore, are parallel.
Apply the Pythagorean Theorem to the right triangle to
find LP.
LP 2 = 2.52 + 17.62
LP 2 = 6.25 + 309.76
LP 2 = 316.01
—
LP = √316.01 ≈ 17.8 inches
A
The distance between the centers of the gears is about
17.8 inches.
B
C
41. Sample answer: Every point is the same distance from the
37. The perimeter of the quadrilateral is
6.4 + 4.5 + 6.4 + 8.3 = 25.6 units.
6.4
X
5.2
8.3
3.1 W
— — —
42. PA ≅ PB ≅ PC by the External Tangent Congruence
Theorem (Thm. 10.2).
V
3.3
Y
5.2
344
4.5
1.2 T 1.2 3.3
center, so the farthest two points can be from each other is
opposite sides of the center.
3.1
6.4
Geometry
Worked-Out Solutions
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Chapter 10
43. Given
Prove
— is a common internal tangent in ⊙A and ⊙B.
RS
AC
BC
Line m is tangent to ⊙Q at point P.
47. a. Given
—
m ⊥ QP
Prove
RC
SC
—=—
Q
R
A
B
C
m
S
STATEMENTS
— is a common
1. RS
REASONS
1. Given
internal tangent in
⊙A and ⊙B.
— ⊥ AR
— and RS
— ⊥ BS
—
2. RS
2. Tangent Line to Circle
Theorem (Thm. 10.1)
P
R
— is not perpendicular to line QP
—. Then the
Assume that mP
perpendicular segment from Q to m intersects m at some
—. The perpendicular segment
other point R. Construct QR
is the shortest distance from Q to m. So, QR < QP, R
must be inside ⊙Q, and m must be a secant line. This is a
contradiction, so m must be tangent to ⊙Q.
—
m ⊥ QP
b. Given
Line m is tangent to ⊙Q.
Prove
3. ∠CRA and ∠CSB are
right angles.
3. Definition of perpendicular
Assume m is not tangent to ⊙Q. Then m must intersect
4. ∠RCA ≅ ∠SCB
4. Vertical Angles Congruence
Theorem (Thm. 2.6)
5. △RAC ∼ △SBC
5. AA Similarity Triangle
Theorem (Thm. 8.3)
— ≅ QR
—. Because m ⊥QP
—, QP< QR. This is a
so QP
contradiction, so m must be tangent to ⊙Q.
AC RC
6. — = —
BC SC
6. Corresponding sides
of similar figures are
proportional.
— and QR
— are both radii of ⊙Q,
⊙Q at a second point R. QP
—
48. Use the Pythagorean Theorem to find the length of AE .
AE 2 + 42 = 122
AE 2 = 144 − 16
—
AE = √144 − 16
—
= √128
44. Yes, it is true.
—
— ≅ AW
—, BW
— ≅ BX
—, CX
— ≅ CY
—, and DY
— ≅ DZ
—. So,
AZ
(AW + WB) + (CY + YD) = (AZ + DZ ) + (BX + CX ).
45. 2x − 5 = x + 8
2x − 5 + 4y − 1 = x + 8 + x + 6
2x + 4y − 6 = 2x + 14
4y − 6 = 14
4y = 20
y=5
— and ST
— are tangent to ⊙P.
SR
— ≅ ST
—
Prove SR
R
P
T
—
= 8√2
B
x = 13
46. Given
= √64 ⋅ 2
S
By the Tangent Line to Circles Theorem (Thm. 10.1),
— ≅ PR
— and PS
— ≅ PS
—, so
m∠PRS = m∠PTS = 90°. PT
△PTS ≅ △PRS by the HL Congruence Theorem (Thm. 5.9).
— ≅ ST
—.
So, SR
8E
C
D
12
P
F
A
12
—, AC
—, and BC
— are all tangents to circle O, and PD
—,
Because AB
—, and PE
— are radii and perpendicular to the tangents at the
PF
— is an altitude to △APB, PE
— is an
points of tangency, PD
— is an altitude to △CPA.
altitude to △BPC, and PF
1
Area of △APB = — r 12 = 6r
2
1
Area of △BPC = — r 8 = 4r
2
1
Area of △CPA = — r 12 = 6r
2
The total area is 6r + 4r + 6r = 16r.
—
—
1
Area of △ ABC = — 8√ 2 8 = 32√2
2
—
16r = 32√2
⋅ ⋅
⋅ ⋅
⋅ ⋅
⋅
⋅
—
—
32√2
r = — = 2√2 ≈ 2.83
16
—
The radius of circle P is 2√2 ≈ 2.83.
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Geometry
Worked-Out Solutions
345
Chapter 10
Maintaining Mathematical Proficiency
49. m∠JKM = m∠JKL + m∠LKM
3. a.
5
50. AC = AB + BC
m∠JKM = 15° + 28°
4
m∠JKM = 43°
2
3 = AB
1
0
+
52
=
=
b2
+
1
0
4
—5
()
5
6
7
−5
c2
c.
5
− 2bc cos A
3
2
1
0
4
5
2
3
4
5
−4
−5
d.
5
C
4
3
2
1
0
A
B
−7 −6 −5 −4 −3 −2 −1
0 1
−1
−48 = −50 cos A
m∠A =
3
−3
2 = 50 − 50 cos A
24
cos−1 —
25
B
2
−2
()
cos A =
A
−7 −6 −5 −4 −3 −2 −1
0 1
−1
3
—5
48
—
50
C
4
( √—2 )2 = 52 + 52 − 2(5)(5) cos A
−2
−3
24
—
25
−4
( ) ≈ 16.26°
−5
d. mBC ≈ 106.26°
Use the Law of Cosines to verify.
10.2 Monitoring Progress (pp. 539 –541)
—
BC = √(4 − (−4))2 + (3 − 3)2 = √ (4 + 4)2 = √82 = 8
——
—
a2 = b2 + c2 − 2bc cos A
is a major arc, so m
2. QRT
QRT = m
QR + m
RS + m
ST
1. TQ is a minor arc, so mTQ = 120°.
= 60° + 100° + 80° = 240°.
(8)2 = 52 + 52 − 2(5)(5) cos A
64 = 50 − 50 cos A
14 = −50 cos A
14
7
—
cos A = −—
50 = − 25
(
7
)
m∠A = cos−1 −—
25 ≈ 106.26°
2. Circular arcs are measured in degrees or radians and are
based on the measure of the central angle.
346
4
−4
a2 = b2 + c2 − 2bc cos A
3
−3
Use the Law of Cosines to verify.
——
—
—
BC = √(4 − 3)2 + (3 − 4)2 = √ 1 + 1 = √2
=
7
B
2
−2
m∠A = cos−1 —35 ≈ 53.1°
c. m
BC ≈ 16.26°
A
−7 −6 −5 −4 −3 −2 −1
0 1
−1
−30 = −50 cos A
=
6
2
20 = 50 − 50 cos A
cos A =
7
3
2
( √—
20 ) = 52 + 52 − 2(5)(5) cos A
30
—
50
6
C
4
Use the Law of Cosines to verify.
——
—
—
—
BC = √(3 − 5)2 + (4 − 0)2 = √ 4 + 16 = √20 = 5√2
a2
7
5
− 2(5)(5) cos A
m∠A = cos−1 —54 ≈ 36.9°
b. m
BC ≈ 53.1°
6
b.
10 = 50 − 50 cos A
cos A =
5
−5
−40 = −50 cos A
40
—
50
4
−4
a2 = b2 + c2 − 2bc cos A
=
3
−3
Use the Law of Cosines to verify.
——
—
—
BC = √(4 − 5)2 + (3 − 0)2 = √ 1 + 9 = √10
( √10 )
B
2
−2
1. a. mBC ≈ 36.9°
52
A
−7 −6 −5 −4 −3 −2 −1
0 1
−1
10.2 Explorations (p. 537)
— 2
C
3
10 = AB + 7
Geometry
Worked-Out Solutions
3. TQR is a semicircle, so mTQR = mTQ + mQR
= 120° + 60° = 180°.
4. QS is a minor arc, so mQS = mQR + mRS
= 60° + 100° = 160°.
5. TS is a minor arc, so mTS = 80°.
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Chapter 10
6. RST is a semicircle, so mRST = mRS + mRS T
= 100° + 80° = 180°.
m
AB = m
CD .
7. AB ≅ CD , because the circles are congruent and
8. Because the two circles have different radii, the circles are
not congruent and, therefore, MN is not congruent to PQ .
10.2 Exercises (pp. 542 –544)
Vocabulary and Core Concept Check
1. If ∠ACB and ∠DCE are congruent central angles of ⊙C,
then AB and DE are congruent arcs.
2. The circle with a diameter of 6 inches does not belong. The
others have a diameter of 12 inches.
Monitoring Progress and Modeling with Mathematics
arc is ADB and its measure is 360° − 135° = 225°.
4. The minor arc is EF and it has a measure of 68°. The major
arc is FGE and its measure is 360° − 68° = 292°.
5. The minor arc is JL and it has a measure of 120°. The major
and its measure is 360° − 120° = 240°.
arc is JKL
and it has a measure of 170°. The
6. The minor arc is MN
and its measure is 360° − 170° = 190°.
major arc is NPM
7. BC is a minor arc and it has a measure of 70°.
8. DC is a minor arc and it has a measure of 65°.
is a minor arc and it has a measure of 45°.
9. ED
10. AE is a minor arc and it has a measure of 70°.
11. EAB is a semicircle and it has a measure of 180°.
12. ABC is a semicircle and it has a measure of 180°.
13. BAC is a major arc and it has a measure of 290°.
is a major arc and it has a measure of 315°.
14. EBD
= mJK
+ mKL
= 53° + 79° = 132°
15. a. mJL
b. mKM = mKL + m
LM = 79° + 68° = 147°
= mJK
+ mKL
+ mLM
= 53° + 79° + 68° = 200°
c. mJLM
d. mJM = 360° − 200° = 160°
16. a. m
RS = m
QRS − m
QR = 180° − 42° = 138°
b. mQRS = 180°
c. m
QST = m
QR + m
RS + m
ST = 42° + 138° + 42° = 222°
d. mQT = mQTS − m
TS = 180° − 42° = 138°
3. The minor arc is AB and it has a measure of 135°. The major
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17. a. mAE = mAH + mHG + mGF + mFE
= 17° + 26° + 28° + 32° = 103°
b. m
ACE = m
AB + m
BC + m
CD + m
DE
= 66° + 55° + 89° + 47° = 257°
c. m
GDC = m
GF + m
FE + m
ED + m
DC
= 28° + 32° + 47° + 89° = 196°
d. m
BHC = 360° − m
BC = 360° − 55° = 305°
= m
e. mFD
FE + m
ED = 32° + 47° = 79°
f. m
FBD = 360° − (m
FE + m
ED )
= 360° − (32° + 47°) = 281°
⋅ 360° = 72°
Soccer: mWX = 30% ⋅ 360° = 108°
= 15% ⋅ 360° = 54°
Volleyball: mXY
Cross-Country: m
YZ = 20% ⋅ 360° = 72°
= 15% ⋅ 360° = 54°
None: mZV
≅ CD
because they are in the same circle and m
19. AB
AB = m
CD .
is not congruent to MN
because the circles are not
20. LP
18. Football: mVW = 20%
congruent.
22. QRS is not congruent to FGH because the circles are not
21. VW ≅ XY because the circles are congruent and mVW = mXY .
congruent.
23. (2x − 30)° + x° = 180°
3x − 30 = 180
3x = 210
x = 70
m
AB = 2 ⋅ 70° − 30° = 110°
24. 6x° + 7x° + 7x° + 4x° = 360°
24x = 360
x = 15
m
RST = 13 15 = 195°
⋅
25. Your friend is correct. The arcs must be in the same circle or
congruent circles.
26. Your friend is not correct. AMB is a semicircle,
so x° + 4x° = 180°.
28. JK and NP are not in the same circle. JK ≅ RQ or
LM ≅ NP.
29. m
ACD = 360° − 20° = 340°
D
m
AC = 180° − 20° = 160°
27. AD is the minor arc. The red arc is the major arc, ABD .
20°
A
P
Geometry
Worked-Out Solutions
B
C
347
Chapter 10
m
AE = 360° − 175° = 185° (major arc)
30. mAE = 60° + 25° + 70° + 20° = 175° (minor arc) or
C 25°B
70°
36. The circles are similar because they are concentric, with
different radii.
37.
60°
D
20°
E
A
P
C
B
D
A
E
∠BAC ≅ ∠DAE
a. Given
360°
20
360°
32. a. Each arc measure is — = 15°.
20
b. The measure of the minor arc from the Tokyo time zone to
the Anchorage time zone is 6 15° = 90°.
Prove
BC ≅ DE
b. Given
BC ≅ DE
c. If two locations differ by 180° on the wheel, then it is 3 p.m.
Prove
31. The measure of each arc is — = 18°.
m
BC = m∠BAC, m
DE = m∠DAE and
m∠BAC = m∠DAE, so m
BC = m
DE . Because BC and
DE are in the same circle, BC ≅ DE .
⋅
⊙O with center O(0, 0) and radius r, ⊙A with
center A with center A(a, 0) and radius s
33. Given
⊙O ∼ ⊙A
Prove
∠BAC ≅ ∠DAE
m
BC = m∠BAC and m
DE = m∠DAE. Because
BC ≅ DE , ∠BAC ≅ ∠DAE.
at one location when it is 3 a.m. at the other location.
38. The circumference of a circle is 2πr. So, the arc length is
θ
πrθ
s = 2πr — = —.
360° 180°
⋅
y
Maintaining Mathematical Proficiency
O
r
A
s
39.
x
c2 = a2 + b2
172 = x2 + 82
289 = x2+ 64
Translate ⊙A left a units so that point A maps to point O.
The image of ⊙A is ⊙A′ with center O, so ⊙A′ and ⊙O are
concentric circles. Dilate ⊙A′ using center of dilation O and
r
scale factor — , which maps the points s units from point O to
s
r
the points — (s) = r units from point O. So, this dilation maps
s
⊙A′ to ⊙O. Because a similarity transformation maps ⊙A to
⊙O, ⊙O ∼ ⊙A.
—
225 = x2
—
x = √225 = 15
Because the side lengths 8, 15, and 17 are integers that satisfy
the equation c2 = a2 + b2, they form a Pythagorean triple.
√2
⋅
√
40. hypotenuse = leg
35.
—
41. c2 = a2 + b2
x2 = 72 + 112
C
x2 = 49 + 121
B
A
—
x = √170 ≈ 13.04
—
Because √ 170 is not an integer, the side lengths do not form
a Pythagorean triple.
D
a. Given
— ≅ BD
—
AC
Prove
⊙A ≅ ⊙B
Translate ⊙B so that point B maps to point A. The image
— ≅ BD
—, this
of ⊙B is ⊙B′ with center A. Because AC
translation maps ⊙B′ to ⊙A. A rigid motion maps ⊙B to
⊙A, so ⊙A ≅ ⊙B.
b. Given
—
x = 13 2 ≈ 18.38
—
Because 13√ 2 is not an integer, the side lengths do not form
a Pythagorean triple.
34. yes; CD is the radius of ⊙C and DC is the radius of ⊙D.
Because the radii are the same, the circles are congruent.
—
⊙A ≅ ⊙B
— ≅ BD
—
Prove AC
42.
c 2 = a2 + b2
142 = x2 + 102
196 = x2 + 100
—
x = √96 ≈ 9.80
—
Because √ 96 is not an integer, the side lengths do not form a
Pythagorean triple.
Because ⊙A ≅ ⊙B, the distance from the center of the
circle to a point on the circle is the same for each circle.
— ≅ BD
—.
So, AC
348
Geometry
Worked-Out Solutions
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Chapter 10
10.3 Explorations (p. 545)
2. Check students’ work. The perpendicular bisector is a
1. a. The diameter with an endpoint of (4, 3) will pass through
the center and the other endpoint of the diameter will be
(−4, −3).
3. Check students’ work.
DF = EF
5
4
The results are the same. If a chord is perpendicular to a
diameter of a circle, then the diameter is a perpendicular
bisector of the chord.
3
2
1
0
−7 −6 −5 −4 −3 −2 −1
0 1
−1
diameter. The results are the same. A perpendicular bisector
of a chord is a diameter of the circle.
2
3
4
5
6
7
−2
4. A chord is a diameter of a circle when it is a perpendicular
bisector of a chord or it passes through the center of the circle.
−3
−4
10.3 Monitoring Progress (pp. 547 –548)
−5
— —
b. The diameter with an endpoint of (0, 5) will pass through
the center and the other endpoint of the diameter will be
(0, −5).
1. Because AB ≅ BC , then by the Congruent Corresponding
Chords Theorem (Thm. 10.6), AB ≅ BC . Therefore,
mBC = 110°.
= 360°
150° + 2mAB
2m
AB = 210°
= 105°
mAB
2. mAC + mCB + mAB = 360°
5
4
3
2
1
0
−7 −6 −5 −4 −3 −2 −1
0 1
−1
2
3
4
5
6
7
⋅ 5 = 10
4. mCD = mDE
−3
9x° = (80 − x)°
−4
−5
10x = 80
c. The diameter with an endpoint of (−3, 4) will pass
through the center and the other endpoint of the diameter
will be (3, −4).
5
x=8
m
CE = 9x° + (80 − x)°
⋅
= 8x + 80 = 8 8 + 80 = 64 + 80 = 144°
5. PN = NO
4
3
3x = 7x − 12
2
−4x = −12
1
0
−7 −6 −5 −4 −3 −2 −1
0 1
−1
3. CE = 2
−2
2
3
4
5
6
7
x=3
(3x) 2 + 122 = (NK)2
−2
⋅
−3
(3 3)2 + 122 = (NK)2
−4
−5
81 + 144 = (NK)2
—
d. The diameter with an endpoint of (−5, 0) will pass
through the center and the other endpoint of the diameter
will be (5, 0).
5
The radius of circle N is 15 units.
10.3 Exercises (pp. 549 –550)
Vocabulary and Core Concept Check
4
1. To bisect a chord is to divide the chord into two segments of
3
equal length.
2
1
0
−7 −6 −5 −4 −3 −2 −1
0 1
−1
NK = √225 = 15
2
3
−2
−3
−4
−5
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4
5
6
7
2. no; One chord does not necessarily bisect another chord.
Monitoring Progress and Modeling with Mathematics
3. Because AB = DE and are in the same circle, then the
corresponding minors arcs are equal. Therefore, mAB = 75°.
Geometry
Worked-Out Solutions
349
Chapter 10
4. mTU = mUV , so TU = UV = 5.
AQ = QB
15.
4x + 3 = 7x − 6
5. By segment addition, WY = WC + CY and
−3x + 3 = −6
ZX = ZC + CX. Because WC = CX and ZC = YC,
WY = ZX. By arc addition, m
XYZ = m
XY + m
YZ ,
m
WZY = m
WZ + m
ZY , and m
XYZ = m
XWZ . By
substitution, m
XY + m
YZ = m
WZ + m
ZY ,
mXY + 60° = 110° + 60°, mXY + 60° = 170°, m
XY = 110°.
Therefore, m
XYZ = 110° + 60° = 170°.
−3x = −9
x=3
QG 2 = QB 2 + BG2
QG 2 = (7x − 6)2 + 82
⋅
QG 2 = (7 3 − 6)2 + 82
6. Because ⊙C ≅ ⊙P and QR = LM, LM = 11. Therefore,
QG 2 = 152 + 82
QR = 11.
QG 2 = 225 + 64
—
7. By the Perpendicular Chord Bisector Theorem (Thm. 10.7),
QG = √289 = 17
JE = EG. Therefore, x = 8.
The radius of circle Q is 17 units.
8. By the Perpendicular Chord Bisector Theorem (Thm. 10.7),
m
RS = m
ST . Therefore, m
RS = 40°.
AD = BC
16.
4x + 4 = 6x − 6
9. By the Perpendicular Chord Bisector Theorem (Thm. 10.7):
−2x + 4 = −6
5x − 6 = 2x + 9
−2x = −10
3x − 6 = 9
x=5
3x = 15
QD2
x=5
5x + 2 = 7x − 12
QD2 = 122 + 52
QD2 =144 + 25
—
QD = √169 = 13
−2x + 2 = −12
The radius of circle Q is 13 units.
−2x = −14
x=7
—
⋅
QD2 = (2 5 + 2)2 + 52
10. By the Perpendicular Chord Bisector Theorem (Thm. 10.7):
—
= (2x + 2)2 + 52
17. The perpendicular bisectors intersect at the center, so the
11. AC and DB are not perpendicular. BC is not congruent to CD .
right triangle with legs of 6 inches and 3.5 inches have a
hypotenuse equal to the length of the radius.
12. Sample answer: Draw the perpendicular bisector of the
r 2 = 62 + 3.52
control panels and find the midpoint; Because the control
panels are parallel and the congruent chords of the cross
section, they are equal distances from the center, and their
perpendicular bisectors form a diameter.
r 2 = 36 + 12.25
— —
The radius is about 6.95 inches. So, the diameter is about
2 6.95 = 13.9 inches.
13. AC ≅ AD and ∠CAB ≅ ∠DAB. By the Reflexive Property
— ≅ AB
—. So, by the SAS Congruence
of Congruence, AB
Theorem, △CAB ≅ △DAB. Because the triangles are
— ≅ BD
— and therefore AB
— is a perpendicular
congruent, BC
—
—
bisector of CD . So, AB is a diameter.
14. (AE)2 + (ED)2 = (AD)2
32 + (ED)2 = 52
9 + (ED)2 = 25
(ED)2 = 16
r 2 = 48.25
—
r 2 = √ 48.25 ≈ 6.95
⋅
—
18. a. AB is a diameter by the Perpendicular Chord Bisector
Converse (Thm. 10.8).
— —
b. AB ≅ CD by the Congruent Corresponding Chords
Theorem (Thm. 10.6).
—
—
c. JH bisects FG and FG by the Perpendicular Chord
Bisector Theorem (Thm. 10.7).
— —
d. NP ≅ LM by the Equidistant Chords Theorem (Thm. 10.9).
ED = 4
— is not a diameter.
Because CE ≠ ED, AB
350
Geometry
Worked-Out Solutions
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Chapter 10
19. a.
23. Given
— is a diameter of the circle L.
Prove QS
A
P
D
T
B
C
P
S
— and —
Given AB
CD are congruent chords.
Prove
L
AB ≅ CD
Prove
AB ≅ CD
— ≅ CD
—
AB
PA = PB = PC = PD, and because AB ≅ CD ,
∠DPC ≅ ∠APB. By the SAS Congruence Theorem
— ≅ CD
—.
(Thm. 5.5), △PDC ≅ △PAB, so AB
— — mAR = 360 − 2m
AC . m
AC is an integer, so 2m
AC
is even and 360 − 2m
AC is even.
20. Sample answer: AC ≅ RC , so AC ≅ RC and
R
24. ∠A ≅ ∠C because they intercept the same arc, BD .
∠APB ≅ ∠CPD, therefore, △APB ∼ △CPD by the
AP CP
AA Similarity Theorem. So, — = —.
BP DP
25.
C
G
A
F
21.
Q
— ≅ PR
—, LP
— ≅ LP
—, and LT
— ≅ LR
—, so △LPR ≅ △LPT
TP
by the SSS Congruence Theorem (Thm. 5.8). Then
∠LPT ≅ ∠LPR, so m∠LPT = m∠LPR = 90°. By definition,
—
—, so L lies on QS
—.
LP is a perpendicular bisector of RT
—
—
Because QS contains the center, QS is a diameter of ⊙L.
Because PA = PB = PC = PD, △PDC ≅ △PAB by the
SSS Congruence Theorem (Thm. 5.8). So, ∠DPC ≅ ∠APB
and AB ≅ CD .
b. Given
— is a perpendicular bisector of RT
—.
QS
D
E
B
— ≅ CD
—, then GC
— ≅ FA
—. Because EC
— ≅ EA
—,
If AB
P
C
8
6
D
△ECG ≅ △EAF by the HL Congruence Theorem
— = EG
—. If EF
— = EG
—, then because
(Thm. 5.9), so EF
— ≅ ED
— ≅ EA
— ≅ EB
—, △AEF ≅ △BEF ≅ △DEG ≅ △CEG
EC
by the HL Congruence Theorem (Thm. 5.9). Then
— ≅ BF
— ≅ DG
— ≅ CG
—, so AB
— ≅ CD
—.
AF
10
A
B
26. yes; The diameter of the tire that is perpendicular to the
( )
( — ) ≈ 53.13°.
6
Using △PDB, m∠DPB = sin−1 —
≈ 36.87°.
10
Using △PCA, m∠CPA = sin−1
8
10
m∠BPA = m∠CPA − m∠DBP ≈ 53.13° − 36.87° = 16.26°
Therefore, m
AB ≈ 16.26°.
— is a diameter of ⊙L, EG
— ⊥ DF
—.
EG
22. Given
— ≅ FC
—, Prove DC
DG ≅ FG
Maintaining Mathematical Proficiency
27. (n − 2)
⋅ 180° = (4 − 2) ⋅ 180° = 360°
360° − (32° + 25° + 44°) = 360° − 101° = 259°
So, m∠M = 259°.
28. (n − 2)
⋅ 180° = (54 − 2) ⋅180° = 540°
540° − (85° + 134° + 97° + 102°) = 540° − 418° = 122°
D
L
G
—
ground is also perpendicular to AB, so it bisects AB .
E
So, m∠T = 122°.
C
10.1–10.3 What Did You Learn? (p. 551)
F
— ≅ LF
— and LC
— ≅ LC
—, so △LDC ≅ △LFC by the
LD
— ≅ FC
—
HL Congruence Theorem (Thm. 5.9). Then DC
and ∠DLC ≅ ∠FLC. So, DG ≅ FG .
1. Sample answer: The External Tangent Congruency Theorem
(Thm. 10.2) could be used.
2. None of the arcs overlapped at more than one point, and
point E was on CD . There are other cases, such as point C
could be on AB .
3. A general conclusion was reached from several specific cases.
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Geometry
Worked-Out Solutions
351
Chapter 10
10.1–10.3 Quiz (p. 552)
1. The circle is P.
—
— —
2. PN and PK are radii.
—
3. KN is a diameter.
4. JL is a chord.
5. ⃖⃗
SQ is a secant.
⃖⃗ is a tangent line.
6. QR
7.
(x + 9)2 = 152 + x2
x2 + 18x + 81 = 225 + x2
8. 6x − 3 = 3x + 18
3x − 3 = 18
18x + 81 = 225
18x = 144
3x = 21
x=7
x=8
9. AE is a minor arc with a measure of 180° − 36° = 144°.
10. BC is a minor arc with a measure of 180° − (67° + 70°) = 43°.
11. AC is a minor arc with a measure of 67° + 43° = 110°.
12. ACD is semicircle with a measure of 180°.
13. ACE is a major arc with a measure of 180° + 36° = 216°.
14. BEC is a major arc with a measure of 360° − 43° = 317°.
15. JM ≅ KL because they are in the same circle and
m
JM = m
KL .
16. PQ and SR are not congruent because the circles are not
17. ⊙P ≅ ⊙Q and BD = EG. Therefore, BAD ≅ EFG and
18.
PG = PJ
x + 5 = 3x − 1
−2x + 5 = −1
−2x = −6
x=3
Apply the Pythagorean Theorem.
c2 = a2 + b2
PD2 = PJ 2 + JD2
PD2 = 82 + 152
PD2 = 64 + 225
PD2 = 289
—
PD = √289 = 17
The radius of ⊙P is 17 units.
360°
12
b. The measure of the minor arc formed by the hour hand
and minute hand at 7:00 is 360° − 210° = 150°.
19. a. Each arc measure is — = 30°.
c. Sample answer: The minor arc formed by the hour and
minute hands when the time is 5:00 measures 150°.
352
Geometry
Worked-Out Solutions
1. a. Check students’ work.
b. m∠CAD = 90° and m∠CBD = 45°.
1
So, m∠CBD = —2 m∠CAD or m∠CAD = 2
⋅
⋅ m∠CBD.
c. Check students’ work. The measure of an inscribed angle
is equal to half of the measure of the intercepted arc.
2. a. Check students’ work.
b. Sample answer: The angle measures of the quadrilateral
sum to 360°. Opposite angles sum to 180°.
c. Check students’ work. The sum of the opposite angles in
an inscribed quadrilateral is 180°.
3. Inscribed angles are one-half of the intercepted arcs. Opposite
angles of an inscribed quadrilateral are supplementary.
4. m∠G + m∠E = 180°
m∠G + 80° = 180°
m∠G = 100°
∠E and ∠G are supplementary.
10.4 Monitoring Progress (pp. 555–557)
1
1. m∠HGF = —2 mHF
m∠HGF =
1
—2
⋅ 90° = 45°
m
TV = 2 ⋅ 38° = 76°
2. mTV = 2∠U
3. m∠Y = m∠X by the Inscribed Angles of a Circle Theorem
(Thm. 10.11). Therefore, m∠X = 72°.
congruent.
m
EFG = 30° + 70° = 100°.
10.4 Explorations (p. 553)
4. By the Inscribed Right Triangle Theorem (Thm. 10.12), x = 90°.
m
LK = 2
⋅ 40°
m
LM = 180° − 80° = 100°
m∠LKM = —2m
LM = —2
1
So, y = 50.
1
⋅ 100° = 50°
5. By the Inscribed Quadrilateral Theorem (Thm. 10.13):
m∠ABC + m∠CDA = 180°
x° + 82° = 180°
x = 98
m∠BAD + m∠DCB = 180°
y° + 68° = 180°
y = 112
6. By the Inscribed Quadrilateral Theorem (Thm. 10.13):
m∠VST + m∠TUV = 180°
c° + (2c − 6)° = 180°
3c − 6 = 180
3c = 186
c = 62
m∠SVU + m∠UTS = 180°
8x° + 10x° = 180°
18x = 180
x = 10
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Chapter 10
7. Draw the circle that has the diagonal from the back left
13. By the Inscribed Quadrilateral Theorem (Thm. 10.13):
corner to the front right corner as a diameter.
m∠QRS + m∠STO = 180°
x° + 80° = 180°
10.4 Exercises (pp. 558 –560)
x = 100
Vocabulary and Core Concept Check
m∠RST + m∠TQR = 180°
1. If a circle is circumscribed about a polygon, then the polygon
y° + 95° = 180°
is an inscribed polygon.
x = 85
2. “Find m∠AGC ” is different.
m∠AGC = 180° − (25° + 25°) = 180° − 50° = 130°.
The other three are right angles and measure 90° by the
Inscribed Right Triangle Theorem (Thm. 10.12).
14. By the Inscribed Quadrilateral Theorem (Thm. 10.13):
m∠DEF + m∠FGD = 180°
m° + 60° = 180°
m = 120
Monitoring Progress and Modeling with Mathematics
3. m∠ABC =
m∠ABC =
1
—2 mBC
1
—2 84°
⋅
m∠EFG + m∠GDE = 180°
2k° + 60° = 180°
= 42°
2k = 120
m
DF + 70° + 120° = 360°
+ 190° = 360°
mDF
m
DF = 170°
m∠G = —12 mDF
4. mDF + mFG + mGD = 360°
m∠G = —12
+ 160° = 180°
mLM
m
LM = 20°
1 m∠N = —2mLM
k = 60
= 360°
110° + 130° + 54° + mJK
294° + m
JK = 360°
= 66°
mJK
1( m∠KLM = —2 mMJ + mJK )
⋅ 170° = 85°
5. mLM + mMN = 180°
1
3a = 60
a = 20
⋅
⋅ m∠RQS
m
RS = 2 ⋅ 67°
= 134°
mRS
8.
3a° = —12(54° + 66°)
m∠JML = —12( m
JK + m
KL )
m∠N = —2 20° = 10°
6. mRS = 2
15. mKL + mLM + mMJ + mJK = 360°
⋅ m∠TUV
m
TV = 2 ⋅ 30°
= 60°
mTV
m
VU = 180° − 60° = 120°
4b° = —12(66° + 110°)
4b = 88
7. mTV = 2
m
WY = 2 m∠WXY
⋅
m
WY = 2 ⋅ 75°
b = 22
16.
m∠XYZ = —12( m
XZ )
3x° = —12(180°)
3x = 90
x = 30
m
YZ = 2 34° = 68°
⋅
mWY = 150°
m
WY + m
YX + m
XW = 360°
m
YZ + m
YX = 180°
260° + m
XW = 360°
m
YX = 112°
68° + m
YX = 180°
150° + 110° + m
XW = 360°
m∠YZX = —2( m
XY )
m
XW = 100°
1
1
9. ∠DBC ≅ ∠DAC and ∠BCA ≅ ∠BDA by the Measure of an
Inscribed Angle Theorem (Thm. 10.10).
10. ∠XWY ≅ ∠YZX and ∠WXZ ≅ ∠WYZ by the Measure of an
Inscribed Angle Theorem (Thm. 10.10).
11. m∠EHF = m∠EGF by
⋅ m∠PRS
m
RS = 2 ⋅ 40°
m
RS = 80°
12. mPS = 2
the Inscribed Angles of a
Circle Theorem (Thm. 10.11).
Therefore, m∠EHF = 51°.
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2y° = —2(112°)
2y = 56
y = 28
17. The inscribed angle was not doubled.
m∠BAC = 2(53)° = 106°
18. Place the right angle of the carpenter’s square on the edge of
the circle and connect the points where the sides intersect the
edge of the circle.
Geometry
Worked-Out Solutions
353
Chapter 10
19. m∠DAB + m∠BCD = 180°
22. yes; ∠PTQ, ∠PSQ, and ∠PRQ are all congruent because
26y° + 2x° = 180°
2x = 180 − 26y
x = 90 − 13y
m∠ABC + m∠ADC = 180°
3x° + 21y° = 180°
3(90 − 13y) + 21y = 180
270 − 39y + 21y = 180
270 − 18y = 180
−18y = −90
y=5
x = 90 − 13 5 = 25
m∠DAB = 26 5 = 130°
m∠ABC = 3x = 3 25 = 75°
m∠BCD = 2 25 = 50°
m∠ADC = 21y = 21 5 = 105°
⋅
⋅
⋅
⋅
⋅
20. m∠ABC + m∠ADC = 180°
4x° + 24y° = 180°
x + 6y = 45
x = 45 − 6y
m∠DAB + m∠BCD = 180°
9y° + 14x° = 180°
9y + 14(45 − 6y) = 180
9y + 630 − 84y = 180
630 − 75y = 180
−75y = −450
y=6
x = 45 − 6 6 = 9
m∠DAB = 9y = 9 6 = 54°
m∠ABC = 4x = 4 9 = 36°
m∠BCD = 14x = 14 9 = 126°
m∠ADC = 24y = 24 6 = 144°
⋅
⋅
⋅
⋅
⋅
1
21. m∠BAC = —2 mBC
2x° =
x=
they all intercept PQ .
23. Construct an equilateral triangle by drawing a segment and
rotating one image of the segment 60° clockwise about one
endpoint of the original segment and the other image of the
segment 60° counterclockwise about the other endpoint of
the original segment. The endpoints of
the images should meet to form the third
vertex of the triangle. Then bisect two of
the angles. Where the angle bisectors
intersect is the center of the circle.
24. Construct a regular hexagon by drawing a segment and
rotating one image of the segment 120° clockwise about one
endpoint of the original segment and then rotate that image
of the segment 120° clockwise about the endpoint of the
image. Continue this pattern until the
endpoints of the images meet to form
a hexagon. Then bisect two of the
angles. The point where the angle
bisectors intersect is the center of
the circle.
25. yes; Opposite angles are always supplementary.
26. yes; Opposite angles are always supplementary.
27. no; Opposite angles are not always supplementary.
28. no; Opposite angles are not always supplementary.
29. no; Opposite angles are not always supplementary.
30. yes; Opposite angles are always supplementary.
31. Moon A and moon C are 220,000 kilometers apart.
A
B
⋅ 6y°
1
—2
3
—2 y
m
AB + m
BC + m
CA = 360°
100,000 km
10,000 km
° = 360°
( )
6y° + 6y° + 4
3
—2 y
12y + 6y = 360
18y = 360
y = 20
x = 30
mAB = 6y = 6 20 = 120°
m
BC = 6y = 6 20 = 120°
m
CA = 4x = 4 30 = 120°
1
1
m∠ BAC = —2 m
BC = —2 120 = 60°
⋅
⋅
⋅
⋅
⋅
m∠ BCA = —m
AB = — ⋅ 120 = 60°
m∠ ABC = —2 m
AC = —2 120 = 60°
354
O
10,000 km
100,000 km
C
1
1
32. You can sit at any point on the circle circumscribed about the
1
2
1
2
triangle that seat F7 makes with the movie screen because
the angle intercepts the same arc.
Geometry
Worked-Out Solutions
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Chapter 10
By Case 1, proved in part (a), m∠DBA = —12m
AD and
m∠DBC = —12m
CD . By the Arc Addition Postulate
(Post. 10.1), m
AC + m
CD = m
AD , so
m
AC = m
AD − m
CD . By the Angle Addition Postulate
(Post. 1.4), m∠DBC + m∠ABC = m∠DBA,
so m∠ABC = m∠DBA − m∠DBC. Then
m∠ABC = —12m
AD − —12 m
CD
33. The length of the hypotenuse of an inscribed right triangle in
a circle is always two times the radius, which is the diameter.
34. ∠AYB is always a right angle for any non-endpoint location
on semicircle AYB .
35. The diagonals of a rectangle inscribed in a circle are
diameters of the circle because each diagonal splits the
rectangle into two right triangles.
= —12(m
AD − m
CD )
= —12m
AC .
36. yes; Every triangle has a circumcenter.
∠ADC ≅ ∠ABC
Prove
A
x°
C
∠ADC and ∠ABC are inscribed angles intercepting AC .
38. Given
37. a. Case 1:
A
B
Q
C
B
Given
Prove
D
∠ABC is inscribed in circle Q. Let m∠ABQ = x°.
—.
Center Q lies on BC
By the Measure of an Inscribed Angle Theorem (Thm. 10.10),
1
1
m∠ABC = —2m
AC and m∠ADC = —2m
AC . By the Transitive
Property of Equality, m∠ABC = m∠ADC.
m∠ABC = —2m
AC
1
— ≅ QA
—, so △ABC is isosceles. By the Base Angles
QB
Theorem (Thm. 5.6), ∠QBA ≅ ∠QAB, so m∠BAQ = x°. By
the Exterior Angles Theorem (Thm. 5.2), m∠AQC = 2x°.
1
1
Then m
AC = 2x°, so m∠B = x° = —2(2x)° = —2m
AC .
39. To prove the conditional, find the measure of the intercepted
arc of the right angle and the definition of a semicircle
to show the hypotenuse of the right triangle must be the
diameter of the circle. To prove the converse, use the
definition of a semicircle to find the measure of the angle
opposite the diameter.
b. Case 2:
A
D
Q
B
40. By the Arc Addition Postulate (Postulate 10.1),
m
EFG + m
EDG = 360° and m
FGD + m
DEF = 360°.
Using the Measure of an Inscribed Angle Theorem,
m
EDG = 2m∠F, m
EFG = 2m∠D, m
DEF = 2m∠G,
and mFGD = 2m∠E. By the Substitution Property of
Equality, 2m∠D + 2m∠F = 360°, so m∠D + m∠F = 180°.
Similarly, m∠E + m∠G = 180°.
C
Given
Prove
— is a diameter.
∠ABC is inscribed in circle Q. DB
m∠ABC = —2m
AC
1
By Case 1, proved in part (a), m∠ABD = —2 m
AD and
1
m∠CBD =
1
—2 mCD .
By the Arc Addition Postulate (Post. 10.1),
m
AD + m
CD = m
AC . By the Angle Addition Postulate
41.
C
(Post 1.4), m∠ABD + m∠CBD = m∠ABC.
J
Then m∠ABC = —2m
AD + —2m
CD
1
=
=
1
( + m
CD )
1
—2 mAD
1
—2 mAC .
c. Case 3:
A
Given
Prove
A
5
x
Q
B
— is a diameter.
∠ABC is inscribed in circle Q. DB
B
—. x is a diameter
Let x be a segment that extends from C to AB
—. Because
of the smallest circle that connects C and AB
— is diameter. By the Right
inscribed ∠C is a right angle, JK
x
4
Triangle Similarity Theorem (Thm. 9.6), —3 = —5 .
4
—3 = —5
C
D
K 4
3
⋅
5x = 3 4
5x = 12
12
x=—
5 = 2.4 units
m∠ABC = —12m
AC
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Geometry
Worked-Out Solutions
355
Chapter 10
—
42. a. Because FH is diameter, ∠FGH is a right angle. △FJG,
4. 180° − 102° = 78°
1
△GJH, and △FGH are similar by the AA Similarity
FG
GJ FJ
Theorem (Thm. 8.3). Therefore, — = — = — because
JH JG GH
corresponding parts of similar triangles are proportional.
b. By the board count, FJ = 6, so JH = 2.
FJ GJ
—=—
JG JH
6 x
—=—
x 2
78° = —2
61 = y
1
5. m∠ J = —2
30° =
1
—2
FG − m
KH )
⋅ ( m
⋅ (a° − 44°)
60 = a − 44
104 = a
x2 = 12
—
—
x = √12 = 2√3
6. m∠N = 180° − m∠KML
—
x° = 180° − 120°
Therefore, JG = 2√ 3 .
⋅
⋅ (95° + y°)
156 = 95 + y
—
—
x = 60
GK = 2 2√ 3 = 4√ 3
Maintaining Mathematical Proficiency
1
43. 3x = 145
44. —2 x = 63
145
x=—
3
x = 126
7.
m
PQ = 2x°
m
PTQ + m
PQ = 360°
m
PTQ + 2x° = 360°
m
PTQ = 360° − 2x°
m∠S = —2( m
PTQ − m
PQ )
1
45. 240 = 2x
46. 75 = —2 (x − 30)
x = 120
1
150 = x − 30
1
50° = —2(360° − 2x° − 2x°)
180 = x
1
50 = —2(360 − 4x)
50 = 180 − 2x
10.5 Explorations (p. 561)
−130 = −2x
1. a. Check students’ work.
65 = x
b. Sample answer: 49° and 131°
c. Sample answer: 98° and 262°
d. Check students’ work.
The measure of each angle between a chord and a tangent
is half of its intercepted arc.
4000
( 4002.73
)
8. m∠ABD = sin−1 — ≈ 87.8836 ≈ 88°
m∠CBD ≈ 2
⋅ 88° = 176°
m∠CBD = 180° − m
CD
176° = 180° − m
CD
2. a. Check students’ work.
−4° = −m
CD
b. Sample answer: 60°
c. Sample answer: 40° and 80°; The angle measure is half of
m
CD = 4°
the sum of the measures of the intercepted arcs.
10.5 Exercises (pp. 566 –568)
d. Check students’ work.
The measure of an angle between two chords inside is half
of the sum of the measure of the arcs intercepted by the
angle and its vertical angle.
3. When a chord intersects a tangent line, the angle formed
is half of the measure of the intercepted arc. When a chord
intersects another chord, the measure of the angle is half of
the sum of the measures of the arcs intercepted by the angle
and its vertical angle.
1
4. m∠1 = —2
⋅ 148° = 74°
⋅ 2 = 110°.
⋅ 210° = 105°
3. m
XY = 2 ⋅ 80° = 160°
1. m∠1 =
356
outside the circle.
2. To find the measure of a circumscribed angle, find the
measure of the central angle that intercepts the same arc and
subtract it from 180°.
⋅ 65° = 130°
4. m
DEF = 2 ⋅ 117° = 234°
3. mAB = 2
10.5 Monitoring Progress (pp. 562 –565)
1
—2
1. Points A, B, C, and D are on a circle, and ⃖⃗
AB intersects ⃖⃗
CD
1
at point P. If m∠APC = —2 ( mBD − mAC ), then point P is
Monitoring Progress and Modeling with Mathematics
5. The sum of the measures of the arcs intercepted by the
two angles is 55
Vocabulary and Core Concept Check
2. m
RST = 2 ⋅ 98° = 196°
Geometry
Worked-Out Solutions
1
5. m∠1 = —2
6. m∠3 =
1
—2
⋅ 260° = 130°
⋅ (360° − 140°) = — ⋅ 220° = 110°
1
2
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Chapter 10
CD + m
BA )
⋅ ( m
x° = ⋅ (145° + 85°)
x = ⋅ 230 = 115
8. 180° − x° = — ⋅ ( m
KL + m
JM )
180° − x° = — ⋅ (2x° − 30° + 30°)
180 − x = — ⋅ (2x)
1
7. x° = —2
1
—2
1
—2
1
2
1
2
1
2
180 − x = x
1
14. m∠G = —2 [ 17x° − (360° − 17x°) ]
1
75° = —2(17x° − 360° + 17x°)
1
75 = —2(34x − 360)
75 = 17x − 180
255 = 17x
x = 15
15. ∠SUT is not a central angle.
m∠SUT = —2( m
ST + m
QR )
1
180 = 2x
90 = x
1
m∠SUT = —2(46° + 37°)
1
9. m∠E = —2 ( mGD − mDF )
1
29° =
1
—2 (114°
⋅ 83°
= 41.5°
− x°)
58 = 114 − x
1
16. The —2 was left out of the equation.
−56 = −x
1
10. m∠S = —2 ( mUV − mTW )
1
34° = —2[ (3x − 2)° − (x + 6)° ]
68 = 3x − 2 − x − 6
180°, solve the equation 90° + 30° + m∠1 = 180°.
90° + 30° + m∠1 = 180°
120° + m∠1 = 180°
m∠1 = 60°
76 = 2x
18. When a chord intersects a tangent line, the angle formed is
38 = x
1
11. m∠T = —( m
PQ − m
RS )
2
1
x
— ° = —[ (x + 70)° − (x + 30)° ]
2
2
x = x + 70 − x − 30
()
half of the measure of the intercepted arc, which in this case
is 120°.
1
m∠2 = —2(120°) = 60°
19. Because the sum of the angles of a triangle always equals
180°, solve the equation 960° + 90° + m∠3 = 180°.
60° + 90° + m∠3 = 180°
x = 40
m∠X = —2 [ (360° − m
YZ ) − m
YZ ]
1
150° + m∠3 = 180°
m∠3 = 30°
1
(6x − 11)° = —2 [ (360° − 125°) − 125° ]
1
6x − 11 = —2 (360 − 250)
1
6x − 11 = —2
⋅
110
6x − 11 = 55
6x = 66
x = 11
⋅ 73° = 146°
m
MLP + m
MP = 360°
+ 146° = 360°
mMLP
m
MLP = 360 − 146 = 214°
1
(
m∠N = —2 m
MLP − m
MP )
13. mMP = 2
1
x° = —2(214° − 146°)
1
x = —2
⋅ 52° = 26°
17. Because the sum of the angles of a triangle always equals
68 = 2x − 8
12.
1
m∠1 = —2(122° − 70°) = —2
56 = x
1
= —2
⋅ 68
x = 34
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20. Because ∠2 and ∠5 sum to 90° and form a straight line with
∠4, the angles are supplementary.
90° + m∠4 = 180°
m∠4 = 180°
21. ∠5 is complementary to ∠2, which measures 60°.
m∠5 + 60° = 90°
m∠5 = 30°
22. The triangle is equiangular, so m∠1 = m∠6 = 60°.
4000
( 4001.2
)
23. m∠ZWX = sin−1 — ≈ 88.5967 ≈ 88.6°
m∠ZWX ≈ 2
⋅ 88.6° = 177.2°
m∠ZWX = 180° − m
CD
177.2° = 180° − m
CD
−2.8 = −m
CD
m
CD = 2.8°
Geometry
Worked-Out Solutions
357
Chapter 10
4000
( 4000.2
)
4000
( 4000.01 ) ≈ 0.1281
24. m∠FCT = cos−1 — ≈ 0.5729
m∠ECT = cos−1 —
30. yes; When the circumscribed angle is 90°, the central angle is 90°.
A
31. a.
C
t
B
m∠FCE ≈ m∠FCT + m∠ECT = 0.5729 + 0.1281 = 0.7010
m∠FCE = m
SB ≈ 0.7
1
25. 40° = —2 (7x° − 3x°)
C
A
t
80 = 4x
B
20 = x
7x = 140
3x = 60
x = 20
m
CD = 360° − 10x°
m
CD = 360° − 10 ⋅ 20°
m
CD = 160°
360° − 2m∠BAC = m
AB
—
c. If AB is a diameter, m∠BAC = 90° and
m
BA = 2m∠BAC = 360° − 2m∠BAC.
32. △ABC is an equilateral and equiangular triangle.
1
Angle = —2(c)
—2 (b − a) =
2
2m∠BAC = m
AB
1
1
1
1
180° − m∠BAC = —2 m
AB
Angle = —2(b − a)
26.
m∠BAC = —2 m
AB
b.
1
—2 (c)
⋅ —(b − a) = 2 ⋅ —(c)
1
2
1
2
A
P
b−a=c
c=b−a
27.
1
m
LJ + m
LK = 180°
= 180° − m
mLJ
LK
1
(
m∠LPJ = —2 180° − m
LK − m
LK )
1
LK )
m∠LPJ = —2( 180° − 2m
m∠LPJ = 90° − m
LK
∠LPJ is less than 90° and greater than 0°.
—
28. AB is not a diameter, so m
AB ≠ 180°. So, x° < 180° and
x° ≠ 90°.
Prove
Y
Z
m∠LPJ = —2 ( m
LJ − m
LK )
⃖⃗
JL and ⃖⃗
NL are secant lines that intersect at point L.
29. Given
B
X
m∠JPN > m∠JLN
C
m∠YXZ = 60°, therefore m
YZ = 2
⋅ 60° = 120°.
m∠XZY = 60°, therefore m
XY = 2 ⋅ 60° = 120°.
= 2 ⋅ 60° = 120°.
m∠XYZ = 60°, therefore mXZ
1
(
m∠A = —2 ⋅ mXZY − mXY ), therefore
− 120°) = —(120)° = 60°.
⋅ (240°
(
m∠B = — ⋅ mXYZ − m
XZ ), therefore
m∠A = — ⋅ (240° − 120°) = —(120)° = 60°.
m∠C = — ⋅ ( m
YXZ − m
YZ ), therefore
m∠A = — ⋅ (240° − 120°) = —(120)° = 60°.
1
2
m∠A = —12
1
2
1
2
1
2
1
2
1
2
1
2
Therefore △ABC is an equiangular and an equilateral triangle.
33. a. By the Tangent Line to Circle Theorem (Thm. 10.1), m∠BAC
is 90°, which is half the measure of the semicircular arc.
J
b.
K
D
B
L
P
M
N
C
By the Angles Inside a Circle Theorem (Thm. 10.15),
m∠JPN = —12 ( m
JN + m
KM ). By the Angles Outside the
Circle Theorem (Thm. 10.16), m∠JLN = —12 ( m
JN − m
KM ).
Because the angle measures are positive,
1
JN + m
KM ) > —1 m
JN > —1 ( m
JN − m
KM ),
—( m
2
2
2
so, m∠JPN > m∠JLN.
358
Geometry
Worked-Out Solutions
A
By the Tangent Line to Circle Theorem (Thm. 10.1),
1
m∠CAD = 90°. m∠DAB = —2 m
DB and by part (a),
1
m∠CAD = —2mAD. By the Angle Addition Postulate
(Post. 1.4), m∠BAC = m∠BAD + m∠CAD.
1
1
1
So, m∠BAC = —2 m
DB + —2 mAD = —2 ( m
DB + m
AD ).
By the Arc Addition Postulate (Post. 10.1),
1
m
DB + m
AD = m
ADB , so m∠BAC = —2 ( m
ADB ).
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Chapter 10
c.
37. Given
D
B
Prove
a secant and a tangent line
m∠1= —2( m
BC − m
AC )
1
B
A
C
A
1
By the Tangent Line to Circle Theorem (Thm. 10.1),
1
m∠CAD = 90°. m∠DAB = —2 m
DB and by part (a),
1 m∠DAC = —2mABD . By the Angle Addition Postulate
(Post. 1.4), m∠BAC = m∠DAC − m∠DAB.
1
1
1
So, m∠BAC = —2m
ABD − —2 m
DB = —2 ( m
ABD − m
DB ).
By the Arc Addition Postulate (Post. 10.1),
1
m
ABD − m
DB = m
AB , so m∠BAC = —2 ( m
AB ).
By the Exterior Angle Theorem (Thm. 5.2),
m∠2 = m∠1 + m∠ABC, so m∠1 = m∠2 − m∠ABC.
By the Tangent and Intersected Chord Theorem (Thm. 10.14),
1
m∠2 = —2m
BC and by the Measure of an Inscribed Angle
1
Theorem (Thm. 10.10), m∠ABC = —2m
AC . By the
Substitution Property,
1
1
1
m∠1 = —2m
BC − —2 m
AC = —2 ( m
BC − m
AC );
34. closer; The smaller arc needs to be 20°.
? 1
30° = —2(80° − 20°)
? 1
30° = —2(60°)
Given
Prove
Prove
— and BD
— intersect inside a circle.
Chords AC
1
m∠1 = —2( m
DC + m
AB )
By the Exterior Angle Theorem (Thm. 5.2),
m∠1 = m∠2 + m∠3, so m∠2 = m∠1 − m∠3. By the
Tangent and Intersected Chord Theorem (Thm. 10.14),
1
1
m∠1 = —2m
PQR and m∠3 = —2m
PR . By the Substitution
1 1 1
Property, m∠2 = —2mPQR − —2mPR = —2( m
PQR − m
PR );
C
STATEMENTS
— and BD
—
1. Chords AC
REASONS
Given
1. Given
Prove
intersect.
2. m∠ACB = —2 m
AB and
1
DC
m∠DBC = —2 m
1
1
1
DC + m
AB )
5. m∠1 = —2 ( m
1
36.
two secant lines
m∠3 = —2 ( m
XY − m
WZ )
1
X
2. Measure of an Inscribed
Angle Theorem
(Thm. 10.10)
W
3
2
3. m∠1 = m∠DBC + m∠ACB 3. Exterior Angle Theorem
(Thm. 5.2)
4. m∠1 = —2 m
DC + —2 m
AB
Q
1
R
1
B
1
2
D
A
two tangent lines
m∠2 = —2 ( m
PQR − m
PR )
P
3
30° = 30°
35. Given
2
C
1
Y
By the Exterior Angle Theorem (Thm. 5.2),
m∠1 = m∠3 + m∠WXZ, so m∠3 = m∠1 − m∠WXZ.
By the Measure of an Inscribed Angle Theorem (Thm. 10.10),
1
1
m∠1 = —2m
XY and m∠WXZ = —2 m
WZ . By the Substitution
1 1 1
Property, m∠3 = —2 mXY − —2 mWZ = —2( m
XY − m
WZ ).
4. Substitution Property of
Equality
5. Distributive Property
A
38.
C
P
A
C
D
B
B
∠PCB, ∠PBC, and ∠CAB intercept BC .
By the Angles Outside the Circle Theorem (Thm. 10.16),
m∠ADB = —2( ( 360° − m
AB ) − m
AB )
1
= —2( 360° − 2m
AB )
1
= 180° − m
AB
= 180° − m∠ACB.
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Geometry
Worked-Out Solutions
359
Chapter 10
39.
m∠P = —2( m
WZ − m
XY )
1
m
WZ + m
ZY = 200°
= mZY
mWX
mWZ + mWX = 200°
m
WX = 200° − m
WZ
mWX + mXY = 160°
m
WX = 160° − mXY
200° − mWZ = 160° − m
XY
= −mXY
40° − mWZ
40° = m
WZ − m
XY
1
m∠P = —2( mWZ − m
XY )
m∠P =
1
—2 (40°)
= 20°
x2 = 12x + 35
42.
x2 − 12x − 35 = 0
Use completing the square.
x2 − 12x = 35
x2
− 12x + 62 = 35 + 62
(x − 6)2 = 71
—
x = 6 ± √71
−3 = x2 + 4x
43.
0 = x2 + 4x + 3
x+1=0
1
130° = 85° + m
EA
10.6 Explorations (p. 569)
m
EA = 45°
= 20°
mEF
m
FA = 45° − 20° = 25°
1
m∠DGB = —2( m
DCB + m
FA )
1
90° = —2( mDCB + 25° )
180° = m
DCB + 25°
mDCB = 155°
1. a. Check students’ work.
B
E
F
D
A
C
⋅ 60° = 120°
m
ED = m
AED − ( m
AF + m
FE )
m
ED = 120° −(25° + 20°) = 75°
m∠EHD = —12( m
ED + m
ABC )
115° = —12( 75° + m
ABC )
230° = 75° + m
ABC
mABC = 155°
m
AB = 155° − 70° = 85°
= 75°.
So, m
AB = 85° and mED
Maintaining Mathematical Proficiency
x2 + x − 12 = 0
b. Sample answer:
BF
CF
7.5
13.2
DF
EF
6.6
15
BF
⋅ CF
99
DF
⋅ EF
99
The products are equal.
c. Check students’ work. If two chords intersect inside a
circle, then the product of the lengths of the segments
of one chord is equal to the product of the lengths of the
segments of the other chord.
2. a. Check students' work.
(x − 3)(x + 4) = 0
x=3
x = −3
The solutions are x = −1 and x = −3.
65° = —2( 85° + m
EA )
x−3=0
x+3=0
or
x = −1
1
41.
—
(x + 1)(x + 3) = 0
m∠DHC = —2( m
DC + m
EA )
m
AED = 2m∠JAD = 2
—
The solutions are x = 6 + √ 71 and x = 6 − √71 .
40. m∠DHC = 180° − 115° = 65°
m∠JAD = 60°
—
x − 6 = ±√71
b. Sample answer:
or
x+4=0
x = −4
The solutions are x = 3 and x = −4.
BE
BC
17.6
35.9
BF
BD
22
28.72
BE
⋅ BC
631.84
BF
⋅ BD
631.84
The products are equal.
360
Geometry
Worked-Out Solutions
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Chapter 10
c. If two secants intersect outside a circle with a common
endpoint, then the product of the lengths of the segments
of one secant is equal to the product of the lengths of the
segments of the other secant.
7. 122 = x(x + 10)
144 = x2 + 10x
0 = x2 + 10x − 144
0 = (x − 8)(x + 18)
3. The products of the lengths of the segments of one chord or
0=x−8
secant is equal to the product of the lengths of the segments
of the other chord or secant.
352
10.6 Monitoring Progress (pp. 570 –572)
⋅3=4⋅6
2. 2(x + 1) = 4
3x = 24
⋅3
1029 = 28r
1029
—
28 = r
1029
The radius is approximately —
≈ 36.75 feet.
28
2x = 10
x=5
3. 6(6 + 9) = 5(5 + x)
10.6 Exercises (pp. 573 –574)
Vocabulary and Core Concept Check
6(15) = 25 + 5x
1. The part of the secant segment that is outside the circle is
called an external segment.
90 = 25 + 5x
2. A tangent intersects a circle at one point and a secant
65 = 5x
segment intersects a circle at two points.
x = 13
4. 3(3 + x + 2) = (x + 1)(x + 1 + x − 1)
3(5 + x) = (x + 1)(2x)
Monitoring Progress and Modeling with Mathematics
3. x
⋅ 12 = 10 ⋅ 6
0 = (2x + 5)(x − 3)
or
0=x−3
5
−—2
x (x + 8) = 8
5.
x=3
x2 + 8x − 48 = 0
(x − 4)(x + 12) = 0
x−4=0
6. 2x
24
x=—
5 = 4.8
⋅ 12 = 15(x + 3)
7. 6
24x = 15x + 45
⋅ (5 + x)
24 = 5x
x = −12
A negative value for x does not make sense. So, the only
solution is x = 4.
x = ±2
49 = 25 + 5x
x + 12 = 0
or
x=4
—
x= 4
=5
⋅6
+ 8x = 48
x2
⋅4
√
6.
9x = 207
x = 23
A negative value for x does not make sense. So, the only
solution is x = 3.
A negative value for x
does not make sense.
So, the only solution is
x = 2.
⋅ 18
9x − 27 = 180
x=5
0 = 2x2 − x − 15
0 = 2x + 5
4. 9(x − 3) = 10
12x = 60
15 + 3x = 2x2 + 2x
72
⋅ CD
= 14 ⋅ (2r + 14)
1225 = 28r + 196
2x + 2 = 12
x=8
5. x2 = 1
x = −18
8. CB2 = CE
72
AF = —
18 = 4
x=
0 = x + 18
A negative value for x does not make sense. So, the only
solution is x = 8.
⋅ AF = AC ⋅ AD
18 ⋅ AF = 8 ⋅ 9
4. AE
1. x
or
x=8
8.
⋅ 16 = 8(8 + x)
96 = 64 + 8x
9x = 45
32 = 8x
x=5
x=4
5
⋅ 12 = x(x + 4)
60 = x2 + 4x
0 = x2 + 4x − 60
(x + 10)(x − 6) = 0
x + 10 = 0
or
x = −10
x−6=0
x=6
A negative value for x does not make sense. So, the only
solution is x = 6.
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Geometry
Worked-Out Solutions
361
Chapter 10
9.
4
⋅ 9 = (x − 2)(x − 2 + x + 4)
36 = (x − 2)(2x + 2)
36 = 2(x − 2)(x + 1)
203,0002 = 83,000(DB)
⋅ 10 ) = (8.3 ⋅ 10 )(DB)
4.1209 ⋅ 10 = (8.3 ⋅ 10 )(DB)
4.1209 ⋅ 10
——
= DB
8.3 ⋅ 10
4.9649 ⋅ 10 = DB
(2.03
The distance from Cassini to Tethys is approximately
496,494 kilometers.
(x + 4)(x − 5) = 0
or x − 5 = 0
x = −4
x=5
17. x
⋅ 45 = 27 ⋅ 50
⋅ x = 188 ⋅ 62
x2 = 11,656
A negative value for x does not make sense. So, the only
solution is x = 5.
x ≈ ±107.96
A negative distance does not make sense. So, the distance
from the end of the passage to either side of the mound is
approximately 108 feet.
45x = 1350
x = 30
18. 4
⋅ 16
√
⋅ CN = 6 ⋅ 8
4CN = 48
—
x = 144 = ±12
A negative value for x does not make sense. So, the only
solution is x = 12.
12. 242 = 12(12 + x)
576 = 144 + 12x
432 = 12x
CN = 12
CN is 12 centimeters. If the sparkle moves from C to D at
a rate of 2 centimeters per second, it will take the sparkle
3 seconds to reach the outer circle. The sparkle will have to
move at 4 centimeters per second from C to N in order to
reach the outer circle at the same time.
19. Prove
36 = x
13.
4
5
0 = x2 − x − 20
11. x2 = 9
4
10
4
18 = x2 − x − 2
10. x
5 2
10
18 = (x − 2)(x + 1)
x+4=0
16.
EA
⋅ EB = EC ⋅ ED
C
(x + 4)2 = x(x + 12)
x2 + 8x + 16 = x2 + 12x
8x + 16 = 12x
x=4
— and CD
— are chords
1. AB
14. ( √ 3 ) = x(x + 2)
REASONS
1. Given
intersecting in the
interior of the circle.
3 = x2 + 2x
0 = x2 + 2x − 3
0 = (x + 3)(x − 1)
or
B
STATEMENTS
— 2
x = −3
E
D
16 = 4x
x+3=0
A
x−1=0
x=1
A negative value for x does not make sense. So, the only
solution is x = 1.
15. The chords were used instead of the secant segments.
⋅ DF = BF ⋅ AF
(CD + 4) ⋅ 4 = 8 ⋅ 3
4CD ⋅16 = 24
4 ⋅ CD = 8
CF
2. ∠AEC ≅ ∠DEB
2. Vertical Angles Congruence
Theorem (Thm. 2.6)
3. ∠CAB ≅ ∠CDB
3. Inscribed Angles of a Circle
Theorem (Thm. 10.11)
4. △EAC ∼ △EDB
4. AA Similarity Theorem
(Thm. 8.3)
EA EC
5. — = —
ED EB
5. Corresponding sidelengths
of similar triangles are
proportional.
6. EA • EB = EC • ED
6. Cross Products Property
CD = 2
362
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Chapter 10
20. Prove
EA
⋅ EB = EC ⋅ ED
23.
A
E
C
D
STATEMENTS
REASONS
— and ED
— are
1. EB
1. Given
secant segments.
2. Inscribed Angles of a Circle
Theorem (Thm. 10.11)
3. ∠E ≅ ∠E
3. Reflexive Property of
Congruence (Thm. 2.2)
4. △EAD ∼ △ECB
4. AA Similarity Theorem (Thm. 8.3)
EA ED
5. — = —
EC EB
5. Corresponding side lengths of
similar triangles are proportional.
⋅
Theorem (Thm. 10.20) to find QR: QR2 = RS RP. Then
use the Pythagorean Theorem (Thm. 9.1) to find PQ.
25. AB2 = BC
122 = 8
E
y
EC = 10
△CDP ~ △CPE by the AA Similarity Theorem (Thm. 8.3).
PD PC
—=—
EP EC
r
4
—=—
r 10
40 = r 2
—
—
r = √40 = 2√10 ≈ 6.3
r
r
D
O r
By the Tangent Line to Circle Theorem (Thm. 10.1), ∠EAO is
a right angle, which makes △AEO a right triangle. By the
Pythagorean Theorem (Thm. 9.1), (r + y)2 = r 2 + x2.
So, r 2 + 2yr + y2 = r 2 + x2. By the Subtraction Property
of Equality, 2yr + y2 = x2. Then y(2r + y) = x2, so
EC ED = EA2.
⋅ (8 + EC)
80 = 8EC
A
C
⋅ BE
144 = 64 + 8EC
6. EA • EB = EC • ED 6. Cross Products Property
x
⋅ AC = AD ⋅ AE
24. Sample answer: Use Segments of Secants and Tangents
2. ∠ABC ≅ ∠ADC
21.
AB
AB(AB + BC) = AD(AD + DE)
AD(AD + DE)
AB + BC = ——
AB
AD2 + (AD)(DE)
BC = —— − AB
AB
AD2 + (AD)(DE) − AB2
BC = ——
AB
B
—
The radius of circle P is 2√10 ≈ 6.3 units.
26. no; The side lengths are not proportional.
⋅
A
22.
E
D
C
STATEMENTS
— is a tangent
1. EA
— is a
segment and ED
REASONS
1. Given
Maintaining Mathematical Proficiency
secant segment.
2. ∠E ≅ ∠E
AC
3. m∠EAC = —2m
1
AC
4. m∠ADC = —2 m
1
2. Reflexive Property of
Congruence (Thm. 2.2)
3. Tangent and Intersected Chord
Theorem (Thm. 10.14)
4. Measure of an Inscribed Angle
Theorem (Thm. 10.10)
5. m∠EAC = m∠ADC 5. Transitive Property of Equality
6. ∠EAC ≅ ∠ADC
6. Definition of congruence
7. △EAC ∼ △EDA
7. AA Similarity Theorem (Thm. 8.3)
EA EC
8. — = —
ED EA
8. Corresponding side lengths of
similar triangles are proportional.
9. EA2 = EC • ED
9. Cross Products Property
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x2 + 4x = 45
27.
x2
+ 4x + 22 = 45 + 22
(x + 2)2 = 49
—
x + 2 = √49
—
x = −2 ± √49
x = −2 + 7 = 5
x = −2 − 7 = −9
The solutions are x = 5 and x = −9.
Geometry
Worked-Out Solutions
363
Chapter 10
28.
2. a. Sample answer:
x2 − 2x = 9
x2 + 2x + 12 = 9 + 12
Center
(x + 1)2 = 10
—
x + 1 = √10
—
x = −1 ± √ 10
—
—
The solutions are x = −1 + √ 10 and x = −1 − √10 .
x2 + 6x = 7
x2 + 6x + 33 = 7 + 32
x2 + y2 = 4
(2,0)
(x − 2)2 + y2 = 4
(0,2)
x2 + (y − 2)2 = 4
(x + 1)2 + (y − 2)2 = 4
(2,1)
(x − 2)2 + (y − 1)2 = 4
(−3,−2)
(x + 3)2 + (y + 2)2 = 4
b. An equation of a circle with center (h, k) and radius 2 is
(x − h)2 + (y − k)2 = 4.
(x + 3)2 = 16
—
x + 3 = √16
c. An equation of a circle with center (h, k) and radius r is
(x − h)2 + (y − k)2 = r 2.
x = −3 ± 4
x = −3 + 4 = 1
x = −3 − 4 = −7
——
3. d = √ (x − h)2 + (y − k)2
d2 =
The solutions are x = 1 and x = −7.
——
( √(x − h)2 + (y − k)2 )2
d 2 = (x − h)2 + (y − k)2
30. −4x2 + 8x + 44 = 16
The equation is the same. The distance from the center to any
point on the circle is the radius; therefore, d would represent
the radius r.
x2 − 2x − 11 = −4
x2 − 2x = 7
x2 − 2x + 13 = 7 + 12
(x −
(0,0)
(−1,2)
29. 2x2 + 12x + 20 = 34
2x2 + 12x = 14
Equation of circle
=8
—
x − 1 = √8
4. The equation of a circle with center (h, k) and radius r in the
coordinate plane is (x − h)2 + (y − k)2 = r 2.
1)2
—
x = 1 ± 2√2
—
—
The solutions are x = 1 + 2√2 and x = 1 − 2√2 .
5. Center: (4, −1), r = 3
(x − h)2 + (y − k)2 = r 2
(x − 4)2 + ( y − (−1) )2 = 32
(x − 4)2 + (y + 1)2 = 9
10.7 Explorations (p. 575)
1. a. Sample answer:
10.7 Monitoring Progress (pp. 576 −578)
Radius
Equation of circle
1
x2 + y2 = 1
2
x2 + y2 = 4
3
x2 + y2 = 9
4
x2 + y2 = 16
5
x2 + y2 = 25
6
x2 + y2 = 36
b. An equation of a circle with center (0, 0) and radius r is
x2 + y2 = r2.
1. (x − h)2 + (y − k)2 = r2
(x − 0)2 + (y − 0)2 = 2.52
x2 + y2 = 6.25
The standard equation of the circle is x 2 + y 2 = 6.25.
2.
(x − h)2 + (y − k)2 = r 2
( x − (−2) )2 + ( y − 5)2 = 72
(x + 2)2 + (y − 5)2 = 49
The standard equation of the circle is (x + 2)2 + (y − 5)2 = 49.
——
3. r = √ (3 − 1)2 + (4 − 4)2
—
= √(2)2 + (0)2
—
= √4
=2
(x − h)2 + (y − k)2 = r 2
(x − 1)2 + (y − 4)2 = 4
The standard equation of the circle is (x − 1)2 + (y − 4)2 = 4.
364
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Chapter 10
x2 + y2 − 8x + 6y + 9 = 0
4.
4. Using the endpoints of the diameter, (−12, −3) and (6, −3),
the center is found by determining the midpoint.
x2 − 8x + y2 + 6y = −9
(x2
− 8x +
42)
+
(y2
+ 6y +
32)
= −9 +
42
+32
(x − 4)2 + (y + 3)2 = −9 + 16 + 9
(x − 4)2 + (y + 3)2 = 16
(x − 4)2 + ( y − (−3) )2 = 42
The center is (4, −3) and the radius is 4.
( −122+ 6, −3 +2 (−3) ) = ( −62 , −62 ) = (−3, −3).
— —
— —
Use the center and an endpoint of the diameter to find the radius.
——
r = √ (x − h)2 + (y − k)2
———
= √( 6 − (−3) )2 + ( −3 − (−3) )2
—
= √92 + 02
y
—
2
4
6
8 x
= √81
=9
−2
(4, −3)
−4
(x − h)2 + (y − k)2 = r2
( x − (−3) )2 + ( y − (−3) )2 = 92
−6
(x + 3)2 + (y + 3)2 = 81
−8
5. The center is (0, 0) and the radius is 7.
——
(x − h)2 + (y − k)2 = r2
——
(x − 0)2 + (y − 0)2 = 72
5. r = √ (x − h)2 + (y − k)2
= √(0 − 0)2 + (1 − 0)2
x2 + y2 = 49
—
= √1
6. The center is (4, 1) and the radius is 5.
=1
The radius of the circle is 1.
(x − h)2 + (y − k)2 = r2
——
—
2
d = √(1 − 0)2 + ( √ 5 − 0 )
(x − 4)2 + (y − 1)2 = 52
—
(x − 4)2 + (y − 1)2 = 25
—
d = √1 + 5 = √6
Because the radius —is 1 and the distance between the center
and the
point ( 1, √ 5 ) is greater than the radius, the point
( 1, √—5 ) does not lie on the circle.
6. Three seismographs are needed to locate an earthquake's
epicenter because two circles can intersect at more than
one point.
10.7 Exercises (pp. 579−580)
Vocabulary and Core Concept Check
1. The standard equation of a circle with center (h, k) and
radius r is r 2 = (x − h)2 + (y − k)2.
2. The radius can be determined by using the distance formula
to find the distance between the center and a point on the
circle.
Monitoring Progress and Modeling with Mathematics
3. The center is (0, 0) and the radius is 2.
(x − h)2 + (y − k)2 = r 2
(x − 0)2 + (y − 0)2= 22
x2 + y2 = 4
7. The center is (−3, 4) and the radius is 1.
(x − h)2 + (y − k)2 = r 2
( x − (−3) )2 + (y − 4)2 = 12
(x + 3)2 + (y − 4)2 = 1
8. The center is (3, −5) and the radius is 7.
(x − h)2 + (y − k)2 = r 2
(x − 3)2 + ( y − (−5) )2 = 72
(x − 3)2 + (y + 5)2 = 49
——
9. r = √ (x − h)2 + (y − k)2
——
= √(0 − 0)2 + (6 − 0)2
—
—
= √62 = √36 = 6
(x − h)2 + (y − k)2 = r 2
(x − 0)2 + (y − 0)2 = 62
x2 + y2 = 36
——
10. r = √ (x − h)2 + (y − k)2
——
= √(4 − 1)2 + (2 − 2)2
—
—
= √32 = √9 = 3
(x − h)2 + (y − k)2 = r2
(x − 1)2 + (y − 2)2 = 32
(x − 1)2 + (y − 2)2 = 9
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Geometry
Worked-Out Solutions
365
Chapter 10
——
11. r = √ (x − h)2 + (y − k)2
16. x2 + y2 + 4y + 22 = 32 + 22
——
x2 + (y + 2)2 = 36
= √(3 − 0)2 + (−7 − 0)2
The center is (0, −2) and the radius is 6.
—
= √32 + (−7)2
—
—
= √9 + 49 = √58
y
8
(x − h)2 + (y − k)2 = r2
— 2
(x − 0)2 + (y − 0)2 = ( √ 58 )
−8
x2 + y2 = 58
−4
4
−4
8 x
(0, −2)
12. The coordinates of the center should be subtracted.
( x − (−3) )2 + ( y − (−5) )2 = 9
(x + 3)2 + (y + 5)2 = 9
13. For the equation
x2
+
y2
= 49, the center is (0, 0) and the
radius is 7.
x2 − 8x + y2 − 2y = −16
17.
) + ( y2 − 2y + (−1)2 ) = −16 + (−4)2 + (−1)2
( − 8x +
( x2 − 8x + (−4)2 ) + ( y2 − 2y + (−1)2 ) = 1
x2
(−4)2
(x − 4)2 + (y − 1)2 = 1
y
8
The center is (4, 1) and the radius is 1.
4
(0, 0)
−8
−4
6
4
y
8 x
4
−4
(4, 1)
2
−8
2
14. For the equation (x + 5)2 + (y − 3)2 = 9, the center is
4
6 x
(−5, 3) and the radius is 3.
8
x2 + 4x + y2 + 12y = −15
18.
y
( + 4x + ) + ( y2 + 12y + (6)2 ) = −15 + (2)2 + (6)2
( x2 + 4x + (2)2 ) + ( y2 + 12y + (6)2 ) = 25
x2
6
(2)2
(x + 2)2 + (y + 6)2 = 25
4
(−5, 3)
The center is (−2, −6) and the radius is 5.
2
y
−8
−6
−4
−2
x
−8
−6
−4
−2
−4
x2 − 6x + (−3)2 + y2 = 7 + (−3)2
(−2, −6)
(x − 3)2 + y2 = 16
The center is (3, 0) and the radius is 4.
4
4 x
−2
x2 − 6x + y2 = 7
15.
2
−6
−8
y
−10
(3, 0)
2
4
6
x
−4
19. r 2 = (x − h)2 + (y − k)2
82
(2 − 0)2 + (3 − 0)2
64
4+9
64 ≠ 13
Because the radius is 8 and the distance between the center
and the point (2, 3) is less than the radius, the point (2, 3)
does not lie on the circle.
366
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Chapter 10
20. r 2 = (x − h)2 + (y − k)2
b. d 2 = 32 + 42
—
32
(4 − 0)2 + ( √ 5 − 0 )
d 2 = 9 + 16
9
16 + 5
d2
2
9 ≠ 21
—
and the point ( 4, √ 5 ) is greater than the radius, the point
( 4, √—5 ) does not lie on the circle.
21.
= (x −
—
h)2
+ (y −
= 25
d 2 = 61
—
d = √ 61 ≈ 7.8
Zone 2
Zone 3
d 2 = 12 + 22
d 2 = 02 + 32
d2 = 5
d2 = 9
—
d = √ 5 ≈ 2.2
k)2
r 2 = ( √6 − 0 ) + (2 − 0)2
r2
d 2 = 36 + 25
d=5
Because the radius is 3 and the distance between the center
r2
d 2 = 62 + 52
2
Zone 1
=6+4
—
d = √9 = 3
Zone 1
d 2 = 12 + 62
r 2 = 10
d 2 = 1 + 36
—
r = √10
2
( √—
10 )
d 2 = 37
(3 − 0)2 + (−1 − 0)2
—
d = √37 ≈ 6.1
9+1
10
10 = 10
Zone 2
—
Because the radius is √10 and the distance between the
center and the point (3, −1) is equal to the radius, the point
(3, −1) lies on the circle.
24. a.
y
6
22. r2 = (x − h)2 + (y − k)2
(0, 5)
4
2
r2 = (5 − 0) + (2 − 0)2
(6, 3)
r2 = 25 + 4
(0, 0)
r2 = 29
−2
—
r = √29
2
( √—
29 )
( √—7 − 0 )2 + (5 − 0)2
7 + 25
29
29 ≠ 32
2
—
center and the point ( √ 7 , 5 ) is greater than the radius, the
—
point ( √ 7 , 5 ) does not lie on the circle.
B
A
Zone 3
4 Zone 2
−4 Zone 1 4
x
6
4
y
−8
8
y
—
8
6
Two of the circles overlap, so there are locations that may
receive calls from more than one tower.
Because the radius is √29 and the distance between the
23. a.
4
−2
−2
8
x
2
4
6
8
x
−2
−4
b. City B has complete coverage. It’s entire radius is within the
−8
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circle with center (6, 3). Parts of City A do not have coverage.
Geometry
Worked-Out Solutions
367
Chapter 10
25.
k−b
h−a
the tangent line is perpendicular to the radius, the slope of
h−a
the tangent line is − — . The equation of the tangent
k−b
line, where d is the y-intercept is:
h−a
y=− — x+d
k−b
Substitute (a, b) into this equation.
h−a
b=− — a+d
k−b
Subtract these two equations.
y
28. The slope of the radius from (h, k) to (a, b) is —. Because
2
−2
2
x
(
−2
(−2, −4)
h−a
h−a
y−b=− — x+ — a+d−d
k−b
k−b
h−a
h−a
y−b= — x− — a
b−k
b−k
h−a
y − b = — (x − a)
b−k
(
The center of the image is (−2, 4) and the radius is 4.
(
(x − h)2 + (y − k)2 = r2
( x − (−2) )2 + (y − 4)2 = 42
(x + 2)2 + (y − 4)2 = 16
The equation of the image is (x + 2)2 + (y − 4)2 = 16. The
equation of the image of a circle after a translation m units to
the left and n units down is (x + m)2 + (y + n)2 = r2.
(x − 4)2 + (6 − 3)2 = 9
(x − 4)2 + (3)2 = 9
(x − 4)2 + 9 = 9
b. D; This circle has a center of (0, 3) and a radius of 2.
(x − 4)2 = 0
—
c. B; This circle has a center of (3, 0) and a radius of 2.
x=4
The line is a tangent because the system has one solution, (4, 6).
27. Determine the midpoint of two of the sides:
The midpoint of X(4, 5) and Y(4, 13) is
4 + 4 5 +13
—, — = (4, 9).
2
2
— is y = 9. The
The equation of the perpendicular bisector to XY
midpoint of Y(4, 13) and Z(8, 9) is
4 + 8 13 + 9
—, — = (6, 11).
2
2
Determine the equation of the line that is the perpendicular
—.
bisector of YZ
4
13 − 9
—= —
Slope of YZ
= — = −1
−4
4−8
The slope of the perpendicular bisector is 1.
)
)
)
y = mx + b
(x + 2)2 + (y − 2)2 = 16, y = 2x − 4
30.
(x + 2)2 + (2x − 4 − 2)2 = 16
(x + 2)2 + (2x − 6)2 = 16
(x + 2)2 + ( 2(x − 3) )2 = 16
(x2 + 4x + 4) + 4(x2 − 6x + 9) = 16
(x2
+ 4x + 4) + (4x2 − 24x + 36) = 16
5x2 − 20x + 40 = 16
5(x2 − 4x + 8) = 16
16
(x2 − 4x + 8) = —
5
16
2
x2 − 4x + (−2)2 = —
5 − 8 + (−2)
16
(x − 2)2 = —
5 −8+4
11 = 6 + b
5=b
16
(x − 2)2 = —
5 −4
The equation is y = x + 5.
Where y = 9 and y = x + 5
intersect is the center of the
circle, which is (4, 9).
Determine the radius:
(x − 4)2 + (y − 9)2 = r2
y
16
4
(x − 2)2 = − —5
10
Because the square root of a negative number does not yield
a real number, the system has no solution. So, the line is
none of these.
Z
8
(4 − 4)2 + (5 − 9)2 = r2
6
(−4)2 = r2
4
16 = r2
2
Geometry
Worked-Out Solutions
20
—
(x − 2)2 = —
5 − 5
Y
12
The equation of the circle
is (x − 4)2 + (y − 9)2 = 16.
368
—
√(x − 4)2 = √0
d. A; This circle has a radius of (0, −3) and a radius of 2.
(
) ( )
) ( )
29. (x − 4)2 + (y − 3)2 = 9, y = 6
26. a. C; This circle has a center of (−3, 0) and a radius of 2.
(
)
)
(
−6
(
)
(
(0, 0)
−6
X
2
4
6
8
x
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Chapter 10
1
5
31. (x − 5)2 + (y + 1)2 = 4, y = —x − 3
(
(
(
(
4
2
3
4
2
2
2
3
16 2
32
9
3
50
25 2
—
—
x
+
3 x + 25
9
50
25 2
—
x +—
3x
9
25 2
50
—
9 —
9x +9
3x
(x2
⋅
⋅
= 25
=0
=9
+ 150x = 0
⋅0
25x(x + 6) = 0
—
−b ± √b2 − 4ac
Use the Quadratic Formula; x = ——, where
2a
a = 26, b = −270, and c = 625.
⋅ ⋅
———
−(−270) ± √ (−270)2 − 4 26 625
x =————
2 26
25x = 0
y=
4
− —3
⋅0+2=2
x+6=0
y=
4
− —3
⋅ (−6) + 2 = 8 + 2 = 10
⇒
x=0
⇒
(0, 2)
⇒
x = −6
⇒
(6, 10)
4
− —3x
——
Because the equation y =
+ 2 contains the center of the
circle (−3, 6), and the system has two solutions, the line is a
secant line.
270 ± √ 72,900 − 65,000
x = ———
52
—
270 ± √ 7900
x = ——
52
33. The center is (0, k) and passes through the points (−1, 0) and
—
270 + √ 7900
x = ——
52
= 6.9, y = 0.2 6.9 − 3 = −1.6
⋅
(1, 0). Find the radius.
⇒
(6.9,−1.6)
—
270 − √ 7900
x = ——
52
= 3.5, y = 0.2 3.5 − 3 = −2.3 ⇒
⋅
——
(1 − 0)2 + (0 − k)2 = r2
—
(3.5,−2.3)
r = √1 + k2
— 2
(x − 0)2 + (y − k)2 = ( √1 + k2 )
x2 + y2 − 2yk + k2 = 1 + k2
—
d = √(3.4)2 + (0.7)2 = √11.56 + 0.49 = √ 12.05
= 3.5 ≠ 8 (the diameter)
The system has two solutions, and (5, −1) is not on the line.
So, the line is a secant line.
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All rights reserved.
(x − 0)2 + (y − k)2 = r2
1 + k2 = r2
———
d = √(6.9 − 3.5)2 + (−1.6 − (−2.3))2
——
⋅
25x2
26x 2 − 270x + 625 = 0
⋅
2
2
)
⋅
) = 25
(x + 3) + ( − —x − 4 ) = 25
(x + 3) + (−1) ( —x + 4 ) = 25
+ 6x + 9) + ( —x + —x + 16 ) = 25
4
(x + 3)2 + − —3 x + 2 − 6
)
)
2
1
(x − 5)2 + —x − 3 + 1 = 4
5
2
1
(x − 5)2 + —x − 2 = 4
5
4
1
(x 2 − 10x + 25) + —x 2 − —x + 4 = 4
5
25
26 2 54
—x − —x + 29 = 4
5
25
26 2 54
— x − —x + 25 = 0
5
25
26 2
54
25 — x − 25 — x + 25 25 = 0
25
5
⋅
4
(x + 3)2 + (y − 6)2 = 25, y = − —3 x + 2
32.
x2 + y2 − 2yk − k2 + k2 = 1
x2 + y2 − 2yk = 1
Yes, the equation of the circle with center (0, k) and passing
through the points (−1,0) and (1,0) is x2 + y2 − 2yk = 1.
Geometry
Worked-Out Solutions
369
Chapter 10
r 2 = (x − h)2 + y2
34.
XY = XZ
9.
(4r)2 = (x −15r)2 + y2
9a2 − 30 = 3a
16r 2 = (63 − 15r)2 + 162
9a2 − 3a − 30 = 0
16r 2 = 3969 − 1890r + 225r 2 + 256
−209
⋅
r2
3a2 − a − 10 = 0
= −1890r + 4225
0 = 209
⋅
(3a + 5)(a − 2) = 0
r 2 − 1890r + 4225
3a + 5 = 0
−5
a=—
3
a−2=0
—
−b ± √ b2 − 4ac
Use the Quadratic Formula; x = ——, where
2a
a = 209, b = −1890, and c = 4225.
———
−(−1890) ± √ (−1890)2 − 4(209)(4225)
x = ————
2 209
—
⋅
1890 ± √40,000 1890 ± 200
x = —— = —
418
418
a=2
XY = XZ
10.
2c2
+ 9c + 6 = 9c + 14
2c2 = 8
x = 5 and x ≈ 4.04
c2 = 4
Because the specification was that all circles have integer
radii, the value of x is 5.
A negative value does not make sense. So, the only solution
is c = 2.
The center of circle A is (15, 0) and the radius is 10. Therefore,
the equation is (x − 15)2 + y2 = 102, or (x − 15)2 + y2 = 100.
Maintaining Mathematical Proficiency
is a minor arc with measure
36. PR
m
PR = m
PQ + m
QR = 65° + 25° = 90°.
37. PRT is a major arc with measure
m
PRT = 53° + 25° + 65° + 90° = 270°.
38. ST is a minor arc with measure 180° − 53° = 127°.
39. RST is a semicircle with measure 180°.
40. QS is a minor arc with measure 25° + 53° = 78°.
35. RS is a minor arc with measure 53°.
—
c = ±√4 = ±2
(r + 3)2 = r 2 + 92
11.
r2
+ 6r + 9 = r 2 + 81
6r + 9 = 81
6r = 72
r = 12
12. If 10 is the radius, then the diameter is 20.
522
2704
482 + 202
2304 + 400
2704 = 2704
— is tangent to circle C.
AB
14. LM is a minor arc, so m
LM = m∠LPM = 60°.
15. KM is a minor arc, so m
KM = m∠KPM = 100° + 60° = 160°.
16. KN is a minor arc, so m
KN = m∠KPN = 80°.
13. KL is a minor arc, so mKL = m∠KPL = 100°.
10.4 –10.7 What Did You Learn? (p. 581)
1. A ruler and a right angle used for mechanical drawing could
be used together to mark right angles. A ruler and a compass
will also work.
2. ∠BAC could be acute or obtuse.
17. The red arcs are not congruent because the two circles are
Chapter 10 Review (pp. 582–586)
—
1. PK is a radius.
—
2. NM is a chord.
3. ⃗
JL is a tangent.
—
4. KN is a diameter.
5. ⃖⃗
NL is a secant.
6. PN is a radius.
—
7. The common tangent is an internal tangent.
8. The common tangent is an external tangent.
not congruent.
18. The red arcs are congruent because the circles are congruent
and m
AB = m
EF .
— —
19. AB ≅ ED by the definition of measure of minor arcs. So, by
the Congruent Corresponding Chords Theorem (Thm. 10.6),
m
ED = m
AB = 61°.
— —
—
20. You are given BE ≅ ED and CA is a perpendicular bisector.
Therefore, by the Congruent Corresponding Chords Theorem
(Thm. 10.6), m
AD = m
AB = 65°.
— —
21. By the Equidistant Chords Theorem (Thm. 10.9), AB ≅ ED .
Therefore, m
AB = m
ED = 91°.
370
Geometry
Worked-Out Solutions
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Chapter 10
22. Using △KNQ: r 2 = (2x)2 + 102
30. By the Angles Inside the Circle Theorem (Thm. 10.15),
1
Using △PLQ: r 2 = (3x − 12)2 + 102
x = —2 (152 + 60)
QN = QP and by the Equidistant Chords Theorem
(Thm. 10.9),
x = 106
4x = 6x − 24
31. By the Angles Outside the Circle Theorem (Thm. 10.16),
2x = 24
x = 12
r 2 = (2
1
x = —2 (212)
1
40 = —2 (96 − x)
⋅ 12)
2
80 = 96 − x
+ 102
−16 = −x
r 2 = (24)2 + 102
x = 16
r 2 = 676
—
r = √676 = 26
32. By the Tangent and Intersected Chord Theorem (Thm. 10.14),
= 2 120° = 240°.
mXYZ
The radius is 26.
23. By the Measure of an Inscribed Angle Theorem
(Thm. 10.10), x° = 2
⋅
JK ⋅ JL = JM ⋅ JN
4 ⋅ (4 + 6) = x ⋅ (x + 3)
33.
⋅ 40° = 80°.
40 = x2 + 3x
24. By the Inscribed Quadrilateral Theorem (Thm. 10.13):
0 = x2 + 3x − 40
q° + 80° = 180°
(x + 8)(x − 5) = 0
q = 100
x+8=0
4r ° + 100° = 180°
4r = 80
25. By the Inscribed Angles of a Circle Theorem (Thm. 10.11),
m∠KJL = m∠KML.
14d ° = 70°
34.
⋅
△NPQ is a right triangle:
⋅
3x(x − 3) = 0
3x = 0
m∠NPQ = 90°
or
x−3=0
x=0
3y° = 90°
x=3
Because 0 does not make sense, the only solution is x = 3.
y = 30
m∠PNQ + m∠PQN = 90°
50° + 4z° = 90°
⋅ WX = XY ⋅ XZ
12 ⋅ 12 = 8 ⋅ (8 + x)
35. WX
144 = 64 + 8x
4z = 40
80 = 8x
z = 10
27. By the Inscribed Angles of a Circle Theorem (Thm. 10.11),
m∠VRT = m∠VST and m∠RTS = m∠RVS.
m∠VRT = m∠VST
m° = 44°
m∠RTS = m∠RVS
n° = 39°
28. By the Measure of an Inscribed Angle Theorem (Thm. 10.10),
⋅ 56° = 28°.
29. By the Angles Outside the Circle Theorem:
x=
⋅
3x2 − 9x = 0
26. By the Inscribed Right Triangle Theorem (Thm. 10.12),
1
—2 (250
⋅
TP PR = QP PS
(x + 3) x = (6 − x) 2x
x2 + 3x = 12x − 2x2
d=5
c° =
x=5
A negative value for x does not make sense. So, the only
solution is x = 5.
r = 20
1
—2
x−5=0
or
x = −8
− 110); x = 70
The value of x is 70.
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x = 10
⋅ AB = AC ⋅ AD
20 ⋅ 20 = 12 ⋅ (12 + 2r)
36. AB
400 = 144 + 24r
256 = 24r
r ≈ 10.7
The radius of the rink is approximately 10.7 feet.
37. The center of the circle is (4, −1) and the radius is 3.
(x − h)2 + (y − k)2 = r 2
(x − 4)2 + ( y − (−1) )2 = 32
(x − 4)2 + (y + 1)2 = 9
Geometry
Worked-Out Solutions
371
Chapter 10
38. The center of the circle is (8, 6) and the radius is 6.
(x −
+ (y −
h)2
k)2
=
(x − 8)2 + (y − 6)2 = 62
(x −
+ (y −
8)2
6)2
x2 + y2 − 12x + 8y + 48 = 0
47.
r2
x2 − 12x + y2 + 8y = −48
x2
= 36
39. The center of the circle is (0, 0) and the radius is 4.
− 12x + (−6)2 + y2 + 8y + 42 = −48 + (−6)2 + 42
(x − 6)2 + (y + 4)2 = 4
The center is (6, −4) and the radius is 2.
(x − h)2 + (y − k)2 = r 2
(x −
0)2
+ (y −
0)2
=
y
42
2
x 2 + y2 = 16
2
40. The center of the circle is (0, 0) and the radius is 9.
(x − h)2 + (y − k)2 = r 2
−4
(x − 0)2 + (y − 0)2 = 92
(x − h)2 + (y − k)2 = r 2
( x − (−5) )2 + (y − 2)2 = 1.32
(x + 5)2 + (y − 2)2 = 1.69
42. The center of the circle is (6, 21) and the radius is 4.
(x − h)2 + (y − k)2 = r 2
(x − 6)2 + (y − 21)2 = 42
(x − 6)2 + (y − 21)2 = 16
43. The center of the circle is (−3, 2) and the radius is 16.
(x − h)2 + (y − k)2 = r2
( x − (−3) )2 + (y − 2)2 = 162
(x + 3)2 + (y − 2)2 = 256
44. The center of the circle is (10, 7) and the radius is 3.5.
(x −
h)2
+ (y −
k)2
=
r2
(x − 10)2 + (y − 7)2 = 3.52
(x − 10)2 + (y − 7)2 = 12.25
45. The center of the circle is (0, 0) and the radius is 5.2.
(x − h)2 + (y − k)2 = r 2
(x − 0)2 + (y − 0)2 = 5.22
x2 + y2 = 27.04
——
46. r = √ (x − h)2 + (y − k)2
8 x
(6, −4)
——
48. r = √ (x − h)2 + (y − k)2
——
= √(−5 − 0)2 + (0 − 0)2
—
= √(−5)2 + 02
—
= √25 = 5
52
(4 − 0)2 + (−3 − 0)2
25
16 + 9
25 = 25
Because the radius is 5 and the distance between the center
and the point (4, −3) is equal to the radius, the point (4, −3)
lies on the circle.
Chapter 10 Test (p. 587)
1
1. ∠1 is an inscribed angle, so m∠1 = —2
∠2 is a central angle, so m∠2 = 145°.
1
m∠2 =
1
—2
⋅ 180° = 90°.
3. ∠1 is outside the circle, so
m∠1 = —12 (96° − 38°) = —12
⋅
4. ∠1 and ∠2 are outside the circle.
m∠1 = —12 (77° − 48°) = —12 29° = 14.5°
⋅
m∠2 = —12(263° − 97°) = —12
+ (1 −
= √02 + (−5)2
—
= √25 = 5
(x − h)2 + (y − k)2 = r2
( x − (−7) )2 + (y − 6)2 = 52
(x + 7)2 + (y − 6)2 = 25
Geometry
Worked-Out Solutions
⋅ 58° = 29°.
∠3 is an inscribed angle, so
m∠3 = 180° − (180° − 66°) − 29° = 37°.
—
6)2
⋅ 180 = 90° and
∠2 is inside the circle, so m∠2 = —12 (96° + 36°) = 66°.
360° − (215° + 48°) = 97°
7)2
⋅ 145 = 72.5°.
2. ∠1 and ∠2 intercept a circle, so m∠1 = —2
——
= √(−7 +
372
6
−6
x2 + y2 = 81
41. The center of the circle is (−5, 2) and the radius is 1.3.
4
−2
5. AG
⋅ GD = BG ⋅ GF
2 ⋅ 9 = 3 ⋅ GF
18 = 3 ⋅ GF
GF = 6
⋅
⋅ 166° = 83°
6. CE ⋅ CD = CB ⋅ CF
CE ⋅ 9 = 3 ⋅ 12
9 ⋅ CE = 36
CE = 4
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Chapter 10
7. CA
— —
⋅ CA = CB ⋅ CF
CA = 3 ⋅ 12
HL Congruence Theorem (Thm. 5.9). Then ∠LCK ≅ ∠JCK,
and by the Congruent Central Angles Theorem (Thm. 10.4),
JM ≅ LM .
2
CA2
= 36
—
CA = √ 36
—
—
14. GE is a perpendicular bisector of DF , so DG = FG by the
CA = 6
Perpendicular Bisector Theorem (Thm. 6.1). By the Congruent
Corresponding Chords Theorem (Thm. 10.6), DG ≅ FG .
8. Sample answer:
A
— —
13. CJ ≅ CL and CK ≅ CK , so △CKJ ≅ △CKL by the
B
15.
y
16
E
C
12
D
a.
1
m∠CDE = —2
1
m∠CAE = —2
8
CAE
⋅ m
m
CDE
⋅
4
4
The sum of the two arcs is 360°. So,
m
CAE + m
CDE = 360°.
CAE + — ⋅ m
CDE
⋅ m
1
m∠CDE + m∠CAE =
m∠CDE + m∠CAE =
( mCAE + mCDE )
1
—2
1
—2
8
12
42
(−2)2
16
⋅ 360°
4 < 16
yes; Actor A is inside the area that is illuminated by the light.
Therefore, ∠CDE and ∠CAE are supplementary.
1
CDE and m∠CAE = — ⋅ m
CDE
⋅ m
Actor B (8, 5):
1
2
m∠CBE = m∠CAE and both angles intercept the same
arc. Therefore, ∠CBE ≅ ∠CAE.
9. 5x − 4 = 3x + 6
(6 +
10.
r)2
=
122
+
(8 − 13)2 + (5 − 4)2
42
+1
16
25 + 1
16
(−5)2
r2
2x = 10
36 + 12r + r 2 = 144 + r 2
x=5
36 + 12r = 144
12r = 108
26 > 16
no; Actor B is outside the area that is illuminated by the light.
Actor C (15, 5):
(15 − 13)2 + (5 − 4)2
42
+1
16
r=9
(2)2
——
5 < 16
11. r = √ (x − h)2 + (y − k)2
r = √(−1 − 0)2 + (4 − 2)2
—
—
r = √1 + 4 = √5
——
x2
x2
+ (y −
2)2
=√
+ (y −
2)2
=5
yes; Actor C is inside the area that is illuminated by the light.
16. a.
—
r 2 = 1602 + (r − 80)2
r 2 = 25,600 + r 2 − 160r + 6400
(5)2
0 = 32,000 − 160r
A circle with center (0, 2) and point (−1, 4) on the circle has
—
a radius of √ 5 and has an equation of x2 + (y − 2)2 = 5.
( 2√—2 )2 + (−1 −2)2
5
8+9
5
—
x
(11 − 13)2 + (4 − 4)2
m∠CDE + m∠CAE = 180°
b. m∠CBE = —2
16
Actor A (11, 4):
1
2
m∠CDE + m∠CAE = —2
(13, 4)
r = 200
The radius is 200 feet.
b. S = 3.87√ fr
—
⋅
—
= 3.87√0.7 200
17 > 5
The point, ( 2√ 2 , −1 ) does not lie on the circle.
≈ 45.79
— —
12. By the Equidistant Chords Theorem (Thm. 10.19), ST ≅ RQ .
By the Congruent Corresponding Chords Theorem
(Thm. 10.6), ST ≅ RQ .
160r = 32,000
The car’s speed is about 45.79 miles per hour.
Chapter 10 Standards Assessment (pp. 588 –589)
—
—
c. AD is a diameter.
1. a. BG is a chord.
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All rights reserved.
—
—
d. FE is a chord.
b. CD is a radius.
Geometry
Worked-Out Solutions
373
Chapter 10
2. Map Circle C to Circle C′ by using the translation
(x, y) → (x − 2, y + 2) so that Circle C ′and Circle D have
the same center at (0, 3). Dilate Circle C′ using a center of
dilation (0, 3) and a scale factor of 4. Because there is a rigid
transformation that maps Circle C to Circle D, Circle C is
similar to Circle D.
3. Given
△JPL ≅ △NPL
—
PK is an altitude of △JPL.
—
PM is an altitude of △NPL.
Prove
△PKL ∼ △NMP
6. m∠ACB = 96°
1
1
m∠JGK = —2(60° + 132°) = —2 (192°) = 96°
m∠JGK = m∠ACB
1
1
m∠LGM = —2 (60° + 132°) = —2 (192°) = 96°
m∠LGM = m∠ACB
1
1
m∠STV = —2(276° − 84°) = —2 (192°) = 96°
m∠STV = m∠ACB
m∠VWU = 96°
m∠VWU = m∠ACB
J
7. a. Inverse
b. Converse
K
c. Biconditional
P
d. Contrapositive
8. No, by the Inscribed Quadrilateral Theorem, A quadrilateral
L
M
N
— and PK
— are altitudes,
m∠JPL = 90°, m∠LPN = 90°, and PM
so △JPL ∼ △PKL and △NPL ∼ △NMP by the Right
Triangle Similarity Theorem (Thm. 9.6). By the Transitive
Property, △PKL ∼ △NMP.
can be inscribed in a circle if and only if its opposite angles
are supplementary. The sum of the measures of the opposite
angles are 70° + 70° = 140° and 110° + 110° = 220°, not
180°. Therefore, this quadrilateral cannot be inscribed in
a circle.
4. B;
x2 + y2 + 14x − 16y + 77 = 0
x2 + 14x + y2 − 16y = −77
(x2
+ 14x +
72)
+ ( y2 − 16y + (−8)2 ) = −77 + 72 + (−8)2
(x2 + 14x + 72) + ( y2 − 16y + (−8)2 ) = −77 + 49 + 64
(x + 7)2 + (y − 8)2 = 36
The center is (−7, 8) and the radius is 6.
−2 − (−6) −2 + 6 4 1
1 − (−7)
1+7
8 2
−6 − (−2) −6 + 2 −4
—
Slope of XY = — = — = — = −2
3−1
2
2
−10 − (−6) −10 + 6 −4 1
—
Slope of YZ = —— = — = — = —
−5 − 3
−8
−8 2
−10
+
6
−4
−10
−
(−6)
Slope of —
ZW = —— = — = — = −2
−5 − (−7)
−5 + 7
2
—
5. Slope of WX = — = — = — = —
— ⊥ XY
—, WX
— ⊥ WZ
—, XY
— ⊥ YZ
—, YZ
— ⊥ ZW
—
WX
There are right angles at W, X, Y, Z. By definition, if a
quadrilateral has four right angles it is a rectangle.
374
Geometry
Worked-Out Solutions
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