CHAPTER 10 Think & Discuss (p. 593) 1. point B
c. m
2. one
09 9 3 9 12
3. The fireworks would not be so high above the Earth. The
ships would be smaller.
x 42 x2 62
264 360 2x
8x 20 0
10.1 Guided Practice (p. 599)
96 2x
8x 20
1.
x 48
3. 15y 15 242
4. 2z2 7 19
15y 225 576
2z2 12
15y 351
intersect a circle at two points. A chord is a line segment having its endpoints on the circle, while a secant is a line that passes through two points on a circle. ↔ 3. mCPX 90; According to Thm. 10.1, XY is perpendicular to CP. Perpendicular lines form right angles. Therefore mCPX 90.
z2 6
2 y 117 5 or 235
z ± 6
xx 12
64 x2 12x x2 12x 64 0
5. No; 52 52 72, so by the Converse of the Pythagorean
Thm., ABD is not a right , so BD is not to AB. ↔ If BD were tangent to C, B would be a right angle. ↔ Thus, BD is not tangent to C.
x40
x 16
x4
x y 18
6.
x 18 y
3x 4y 64
6. 4
x 18 10
318 y 4y 64
x8
8. 5
9. r 7.5 cm
y 10
12. r 4 cm
Solution: 8, 10
JL 145
7. 2
10.1 Practice and Applications (pp. 599–602)
54 3y 4y 64
7. 82 92 JL2
radius
4. 6.5 cm
x 16x 4 0 x 16 0
2. Both a chord and a secant
chord diameter
x 52
5.
d. A: 7, 0; B: 5, 9
132 12360 x x
2.
x2 8x 16 x2 36
82
9 b 4
3 4
3 9 equation: y x 4 4
Chapter 10 Study Guide (p. 594) 1.
433 b
0
sin L
8 145
mL 41.6
sin J
9 145
mJ 48.4
12.0 8. A3, 0, B9, 9 a. AB 3 92 0 92
10. r 3.35 in. 13. d 52 in.
15. d 17.4 in.
11. r 1.5 ft
14. d 124 ft
16. d 8.8 cm
17. C and G are congruent because they have the same
radius, 22.5. 18. B
19. E
25. G
26. external
20. F
21. D
22. A
27. external
29.
23. C
24. H
28. internal
30.
15 b. mdpt AB
32 9, 0 2 9
3, 4
1 2
4 common tangents
204
Geometry Chapter 10 Worked-out Solution Key
no common tangents
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued 32. center is 2, 2
31.
52. The statements QR > QP and QP > QR cannot both be
true at once. Therefore, the assumption that R exists must be false. Then QP.
radius is 2 units
53. Assume that is not tangent to P, that is, there is another point X on that is also on Q. X is on Q, so QX
2 common tangents
QP. But the perpendicular segment from Q to is the shortest such segment, so QX > QP. QX cannot be both equal to and greater than QP. The assumption that such a point X exists must be false. Then is tangent to P.
33. center is 6, 2; radius is 2 units 34. The two circles intersect at one point, 4, 2. 35. The two circles have three common tangents, lines x 4,
y 0, y 4.
54.
A
D
36. No; 52 142 152, so by the Converse of the
Pythagorean Thm., ABC is not a right , so AB is not ↔ to AC. Then AB is not tangent to C.
B
C
37. No; 52 152 172, so by the Converse of the
Pythagorean Thm., ABC is not a right , so AB is not ↔ to AC. Then, AB is not tangent to C.
38. Yes;
162
122
55. Square; BD and AD are tangent to C at A and B,
respectively, so A and B are right angles. Then, by the Interior Angles of a Quadrilateral Thm., D is also a right angle. Then CABD is a rectangle. Opposite sides of a rectangle are congruent, so CA CB and AD CB. But CA and CB are radii, so CA CB and by the Transitive Prop. of Congruence, all 4 sides of CABD are congruent. CABD is both a rectangle and a rhombus, so it is a square by the Square Corollary.
202,
so by the Pythagorean Thm. ↔ ABC is a right , so AB AC. Therefore, AB is tangent to C.
39. Yes; 202 212 292, so by the Pythagorean Thm.,
↔ ABC is a right , so AB AC. Therefore, AB is tangent to C.
40.
282 r2 r 82 784
r2
r2
41. d 45 8
53 ft
16r 64
56. a. m
16r 720 0
53 1 48 2
b. The slope of j is 2. Since j is tangent to C at P,
16r 720
CP j. The slopes of 2 perpendicular lines are nega-
r 45 ft
tive reciprocals of each other. The slope of CP is 12,
↔ ↔ 42. AF, BE 43. GD, HC, FA, or EB
so the slope of j is 2.
44. Yes; HC is a chord. The diameter of a circle is the great-
est distance across the circle.
c.
46. 2x 7 5x 8
45. JK
13 b
15 3x
y 2x 13
x5 47.
5x2
9 14
48.
2x 5
3 82 b
d. Choose any point on the circle, determine the slope of
3x2
5x2 5
3x2 12 0
x2 1
x2 4
the radius to that point and use its negative reciprocal along with the chosen point to find the equation.
2x 7 57.
x ±1 x ±2 ↔ ↔ ↔ 49. PS is tangent to X at P, PS is tangent to Y at S, RT ↔ is tangent to X at T, and RT is tangent to Y at R. Then, PQ TQ and QS QR. (2 tangent segments with the same exterior endpoint are .) By the def. of congru ence, PQ TQ and QS QR, so PQ QS TQ QR by the addition prop. of equality. Then, by the Segment Addition Post. and the Substitution Prop., PS RT or PS RT. 50. QR > QP
51. QP > QR
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 10 Worked-out Solution Key
205
Chapter 10 continued 9. BC is a diameter; a chord that is the perpendicular bisec-
tor of another chord is a diameter.
mBC ; The measure of an arc formed 10. mABC mAB by two adjacent arcs is the sum of the measures of the two arcs.
DA ; If the diameter of a circle is 11. BC CA, BD
10.1 Mixed Review (p. 602)
perpendicular to a chord, then the diameter bisects the chord and its arc.
58. let x represent the length of the third side,
4 10 > x
4 x > 10
14 > x
x > 6
10.2 Practice and Applications (pp. 607–611) 12. minor arc
6 < x < 14 59. Since the slope of PS
3 8
slope of QR, PS QR. Since the slope of PQ 3 slope of SR, PQ SR. Then, PQRS is a parallelogram by def.
61.
x 3 11 5
16. major arc
17. semicircle
18. major arc
19. major arc
20. 60 21. 55
J
2x 54
3x 84
3 5
x 27
x 28
67.
33 18 x 42
10 8 3 x
65.
18x 1386
10x 24
x 77
x2
3 2 x3 x
66.
69. 142 62 C 2
C 258
6 tan A 14
tan C 14 6
mA 23.2
mC 66.8
15.2 10 70. sin 43 AB
mA 90 43
AB 14.7
10 tan 43 CB
47
71. 82 CB2 142
CB 11.5
33. 145
34. 145
mML 130 85 215 by the mKML mKM Arc Add. Post. Since D N, ABC KML .
8 cos A 14
mB 34.8
mA 55.2
270 30 mBC
3x 210 x 70 37. 4x x 180
110
436 mMB
5x 180
144
x 36 38. 4x 6x 27x 360
mRST 6x 7x 1315
24x 360
CB 10.7
8 sin B 14
29. 65
31. 120
36. x 2x 30 180
x 9
28. 60
30. 115
26. 60
(both have radius 4). By the Arc Add. Post., mAC mAE mEC 70 75 145. mKL 145 and KL; mABC 360 mAC since D N, AC 360 145 215.
10x 9x 9
9x
25. 180
KL and ABC KML ; D and N are congruent 35. AC
8 x 5 9 x 1 2x
27. 65
23. 305
L
32. 145
3x 4x 8
2x 3x 9
K
60
M
3 4 x2 x
2 5
68.
Q 55
x 12 7 3
63.
22. 300
24. 180 N
5x 33 x6 64.
x 9 6 2
62.
14. semicircle
15. minor arc
60. The length of PQ 125 SR so PQ SR. The length of PS 41 QR so PS QR. Then PQRS
is a parallelogram by Thm. 6.2.
13. minor arc
39.
AB
x 15
195
; 2 arcs are congruent if and only if their corre CB
sponding chords are congruent. 40. AB CD; if 2 arcs are congruent, then their correspond-
ing chords are congruent. Lesson 10.2
41. AB AC; in a circle, 2 chords are congruent if and only
if they are equidistant from the center.
10.2 Guided Practice (p. 607) 1. minor arc
72; mMN 72; No, KL and MN are not arcs 2. mKL of the same nor of s. 3. 60
206
4. 300
5. 180
6. 100
7. 220
8. 40
Geometry Chapter 10 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued 42. ED 10; in a circle, 2 chords are congruent if and only
59. Draw radii LG and LH. LG LH, LJ LJ, and since
EF GH, LGJ LHJ by the HL Congruence Thm. Then, corresponding sides GJ and JH are congruent as are corresponding angles GLJ and HLJ. By the def. of EH . congruent arcs, GE
if they are equidistant from the center. 43. 40; a diameter that is perpendicular to a chord bisects
the chord and its arc. 44. 170; 2 minor arcs are the same measure if their chords
are the same measure.
60. Assume center L is not on EF. GL HL because both
are radii. GJL HJL and GJ HJ because EF is the perp. bisector of GH. GLJ HLJ by the HL Congruence Thm. GJL HJL, because corresponding parts are congruent. Since GJL and HJL form a linear pair and are congruent, both must be right angles. Therefore,JL GH. According to the Perpendicular Postulate, the center must be on EF. Therefore, EF is a diameter of L.
45. 15; in a circle, 2 chords are congruent if and only if they
are equidistant from the center. 46. 7; If a diameter of a circle is perpendicular to a chord,
then the diameter bisects the chord and its arc. 47. 40; Vertical Angles Thm., def. of minor arc
farther from the center
48.
61. Draw radii PB and PC. PB PC and PE PF. Also,
49. 15
50. 90
since PE AB and PF CD, PEB and PFC are right triangles and are congruent by the HL Congruence Thm. Corresponding sides BE and CF are congruent, so BE CF and by the multiplication prop. of equality, 2BE 2CE. By Thm. 10.5 PE bisects AB and PF bisects CD, so AB 2BE and CD 2CF. Then by the Substitution Prop., AB CD or AB CD.
51. 2 A.M.
52. During step 1, the ski patrol marks off a chord of a circle.
During steps 2 and 3, the patrol marks off a diameter of the circle, because this line is the perpendicular bisector of the chord from step 1. The middle of the diameter is the center of the circle.
62. Draw radii PB and PC. PB PC. By Thm. 10.5 PE
bisects AB and PF bisects DC. By the multiplication property of equality, 12AB BE and 12DC CF. Since AB DC, by the Substitution Prop., BE CF. Since PE AB and PF DC, PEB and PFC are right triangles and are congruent by the HL Congruence Thm. Therefore, PE PF because they are corresponding sides.
53. This follows from the definition of the measure of a
minor arc. (The measure of a minor arc is the measure of its central angle.) If 2 minor arcs in the same circle or congruent circles are congruent, then their central angles are congruent. Conversely, if 2 central angles of the same circle or congruent circles are congruent, then the measures of the associated arcs are congruent.
63.
90 120 150
54. Circles may vary; to find the center, draw a chord with
the straight edge and construct its perpendicular bisector. Extend the perpendicular bisector so it touches 2 points on the circle. Find the midpoint of this bisector; this will be the center.
56. AB DC; AP, PB, PC and PD are all radii, so they are
all congruent. Therefore, APB DPC by SSS. Since the 2 triangles are congruent, APB DPC; because corresponding parts are congruent. By def. of minor arc, . AB DC
57.
Since AB
, APB CPD by the def. of DC
congruent arcs. PA, PB, PC, and PD are all radii of P, so PA PB PC PD. Then APB CPD by the SAS Congruence Post., so corresponding sides AB and DC are congruent. 58. You would have to state that by the def. of congruent
circles, the radii were congruent. Then proceed with the proof.
Copyright © McDougal Littell Inc. All rights reserved.
30
A(2, 30°) E(2, 150°) 180
210
55. Yes; construct the perpendiculars from the center of the
circle to each chord. Use a compass to compare the lengths of the segments.
60
B(4, 120°)
0
2
C(4, 210°)
D(4, 330°)
240
330
300 270
64. 120
65. 90
66. 150
67.
210
68. a. Construct the perpendicular bisector of each chord.
The point at which the bisectors intersect is the center. Connect the center with any point on the circle. b. Extend the tangent lines until they intersect. Construct
the bisector of the angle formed by the tangent lines. This bisector is a diameter of the circle. Half of the diameter is a radius. c. Large objects, you are able to make more accurate
constructions. 69. 81 36 x2
x 35
Geometry Chapter 10 Worked-out Solution Key
207
Chapter 10 continued Lesson 10.3
10.2 Mixed Review (p. 611) 70.
interior: 3, 1
y
exterior: 2, 4
(2, 4) B(0, 2)
1
1.
A(4, 2) (3, 1)
1
Activity 10.3: Investigating Inscribed Angles (p. 612) Exploring the Concept
Circle 1
T R
x
C(3, 0)
P
71.
1, 1 lies in the interior
y
B(0, 0)
(1, 1)
1
U
V
Investigate
1 1
x
1.
C(4, 1) (1, 2)
72.
1, 5 lies in the exterior
(2, 1)
1
B(0, 1)
mRPS 90 70 55
Circle 1 Circle 2 Circle 3
2, 1 lies in the interior
y
1
S
1, 2 lies in the exterior
A(2, 3)
2.
mRTS 40 30 26 3.
U
x
mRUS 44 30 27
mRVS 46 36 28
S
T
R
A(2, 3)
V C(2, 3) R
P
(1, 5)
P
S
73. A(3, 2) 1
B(0, 0)
y
0, 1 lies in the interior
C(3, 2) (0, 1)
1, 2 lies in the exterior
U
T
Circle 2
V
Circle 3
Make a Conjecture 1
x
3. The measure of an inscribed angle is approximately equal 1
to 2 the measure of the corresponding central angle.
(1, 2)
Extension 74. rhombus; PQ QR RS PS 10, so PQRS is a
rhombus by the Rhombus Corollary. 75. square; PQ QR RS PS 32, so PQRS is a
rhombus by the Rhombus Corollary; PR QS 6, so PQRS is a rectangle. (A parallelogram is a rectangle if and only if its diagonals are congruent.) Then, PQRS is a square by the Square Corollary.
76.
78.
9 x x 16
77.
8 x x 32
x2 144
x2 256
x 12
x 16
4 x x 49 x2
196
x 14
79.
About 77; the star divides the into 7 arcs, so each has 360 . Each inscribed corresp. to an arc with measure 7 measure 3
360 1080 1 . Then x 7 7 2
324
x 18
10.3 Guided Practice (p. 616) 1. B
2. No; according to Thm.
A
10.11, since the opposite angles are not supplementary, the quadrilateral cannot be inscribed in a circle.
C
intercepted arc AC
40 3. mKL 6. x 115
4. mKML 180 7. y 150
z 75
208
Geometry Chapter 10 Worked-out Solution Key
1080 7
540 1 77 77. 7 7
9 x x 36 x2
5. mLMK 210 8. x 95
y 100
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued 10.3 Practice and Applications (pp. 617–619)
64 9. mCB
156 10. mBC
12. mABC 55
228 11. mBC
13. mABC 109
mentary, so according to Thm. 10.11, it cannot always be inscribed in a circle. 28. No; the opposite angles are not always supplementary, so
according to Thm. 10.11, it cannot always be inscribed in a circle.
14. mABC 90 15. x 47; EGH and EDH intercept the same arc so
29. Yes; both pairs of opposite angles of an isosceles trape-
their measures must be the same.
zoid are supplementary.
16. x 90, y 50; MLK intercepts an arc whose chord
According to Thm. 10.1, a line that is perpendicular to a radius of the circle is tangent to the circle.
30.
is a diameter so the measure of the arc is 180 and x 90; mLKM can be found using the Triangle Sum Thm.
A C
17. x 45, y 40; QPR and QSR intercept the same
arc and PQS and PRS intercept the same arc so mPQS 40 and mQSR 45.
18. x 90
19. x 80
20. x 65
y 90
y 78
y 90
z 112
z 160
z 180
21.
2x 26y 1
→
6y 6y 4x 360
AC is a diameter of M.
31. A M
x 32y
C
x 30
12y 4x 360 3y x 90
32.
3
3y 2y 90 9 2y
B
90
A M
y 20
C
mCBA 90; since it intercepts AC which is a diameter, the intercepted arc is 180. CBA is an inscribed angle so its measure is 90.
mA 60, mB 60, mC 60 22. 2x 26y 180 Mult. by 3 →
6x 78y 540
3x 21y 180 Mult. by 2 → 6x 42y 360 36y 180 mA 130, mB 75
y5
mC 50, mD 105
x 25
23. 4x 24y 180 Mult. by 7 →
28x 168y 1260
14x 9y 180 Mult. by 2 → 28x 18y 360 150y 900 mA 54, mB 36
y6
mC 126, mD 144
x9
24. Yes; every angle of a square measures 90 so both pairs
of opposite angles are always supplementary. Therefore the square can always be inscribed in a circle, according to Thm. 10.11 25. Yes; every angle of a rectangle is a right angle so both
pairs of opposite angles are always supplementary. Therefore, the rectangle can always be inscribed in a circle, according to Thm. 10.11. 26. No; the opposite angles are not always supplementary, so
according to Thm. 10.11, it cannot always be inscribed in a circle. 27. No; the opposite angles of a kite are not always supple-
Copyright © McDougal Littell Inc. All rights reserved.
33. BA; a line perpendicular to a radius of a circle at its end-
point is tangent to the circle. 34. Answers may vary. Sample answer: mC 90 when
↔ ↔ CQ AB, mA mB 45. As you drag C toward A, mA increases and mB decreases. As you drag C away from A, mA decreases and mB increases.
35. QB; isosceles; base angles; A B; Exterior Angle;
; 12mAC 2x; 2x; 2; 12mAC
36. Draw diameter containing QB, intersecting the circle
and at D. By the proof in Ex. 35, mABD 12mAD 1 mDBC 2mDC . By the Arc. Add. Post., mDC mAC . By the Angle Add. Post., mAD mABD mDBC mABC. By repeated applica. tion of the Substitution Prop., mABC 12mAC
37. Draw the diameter containing QB, intersecting the circle
at point D. By the proof in Ex. 35, mABD 12mAD 1 and mDBC 2mDC . By the Arc Add. Post., mAD mCD , so mAC mAD mCD by the submAC traction prop. of equality. By the Angle Addition Post., mABD mABC mCBD, so mABC mABD mCBD by the subtraction prop. of equality. Then, by repeated application of the Substitution Prop., . mABC 12mAC
Geometry Chapter 10 Worked-out Solution Key
209
Chapter 10 continued 38. Given: X with inscribed
A
angles ACB and ADB and diameter AB.
46. D
Prove: ADB ACB
Proof: ADB and ACB intercept diameter AB. Therefore the meaB sure of the intercepted arc is 180. So mADB 90 and mACB 90. By the Substitution Prop. mADB mACB so ADB ACB. A
ABC, AC is a diameter of O.
GJ 23 3.46 in. 47.
7 LX2
O
Given: O with inscribed ABC, B is a right angle Prove: AC is a diameter of O. Use the Measure of an Inscribed Angle Thm., to show the inscribed right angle intercepts an arc with measure 290 180. Since AC intercepts an arc that is half the measure of the circle, it must be a diameter. 40. Given: DEFG is inscribed
10.3 Mixed Review (p. 620) 48. 6 12 b
y x 8
1 2
1 2
y3
y 43x 7
52. 4
8
16 b
12x
y 45x 16 55.
y
G
mD mF 180
4
0b
P
1 2
53. 12 5 5 b
12
54.
1 2
4
7b
1 2 mDGF ;
1 2
y 2x 9 51. 7 3 0 b
3b
R
cle. (Position the vertex of the tool on the circle and mark the 2 points where the sides intersect the . Repeat, placing the vertex at a different point on the circle. The center is the point where the diameters intersect.)
R Q
Q
2 2
2
Q
2 2
x
R
56.
P
57.
y
P
R
y
P
R 2 2
x
2
P
Q
41. Use the carpenter’s square to draw 2 diameters of the cir-
y
P
Q
1 2
1 2
1 25 b 9 b
50. 3 30 b
C
D mG mDEF ; mDGF mDEF 360; so mDEF 180 By the Sub. Prop. mDGF and mE mG 180; Likewise mD mGFE ; mGFE mGDE 360; so mF mGDE mGDE 180; By Sub. Prop. mGFE
49.
8 b
y
E
Prove: mD mF 180, mE mG 180
5.3 in.
2.65 in.
F
in a circle
LM 2LX
LX 7
Use the Arc Addition Post. to show C that mAEC mABC and thus mABC 180. Then use the Measure of an Inscribed Angle Thm. to show mB 90, so that B is a rt. and ABC is a rt. .
Proof: mE
1 LX LX 7
B
6.92 in.
12 GJ2
C
E
Prove: ABC is a right triangle.
GK 23.46
2 GJ GJ 6
X
39. Given: O with inscribed
JH GJ GJ FJ
R Q
x
2
R
P
2 Q 2
R
x
P Q
43. 27x 16 18x 32
42. B
14x 32 18x 32 64 4x 16 x C 44. a right triangle 45. GJ is the geometric mean of FJ and JH; Thm. 9.2 justi-
fies this answer.
210
Geometry Chapter 10 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued 58. 136 61.
3
2
59.
23. m1 2 120 46 37 1
3 1 or 0.5 60. or 0.8660 2 2
24. m1 2 235 125 55 1
25. m1 2 235 125 55 1
or 0.8660 62. 3 or 1.7320
27. 15a 2 255 105 1
26. 8a 10 130
Quiz 1 (p. 620) 1. x 90; Thm. 10.1 4. 133
5. 227
2. x 12; Thm. 10.3
6. 313
7. 180
8. 47
3. 47 9. 85.2
28.
8a 120
15a 75
a 15
a5
a 1 a 70 a 30 2 2
Lesson 10.4
a a 70 a 30
10.4 Guided Practice (p. 624)
a 40
1. Each angle is equal to one-half the measure of the inter-
cepted arc. 2. mSTU 2 105 210
29. 60
3. m1 2 55 65 60
30. 60
4. mDBR 2 190 60 65
31. 30
1
1
5. mRQU 6. 7.
1 2 270
90 90
35. 4000
1 mN 280 35 22.5 1 m1 288 88 88
mTCE tan1
TF2 1600.04
1 2 180
90 236 72
TF 40.0005
mTCF tan1
12. mABC 2126 252 1
0.701
15. 144 5x 17
0.7
127 5x x 25.4 16. 28x 29 10x 50
16x 58 10x 50 6x 108 x 18 18. m1 19. m2
1 2 130 95 112.5 1 2 25 75 50 1 2 32 122 77
m1 180 77 103
0.573 40.0005 4000
mFCB 0.128 0.573 mSB
13. m3 2 220 110
17. m1
0.128 8.944 4000
40002 TF2 4000.22
9. mGHJ 2140 280
284 168 14. mAB
34. 60
4000.012
TE 8.944
1
11.
33. 30
TE 80.0001
8. m1 2 220 110
mDE
TE2 2
10.4 Practice and Applications (pp. 624–627)
10. m2
32. 90 2
36. The measure of BAC is equal to one-half the measure
; Theorem 10.2. of AC
37. Diameter; 90; a tangent line is to the radius drawn to
the point of tangency.
↔
38. Draw the diameter containing PQ perpendicular to AB ;
By Thm. 10.8 mPBC By the proof of Case 1 . By the Arc Add. Post., mBP mPC mPBA 12mBP mCPB . By the Angle Add. Post., mABP mPBC mABC. By repeated application of the Substitution Prop., mABC 12mAPC . 1 2 mPC .
20. m1 2 105 51 27 1
21. m1 2 122 70 26 1
22. m1 2 142 52 45 1
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Geometry Chapter 10 Worked-out Solution Key
211
Chapter 10 continued r SV r QS and QR RS QS, so QR QS RS QT r SV r QS. 2r QR RS QT SV. By substitution 2r QR RS QU US. QU US QS. Therefore r 12QR RS QS.
39. The proof would be similar, using the Angle Addition
and Arc Addition Postulates, but you would be subtract instead of adding. ing mPBC and mPC 40. Reasons 1. Given 2. Through any 2 points there is exactly 1 line. 3. ACB; Exterior Angle Thm. 4. Def. of the measure of a minor arc 5. Def. of the measure of a minor arc 6. Substitution Property 7. Distributive Property 41.
Case 1: Draw BC. Use the Exterior Angle Thm. to show that m2 m1 mABC, so that m1 m2 mABC. Then use Thm. 10.12 to show that m2 12mBC and the Measure of an Inscribed Angle Thm. to show that 1 . Then mABC 2mAC mAC . m1 12mBC
B A 1 2 C
P 4 2 3
Q
R
Case 2: Draw PR. Use the Exterior Angle Thm. to show that m3 m2 m4, so that m2 m3 m4. Then use Thm. 10.12 to show that m3 12mPQR and m4
W 3
4 Z Y
42. E
Case 3: Draw XZ. Use the Exterior Angle Thm. to show that m4 m3 mWXZ, so that m3 m4 mWXZ. Then use the Measure of an Inscribed Angle Thm. to and show that m4 12mXY 1 mWXZ 2mWZ . Then, mWZ . m3 12mXY
43. C
44. R is a right angle; circle P is inscribed i nQRS; T, U,
and V are points of tangency. By Thm. 10.1, PT QR and PV RS, so RTP and RVP ar right angles. Using the Interior Angles of a Quadrilateral Thm. TPV is a right angle. By the Rectangle Corollary, quad. TPVR is a rectangle; TP PV RV TR, so by the Rhombus Corollary quad. TPVR is a rhombus. By the Square Corollary, quad TPVR is a square. According to Thm. 10.3, QT QU and SU SV. Using the Segment Addition Postulate, QT TR QR and SV VR SR. Since RV r TR, by Substitution Prop. SV r SR and QT r QR. The perimeter of QRS QT
212
1
r1 10.4 Mixed Review (p. 627) 46.
9 12 12 LM
Geometry Chapter 10 Worked-out Solution Key
LP 25 LP 16
9LM 144
LP2 400
LM 16
LP 20
LP 9 47. LP 4 36 LP2 LP 6
r 102 r2 222
48.
49. x 25
r2 20r 100 r2 484 20r 384 r 19.2 ft 50. 2x 5 x 3
51. 6x 12 10x 4
x8
1 2 mPR . Then, m2 1 2 mPQR mPR .
X
45. r 2 3 4 5
8 4x 2x
Lesson 10.5 Activity 10.5 Investigating Segment Lengths (p. 628) Investigate 1. yes; yes; yes 3. The products are equal. 4. The products remain equal. Conjecture 4. The product of the segments of intersecting chords are
equal. Investigate 5 The products are equal. Conjecture 6. The product of the length of the external segment of a
secant and the total length of the secant is equal to the product of the length of the other secant and the length of its external segment.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued 1. A
2. HF
B
x 4313
36 x2
HJ HG HK
6x 19. 422 66 x
C D
x2 576
823
x 24 64 x12 x
21.
External segment is AB
64 12x x2
3. 15; 18
4. 12; 15
x 15 10 18
12 x 15 40
15x 180
12x 600
x 12
x 50
x2 12x 64 0
x 16x 4 0 x 16
6. 2
6 16 8 x 8
42 2 2 x
96 8x 64
16 4 2x
32 8x
12 2x
4x
144 x10 x x2 10x 144 0
x 18x 8 0 x 18
x2
x x 3 22
x2 36
x2 3x 4 0
x 4x 1 0 x 4
x1
Solution: x 1 9. The segment from you to the center of the aviary is a
secant segment that shares an endpoint with the segment that is tangent to the aviary. Let x be the length of the internal secant segment (twice the radius of the aviary) and use Thm. 10.7. Since 4040 x 602, the radius is about 50 2 , or 25 ft. 10.5 Practice and Applications (pp. 632–634) 11. 45; 27
12. 16
9x 180
45x 1350
x2 144
x 20
x 30
x 12
35 155 x 420 225 15x 195 15x
14. 4x 16
7
9x 121 0
x
x2 3x 4
x6
121 xx 9
23.
8. x 3; 22
x2 4 9
x8
Solution: x 8
6x
7. 9
x4
Solution: x 4 22.
5. 16; x 8
10. 9; 15
20. x2 1248
88 36 6x x
13. 12
18. 72x 4078
17. 72 2x2
10.5 Guided Practice (p. 632)
9 ± 81 41121 21 9 565 7.38 2 xx 29 1550
24.
x 29x 750 0 2
x
29 ± 841 41750 21 29 3841 16.49 2
25. 400 88 x
400 64 8x 336 8x x 42
400 y30 y y2
30y 400 0
y 40y 10 0 y 40
y 10
Solution: y 10 26. 818 12x
x 12
y2 226 y 213
4x 112 x 28
13 x 15. 24x 12
17
24x 204 x 8.5
16. 15x 10x 1
15x 10x 10 5x 10 x2
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 10 Worked-out Solution Key
213
Chapter 10 continued 27. 314 6x
10.5 Mixed Review (p. 635)
x7
40. AB 25 4
8 yy 13 2
0 y2 13y 64
Midpoint:
13 ± 425 2
13 517 3.81 2
Midpoint: 2 , 1 11
45. AB 196 0
15.26
14
Midpoint: 4, 46. 1
92
1 2 2
Midpoint: 2, 2
b
47.
0b
29. 4.875 ft; the diameter through A bisects the chord into
two 4.5 ft segments. Use Thm. 10.15 to find the length of the part of the diameter containing A. Add this length to 3 and divide by 2 to get the radius. 30. BA2 12,50020,500
y
49.
y 13x 10 3
yx9 1 510 b
51.
52.
y 37x 81 7
A B
D
B
D
C 2 2
C x
2
A6, 8, B1, 4, C2, 2, D7, 3
32. EA is tangent to a circle; ED is a secant of the same cir-
cle. Draw AC and AD. EAD is a right triangle, ECA is similar to EAD by Thm. 9.1. Since corresponding sides of similar triangles are proportional,
A1, 8, B6, 4, C5, 2, D0, 3 53.
Using cross-products, EA2 EC ED. 34.
1 8000
5280 1
b
y
A
By the Cross Product Prop., EA EB EC ED.
9 376 b 81 7
y 5x 51
EA ED . EC EB
EA ED . EC EA
4 132 b 10 3 b
51 b
E E by the Reflexive Prop. of Congruence, so BCE ~ DAE by the AA Similarity Thm. Then, since lengths of corresponding sides of similar triangles are proportional,
3
3 y 2 17
9b
BC 16,007.8 mi 31. B and D intercept the same arc, so B D.
8 26 b 17 b
1 2x
48. 9 10 b
50.
BA 16,007.8 mi
y
A
12 1
A
8 in.
1 x 8000 mi
D
looking toward the horizon, it appears to be flat. 37. AE ACAD 2
38. AB AE; By applying the substitution property and
using the results from Exercises 36 and 37, the two are equal. 39. Tangents to intersecting circles are equal. No counter-
examples should be found.
Geometry Chapter 10 Worked-out Solution Key
B C
35. They probably thought the Earth was flat because, 36. AB ACAD
15
72, 32
44. AB 64 169
CN 18
2
43. AB 81 144
17.49
2
33. 11 8000x
Midpoint 3, 0
42. AB 81 225
13 ± 132 4164
28. 1512 10CN
214
10
1 Midpoint 2, 4
64 y2 13y
y
41. AB 36 64
29 5.39
D
B C
2 2
x
A6, 8, B1, 4, C2, 2, D7, 3 A8, 11, B3, 7, C4, 5, D9, 6
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued 54.
y
A
12. center:
13. center: 3, 2; radius: 2; x 32 y 22 4
B
14. center: 0, 1; radius: 2; x2 y 12 4
D C
A
15. center: 3, 3; radius: 1; x 32 y 32 1
x
2
B C
D
16. center: 0.5, 1.5; radius: 2.5;
6
x 0.52 y 1.52 6.25
17. center: 2, 2; radius: 4; x 22 y 22 16
A6, 8, B1, 4, C2, 2, D7, 3
1 1 1 1 A6, 22 , B1, 12 , C2, 32 , D7, 22
Quiz 2 (p. 635) 2. x
1. 202
1 2 110
168
139 3. 28 2 82 x 1
18. center: 0, 0; radius: 6; x2 y2 36 21. x 32 y 22 4 22. x 12 y 32 36
818 xx 18
3
5 9
x 12 y 22 25
25. r 4 0
26. x 52 y 32 16
x2
y2
2
x2 18x 144 0
x 32 y 22 4
x 24x 6 0 x 24
27.
28.
y
y
x6
Solution: x 6
2 2 2
100 xx 15
6.
x2
24. r 9 16
23. r 9
x 6.25
x 26
20. x 42 y2 16
19. x2 y2 1
4. 16x 100
56 82 x 5.
12, 34 ; radius: 12
x
2
1
15x 100 0
1
x 20x 5 0 x 20
29.
Solution: x 5
r 20 2
r2
492
30.
y
7. Solve 202r 20 492 (Thm. 10.17) or solve
y
2
(the Pythagorean Thm.); 50.025 ft. 1
x 2 2
Lesson 10.6 10.6 Guided Practice (p. 638) 1. x h y k 2
x
x2 y 42 1
x2 y2 25
x5
1
2
x 32 y2 9
r2
2. The center is 3, 4 and the radius is 9 3 units. Plot
31.
3. Center: 0, 0; radius: 2;
x 32 y 42 16 32.
y
y 1
points 3 units above, below, to the right, and to the left of the center. x2
x
2
2
y 4 2
2 2
4. Center: 2, 0; radius: 4; x 22 y2 16
1 x
x 1
5. Center: 2, 2; radius: 2; x 22 y 22 4 6. r 1 9
x 52 y 12 49
10 x2 y2 10
33. exterior
34. interior
37. interior
38. on the circle
x 12 2 y 12 2 14
35. on the circle 39. exterior
36. exterior 40. interior
10.6 Guided Practice and Applications (pp. 638–640) 7. center: 4, 3; radius: 4
8. center: 5, 1; radius: 5
9. center:0, 0; radius: 2
10. center: 2, 3; radius: 6
11. center: 5, 3; radius: 1 Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 10 Worked-out Solution Key
215
Chapter 10 continued 41. A: x2 y2 9, B: x 52 y 32 6.25,
2. m moves farther away from k; circle P gets bigger; yes
C: x 22 y 52 4
3. parabola Conjecture 4. The set of points in a plane that are equidistant from a line and a point in the plane is a parabola.
y
C B 1
10.7 Guided Practice (p. 645) 1. exterior
A
1 1
A
42. A; B; B and C; none; C 44. x 32 y2 49
45. Answers may vary. 3. B
image: x 22 y 42 16
y 2 2 2
2
4. A
2 2
49. D
b2
53.
A
B 2
The two points on the intersection of the bisector of AB and B with radius 5
x
8. The intersection of C with radius 3 and D with radius
50. C
5 25 b 22
5, 2, 1, or 0 points depending on the distance between C and D.
25 25 b 2
10.7 Practice and Applications (pp. 645–647)
2
0 b 22
b0
6. C
y
2
51. Yes; find the point r units (r is the radius of B) from the point of tangency and that is the center of B. 52. 3 32 b 02
5. D
7.
x
48. x p2 y q2 q2
99
B
43. x 32 y2 1
46. Any way it is rolled, the width is the same. 47.
The locus of points that are equidistant from A and B form the perpendicular bisector of AB.
2.
x
1
9.
10.
m
1 in. 1 in.
b 2
P
k
1 in.
10.6 Mixed Review (p. 640) n
54. kite, rhombus, rectangle, parallelogram
The points on P with radius 1 inch
55. parallelogram, rectangle, rhombus, kite, isosceles
trapezoid
11.
The 2 lines parallel to k and 1 in. away 12.
56. kite, rhombus, rectangle, parallelogram 57. 6, 7; 9.2
m
1 in.
1 in.
C
58. 8, 2; 8.2
j
1 in.
59. 3, 11; 11.4
n
60. 2, 13; 13.2 61. No; P is not equidistant from the sides of A. 62. Yes; P is equidistant from the sides of A.
The interior of C with radius 1 inch
The points on and beyond 2 lines parallel to j and 1 in. from j.
Lesson 10.7 Activity 10.7 (p. 641)
Investigate 1. Yes; it is AB units from each because that is the radius of both circles and y is on both circles and k goes through the center of the first circle; Yes; same reason as y.
216
Geometry Chapter 10 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued 13.
14.
25.
y
j
k
C
A
1
k
1 1
All points on line m
The points on the bisector of A
15.
16.
x
1
D
2 points, 2, 2 and 4, 4, the intersections of y x with y 2 and y 4
r
26. C
s
R R
The points on C with radius 12 of the original circle
17.
The points on the bisector of the vertical angles and the points on the line to the bisector through the vertex
k
d1
d<1
1
R R
18.
M
k
R
k
A
30 30
k
A
B
B q
N
→ → Any point on AM or AN 19. x 3 21. M1, 1
K5, 5
d3
d>3
Let d be the distance from R to k; the locus of points is 4 points if d < 1, 3 points if d 1, 2 points if 1 < d < 3, 1 point if d 3, and 0 points if d > 3.
Any point on line q 20. y 3
27.
y
y
d4
d<4
22. x 5 y 5 9 2
2
point3, 3 m
k
R
R 2
51 1 51
P
2 2
2
x
Q
P
2 2
2
Q
x
3 13 b 6b
y
y x 6
d>4
23. y 4, y 2 24.
Let d be the distance from R to the bisector of PQ; the locus of points is 2 points if d < 4, 1 point if d 4 and 0 points if d > 4.
R 2
y
P A
2 2
2
x
Q
1 1 1
x
1
B
28.
29. 0, 6
y
12
A
points on the x-axis such that 4 < x < 4
C
B 12
x
30. d 9 676
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 10 Worked-out Solution Key
217
Chapter 10 continued 26.2 mi
3.
No; your friend is outside a 14 mi radius from the epicenter.
2 2
locus is 2 points. If d 4, the locus is 1 point. If d > 4, the locus is 0 points. 33. D
x
2
2 2
1 2 AB
35. 1 ft 0.5 cm
34. C
y
2
31. Let d be the distance from P to K. If 0 < d < 4, the
32. the with center at the midpoint of AB and radius
4.
y
x 12 y2 36
2 x
x 42 y 72 25
5. r 25 49
74
6 ft
x 22 y 22 74
2 ft
The points that are in both the exterior of the with center P and radius of 6 units and the interior of the circle with center P and radius 9 units
6.
9 ft 5 ft
9 6 P
10.7 Mixed Review (p. 647) 36. 22
37. 69
38. 70
39. 12x 210
x 17.5 40. 930 1010 x
7. 4 cm
41. x2 1636
270 100 10x
A
x 24
x 17 42.
43.
y
→ A set of points formed by 2 rays on opposite sides of AB, each → to AB and 4 cm from it, and a semicircle with center A and radius 4 cm
y
3 3 3
2
x
3
2 2
B
4 cm
8.
2 x
100 yd 10 yd
44.
45.
y
y
The points that are on the field and on or outside the circle whose center is the center of the field and whose radius is 10 yd
4 8
50 yd
x 62 y 42 9
x2 y2 81
8
4
x
1
x2 y 72 100
Math & History (p. 648)
x
1
x 42 y 52 1
1. June 21: 75 min
Dec. 21: 46 min
Quiz 3 (p. 648) 1.
2.
y
y 2
4 4 4
4
x
2 2
2
x
x2 y2 100 x 32 y 32 49
218
Geometry Chapter 10 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued 28. x 42 y 12 16
Chapter 10 Review (pp. 650–652)
y 4
10.1 Tangents to Circles 1. BN
2. C, D, or R 3. BF or BN 4. P 5. QE ↔ BC ↔ 7. BF 8. R ↔ ↔ 9. yes; BC is tangent to P at point B so BC PB.
6.
2 x
10. According to Thm. 10.3, since the segments from point S are tangent to Q, they are congruent.
29. x 62 y2 10
y
1 x 1 1
10.2 Arcs and Chords 11. 62
12. 118
13. 239
14. 85
15. 275
16. 324 10.3 Inscribed Angles
10.7 Locus
17. True; the sides of the triangles opposite the inscribed
30.
The perpendicular bisector of RS contains the points equidistant from R and S.
S
angles are diameters, so the inscribed angles are right angles. R
18. False; as AC is a diameter, it is perpendicular to DB and
AED is a right angle. ACD cannot be a right angle because in ADC, D is a right angle.
T
19. True; ABCD is inscribed in a circle, so opposite angles 31.
are supplementary.
m
4 in.
10.4 Other Angle Relationships in a Circle
l
Parallel lines 4 in. from and every point between and each line
4 in.
20. 112
21. 55
22. 64
23. 94
n
10.5 Segment Lengths in Circles
32.
25. 10x 10 1237
24. 16x 80
x5
10x 100 444 A
x 34.4
B
Two points where the circles with center B and radius 4 cm and center A and radius 3 cm intersect.
400 x30 x
26.
x2
30x 400 0
x 40x 10 0 x 40, x 10 Solution: x 10 Chapter 10 Test (p. 653)
10.6 Equations of Circle
1. Thm. 10.3, converse of Base Angles Thm.
27. x 22 y 52 81
y
2 2 2
Copyright © McDougal Littell Inc. All rights reserved.
2
x
Geometry Chapter 10 Worked-out Solution Key
219
Chapter 10 continued 18. x 302 y 302 900
2. 42 JK2 82
JK 6.9
19.
mP 4 5
mK 4 8
mP2 9
mK2 48
mP 3
mK 43
2
2
2
2
2
3.5 ft
2
A
PK 43 3 3. KJH and KMH are rt s. HJ and HM are radii of H so they are . HK HK by the Reflexive Prop.
of Congruence. So KJH KMH by HL Thm. KHM KHJ because they are corresponding angles. By the Converse of Thm. 10.9 LM JL .
4. HMK is a 30 60 90, short leg: 4; long leg:43; hypotenuse: 8.
B
Let A and B be the ends of the cable. The locus consists of the points on or inside a region bounded by two semicircles with centers A and B and radius 3.5 ft and two segments of opp. sides of AB, both to AB and 3.5 ft from AB. Chapter 10 Standardized Test (pp. 654–655) 1. D
2. C
3. E
5. According to Thm. 10.5, since AD FB, FH BH and
FA BA .
9. D
A 7x2 5 180
mF 135
2
7x 175
mG 140
x5
mH 45
EC 7. FB EC and FB 8. m1 72.5
9. m1 90
m2 145
m2 90
12.
m3 45 11. m1 29
m2 75
11. mE 40
10. x2 15 6x2 10 180
. FE BC 6. FE BC so by a converse of Thm. 10.4
50 mFG 40 mEF 240 mEH
13. ABC is inscribed and
AB is the hypotenuse. The hypotenuse of an inscribed right triangle
m2 66
is a diameter.
m3 37
14. P:
a. supplementary
2
3 2,
15.
x 32 2 y 22 254
d 9 16
b. congruent
D
7. A
x 12 y 52 34
PHJ 156.9 JN
C
6. C
Mdpt 1, 5
MHP tan134 36.9
B
5. B
34
MHJ 120 JM
12.
4. B
8. d 25 9
MHK 60 JHK
10. m1 120
3.5 ft
5 r 52
E
32, 2
A
16. the point
mCDE 2mCAE
17. a. 10x 30
x3
mCBE mCAE 14. 312 x9
13. 3x 18
x6 16.
15.
x4
312
d. 90
x6
17.
y
x2
18.
2
BC
x 24 e. 45
f. 45
; Their corresponding chords are congruent so CD
19. 120 2 180 AG 1
x
240 180 AG 1 1
c. 180
by Thm. 10.4 the arcs are congruent.
y
2 2
b. x2 1832
x
60 AG
180 x 20. mBH x mHD
16 2180 x x 1
32 180 2x 2x 148
x 42 y 62 64
220
x 74
Two points where the line y x and circle with center 4, 0 and radius 4 units intersect.
Geometry Chapter 10 Worked-out Solution Key
106, mHD 74 mBH
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 10 continued 21.
22. 2 in. A A
2 in.
B
B
all points on the perpen-
all points on the lines
dicular bisector of AB.
parallel to and 2 in. from ↔ AB and all points less ↔ than 2 in. from AB. All points on the bisector of AB joining 2 lines parallel to AB and 2 in. from AB.
23. 2 in.
A
B 2 in.
25. m1 80; m2 40;
24. 100
m3 25 26. Farther away; this will increase the measure of the
smaller intercepted arc and thereby decrease the mB. 28. 55 d 100
27.
25 5d 100 d 15 ft r 7.5 ft Use Thm. 10.17. 10 ft
5 ft
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 10 Worked-out Solution Key
221
Chapter 10 continued Algebra Review (pp. 656–657) 1. A lw
3
1 A bh 2
A r
6.
8.
3 VS
9.
V lwh
10.
V h lw
13. 5 x
17. x 2 9x
44.
19. 5x 7 13
20. 2x 16 10
5x 20
2x 26
21. 2x 14x 48
x 13 22.
16x 48
46.
x3 6x 1
7d 1 3d 4
2y 12 2y 6 24 2y 212 y
43.
36s2 4s 4s9s 1 4s2 12s 4ss 3
y6 12 y
5h 1 h1
45.
t2
9s 1 s3
t2 1 t 1t 1 2t 1 t 1t 1
1
1 2x
16. 3x 6
18. 2 x 3x
x4
5x2 15x 5xx 3 30x2 5x 5xx 1
a2 a8
42.
40.
14d2 2d 2d7d 1 6d2 8d 2d3d 4
b c2 a2 15. 2x 14
27w 9ww2 3 9w 3ww2 3
12. 92 b2 c2
14. x2 2
3w3
3
P 2l 2w P 2w l 2 P wl 2
S s 6
9w3
5a 10 5a 2 39. 5a 40 5a 8
V h r2
S 6s2
11.
38.
41.
V r2h
5x2 x x5x 1 5x 1 5x 1 x
C 2r C r 2
A r
V S3
7.
1 A hb1 b2 2 2A b2 b1 h
2
4V r 3
4.
2A h b 5.
3 V r3 4
2.
A w l 3.
37.
t1 t1
m2 4m 4 m 22 m2 4 m 2m 2
m2 m2
3x 5
x 6x 30
x3
5x 30 x 6
23. x 0.30120
24. x 0.15340
x 36
x 51 miles
25. x 0.71200
26.
x 142 27. 34 x136
$12.50 x 28. 11 x50
x 25% 29. 200 x50
x 22% 30. 8 x52
x 400% 31. 3 0.30x
x 0.5025
x 15.38% 32. 16 0.64x
x 10
x 25 meters
33. 25.950.08 $2.08
34. 3 x18
x 16.67% 35.
5x 1 10x2 2x
222
36.
16a3 2a2 8a
Geometry Chapter 10 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
