Document 1807083

Chapter 10
Prerequisite Skills (p. 648)
4. 2 common tangents
5. 1 common tangent
1. Two similar triangles have congruent corresponding
angles and proportional corresponding sides.
2. Two angles whose sides form two pairs of opposite rays
are called vertical angles.
3. The interior of an angle is all of the points between the
sides of the angle.
4.
c 2 ? a 2 1 b2
5.
0.92 ? 0.62 1 0.82
172 ? 112 1 122
6. 0 common tangents
0.81 ? 0.36 1 0.64
289 ? 121 1 144
7.
0.81 < 1
289 > 265
The triangle is acute.
6.
c 2 ? a 2 1 b2
c 2 ? a 2 1 b2
2
2.5
25 0 9 1 16
The triangle is obtuse.
25 5 25
7. 6x 2 8 5 5x
? 1.5 1 2
2
CE 2 0 CD 2 1 DE 2
(3 1 2)2 0 32 1 42
}
Yes, DE is tangent to (C.
x58
2
6.25 ? 2.25 1 4
QT 2 5 QS 2 1 ST 2
8.
(r 1 18)2 5 r 2 1 242
6.25 5 6.25
r 1 36r 1 324 5 r 2 1 576
2
The triangle is right.
36r 5 252
8. (8x 2 2) 1 (2x 1 2) 5 180
r57
10x 5 180
x 5 18
9. x 5 9
2
x 5 63
9. 7x 5 5x 1 40
2x 5 40
x 5 20
Lesson 10.1
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Investigating Geometry Activity 10.1 (p. 650)
1. Answers will vary.
10.1 Exercises (pp. 655–658)
Skill Practice
}
then AB is a diameter.
2. When referring to a segment, “a radius” and “a diameter”
is used. When referring to a length, “the radius” and “the
diameter” is used.
2. Tangent segments from a common external point are
congruent.
3. MQ 5 MP 5 5.5
MN 5 LM 5 7
LQ 5 LM 1 MQ 5 7 1 5.5 5 12.5
PN 5 PM 1 MN 5 5.5 1 7 5 12.5
4. Because AC 5 EC and BC 5 DC, it follows that
} }
AB 5 ED which makes AB > ED.
10.1 Guided Practice (pp. 651–654)
3. G; B is a point of tangency.
@##$ is a common tangent.
4. H; BH
}
5. C; AB is a chord.
7. F; @##$
AE is a tangent.
}
CB is a radius because C is the center and B is a point
on the circle.
@##$ and a tangent segment is }
2. A tangent is DE
DB.
6. E; @##$
AB is a secant.
8. A; G is the center.
}
}
9. B; CD is a radius.
10. D; BD is a diameter.
}
11. The error is that AB is not a secant, but rather a chord.
}
The length of chord AB is 6 units.
12. The radius of (C is 9 units.
The diameter of (C is 18 units.
}
1. AG is a chord because its endpoints are on the circle.
}
1. The points A and B are on (C. If C is a point on AB,
13. The radius of (D is 6 units.
The diameter of (D is 12 units.
14.
y
3. The radius of (C is 3 units.
The diameter of (C is 6 units.
C
D
The radius of (D is 2 units.
The diameter of (D is 4 units.
3
23
x
Geometry
Worked-Out Solution Key
305
Chapter 10,
continued
15. 4 common tangents
28. The common tangents are internal because they intersect
the segment that joins the centers of the circles.
} }
}
parallel segment XP as shown.
29. C; Using Theorem 10.1, RS > QR. Draw a congruent,
R
5
S
x
3
Q
P
2
Using Pythagorean Theorem,
(XQ)2 1 (XP)}
5 (QP)2,
}
2
2
2
2 1 (XP) 5 8 . XP 5 2Ï15 , so RS 5 2Ï 15 .
} }
} }
30. Using Theorem 10.2, PA > PB and PB > PC. Using
the Transitive Property of Segment Congruence,
} } }
PA > PB > PC.
16. 0 common tangents
17. 1 common tangent
31. Sample Answer: Two lines tangent to the same circle
will not intersect when the lines are tangent at opposite
endpoints of the same diameter. Using Theorem 10.1, the
two lines are perpendicular to the same line, so they are
parallel.
18. Use the Converse of the Pythagorean Theorem. Because
19. Use the Converse of the Pythagorean Theorem. Because
}
}
}
9 2 1 152 Þ 182, AB is not perpendicular to BC and AB is
not tangent to (C.
20. The diameter of (C is 20. Using the Converse of the
Pythagorean Theorem, 202 1 482 5 522 and the triangle
} }
is a right triangle. This implies that CB > BA. Using
}
Theorem 10.1, AB is tangent to (C.
(r 1 16) 5 r 1 24
2
21.
2
2
r 2 1 32r 1 256 5 r 2 1 576
32r 5 320
r 5 10
(r 1 6)2 5 r 2 1 92
22.
r 1 12r 1 36 5 r 2 1 81
2
12r 5 45
32. C is in the interior of ŽABD and AC 5 DC. By
###$ bisects ŽABD.
Theorem 5.6, BC
33. For any point outside of a circle, there is never only
one or more than two tangents to the circle that passes
through the point. There will always be two tangents.
34.
Statements
1. AB 5 AC 5 12, BC 5 8
1. Given
2. radius
r 5 PD 5 PE 5 PF
2. Def. of radius
3. AB 5 AD 1 BD,
AC 5 AF 1 CF,
BC 5 BE 1 CE
3. Postulate 2,
Segment Addition
4. AB 5 AC
4. Given
5. AD 1 BD 5 AF 1 CF
5. Substitution for AB
and AC
6. BE 5 BD, CE 5 CF,
AD 5 AF
6. Theorem 10.2
7. AD 1 BE 5 AD 1 CE
7. Substitution for BD, AF,
and CF
8. BE 5 CE
8. Subtract AD
9. E and P are on the angle
bisector of ŽA.
9. Theorem 5.6
r 5 3.75
(r 1 7)2 5 r 2 1 142
23.
r 1 14r 1 49 5 r 1 196
2
2
14r 5 147
r 5 10.5
24. 3x 1 10 5 7x 2 6
25. 2x 2 1 5 5 13
16 5 4x
2x 2 5 8
45x
x 54
2
x 5 62
26. 3x 2 1 4x 2 4 5 4x 2 1
3x 2 5 3
x2 5 1
x 5 61
When x 5 21, 4x 2 1 5 25. Because length cannot be
negative, x Þ 21 in the figure. So, the answer is x 5 1.
27. The common tangents are external because they do
not intersect the segment that joins the centers of the
two circles.
306
Geometry
Worked-Out Solution Key
Reasons
10. AE 5 AP 1 PE
10. Postulate 2,
Segment Addition
11. BC 5 8
11. Given
12. BE 1 CE 5 8
12. Substitution for BC
13. CE 1 CE 5 8
13. Substitution for BE
14. CE 5 4
} }
15. PE > CE
14. Divide by 2.
15. Theorem 10.1
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
} }
32 1 42 5 52, nABC is a right triangle and AB > AC.
}
By Theorem 10.1, AB is tangent to (C.
Chapter 10,
continued
Statements
Reasons
16. AC 5 AF 1 CF
16. Statement 3
17. 12 5 AF 1 4
17. Substitution for AC
and CF
18. AF 5 8
18. Subtract 4
19. (CE)2 1 (AE)2 5 (AC)2
19. Pythagorean Theorem
20. 42 1 (AE)2 5 122
20. Substitution for CE
and AC
}
21. AE 5 8Ï 2
} }
22. PF > AF
21. Solve for AE.
23. (PF)2 1 (AF)2 5 (AP)2
23. Pythagorean Theorem
24. (PF)2 1 (AF)2
5 (AE 2 PE)2
24. Substitution for AP
22. Theorem 10.1
}
25. r 2 1 82 5 (8Ï 2 2 r)2
25. Substitution for PF, AF,
AE, and PE
}
26. r 5 2Ï 2
26. Solve for radius, r.
41.
Statements
}
1. SR and ST are tangent
to (P.
} } } }
2. SR > RP, ST > TP
}
3. RP 5 TP
} }
4. RP > TP
} }
5. PS > PS
Reasons
1. Given
2. Tangent and radius are
perpendicular.
3. Def. of circle
4. Def. of congruence
5. Reflexive Property
6. nPRS > nPTS
} }
7. SR > ST
6. HL Congruence Theorem
7. Corresponding parts of
congruent triangles are
congruent.
42. a. The slope of the line perpendicular to line * through C
3
4
is 2 }4, so the slope of line * is }3.
b. y 5 mx 1 b
4
3 5 }3 (24) 1 b
Problem Solving
16
35. The wheel has radial spokes because each spoke has
endpoints that are the center and a point on the circle.
36. The wheel has tangential spokes because each spoke
intersects the circle in exactly one point.
37. BE 2 5 EC 2 1 CB 2
25
3
}5b
4
(11,000 1 3959) 5 3959 1 CB
2
25 5 r 2
2
14,9592 5 39592 1 CB 2
208,098,000 5 CB 2
14,426 ø CB
BA 5 BC ø 14,426 miles
38. mŽARC 5 mŽBSC 5 908, so ŽARC > ŽBSC. Also,
ŽRCA > ŽSCB because vertical angles are congruent.
Therefore, nARC , nBSC by the AA Similarity
AC
25
The equation for * is y 5 }3 x 1 }
.
3
c. 32 1 (24)2 5 r 2
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3 5 2}
1b
3
RC
Postulate. So, }
5}
because the ratio of corresponding
BC
SC
sides is the same.
55r
25
d. The y-intercept is } and the radius is 5. So, the
3
25
10
255}
.
distance from the y-intercept to (C is }
3
3
Mixed Review for TAKS
43. D; After nLMN is dilated by a scale factor of 2 with the
origin as the center of dilation, the coordinates of vertex
M(6, 2) are (6 + 2, 2 + 2) 5 (12, 4).
44. H; 12x 1 6y 5 18
39. a. Because R is exterior to (Q and P is on (Q,
QR > QP.
}
b. Because QR is perpendicular to line m it must be the
shortest distance from Q to line m. Thus, QR < QP.
}
c. It was assumed QP was not perpendicular to line m but
}
QR was perpendicular to line m. Since R is outside of
(Q you know that QR > QP, but part (b) tells you that
QR < QP which is a contradiction. Therefore, line m is
}
perpendicular to QP.
40. Assume line m is not tangent to (Q. So, there is another
point X on line m that is also on (Q. X is on (Q, so
QX 5 QP. But the perpendicular segment from Q to line
m is the shortest such segment, so QX > QP.
QX cannot be both equal to and greater than QP. The
assumption that point X exists must be false. Therefore,
line m is tangent to (Q.
6y 5 212x 1 18
y 5 22x 1 3
y 5 3 2 2x
The expression 3 2 2x can be correctly substituted for y.
y2 2 y1
25 2 1
26
2
45. A; Slope: m 5 } 5 } 5 } 5 } 5 0.4
5
x2 2 x1
210 2 5
215
y 5 mx 1 b
1 5 0.4(5) 1 b
1521b
21 5 b
So, y 5 0.4x 2 1 describes the line that contains the
points (5, 1) and (210, 25).
Geometry
Worked-Out Solution Key
307
continued
Lesson 10.2
C C
13. LP and MN have the same measure but they are not
congruent because they are arcs of circles that are not
congruent.
10.2 Guided Practice (pp. 660–661)
C
CQ 5 1208
mT
C
2. Q
RT is a major arc.
C
QR 1 mC
RT 5 608 1 1808 5 2408
mQRT 5 mC
C
3. TQR is a semicircle.
C
QR 5 1808
mT
CS is a minor arc.
4. Q
QS 5 mC
QR 1 mC
RS 5 608 1 1008 5 1608
mC
C
5. TS is a minor arc.
CS 5 808
mT
C
6. R
ST is a semicircle.
RST 5 1808
mC
C
7. AB > C
CD because they are in congruent circles and
CB 5 mC
mA
CD .
CN and PCQ have the same measure but they are not
8. M
1. TQ is a minor arc.
congruent because they are arcs of circles that are not
congruent.
C C
C C
14. VW > XY because they are in congruent circles and
mVW 5 mXY .
15. The statement is incorrect because you can tell that the
circles are congruent. The circles have the same radius,
}
CD.
16.
C B
P
A D
20°
C
C
CC 5 1808 2 mC
AD 5 1808 2 208 5 1608
mA
mACD 5 3608 2 mAD 5 3608 2 208 5 3408
17. A; nAPB is a right triangle.
AP 2 1 PB 2 5 AB 2
32 1 32 5 AB 2
18 5 AB 2
}
3Ï2 5 AB
18.
E
100°
10.2 Exercises (pp. 661–663)
Skill Practice
C C
1. If ŽACB and ŽDCE are congruent central angles of
(C, then AB and DE are congruent.
2. You need to know that the radii of two circles are the
same in order to show that the two circles are congruent.
C
CC 5 708
mB
CC is a minor arc.
4. D
CC 5 1808 2 mECD 2 mC
BC
mD
5 1808 2 458 2 708 5 658
CB is a minor arc.
5. D
CB 5 mDCC 1 mCCB 5 658 1 708 5 1358
mD
CE is a minor arc.
6. A
CE 5 mBCC 5 708
mA
CD is a minor arc.
7. A
CD 5 mACE 1 mECD 5 708 1 458 5 1158
mA
C
8. A
BC is a semicircle.
C
BC 5 1808
mA
C
9. A
CD is a major arc.
C
C
CD 5 1808 1 658 5 2458
CD 5 mA
BC 1 mC
mA
C
10. E
AC is a major arc.
C
CC 5 1808 1 708 5 2508
AC 5 mC
EB 1 mB
mE
}
C
11. C; Q
RS is a semicircle because you know that QS is
a diameter.
CD 5 1808 2 708 2 408 5 708
12. mC
C
AB > C
CD because they are in the same circle and
CB 5 mC
mA
CD .
3. BC is a minor arc.
308
Geometry
Worked-Out Solution Key
C
G
F
120°
C
C
mGE 5 3608 2 1008 2 1208 5 1408
C
C
If mGH 5 1508, point H must be 108 beyond point E,
placing it on EF . Or, point H is 308 beyond point F,
placing it on EF .
19. Sample answer:
A 60° 25°
B
C
R
70°
E D
20°
C C C C C
mAE 5 mAB 1 mBC 1 mCD 1 mDE
5 608 1 258 1 708 1 208 5 1758
20°
15°
D E A 35°
C
B 25°
R
C
mAE 5 158
C
Two possible values for mAE are 1758 and 158.
20. nPAQ is a right triangle with mŽAPQ 5 308 and
mŽAQP 5 608. nPBQ is a right triangle with
mŽBPQ 5 308 and mŽBQP 5 608. So,
mŽAQB 5 mŽAQP 1 mŽBQP 5 608 1 608 5 1208.
Therefore, mAUB 5 1208.
C
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 10,
Chapter 10,
continued
3
21. a. tan X 5 }
4
mŽX ø 36.98
C
mBD ø 36.98
4
b. tan Y 5 }
3
mŽY ø 53.18
C
CB 5 mC
c. mA
AD 2 mC
BD
mAD ø 53.18
5 53.18 2 36.98 5 16.28
Problem Solving
22. The measure of the arc is 608.
23. 3608 4 20 5 188
The measure of each arc in the outermost circle of the
dartboard is 188.
24. a. 3608 2 908 5 2708
The measure of the arc surveyed by the camera is 2708.
b. 2708 4 108 per minute 5 27 minutes
It takes the camera 27 minutes to survey the area once.
c. The camera must go 2708 2 858 5 1858
counterclockwise and another 1858 clockwise to return
to the same position. So, 1858 1 1858 5 3708.
3708 4 108 per minute 5 37 minutes. It will take the
camera 37 minutes.
d. In 15 minutes, the camera can go 108(15) 5 1508.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The camera will go 508 counterclockwise and then
1008 clockwise. So, the camera will be 1008 from wall
A after 15 minutes.
3608 5 68 per minute
25. a. }
60 min
After 20 minutes, the minute hand will move
20(68) 5 1208.
3608
} 5 308 per hour
12 h
The hour hand moves 308 per hour.
308
} (20 min) 5 108 in 20 minutes
60 min
At 1:20, the hour hand is 308 1 108 5 408 from the
12. The minute hand is 1208 from 12. The minor arc
between the hour and minute hand is
1208 2 408 5 808.
b. You want the arc between the hour and minute hand
to be 1808. Use guess, check, and revise. At 1:40, the
minute hand is 40(68) 5 2408 from 12. The hour hand
308
60 min
moves } (40 min) 5 208 in 40 minutes. The hour
hand is 308 1 208 5 508 from 12. The arc is
2408 2 508 5 1908. At 1:38, the minute hand is
38(68) 5 2288 from 12. The hour hand moves
308
60 min
} (38 min) 5 198 in 38 minutes. The hour hand is
308 1 198 5 498 from 12. The arc is 2288 2 498 5 1798.
Mixed Review for TAKS
26. C; The volume of the cone depends on both the radius of
the base r and the height h of the cone.
Lesson 10.3
10.3 Guided Practice (pp. 664–666)
C
2. 2mC
AB 1 mC
AC 5 3608
CB 1 1508 5 3608
2mA
AB 5 2108
2mC
CB 5 1058
mA
1. mBC 5 1108
3. 9x8 5 (80 2 x)8
10x 5 80
x58
C
mCD 5 9x8 5 9(88) 5 728
4. 9x8 5 (80 2 x)8
10x 5 80
x58
C
5. mC
CE 5 mC
CD 1 mC
DE 5 728 1 728 5 1448
mDE 5 (80 2 x)8 5 (80 2 8)8 5 728
6. QR 5 ST 5 32
1
1
7. QU 5 } QR 5 }(32) 5 16
2
2
8. QC 2 5 QU 2 1 UC 2
QC 2 5 162 1 122
QC 2 5 400
QC 5 20
10.3 Exercises (pp. 667–670)
Skill Practice
C C C
1. Sample answer: Point Y bisects XZ if XY > YZ .
2. If two chords of a circle are perpendicular and congruent,
one of them does not have to be a diameter. A square
inscribed in a circle is an example of two chords that are
congruent and perpendicular.
C C
CB 1 mC
4. 2mA
AD 5 3608
CB 1 1288 5 3608
2mA
CB 5 2328
2mA
CB 5 1168
mA
3. mAB 5 mED 5 758
5. EG 5 EJ 5 8
}
}
6. By Theorem 10.5, BD bisects AC.
4x 5 3x 1 7
x57
}
}
7. By Theorem 10.5, LN bisects PM.
5x 2 6 5 2x 1 9
3x 5 15
x55
Geometry
Worked-Out Solution Key
309
}
continued
}
8. Because SQ and TQ are radii, they have the same measure.
6x 1 9 5 8x 2 13
22 5 2x
11 5 x
}
}
9. AB and CD are equidistant from the center, so by
Theorem 10.6, they are congruent.
5x 2 7 5 18
5x 5 25
x55
}
}
10. AD and BC are equidistant from the center, so by
Theorem 10.6, they are congruent. By Theorem 10.5,
AD 5 2(3x 1 2) 5 6x 1 4.
6x 1 4 5 22
22. a. The converse of Theorem 10.5 is: If a diameter of a
circle bisects a chord and its arc, then the diameter is
perpendicular to the chord. The converse is different
from Theorem 10.4 in that in the hypothesis, it is
known that the first chord is a diameter.
b.
P
C
S
R
Statements
}
1. PT > TR, PS > SR
} }
2. CT > CT
} }
3. PC > CR
x53
}
11. EF and HG are congruent, so by Theorem 10.6, they are
equidistant from Q.
}
4x 1 1 5 x 1 8
Reasons
C C
}
6x 5 18
Q
T
1. Given
2. Reflexive Property
3. All radii in a circle are
congruent.
4. nPCT > nRCT
4. SSS Congruence
Postulate
5. ŽCPT > ŽCTR
5. Congruent parts of
congruent ns are
congruent.
6. mŽCTP 1 mŽCTR
5 1808
6. Linear Pair Postulate
7. mŽCTP 5 908
} }
8. CT > PR
7. Substitution Property
3x 5 7
7
x 5 }3
}
}
12. Because AB is a perpendicular bisector of CD, it is a
diameter by Theorem 10.4.
}
}
}
bisects FG and FG by Theorem 10.5.
}
}
14. Because NP and LM are equidistant from the center,
} }
NP > LM by Theorem 10.6.
C
13. Because JH is a diameter and it is perpendicular to FG, it
}
}
15. D; Choice A is true because they are radii of the circle.
Theorem 10.6, CD 5 EF.
} }
17. You do not know that AC > BD, therefore you cannot
show that BC > CD .
}
}
18. AB is a perpendicular bisector of CD, so by Theorem 10.4,
it is a diameter.
C
C
C C
C C
CB 1 mC
BC 1 mC
CA 5 3608
mA
23. From the diagram, mAB 5 x8. By Theorem 10.3,
mBC 5 mCA . Let y8 5 mBC 5 mCA . Then
x8 1 y8 1 y8 5 3608
x 5 360 2 2y
x is an even number because (360 2 2y) is even.
19. The two triangles are congruent by the SAS Congruence
C
21. Using the facts that nAPB is equilateral which makes it
equiangular and that mAC 5 308, you can conclude that
mŽAPD 5 mŽBPD 5 308. You now know that
} }
mBC 5 308, which makes AC > BC. nAPD > nBPD
} }
by the SAS Congruence Postulate because BP > AP and
} }
PD > PD. Because corresponding parts of congruent
} }
triangles are congruent, AD > BD. Along with
} }
DC > DC, you have nADC > nBDC by the SSS
Congruence Postulate.
C
310
Geometry
Worked-Out Solution Key
}
and PR , SQ is a perpendicular bisector of PR.
}
Therefore, any point on SQ is equidistant from the
endpoints of the segment, so QP 5 QR.
C C
}
}
Postulate. So, AB is the perpendicular bisector of CD.
}
By Theorem 10.4, AB is a diameter.
}
20. By the Pythagorean Theorem, ED 5 4. CE Þ ED, so AB
}
}
is not a perpendicular bisector of CD. Therefore, AB is
not a diameter.
}
c. Since SQ is a diameter of (C and it bisects both PR
By Theorem 10.5 choice C is true. Choice B is true
because the two triangles are congruent by the SAS
Congruence Postulate.
16. From the diagram, CD 5 12 and EF 5 14. However, by
8. Def of >
24.
P
T
R
A
B
In nPAT, PA 5 10 and AT 5 8.
mŽ APT 5 sin211 }
5 53.138.
10 2
8
In nPBR, PB 5 10 and BR 5 6.
mŽBPR 5 sin211 }
5 36.868.
10 2
6
C C C
mAB 5 mAR 2 mBR 5 53.138 2 36.868 5 16.268
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 10,
Chapter 10,
continued
Problem Solving
C C
30. a.
}
}
r 2 5 25,600 1 r 2 2 160r 1 6400
25. In order for AB > BC , AB should be congruent to BC.
160r 5 32,000
26. To find the center of the cross section, you can (1)
construct the perpendicular bisector of the control panel,
extend this segment to the other control panel, and
then find the midpoint of the segment. Or you can (2)
draw two diagonals from the top of one control panel
to the bottom of the other. The center will be at their
intersection.
27.
Statements
}
1. AB > CD
} } }
}
2. PA, PB, PC, and PD
are radii of (P.
} } } }
3. PA > PB > PC > PD
Reasons
2. Given
4. nPCD > nPAB
4. SSS Congruence Postulate
5. ŽCPD > ŽAPB
5. Corr. parts of > ns are
>.
6. mŽCPD 5 mŽAPB
6. Def. of >
7. Ž CPD and ŽAPB
are central Žs.
7. Given
}
1. Given
3. All radii of same circle
are congruent.
C C
8. Def. of arc measure
CD > C
9. C
AB
9. Def. of >
CB > C
28. Because A
CD , Ž APB > ŽCPD by the definition of
} } }
}
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
8. mCD 5 mAB
congruent arcs. PA, PB, PC, and PD are all radii of (P,
} } } }
so PA > PB > PC > PD. Then nAPB > nCPD by the
}
SAS Congruence Postulate, so corresponding sides AB
}
and CD are congruent.
29. a.
B
C
A
D
The longer chord is closer to the center.
b. The length of a chord in a circle increases as the
distance from the center of the circle to the chord
decreases.
Let c be the radius.
c.
d
Let x and y be the distances to the
center of the circle from the chords
shown, with x < y.
y
b x
By the Pythagorean Theorem,
1 2
1 2
b 2
d 2
1 x 2 and c 2 5 }2 1 y 2
2
c 5 }
2
1 2
b
So, }
2
2
1 2
d
1 x 5 }2
2
2
b d
1 y . Because x < y, }
> },
2 2
2
so b > d. Therefore, the length of a chord in a circle
increases as the distance from the center decreases.
r 2 5 1602 1 (r 2 80)2
r 5 200
The radius of the circle is 200 feet.
}
}
b. S 5 3.86Ï fr 5 3.86Ï (0.7)(200) ø 45.67
The car’s speed is about 45.7 miles per hour.
}
}
}
}
}
Suppose center L is not on QS . Since LT and LR are
} }
radii of the circle, they are congruent. With PL > PL by
the Reflexive Property, n RLP > nTLP by SSS. So,
corresponding angles ŽRPL and ŽTPL are congruent
and they form a linear pair. This makes them right angles
}
}
and leads to PL being perpendicular to RT . By the
}
}
Perpendicular Postulate, L must be on QS and thus QS
must be a diameter.
}
}
}
} } }
32. Draw radii LD and LF. So, LD > LF , LC > LC
} }
(Reflexive Property) and since EG > DF ,
nLDC > nLFC by the HL Congruence Theorem.
}
}
Then, corresponding sides DC and FC are congruent,
as are corresponding angles ŽDLC and ŽFLC. By the
definition of congruent arcs, DG > FG .
} } } } } }
33. Case 1: Given: EF > AB, EG > DC , EG > EF
} }
Prove: AB > DC
} } }
}
} }
Draw radii EB and EC . EB > EC and EF > EG. Also,
} }
} }
since EF > AB and EG > DC, nEFB and nEGC are
right triangles and are congruent by the HL Congruence
}
}
Theorem. Corresponding sides BF and CG are congruent,
} }
so BF 5 CG and, by the Multiplication Property of
}
}
Equality, 2BF 5 2CG. By Theorem 10.5, EF bisects AB
}
}
and EG bisects CD , so AB 5 2BF and CD 5 2CG. Then
} }
by the Substitution Property, AB 5 CD or AB > CD.
} } } } } }
Case 2: Given: EF > AB, EG > DC, AB > DC
} }
Prove: EF > EG
} } }
}
}
Draw radii EB and EC, EB > EC. By Theorem 10.5, EF
}
}
}
bisects AB and EG bisects CD, so AB 5 2BF and
CD 5 2CG. We know AB 5 CD, so by the Substitution
Property, 2BF 5 2CG. By the Division Property of
} }
Equality, BF 5 CG, or BF > CG. nEFB and nEGC are
right triangles and are congruent by the HL Congruence
}
}
Theorem. It follows that corresponding sides EF and EG
are congruent.
31.Given: QS is the perpendicular bisector of RT in (L.
C C
Q
34.
T
A
B
P
The point where the tire touches the
ground is a point of tangency. By
Theorem 10.1, line g is perpendicular to
}
}
radius of the circle, TP. Because AB i g,
}
} }
PT > AB. By Theorem 10.5, PQ bisects
}
AB and AB .
C
Mixed Review for TAKS
35. C; The equation that best represents the area A of the
rectangle is A 5 x(x 1 b).
Geometry
Worked-Out Solution Key
311
Chapter 10,
continued
Quiz 10.1–10.3 (p. 670)
1. CA 2 0 CB 2 1 BA 2
6.
mŽS 1 mŽU 5 1808
c8 1 (2c 2 6)8 5 1808
152 0 92 1 122
225 0 81 1 144
3c 5 186
c 5 62
mŽT 1 mŽV 5 1808
225 5 225
}
AB is tangent to (C at B because mŽABC 5 908 and
}
}
radius CB is perpendicular to AB.
10x8 1 8x8 5 1808
18x 5 180
2. CB 2 5 CA 2 1 AB 2
x 5 10
142 0 52 1 122
196 0 25 1 144
10.4 Exercises (pp. 676–679)
196 Þ 169
}
AB is not tangent to (C at A because mŽBAC Þ 908
}
}
and radius CA is not perpendicular to AB.
C C C
FG
1958 5 808 1 mC
CG
1158 5 mF
mC
EG 5 3608 2 mC
EFG 5 3608 2 1958 5 1658
3. mEFG 5 mEF 1 mFG
Theorem 10.9 tells you the hypotenuse of each of these
triangles is a diameter of the circle.
C
1
1
3. mŽA 5 } mBC 5 }(848) 5 428
2
2
C
1
C
D
C
1
5. mLM 5 1808 2 1608 5 208
C C C
ABD 5 mC
AB 1 mC
AB
mC
C
BD 5 2mC
AB
mA
1948 5 2mC
AB
CB
978 5 mA
1
Investigating Geometry Activity 10.4 (p. 671)
2. Answers will vary.
3. The measure of an inscribed angle is one half the
measure of the corresponding central angle.
10.4 Guided Practice (pp. 673–675)
C
1
1
1. mŽHGF 5 } mHF 5 }(908) 5 458
2
2
2. mTV 5 2mŽTUV 5 2(388) 5 768
3. mŽZXW 5 mŽZYW 5 728
4. To frame the front and left side of the statue in your
picture, make the diameter of your circle the diagonal of
the rectangular base. This diagonal connects the upper
left corner to the bottom right corner.
5. mŽB 1 mŽD 5 1808
x8 1 828 5 1808
x 5 98
mŽC 1 mŽA 5 1808
688 1 y8 5 1808
y 5 112
Geometry
Worked-Out Solution Key
1
C
CV 5 2mŽU 5 2(308) 5 608
7. mT
UV 5 1808 2 mC
TV 5 1808 2 608 5 1208
mC
CW 5 2mŽX 5 2(758) 5 1508
8. mY
WX 5 3608 2 m C
X Y 2 mC
YW
mC
5 3608 2 1108 2 1508 5 1008
9. From the diagram, the measure of C
RS is 908. So, the
6. mRS 5 2mŽQ 5 2(678) 5 1348
Lesson 10.4
1. Answers will vary.
C
mŽN 5 }2 mLM 5 }2(208) 5 108
mABD 5 mAB 1 mBD
312
2. The diagonals of a rectangle create two right triangles.
mŽG 5 }2 mFD 5 }2(1708) 5 858
C
C
polygon is inscribed in the circle.
4. mFD 5 3608 2 1208 2 708 5 1708
B
A
1. If a circle is circumscribed about a polygon, then the
measures of the arcs add up to 3708. You can either
change the measure of ŽQ to 408 or change the measure
of QS to 908.
C
C
10. ŽADB > ŽACB because they intercept the same arc, AB .
C
ŽCBD > ŽDAC because they intercept the same arc,
CD .
C
11. ŽJMK > ŽKLJ because they intercept the same arc, JK .
C
C
ZW . ŽXWY > ŽYZX because they intercept the same
CY .
arc, X
ŽMKL > ŽLJM because they intercept the same arc,
LM .
12. ŽWXZ > ŽZYW because they intercept the same arc,
13. mŽR 1 mŽT 5 1808
x8 1 808 5 1808
x 5 100
mŽS 1 mŽQ 5 1808
y8 1 958 5 1808
y 5 85
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4.
Skill Practice
Chapter 10,
continued
14. mŽD 1 mŽF 5 1808
23. A kite cannot always be inscribed in a circle because its
opposite angles are not always supplementary.
608 1 2k8 5 1808
2k 5 120
k 5 60
mŽE 1 mŽG 5 1808
m8 1 608 5 1808
m 5 120
24. A rhombus cannot always be inscribed in a circle because
its opposite angles are not always supplementary.
25. An isosceles trapezoid can always be inscribed in a circle
because its opposite angles are supplementary.
26.
C
A
alt
alt
12
Problem Solving
27.
B
4b 5 88
C
4
. So, }5 5 }
l alt 5 }
.
the smaller triangle, sin A 5 }
5
3
3
12
}
.
Therefore, the diameter JK is also }
5
a 5 20
b 5 22
B
diameter. nABC is a right triangle, so sin A 5 }5. Using
3a 5 60
928 1 4b8 5 1808
5
4
1208 1 3a8 5 1808
mŽK 1 mŽM 5 1808
4
}
JK is a diameter. The altitude from C to AB is also a
}
mŽ J 1 mŽL 5 1808
C
K
J
1
1
15. mŽJ 5 } mKLM 5 }(1308 1 1108) 5 1208
2
2
1
1
mŽK 5 }2 mJML 5 }2 (548 1 1308) 5 928
C
3
20,000 km
A
C
16. B; mAC 5 608
1
C
1
mŽB 5 }2 mAC 5 }2(608) 5 308
17. a. There are 5 congruent arcs. 3608 4 5 5 728. The
1
measure of each inscribed angle is }2 the measure of
1
the arc. }2 (728) 5 368. There are 5 inscribed angles.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Sum 5 5(368) 5 1808.
b. There are 7 congruent arcs. 3608 4 7 ø 51.48. The
1
measure of each inscribed angle is }2 the measure of
1
the arc. }2 (51.48) ø 25.78. There are 7 inscribed angles.
Sum ø 7(25.78) ø 1808.
c. There are 9 congruent arcs. 3608 4 9 5 408. The
1
measure of each inscribed angle is }2 the measure of
1
the arc. }2 (408) 5 208. There are 9 inscribed angles.
Sum 5 9(208) 5 1808.
18.
C
A; 2mŽEFG 5 mEG
2(8x 1 10)8 5 (12x 1 40)8
16x 1 20 5 12x 1 40
4x 5 20
x55
19. In a parallelogram, the opposite angles are congruent.
In an inscribed parallelogram, opposite angles are
supplementary. A rectangle is a parallelogram with
congruent supplementary opposite angles. The mŽR
is 908.
20. A square can always be inscribed in a circle because its
opposite angles are 908 and thus are supplementary.
21. A rectangle can always be inscribed in a circle because
its opposite angles are 908 and thus are supplementary.
22. A parallelogram cannot always be inscribed in a circle
because its opposite angles are not always supplementary.
100,000 km
AC 5 100,000 1 20,000 1 100,000 5 220,000
Moon A is 220,000 km from moon C.
28. Place the carpenter’s square so the endpoints of
the square and the vertex of the square are on the
circumference of the circle, then connect the endpoints.
29. The hypotenuse of the right triangle inscribed in the
circle is the diameter of the circle. So, double the length
of the radius to find the length of the hypotenuse.
C C
C C
C
C
C
C
30. By the Arc Addition Postulate, mEFG 1 mGDE 5 3608
and mFGD 1 mDEF 5 3608. Using the Measure of an
Inscribed Angle Theorem, mEDG 5 2mŽF,
mEFG 5 2mŽ D, mDEG 5 2mŽG, and
mFGD 5 2mŽ E. By the Substitution Property,
2mŽD 1 2mŽF 5 3608, so mŽ D 1 mŽF 5 1808.
Similarly, mŽE 1 mŽG 5 1808.
}
}
31. Let mŽB 5 x8. Because QA and QB are both radii of
} }
(Q, QA > QB and nAQB is isosceles. Because ŽA and
ŽB are base angles of an isosceles triangle, ŽA > ŽB.
So, by substitution, mŽA 5 x8. By the Exterior Angles
Theorem, mŽ AQC 5 mŽ A 1 mŽB 5 2x8. So, by the
definition of the measure of a minor arc, mAC 5 2x8.
C
C
1
Divide each side by 2 to show that x8 5 }2 mAC . Then,
1
by substitution, mŽB 5 }2 mAC .
C
32. Given: ŽABC is inscribed in (Q. Point Q is in the
interior of ŽABC.
1
C
Prove: mŽABC 5 }2 mAC
}
Plan for Proof: Construct the diameter BD of (Q and
C
the Arc Addition Postulate and the Angle Addition
CC .
AD 1 mD
Postulate to show 2mŽABC 5 mC
1
C
1
show mŽABD 5 }2 mAD and mŽDBC 5 }2 DC . Use
Geometry
Worked-Out Solution Key
313
continued
33. Given: ŽABC is inscribed in (Q. Point Q is in the
39.
A
2
exterior of ŽABC.
C
1
D
C
Prove: mŽABC 5 }2 mAC
B
C
C
the Arc Addition Postulate and the Angle Addition
Postulate to show 2mŽABC 5 mC
AD 2 mC
DC .
1
1
show mŽABD 5 }2 mAD and mŽDBC 5 }2 mDC . Use
34. Given: ŽACB and ŽADB are
A
inscribed angles.
D
Prove: ŽADB > ŽACB
O
Paragraph Proof:
By Theorem 10.7,
C
C
B
C
1
1
mŽ ADB 5 }2 mAB and mŽACB 5 }2 mAB . By
mŽPAD 5 cos21 0.375 ø 688
Since nAPB is isosceles, the base angles are equal.
B
T
A
Plan for Proof:
Use the Arc Addition Postulate to show
C
C
Case 2: Given: (T with inscribed nABC. ŽB is a
right angle.
}
Prove: AC is a diameter of (T.
Plan for Proof: Use the Measure of an Inscribed Angle
Theorem to show the inscribed right angle intercepts an
}
arc with measure 2(908) 5 1808. Since AC intercepts an
arc that is half of the measure of the circle, it must be a
diameter.
36. In the figure, nABC is a right triangle with ŽABC
being the right angle. Using Theorem 10.1, since @##$
AB is
}
perpendicular to radius BC, it is tangent to (C at point B.
HJ
GJ
37. } 5 } ; in a right triangle, the altitude from the right
FJ
GJ
angle to the hypotenuse divides the hypotenuse into two
segments. The length of the altitude is the geometric
mean of the lengths of these two segments.
So mŽAPB is less than 458.
Mixed Review for TAKS
40. C;
The students paid 0.3($120) 5 $36 for music, 0.25($120)
5 $30 for decorations, and $120 2 ($36 1 $30) 5
$120 2 $66 5 $54 for food. The bar graph shown in
choice C best represents the amounts spent on food,
music, and decorations.
Lesson 10.5
10.5 Guided Practice (pp. 680–682)
1
1. mŽ1 5 } (2108) 5 1058
2
C
CY 5 2mŽ X 5 2(808) 5 1608
3. m X
2. mRST 5 2mŽT 5 2(988) 5 1968
4. 1808 2 1028 5 788
1
788 5 }2 (958 1 y8)
156 5 95 1 y
61 5 y
C C
1
5. mŽFJG 5 } (mFG 2 mKH )
2
1
308 5 }2 (a8 2 448)
2
60 5 a 2 44
}
x 5 2Ï 3
104 5 a
}
}
}
GK 5 2GJ 5 2(2Ï 3 ) 5 4Ï 3 in.
314
mŽAPB 5 1808 2 mŽPAB 2 mŽPBA
5 1808 2 688 2 688 5 448
E
that mAEC 5 mABC and thus mABC 5 1808. Then use
the Measure of an Inscribed Angle Theorem to show
mŽB 5 908, so that ŽB is a right angle and nABC is
a right triangle.
GJ 5 2Ï 3 in.
Geometry
Worked-Out Solution Key
0.563
mŽPAB 5 mŽPBA 5 688
Prove: nABC is a right triangle.
38. FJ 5 6 in., JH 5 2 in.
GJ
HJ
}5}
FJ
GJ
x
2
}5}
x
6
AD
sin 16.358 5 }
l AD ø 0.563
2
AD
}
nABC. AC is a diameter of (T.
C C
Show that mŽAPB is less than or equal to 458. The
1
diagram shown represents }
of the fuel booster design.
11
}
1
of 3608 or about 32.78. CP bisects
ŽACB is equal to }
11
}
ŽACB and Ž APB, so mŽ ACD ø 16.358. CP is also
}
the perpendicular bisector of AB, and contains point D.
nACD and nAPD are right triangles with ŽD 5 908 in
both triangles.
cos ŽPAD 5 }
5}
ø 0.375
1.5
1.5
substitution, mŽADB 5 mŽACB. By the definition of
congruence, ŽADB > ŽACB.
35. Case 1: Given: (T with inscribed
P
1.5
2
}
Plan for Proof: Construct the diameter BD of (Q and
x 5 12
1.5
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 10,
Chapter 10,
continued
3
6. sin TQS 5 }
5
1
11. 298 5 } (1148 2 x8)
2
mŽTQS ø 36.878
58 5 114 2 x
mŽTQR 5 2(mŽTQS) ø 2(36.878) ø 73.748
x 5 56
C C
1
mŽTQR 5 }2 (mTUR 2 mTR )
1
12. 348 5 } [(3x 2 2)8 2 (x 1 6)8]
2
1
68 5 (3x 2 2) 2 (x 1 6)
73.748 5 }2 (x8 2 (360 2 x8))
68 5 2x 2 8
147.48 5 x 2 360 1 x
76 5 2x
507.48 5 2x
38 5 x
253.74 ø x
10.5 Exercises (pp. 683–686)
1
1
13. D; mŽ4 5 }(808 1 1208) 5 } (2008) 5 1008
2
2
Skill Practice
14. The error is that the given measurements imply two
1. The points A, B, C, and D are on a circle and @##$
AB
1
@##$
}
(
intersects CD at P. If mŽAPC 5 2 mBD 2 mAC )
C C
then P is outside the circle.
C
2. If mAB 5 08, then the two chords intersect on the circle.
1
By Theorem 10.12, mŽ1 5 }2 (mDC 1 mAB )
C C
1
CC 1 08) 5 12 mC
5 2 (mD
DC .
}
}
This is consistent with the measure of an Inscribed Angle
Theorem (Lesson 10.4).
C
1
CB
658 5 2 mA
AB
1308 5 mC
1
3. mŽA 5 } mAB
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
C
1
C
1178 5 2 mD
EF
2348 5 mC
DEF
1
4. mŽD 5 } mDEF
2
1
5. mŽ1 5 } (2608) 5 1308
2
}
}
C
6. D; Because AB is not a diameter, mAB Þ 1808.
1
1
mŽA 5 }2 mAB Þ }2 (1808) Þ 908
C
1
1
7. x8 5 } (1458 1 858) 5 } (2308) 5 1158
2
2
8. 1808 2 122.58 5 57.58
1
57.58 5 }2 (x8 1 458)
115 5 x 1 45
70 5 x
1
9. (180 2 x)8 5 } (308 1 (2x 2 30)8)
2
2(180 2 x) 5 2x
360 2 2x 5 2x
360 5 4x
90 5 x
10. 3608 2 2478 5 1138
1
x8 5 }2 (2478 2 1138)
1
x 5 }2 (134)
x 5 67
C
different measures for BE . Using Theorem 10.12,
C
1008 5 mC
BE 1 608
BE
408 5 mC
1
508 5 }2 (mBE 1 608)
Using Theorem 10.13,
C
CE
308 5 608 2 mB
CE 5 308
mB
1
158 5 }2 (608 2 mBE )
}
15. If ###$
PL is perpendicular to KJ at K, then mŽLPJ 5 908,
otherwise it would measure less than 908. So, mŽLPJa908.
1
16. a. } (1108) 5 558
2
1
558 5 }2 (x8 2 408)
1108 5 x8 2 408
150 5 x
1
1
b. } c8 5 } (b8 2 a8)
2
2
c5b2a
C C
1
17. mŽQ 5 } (mEGF 2 mEF )
2
1
608 5 }2 [(360 2 x)8 2 x8]
120 5 360 2 2x
2x 5 240
x 5 120
C
So, mEF 5 1208.
1
C C
mŽR 5 }2 (mGEF 2 mFG )
1
808 5 }2 [(360 2 x)8 2 x8]
160 5 360 2 2x
2x 5 200
x 5 100
C
So, mFG 5 1008.
C
C C
mGE 5 3608 2 mEF 2 mFG
5 3608 2 1208 2 1008 5 1408
C
So, mGF 5 1408.
Geometry
Worked-Out Solution Key
315
Chapter 10,
continued
C C
1
18. mŽB 5 } (mAD 2 mAC )
2
21. mŽCHD 5 1808 2 1158 5 658
C C
1
CA )
658 5 2 (858 1 mE
CA
1308 5 858 1 mE
458 5 mC
EA
CF 5 mECA 2 mECF 5 458 2 208 5 258
mA
1
CB 2 mACF )
mŽJ 5 2 (mA
1
CB 2 258)
308 5 2 (mA
CB 2 258
608 5 mA
AB
858 5 mC
1
CB 1 mFCD )
mŽFGH 5 2 (mA
1
CD ))
908 5 2 (858 1 (208 1 mE
CD
1808 5 1058 1 mE
CD
758 5 mE
1
mŽCHD 5 }2 (mCD 1 mEA )
1
408 5 }2 (7x8 2 3x8)
}
80 5 4x
20 5 x
C
mC
AC 5 3x8 5 3(20)8 5 608
CD 5 3608 2 mC
AD 2 mC
AC
mC
mAD 5 7x8 5 7(20)8 5 1408
}
5 3608 2 1408 2 608 5 1608
19. a.
t
t
}
C
A
A
}
C
}
B
C
CB
2mŽBAC 5 mA
Problem Solving
For the diagram on the right,
22. mŽ A 5 808
1
mŽBAC 5 }2 mAB
C
CA
2mŽBAC 5 3608 2 mB
CA 5 3608 2 2mŽBAC
mB
CA 5 2(1808 2 mŽBAC)
mB
1
mŽ BAC 5 }2 (3608 2 mBA )
1
1
mŽB 5 }2 (808) 5 408
1
1
23. x8 5 }(1808 2 808) 5 }(1008) 5 508
2
2
c. 2mŽBAC 5 2(1808 2 mŽBAC)
1
24. mŽB 5 }(808 2 x8)
2
mŽBAC 5 1808 2 mŽBAC
2mŽBAC 5 1808
mŽBAC 5 908
1
308 5 }2 (808 2 x8)
C
608 5 808 2 x8
So, these equations give the same value for mAB when
}
AB is perpendicular to t at point A.
x8 5 208
C C
C
Camera B will have a 308 view of the stage when the arc
measuring 308 is reduced to an arc measuring 208. You
should move the camera closer to the stage.
20. Let x8 5 mX W 5 mZ Y .
C
Then mWZ 5 (200 2 x)8 and
mX Y 5 (3608 2 (200 1 x)8) 5 (160 2 x)8
C C
1
mŽP 5 }2 (mWZ 2 mX Y )
25.
C
B
1
D
4001.2 mi
1
5 }2 [(200 2 x)8 2 (160 2 x)8]
5 }2 (40) 5 208
1
mŽ B 5 }2 (808 2 308) 5 }2 (508) 5 258
A
4000 mi
E
Not drawn to scale
4000
sin BCA 5 }
lmŽBCA ø 88.68
4001.2
mŽBCD ø 2(88.68) ø 177.28
C
Let mBD 5 x8.
1
C C
mŽBCD 5 }2 (mDEB 2 mBD )
1
2
177.28 ø } [(3608 2 x8) 2 x8]
x ø 2.8
The measure of the arc from which you can see is about 2.88.
316
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
B
b. For the diagram on the left,
Chapter 10,
continued
}
}
28. Given: Chords AC and BD
26.
intersect.
14°
(180 2 x)°
x°
1
28 5 180 2 2x
2x 5 152
x 5 76
(180 2 x)8 5 (180 2 76)8 5 1048
The measures of the arcs between the ground and the cart
are 768 and 1048.
}
27. Case 1: Given: tangent @##$
AC intersects chord AB at point A
}
on (Q. AB contains the center of (Q.
C
Prove: mŽ CAB 5 }2 mAB
3. mŽ1 5 mŽDBC 1
mŽACB
3. Exterior Angle Theorem
2. Given
C
1
CB
5. mŽACB 5 2 mA
1
CC 1 12 mC
6. mŽ1 5 2 mD
AB
1
CC 1 mACB )
7. mŽ1 5 2 (mD
1
4. mŽDBC 5 }2 mDC
4. Theorem 10.7
5. Theorem 10.7
}
29.
6. Substitution Property
7. Distributive Property
B
}
Paragraph proof: By Theorem 10.1, @##$
CA > AB. By the
definition of perpendicular, mŽCAB 5 908. Because
}
1
AB is a diameter, mAB 5 1808. So, mŽCAB 5 }2 mAB .
C
C
}
Case 2: Given: tangent @##$
AC intersects chord AB at point A
on (Q. The center of the circle, Q, is in the interior
of Ž CAB.
C
1
Prove: mŽ CAB 5 }2 mAPB
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
B
Reasons
}
A
Case 1
P
A
1
C
2
Case 1
}
Case 1: Draw BC. Use the Exterior Angle Theorem to
show that mŽ2 5 mŽ1 1 mŽABC, so that
mŽ1 5 mŽ2 2 mŽABC. Then use Theorem 10.11
1
C
to show that mŽ2 5 }2 mBC and the Measure of an
B
1
C
Inscribed Angle Theorem to show that mŽ ABC 5 }2 mAC .
C C
Q
1
Then, mŽ1 5 }2 (mBC 2 mAC ).
C
C
1. Given
}
Q
1
A
Statements
}
}
1. Chords AC and BD
intersect.
}
2. Draw BC.
}
B
C
C C
1
Prove: mŽ1 5 }2(mDC 1 mAB )
148 5 }2 [(180 2 x)8 2 x8]
1
D
A
Case 2
P
}
Plan for proof: Draw diameter AP. Use the Angle
Addition Postulate and Theorem 10.7 to show that
C
2
1
mŽ CAB 5 908 1 }2 mPB . Use the Arc Addition Postulate
C
C
3
to show that mAPB 5 1808 1 mPB .
}
AC intersects chord AB at point A
Case 3: Given: Tangent @##$
on (Q. The center of the circle, Q, is in the exterior of
Ž CAB.
1
C
Q
4
R
Case 2
}
Case 2: Draw PR. Use the Exterior Angle Theorem to
show that mŽ3 5 mŽ2 1 mŽ4, so that mŽ2 5
mŽ3 2 mŽ4. Then use Theorem 10.11 to show that
C
1
mŽ2 5 2 (mC
PQR 2 mC
PR ).
Prove: mŽCAB 5 }2 mAB
1
1
C
mŽ3 5 }2 mPQR and mŽ4 5 }2 mPR . Then,
P
}
B
Q
X
C
A
Case 3
}
Plan for Proof: Draw diameter AP. Use the Angle
Addition Postulate and Theorem 10.7 to show that
1
C
mŽCAB 5 908 2 }2 mPB . Use the Arc Addition Postulate
C
W
3
Z
4
Y
Case 3
C
to show that mAB 5 1808 2 mPB .
Geometry
Worked-Out Solution Key
317
Chapter 10,
continued
}
Case 3: Draw XZ. Use the Exterior Angle Theorem to
show that mŽ4 5 mŽ3 1 mŽWXZ, so that
mŽ3 5 mŽ4 2 mŽWXZ. Then use the Measure of an
1
Inscribed Angle Theorem to show that mŽ4 5 }2 mXY and
33. J;
Number
Cost
Cost
Number
of bags
of one
of one
Maximum
a
+
+ of hot 1
of
bag of
hot
total cost
dogs
peanuts peanuts
dog
C
1
1
CZ ).
WZ . Then, mŽ3 5 2(mC
XY 2 mW
mŽWXZ 5 2 mC
}
}
Q
30.
3h 1 2.5b a10
T
P
Quiz 10.4–10.5 (p. 686)
1.
R
}
}
x8 1 858 5 1808
Given: PQ and PR are tangents to a circle.
}
Prove: QR is not a diameter.
}
Paragraph Proof: Assume QR is a diameter. Then
mQTR 5 mQR 5 1808. By Theorem 10.13,
x 5 95
mŽC 1 mŽA 5 1808
C C
1
1
C
TR 2 mC
QR ) 5 2(1808 2 1808) 5 08
mŽP 5 2 (mQ
}
y8 1 758 5 1808
y 5 105
}
ABC 5 2mŽD 5 2(958) 5 1908
mC
}
The mŽP cannot equal 08, so QR cannot be a diameter.
DE 5 113, EC 5 15
15 cm
D
DE 2 5 DC 2 1 EC 2
So, z 5 190.
2. mŽE 1 mŽG 5 1808
x8 1 1128 5 1808
113 5 DC 1 15
2
2
2
x 5 68
12,544 5 DC 2
113 cm
mŽF 1 mŽH 5 1808
112 5 DC
2mŽF 5 180
C
B
mŽF 5 908
GHE 5 2mŽF 5 2(908) 5 1808
mC
E
15 cm
mŽDEC 5 sin211 }
ø 82.48
113 2
112
C
So, z 5 180.
mBC 5 mŽBED 1 mŽDEC
3.
5 908 1 82.48 5 172.48
mŽK 1 mŽM 5 1808
7x8 1 1318 5 1808
172.48
3608
} ø 48%
7x8 5 49
About 48% of the circumference of the bottom pulley is
not touching the rope.
x57
mŽL 1 mŽJ 5 1808
(11x 1 y)8 1 998 5 1808
Mixed Review for TAKS
32. C; The graph of y 5 26x 2 has points with y-coordinates
that are the opposite of those of y 5 6x for each
x-coordinate.
(11(7) 1 y) 1 99 5 180
77 1 y 5 81
2
y 5 6x 2
22
1
1
4. x8 5 } (1078 1 838) 5 } (1908) 5 958
2
2
(1,6
)
(1, 26)
y5
x
26x 2
(2, 224)
,0
(0
)
So, the graph of y 5 26x2 is a reflection of y 5 6x2
across the x-axis.
318
y54
So, z 5 262.
,2
(2
)
4
6
C
mJKL 5 2mŽM 5 2(1318) 5 2628
y
Geometry
Worked-Out Solution Key
1
1
5. x8 5 } (748 2 228) 5 }(528) 5 268
2
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
31.
mŽD 1 mŽB 5 1808
Chapter 10,
continued
1
6. 618 5 } (x8 2 878)
2
3. B;
The measure of the arc between any two blades is
122 5 x 2 87
3608
3
209 5 x
} 5 1208. So, an isosceles triangle may be formed
C
with two blades and the distance y from the tip of one
blade to the tip of another blade.
7.
B
D
4001.37 mi
1208
38.5 m
A
4000 mi
38.5 m
308
308
y
E
Not drawn to scale
4000
38.5 m
sin BCA 5 }
l mŽBCA ø 88.58
4001.37
308
mŽBCD ø 2(88.58) ø 1778
C
Let mBD 5 x8.
1
1
y
2
C C
1
2
}y
mŽBCD 5 }2 (mDEB 2 mBD )
cos 308 5 }
38.5
1
1
1778 ø }2 [(3608 2 x8) 2 x8]
38.5 + cos 308 5 }2 y
354 ø 360 2 2x
2 + 38.5 + cos 308 5 y
26 ø 2x
66.7 ø y
xø3
The measure of the arc from which you can see is
about 38.
Mixed Review for TEKS (p. 687)
C
1. A; If mŽ ACB 5 338, then mŽ AB 5 338.
C C
CC 2 mC
408 5 (mB
AD )
C
C
mBC 5 408 1 mAD
1
CC 1 mACD )
608 5 2 (mB
CC 1 mACD
1208 5 mB
CC 5 1208 2 mACD
mB
CD 5 1208 2 mACD
408 1 mA
AD 5 808
2mC
CD 5 408
mA
AD is 408.
The measure of C
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
2. G; 208 5 }(mBC 2 mAD )
2
}
The approximate distance from the tip of one blade to the
tip of another blade is 66.7 meters.
4. H;
By Theorem 10.1, the 6 foot distance forms a right angle
with a radius r of the circle at the point of tangency. So,
(r 1 3)2 5 r 2 1 62
r 1 6r 1 9 5 r 2 1 36
6
2
6r 5 27
r
r13
r 5 4.5
The official is approximately 4.5 1 3 5 7.5 feet from the
center of the discus circle.
C C
1
5. C; x8 5 } (mNM 1 mLK )
2
1
5 }2 (358 1 938)
1
5 }2 (1288) 5 648
The measure of Ž MDN is 648.
C
C
6. mXYZ 5 3608 2 m XQZ
5 3608 2 1998 5 1618
Using Theorem 10.3,
C
1
C
1
mYZ 5 }2 mXYZ 5 }2 (1618) 5 80.58.
7. AB 5 16 cm, so AD 5 8 cm.
10 5 8 1 x
2
2
100 5 64 1 x 2
36 5 x 2
B
2
D
10 cm
A
x
10 cm
C
65x
The distance from the center
}
point of (C to AB is 6 centimeters.
Geometry
Worked-Out Solution Key
319
Chapter 10,
continued
Lesson 10.6
8. Use Theorem 10.14.
x + 18 5 9 + 16
Investigating Geometry Activity 10.6 (p. 688)
18x 5 144
1. The products AE + CE and BE + DE are the same.
x58
2. The products AE + CE and BE + DE are the same.
9. Use Theorem 10.15.
3. AE + CE 5 BE + DE
18 + (18 1 22) 5 x + (x 1 29)
4. PT + QT 5 RT + ST
720 5 x 2 1 29x
9 + 5 5 15 + ST
x 2 1 29x 2 720 5 0
45 5 15ST
(x 2 16)(x 1 45) 5 0
3 5 ST
x 5 16 (x 5 245 is extraneous.)
10.6 Guided Practice (pp. 690–692)
1. 6 + (6 1 9) 5 5 + (5 1 x)
2. 3 + x 5 6 + 4
90 5 25 1 5x
65 5 5x
3x 5 24
tangent to the circle. EC must be less than EA.
11. It is appropriate to use the approximation symbol in the
solution to Example 4 because it is stated in the problem
that each moon has a nearly circular orbit.
x58
13 5 x
3. 3 + [3 1 (x 1 2)] 5 (x 1 1) + [(x 1 1) 1 (x 2 1)]
3(x 1 5) 5 (x 1 1)(2x)
3x 1 15 5 2x 2 1 2x
10.6 Exercises (pp. 692–695)
Skill Practice
1. The part of the secant segment that is outside the circle is
2x2 2 x 2 15 5 0
(2x 1 5)(x 2 3) 5 0
x 5 3 1 x 5 2}2 is extraneous. 2
called an external segment.
2. A tangent segment intersects the circle in only one point
while the secant segment intersects the circle in two points.
5
3. 12 + x 5 10 + 6
4. x 2 5 1(3 1 1)
12x 5 60
x2 5 4
x55
4. 9 + (x 2 3) 5 10 + 18
9x 2 27 5 180
49 5 25 1 5x
9x 5 207
24 5 5x
x 5 23
24
5
x + (x 1 8) 5 6 + 8
5.
}5x
x 2 1 8x 5 48
122 5 x + (x 1 10)
x 1 8x 2 48 5 0
2
144 5 x 2 1 10x
(x 2 4)(x 1 12) 5 0
x2 1 10x 2 144 5 0
(x 2 8)(x 1 18) 5 0
x 5 8 (x 5 218 is extraneous.)
x 5 4 (x 5 212 is extraneous.)
6. 8 + (8 1 x) 5 6 + (6 1 10)
64 1 8x 5 96
7. Use Theorem 10.16.
8x 5 32
152 5 x + (x 1 14)
225 5 x 2 1 14x
x 2 1 14x 2 225 5 0
}}
214 6 Ï142 2 4(1)(2225)
2(1)
x 5 }}}
}
214 6 2Ï274
x 5 }}
2
}
x 5 27 6 Ï274
}
x 5 27 1 Ï274
}
(x 5 27 2 Ï274 is extraneous.)
x54
x + (x 1 4) 5 5 + (5 1 7)
x 2 1 4x 5 60
2
x 1 4x 2 60 5 0
(x 2 6)(x 1 10) 5 0
x 5 6 (x 5 210 is extraneous.)
8. (x 2 2) + [(x 2 2) 1 (x 1 4)] 5 4 + (4 1 5)
(x 2 2)(2x 1 2) 5 36
2x 2 2 2x 2 4 5 36
2x 2 2 2x 2 40 5 0
7.
x 2 2 x 2 20 5 0
(x 2 5)(x 1 4) 5 0
x 5 5 (x 5 24 is extraneous.)
320
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x 5 2 (x 5 22 is extraneous.)
5. 72 5 5 + (5 1 x)
6.
}
10. EC cannot be equal to EA because that would make EC
Chapter 10,
continued
9. x 2 5 9 + (9 1 7)
19. AB 2 5 BC + (BC 1 CD 1 DE)
x 5 144
x 5 12 (x 5 212 is extraneous.)
10. 242 5 12 + (12 1 x)
576 5 144 1 12x
432 5 12x
36 5 x
11.
(x 1 4)2 5 x + (x 1 12)
2
x 1 8x 1 16 5 x 2 1 12x
16 5 4x
45x
2
122 5 8 + (8 1 CD 1 6)
144 5 112 1 8CD
4 5 CD
r
P
4 4
6
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
13. 15 + (x 1 3) 5 2x + 12
CD + DE 5 (r 1 PD) + (r 2 PD)
4 + 6 5 (r 1 4) + (r 2 4)
24 5 r 2 2 16
122 5 6 + (6 1 PQ)
144 5 36 1 6PQ
108 5 6PQ
18 5 PQ
18. RQ 5 RS + (RS 1 SP)
2
RQ 2 5 14 + (14 1 12)
RQ 5 364
40 5 r 2
}
2Ï10 5 r
Problem Solving
20. x + x 5 62 + (250 2 62)
x 2 5 11,656
x ø 107.96
14. x + 45 5 27.50
15x 1 45 5 24x
45x 5 1350
45 5 9x
x 5 30
55x
}
(Ï3 )2 5 x + (2 1 x)
15.
3 5 2x 1 x 2
2
x 1 2x 2 3 5 0
(x 2 1)(x 1 3) 5 0
x 5 1 (x 5 23 is extraneous.)
16. D;
x + x 5 2 + (2x 1 6)
x 2 5 4x 1 12
2
x 2 4x 2 12 5 0
(x 2 6)(x 1 2) 5 0
x 5 6 (x 5 22 is extraneous.)
17. MN 2 5 PN + (PN 1 PQ)
D
E
12. The error is that the wrong segment lengths are being
multiplied. Use Theorem 10.15.
FD + FC 5 FA + FB
4 + (4 1 CD) 5 3 + (3 1 5)
16 1 4CD 5 24
4CD 5 8
CD 5 2
C
The distance from the end of the passage to either side of
the mound is about 108 feet.
}
}
C
21. Given: AB and CD are chords
that intersect at E.
Prove: EA + EB 5 EC + ED
A
E
B
D
Statements
}
1. AB and CD are chords
that intersect at E.
}
}
2. Draw BD and AC.
}
Reasons
1. Given
2. Two points determine a
line.
3. Ž C > Ž B
3. ŽC and ŽB intercept
the same arc.
4. ŽA > ŽD
4. ŽA and ŽD intercept
the same arc.
5. nAEC , nDEB
5. AA Similarity Postulate
EC
EA
5}
6. }
EB
ED
6. Corresponding sides are
proportional.
7. EA + EB 5 EC + ED
7. Cross Product Property
2
PQ 2 1 RQ 2 5 RP 2
PQ 2 1 364 5 262
PQ 2 5 312
PQ ø 17.7
AD + (AD 1 DE) 5 AB + (AB 1 BC)
22.
(AD)2 1 (AD)(DE) 5 (AB)2 1 (AB)(BC)
(AD)2 1 (AD)(DE) 2 (AB)2 5 (AB)(BC)
(AD)2 1 (AD)(DE) 2 (AB)2
AB
}}} 5 BC
}
23. Given: EA is a tangent
}
segment and ED is
E
a secant segment
passing through the center.
A
T
C
D
Prove: (EA)2 5 EC + ED
Geometry
Worked-Out Solution Key
321
Chapter 10,
continued
}
} }
Proof: Draw radius AT. By Theorem 10.1, EA > AT. By
2
2
the Pythagorean Theorem, EA 1 AT 5 ET 2. By the
Segment Addition Postulate, ET 5 EC 1 CT. So,
EA 2 1 AT 2 5 (EC 1 CT)2 by substitution. Simplify the
equation.
EA2 1 AT 2 5 (EC 1 CT)2
C
1
1
27. a. mŽCAB 5 } mBD 5 } (1208) 5 608
2
2
b. mŽCAB 5 mŽEFD 5 608, so ŽCAB > ŽEFD.
Also, ŽACB > ŽFCE by the Vertical Angles
Theorem. So, nABC , nFEC by the AA Similarity
Theorem.
FC
EF
c. } 5 }
BA
AC
EA2 1 AT 2 5 EC 2 1 (EC)(CT) 1 CT 2
AT and CT are both radii, so they are equal. Substitute
CT for AT.
x 1 10
6
y
3
}5}
EA 2 1 CT 2 5 EC 2 1 2(EC)(CT) 1 CT 2
x 1 10
y 5 31 }
6 2
EA 2 5 EC(EC 1 2CT)
x 1 10
EA 2 5 EC(EC 1 CD)
y5}
2
EA 2 5 EC + ED
d. Use Theorem 10.16.
24. 4 + CN 5 6 + 8
EF 2 5 FD + FA
CN 5 12
y 2 5 x + (x 1 10 1 6)
The length CN is 12 centimeters. Sparkles travel from
6 cm
2 cm/sec
C to D in } 5 3 seconds. So, sparkles must travel
y 2 5 x(x 1 16)
y 2 5 x(x 1 16)
e.
from C to N in 3 seconds. The sparkles must travel at a
x 1 10
1}
2 2
2
12 cm
rate of }
5 4 cm/sec.
3 sec
}
x2
4
} 1 5x 1 25 5 x 2 1 16x
25. Given: EB and ED are
secant segments.
x 2 1 20x 1 100 5 4x 2 1 64x
Prove: EA + EB 5 EC + ED
3x 2 1 44x 2 100 5 0
B
(3x 1 50)(x 2 2) 5 0
A
x2250
E
or 3x 1 50 5 0
C
x52
D
}
}
Paragraph proof: Draw AD and BC. ŽB and ŽD
intercept the same arc, so ŽB > ŽD. ŽE > ŽE by the
Reflexive Property of Congruence, so nBCE , nDAE
by the AA Similarity Theorem. Then, since lengths of
corresponding sides of similar triangles are proportional,
EA
EC
ED
EB
} 5 }. By the Cross Product Property,
Length cannot be negative, so x 5 2.
x 1 10
Prove: EA2 5 EC + ED
2 1 10
y5}
5}
56
2
2
6
EF
2
f. } 5 } 5 }, so the ratio of nFEC to n ABC is 2 to 1.
3
1
AB
CF
EA + EB 5 EC + ED.
}
}
26. Given: EA is a tangent segment. ED is a secant segment.
50
x 52}
3
2
5 }1. Let CE 5 2x and CB 5 x.
So, }
CB
By Theorem 10.14,
CE + CB 5 AC + CD
2x + x 5 6 + 10
A
2x 2 5 60
x 2 5 30
E
D
}
}
C
1
Paragraph proof: Draw AC and AD. mŽD 5 }2 mAC and
1
mŽEAC 5 }2 mAC . So, ŽD > ŽEAC. ŽE > ŽE by the
C
Reflexive Property of Congruence, so nDAE , nACE
by the AA Similarity Theorem. Then, since lengths of
corresponding sides of similar triangles are proportional,
EA
EC
ED
EA
} 5 }. By the Cross Product Property,
(EA)2 5 EC + ED.
322
}
x 5 6Ï30
C
Geometry
Worked-Out Solution Key
}
Length cannot be negative, so x 5 Ï30 .
}
CE 5 2x 5 2Ï30
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
5 x(x 1 16)
Chapter 10,
28.
N
continued
So, mŽAEB 5 mŽACD or ŽAEB > ŽACD. Also
ŽA > ŽA by the Reflexive Property of Congruence.
So, nAEB , nACD by the AA Similarity Theorem.
P
2
P9
5
O
b.
The distance from the origin to P9 is 5 (Pythagorean
}
Theorem). ON is the diameter, so its length is 2.
2
2
2
AD
AB
15 1 5
12
12 1 DE
15
20
12
12 1 DE
15
}5}
(NP9) 5 (ON) 1 (OP9) 5 2 1 5 5 29
2
AC
AE
}5}
2
}
}5}
NP9 5 Ï29 . By Theorem 10.16,
(OP9)2 5 PP9 + NP9
300 5 144 1 12DE
}
52 5 PP9 + Ï 29
156 5 12DE
13 5 DE
25
}
} 5 PP9
Ï29
3. 72 5 5(5 1 x)
}
25Ï 29
29
} 5 PP9
49 5 25 1 5x
}
25Ï 29
units.
The distance d is }
29
24 5 5x
24
5
}5x
Mixed Review for TAKS
29. B;
4. x(x 1 w) 5 y( y 1 z)
5, 10, 12, 12, 15
54
y(y 1 z)
x1w5}
x
5 10.8
Mean 5 }
5
Median 5 12
Mode 5 12
Range 5 15 2 5 5 10
y(y 1 z)
w5}
2x
x
The median and mode are the same.
30. H; Based on the results in the graph, if Company X
spends $2100 on advertising, $280,000 is the best
estimate of the company’s sales.
10.6 Extension (pp. 697–698)
1.
1in.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1in.
P
Problem Solving Workshop 10.6 (p. 696)
RQ
RS
1. } 5 }
RT
RQ
2.
1 in.
l
k
m
3.
RQ
4
}5}
419
RQ
4
RQ
1 in.
RQ
13
}5}
C
RQ 2 5 52
}
RQ 5 Ï 52
}
RQ 5 2Ï 13
4.
2. a.
C
1in.
l
1in.
k
B
m
D
A
5. The locus of points consists of two points on * each
E
3 centimeters away from P.
1
C
3 cm
mŽBED 5 }2 mBCD
3 cm
P
l
C
1
mŽAEB 5 1808 2 mŽBED 5 1808 2 }2 mBCD
1
C
mŽACD 5 }2 mBED
1
C
1
C
5 }2 (3608 2 mBCD ) 5 1808 2 }2 mBCD
Geometry
Worked-Out Solution Key
323
Chapter 10,
continued
}}
6. The locus of points consists of four points on a circle
with center at Q and a radius of 5 centimeters. The four
points are the intersections of the circle and two lines
parallel to and 3 centimeters away from m.
4. r 5 Ï (21 2 2)2 1 (2 2 6)2
}}
5 Ï(23)2 1 (24)2 5 5
(x 2 h)2 1 ( y 2 k)2 5 r 2
(x 2 2)2 1 (y 2 6)2 5 52
(x 2 2)2 1 ( y 2 6)2 5 25
3 cm
m
3 cm
5. (x 2 4)2 1 ( y 1 3)2 5 16
5 cm
Q
(x 2 4)2 1 ( y 2 (23))2 5 42
The center is (4, 23) and the radius is 4.
y
1
7. The locus of points consists of a semi-circle centered at
R with radius 10 centimeters. The diameter bordering the
semi-circle is 10 centimeters from k and parallel to k.
8. The portions of line * and m that are no more than
8 centimeters from point P and the portion of the circle,
including its interior, with center P and radius
8 centimeters that is between lines * and m.
(4, 23)
6. (x 1 8)2 1 ( y 1 5)2 5 121
(x 2 (28))2 1 ( y 2 (25))2 5 112
The center is at (28, 25) and the radius is 11.
y
6
m
5
cm
P
x
k
10 cm
10 cm R
8 cm
21
23
8 cm
5
cm
x
(28, 25)
l
9.
2 ft dog 3 ft
house
5 ft
4 ft
epicenter because two circles intersect in two points. You
would not know which point is the epicenter. You need
the third circle to determine which point is the epicenter.
9 ft
10.7 Exercises (pp. 702–705)
dog
Skill Practice
1. The standard equation of a circle can be written for any
circle with known center and radius.
Lesson 10.7
2. The location of the center and one point on a circle is
10.7 Guided Practice (pp. 700–701)
1. x 1 y 5 r
2
2
2
x 2 1 y 2 5 2.52
3. The radius is 2.
x 2 1 y 2 5 6.25
(x 2 h) 1 ( y 2 k) 5 r
2
2.
2
2
(x 2 (22))2 1 ( y 2 5)2 5 72
(x 1 2) 1 ( y 2 5) 5 49
2
2
}}
3. r 5 Ï (3 2 1) 1 (4 2 4)
2
5
2
}
Ï22 1 02 5 2
(x 2 h)2 1 ( y 2 k)2 5 r 2
(x 2 1)2 1 ( y 2 4)2 5 22
(x 2 1)2 1 ( y 2 4)2 5 4
324
Geometry
Worked-Out Solution Key
enough information to draw the rest of the circle because
the distance from the center to the known point is the
radius of the circle.
x2 1 y 2 5 4
4. The center is (2, 3). The radius is 2.
(x 2 2)2 1 (y 2 3)2 5 4
5. The radius is 20.
x 2 1 y 2 5 202
x 2 1 y 2 5 400
6. The center is (5, 0). The radius is 10.
(x 2 5)2 1 ( y 2 0)2 5 102
(x 2 5)2 1 y 2 5 100
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
7. Three seismographs are needed to locate an earthquake’s
6 ft
2 ft
Chapter 10,
continued
22. x 2 1 ( y 1 2)2 5 36
7. The center is (50, 50). The radius is 10.
(x 2 50) 1 ( y 2 50) 5 10
2
2
The center is (0, 22). The radius is 6.
2
(x 2 50) 1 ( y 2 50) 5 100
2
2
y
8. The center is (23,23). The radius is 9.
2
(x 2 (23)) 1 ( y 2 (23)) 5 9
2
2
2
22
(x 1 3) 1 ( y 1 3) 5 81
2
2
x
(0, 22)
9. x 1 y 5 72
2
2
x 2 1 y 2 5 49
23. (x 2 4)2 1 ( y 2 1)2 5 1
10. (x 2 (24))2 1 ( y 2 1)2 5 12
The center is (4, 1). The radius is 1.
(x 1 4)2 1 ( y 2 1)2 5 1
11. (x 2 7)2 1 ( y 2 (26))2 5 82
y
(x 2 7) 1 ( y 1 6) 5 64
2
(4, 1)
12. (x 2 4)2 1 ( y 2 1)2 5 52
1
(x 2 4) 1 ( y 2 1)2 5 25
1
x
13. (x 2 3)2 1 ( y 2 (25))2 5 72
(x 2 3)2 1 ( y 1 5)2 5 49
24. (x 1 5)2 1 ( y 2 3)2 5 9
14. (x 2 (23))2 1 ( y 2 4)2 5 52
The center is (25, 3). The radius is 3.
(x 1 3)2 1 ( y 2 4)2 5 25
y
15. If (h, k) is the center of a circle with radius r, the
equation of the circle should be
(x 2 h)2 1 (y 2 k)2 5 r 2 not (x 1 h)2 1 (y 1 k)2 5 r 2.
(25, 3)
(x 2 (23))2 1 ( y 2 (25))2 5 32
1
(x 1 3)2 1 (y 1 5)2 5 9
21
16. C; r 2 5 16 so r 5 4.
d 5 2r 5 2(4) 5 8
}}
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
25. (x 1 2)2 1 ( y 1 6)2 5 25
}
17. r 5 Ï (0 2 0) 1 (0 2 6) 5 Ï (26) 5 6
2
2
The center is (22, 26). The radius is 5.
2
x 1y 56
2
2
2
y
2
x 1 y 5 36
2
x
2
28
}
}}
18. r 5 Ï (1 2 4)2 1 (2 2 2)2 5 Ï (23)2 5 3
(x 2 1)2 1 (y 2 2)2 5 32
x
(22, 26)
(x 2 1)2 1 (y 2 2)2 5 9
}}
19. r 5 Ï (23 2 1)2 1 (5 2 8)2
}}
5 Ï(24)2 1 (23)2 5 5
26. D; (x 1 2)2 1 ( y 2 4)2 5 25
(0 1 2)2 1 (5 2 4)2 0 25
22 1 12 0 25
5 Þ 25
(x 2(23)) 1 (y 2 5) 5 5
2
2
2
(x 1 3)2 1 (y 2 5)2 5 25
20. x 2 1 y 2 5 49
21. (x 2 3)2 1 y 2 5 16
The center is (0, 0).
The radius is 7.
y
2
22
2
y
1
(0, 0)
x
(3, 0)
1
x 2 1 y 2 2 6y 1 9 5 4
x 1 ( y 2 2 6y 1 9) 5 4
x 2 1 (y 2 3)2 5 4
The equation is a circle.
28.
x 2 2 8x 1 16 1 y 2 1 2y 1 4 5 25
2
(x 2 8x 1 16) 1 ( y 2 1 2y 1 1) 5 25 2 3
(x 2 4)2 1 ( y 1 1)2 5 22
The equation is a circle.
27.
The center is (3, 0).
The radius is 4.
x
Geometry
Worked-Out Solution Key
325
Chapter 10,
continued
x 2 1 y 2 1 4y 1 3 5 16
x 2 1 ( y 2 1 4y 1 3 1 1) 5 16 1 1
x 2 1 ( y 2 1 4y 1 4) 5 17
x 2 1 ( y 1 2)2 5 17
The equation is a circle.
30.
x 2 2 2x 1 5 1 y2 5 81
(x 2 2 2x 1 5 2 4) 1 y 2 5 81 2 4
(x 2 2 2x 1 1) 1 y 2 5 77
(x 2 1)2 1 y 2 5 77
The equation is a circle.
31. (x 2 4)2 1 ( y 2 3)2 5 9
The center is (4, 3).
y 5 23x 1 1
3 0 23(4) 1 6
3 Þ 26
The line does not contain the center of the circle. Solve
for y in both equations and set them equal to find points
of intersection.
(x 2 4)2 1 (y 2 3)2 5 9
29.
( y 2 3) 5 9 2 (x 2 4)
2
y 2 3 5 Ï 9 2 (x 2 4)
2
y 5 3 1 Ï9 2 (x 2 4)
2
}}
23x 1 6 5 3 1 Ï9 2 (x 2 4)2
}}
23x 1 3 5 Ï 9 2 (x 2 4)2
(23x 1 3) 5 9 2 (x 2 4)
2
9x 2 2 18x 1 9 5 9 2 x2 1 8x 2 16
10x 2 26x 1 16 5 0
2
5x2 2 13x 1 8 5 0
Use the quadratic equation to solve for x.
}}
13 6 Ï(213)2 2 4(5)(8)
2(5)
x 5 }}
13 6 3
10
x 5 } l x 5 1 or x 5 1.6
Because x has two solutions, the line intersects the circle
at two points. The line is a secant.
32. (x 1 2)2 1 ( y 2 2)2 5 16
The center is (22, 2).
y 5 2x 2 4
2 0 2(22) 2 4
2 Þ 28
The line does not contain the center of the circle. Solve
for y in both equations and set them equal to find points
of intersection.
Geometry
Worked-Out Solution Key
}}
y 5 2 1 Ï16 2 (x 1 2)2
}}
2x 2 4 5 2 1 Ï16 2 (x 1 2)2
}}
2x 2 6 5 Ï16 2 (x 1 2)2
(2x 2 6)2 5 16 2 (x 1 2)2
4x 2 24x 1 36 5 16 2 x 2 2 4x 2 4
2
5x2 2 20x 1 24 5 0
Use the quadratic equation to solve for x.
}}
20 6 Ï(220)2 2 4(5)(24)
2(5)
x 5 }}
}
20 6 Ï280
x5}
10
The square root of a negative number does not exist.
There are no solutions, so the line does not intersect the
circle. The line is not a secant or a tangent.
33. (x 2 5)2 1 ( y 1 1)2 5 4
1
}}
326
}}
y 2 2 5 Ï16 2 (x 1 2)2
The center is (5, 21).
}}
2
( y 2 2)2 5 16 2 (x 1 2)2
y 5 }5 x 2 3
21 0 }5 (5) 2 3
1
21 Þ 22
The line does not contain the center of the circle.
At x 5 5,
1
1
y 5 }5 x 2 3 5 }5 (5) 2 3 5 22.
The point (5, 22) is in the interior of the circle, so the
line is a secant.
34. (x 1 3)2 1 ( y 2 6)2 5 25
The center is (23, 6).
4
y 5 2}3 x 1 2
4
6 0 2}3(23) 1 2
656
The line contains the center of the circle, so it is a secant
that contains a diameter.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
(x 1 2)2 1 ( y 2 2)2 5 16
Chapter 10,
continued
35. Let radius (O be k, then center (C: (15k, 0),
c. City B has better cell phone coverage because it is
point: (63, 16).
located entirely within the range of Tower Z. City A is
only partially covered by Tower X and Tower Y.
}}}
r 5 Ï(15k 2 63) 1 (0 2 16)
2
2
}}
r 5 Ï225k2 2 1890k 1 4225
43. a. The center C can be found at the intersection of the
lines perpendicular to the tangents.
}}
4k 5 Ï225k2 2 1890k 1 4225
4
y 5 2}3 x 1 b
16k2 5 225k2 2 1890k 1 4225
4
5 5 2}3(4) 1 b
0 5 209k2 2 1890k 1 4225
31
3
}}
}5b
Ï18902 2 4(209)(4225)
k 5 }}
2(209)
1890 6 200
k5}
l k 5 5 or k 5 4.04
418
4
y 5 }3 x 1 b
k 5 5 because k is an integer.
4
13 5 2}3(4) 1 b
(x 2 15)2 1 y 2 5 100
23
3
}5b
Problem Solving
36. a.
(6, 5) is in Zone 3.
City
Center
x
2
4
8 5 8x
15x
The center C is at (1, 9).
}}
}
r 5 Ï(1 2 4)2 1 (9 2 5)2 5 Ï(23)2 1 42 5 5
b.
x 2 1 y 2 5 5.76
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
23
4
5 }3 (1) 1 }
59
y 5 }3 x 1 }
3
3
x 2 1 y 2 5 2.42
center: (1, 9)
radius: 5
1
r 5 }2 d 5 }2 (0.6) 5 0.3
(x 2 1)2 1 ( y 2 9)2 5 25
x 1 y 5 0.3
2
23
4
Zone 3
1
1
37. r 5 } d 5 }(4.8) 5 2.4
2
2
2
23
4
24x 1 31 5 4x 1 23
(0, 3) is in Zone 1.
(1, 6) is in Zone 2.
Zone 2
1
31
2}3 x 1 }
5 }3 x 1 }
3
3
(1, 2) is in Zone 1.
2
23
4
y 5 }3 x 1 }
3
b. (3, 4) is in Zone 2.
y
Zone 1
31
4
y 5 2}3 x 1 }
3
2
y
x 1 y 5 0.09
2
2
38. center: (23, 0)
(1, 9)
radius: AD 5 1
(x 2 (23))2 1 ( y 2 0)2 5 12
2
(x 1 3)2 1 y 2 5 1
22
39. center: (3, 0)
radius: BD 5 7
44. Given: A circle passing through the points (21, 0) and
(x 2 3)2 1 (y 2 0)2 5 72
(1, 0).
Prove: The equation of the circle is x 2 2 2yk 1 y 2 5 1
with center at (0, k).
(x 2 3) 1 y 5 49
2
2
40. Answers will vary.
41. The height (or width) always remains the same as the
figure is rolled on its edge.
42. a.
y
4
24
X
x
Y
Z
x
Yes, cell towers X and Y
cover part of the same area,
so this area may receive
calls from Tower X or
Tower Y.
b. Your home is within range of Tower Y, but your school
is not in the range of any tower. So, you can use your
phone at home.
Construct the perpendicular bisector of chord with the
endpoints (21, 0) and (1, 0). Using Theorem 10.4, this
new chord is a diameter of the circle. Since the new chord
is a segment of the y-axis, the center of the circle is located
at some point (0, k) which makes the equation of the circle
x 2 1 (y 2 k)2 5 r 2 or x 2 1 y 2 2 2yk 5 r 2 2 k2. Now
consider the right triangle whose vertices are (0, 0),
(0, k), and (1, 0) with the distance from (0, k) to (1, 0)
being r, the radius of the circle. Using the Pythagorean
Theorem, you get k2 1 12 5 r 2 or r 2 2 k2 5 1.
Substituting you get x 2 2 2yk 1 y 2 5 1.
Geometry
Worked-Out Solution Key
327
continued
45. a. If r 5 2, there are two possible points of intersection.
b. The locus of intersection points of m and n for all
possible values of r for (C is the intersection of all
circles centered (8, 6) and (10, 8).
Use Theorem 10.1. If the center of the circle is
(8, 8), then the point of intersection for m and n is
(10, 6). If the center of the circle is (10, 6), then the
point of intersection is (8, 8). If r 5 10, these are
also two possible points of intersection. (C has two
possible centers. To find these centers, set the equation
of the circle with radius 10 and center (8, 6) equal to
the equation of the circle with radius 10 and center
(10, 8).
(x 2 8)2 1 ( y 2 6)2 5 (x 2 10)2 1 ( y 2 8)2
x 2 2 16x 1 64 1 y 2 2 12y 1 36
5 x 2 2 20x 1 100 1 y 2 2 16y 1 64
4x 1 4y 2 64 5 0
x 1 y 2 16 5 0
y
rm 5 10 center (8, 6) (x 2 8)2 1 (y 2 6)2 5 100
}}
l y 5 6 1 Ï100 2 (x 2 8)2
rn 5 10
center (10, 8) (x 2 10)2 1 (y 2 8)2 5 100
}}
l y 5 8 1 Ï100 2 (x 2 10)2
}}
(8, 6)
}}
6 1 Ï 100 2 (x 2 8)2 5 8 1 Ï100 2 (x 2 10)2
}}
}}
24x 1 32
4
}}
} 5 Ï 100 2 (x 2 10)2
(2x 1 8)2 5 100 2 (x 2 10)2
Mixed Review for TAKS
46. C; All the triangles shown are equilateral, equiangular, and
isosceles. So, it is not true that all the triangles are scalene.
47. H; By Theorem 5.10, the largest angle is opposite the
largest side. So, the rosebush is planted in the corner
opposite the side that is 15 feet.
x 2 18x 1 32 5 0
(x 2 2)(x 2 16) 5 0
Possible centers for (C are (2, 14) and (16, 0). If the
Quiz 10.6–10.7 (p. 705)
1. 6 + x 5 8 + 9
2. 7 + (7 1 5) 5 6 + (6 1 x)
6x 5 72
4
14
and
center of (C is (16, 0), then line m is y 5 }3 x 2 }
3
3
1
line n is y 5 }4 2 x 1 }2.
7(12) 5 36 1 6x
x 5 12
48 5 6x
85x
3. 162 5 12(x 1 12)
3
4
14
1
}x 2 } 5 }x 1 }
3
3
4
2
16x 2 56 5 9x 1 6
6
1
x 5 8 }7 l y 5 7 }7
4. center: (1, 4)
256 5 12x 1 144
radius: 6
112 5 12x
(x 2 1)2 1 (y 2 4)2 5 62
28
3
}5x
The point of intersection is 1 8 }7, 7 }7 2.
6
1
(x 2 1)2 1 (y 2 4)2 5 36
5. center: (5, 27)
point: (5, 23)
3
If the center of (C is (2, 14), then line m is y 5 }4 x
16
4
and line n is y 5 }3 x 2 }
.
3
4
3
x
m
2
x 5 2 or x 5 16
n
C
2
100 2 (x 2 8)2 5 4 1 4Ï100 2 (x 2 10)2
1 100 2 (x 2 10)2
}}
(x 2 10)2 2 (x 2 8)2 2 4
}} 5 Ï 100 2 (x 2 10)2
4
(10, 8)
r52
2
100 2 (x 2 8)2 5 (2 1 Ï 100 2 (x 2 10)2 )2
3
4
r 5 10
}}}
(x 2 5)2 1 (y 2 (27))2 5 42
(x 2 5)2 1 (y 1 7)2 5 16
16
3
}x 5 }x 2 }
6.
center of tire: (24, 3)
y
9x 5 16x 2 64
radius of tire: 12.1
6
1
x 5 9 }7 l y 5 6 }7
4
The point of intersection is 1 9 }7, 6 }7 2.
1
6
The points of intersection and the center of (C are
collinear.
24
x
(x 2 (24))2 1 (y 2 3)2 5 12.12
(x 1 4)2 1 ( y 2 3)2 5 146.21
center of rim: (24, 3)
radius of rim: 7
(x 2 (24))2 1 ( y 2 3)2 5 72
(x 1 4)2 1 ( y 2 3)2 5 49
328
Geometry
Worked-Out Solution Key
}
r 5 Ï(5 2 5)2 1 (27 2 (23))2 5 Ï(24)2 5 4
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 10,
Chapter 10,
continued
Mixed Review for TEKS (p. 706)
5. B;
182 5 6(6 1 2r)
1. C;
324 5 36 1 12r
Center: (20, 30)
288 5 12r
Radius: 20
24 5 r
(x 2 20) 1 ( y 2 30) 5 20 5 400
2
2
2
The equation (x 2 20) 1 ( y 2 30) 5 400 represents
the broadcast boundary of the radio station.
2
2
The radius of the tank is 24 feet.
6. 6 + x 5 9 + 8
6x 5 72
2. F;
x + 43 5 41 + 42
x 5 12
So, the expression 41 + 42 4 43 gives the value x must
be for the track to be a perfect circle.
132 5 y( y 1 6 1 x)
169 5 y( y 1 6 1 x)
3. G; The standard equation of a circle with radius 5 and
169 5 y( y 1 6 1 12)
center (5, 6) is:
169 5 y( y 1 18)
(x 2 5)2 1 ( y 2 6)2 5 52
169 5 y 2 1 18y
(x 2 5)2 1 ( y 2 6)2 5 25
(x 2 5) 1
2
F1
4
3
2
3
2
}x 2 } 2 6
0 5 y 2 1 18y 2 169
G 5 25
2
}}
218 6 Ï(18)2 2 4(1)(2169)
2(1)
y 5 }}}
5 25
(x 2 5)2 1 1 }3 x 2 }
32
4
16
160
20 2
}
218 6 Ï 324 1 676
5 }}
2
400
x2 2 }
x1}
5 25
x 2 2 10x 1 25 1 }
9
9
9
25
9
250
9
625
9
25
9
250
9
400
9
}
218 6 Ï 1000
5 }}
2
} x 2 2 } x 1 } 5 25
}
218 1 Ï1000
ø 6.8 and
The solutions are }}
2
}x2 2 }x 1 } 5 0
}
218 2 Ï1000
2
}} ø 224.8.
25
9
} (x 2 2 10x 1 16) 5 0
The value of y is about 6.8 units.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
25
} (x 2 2)(x 2 8) 5 0
9
x2250
or
x2850
x52
or
x58
4
2
4
Chapter 10 Review (pp. 708–711)
1. If a chord passes through the center of a circle, then it is
called a diameter.
2
y 5 }3 (2) 2 }3
y 5 }3 (8) 2 }3
y52
y 5 10
2.
A
O
The two points of intersection of the line and the circle
are (2, 2) and (8, 10).
4. D;
Center: (6, 8)
Radius: 5
B
C
An inscribed angle is an angle whose vertex is on the
circle and whose sides contain chords of the circle. The
arc that lies in the interior of an inscribed angle and has
endpoints on the angle is called the intercepted arc of the
angle.
Equation of the boundary of the receiver:
3. The measure of the central angle and the corresponding
(x 2 6) 1 ( y 2 8) 5 5
minor arc are the same. The measure of the major arc is
3608 minus the measure of the minor arc.
}
4. B; KL is a tangent segment.
}
5. C; LN is a secant segment.
}
6. A; LM is an external segment.
2
2
2
(x 2 6)2 1 ( y 2 8)2 5 25
(11, 10):
(11 2 6)2 1 (10 2 8)2a25
52 1 22a
1a
÷
The point (11, 10) is not within range of the receiver.
Geometry
Worked-Out Solution Key
329
continued
1
21. 408 5 }(968 2 x8)
2
XY 5 XZ
7.
9a 2 30 5 3a
2
23. 20 5 12 + (12 1 2r)
400 5 144 1 24r
a52
256 5 24r
Length cannot be negative, so a 5 2.
2
10 }3 5 r
XY 5 XZ
2
The radius of the rink is 10 }3 feet.
2c2 1 9c 1 6 5 9c 1 14
2c2 2 8 5 0
24. center: (4, 21)
2(c2 2 4) 5 0
radius: 3
(c 2 2)(c 1 2) 5 0
(x 2 4)2 1 (y 2 (21))2 5 32
c 2 2 5 0 or c 1 2 5 0
(x 2 4)2 1 (y 1 1)2 5 9
c 5 22
25. center: (8, 6)
Length cannot be negative, so c 5 2.
radius: 6
9. WZ 2 1 XZ 2 5 WX 2
(x 2 8)2 1 (y 2 6)2 5 62
r 2 1 92 5 (r 1 3)2
(x 2 8)2 1 (y 2 6)2 5 36
r 1 81 5 r 1 6r 1 9
2
2
26. center: (0, 0)
72 5 6r
radius: 4
12 5 r
C
CM 5 1808 2 mC
11. mL
MN 5 1808 2 1208 5 608
12. mC
KM 5 mC
KL 1 mC
LM 5 1008 1 608 5 1608
CN 5 1808 2 mKCL 5 1808 2 1008 5 808
13. mK
CD 5 mŽECD 5 618
14. mE
CB 5 mECD 5 618
mA
CB 5 mACD 5 658 16. mACB 5 mECD 5 918
15. mA
1
1
CC
CZ
17. mŽYXZ 5 mY
18. mŽ BAC 5 mB
2
2
x 2 1 y 2 5 42
10. mKL 5 mŽKPL 5 1008
}
}
1
c8 5 }2 (568)
c 5 28
19. mŽE 1 mŽG 5 1808
1
408 5 }2 x8
x 2 1 y 2 5 16
27. x 2 1 y 2 5 92
x 2 1 y 2 5 81
28. (x 2 (25))2 1 ( y 2 2)2 5 1.32
(x 1 5) 1 ( y 2 2)2 5 1.69
29. (x 2 6)2 1 (y 2 21)2 5 42
(x 2 6)2 1 (y 2 21)2 5 16
30. (x 2 (23))2 1 (y 2 2)2 5 162
(x 1 3)2 1 ( y 2 2)2 5 256
31. (x 2 10)2 1 (y 2 7)2 5 3.52
80 5 x
(x 2 10)2 1 (y 2 7)2 5 12.25
32. x 2 1 y 2 5 5.22
x 2 1 y 2 5 27.04
q8 1 808 5 1808
q 5 100
mŽF 1 mŽD 5 1808
1008 1 4r8 5 1808
4r 5 80
4 5 20
1
20. x8 5 } [2508 2 (3608 2 2508)]
2
1
x 5 }2 (250 2 110)
1
x 5 }2 (140)
x 5 70
330
Geometry
Worked-Out Solution Key
x 5 106
2
3a 1 5 5 0 or a 2 2 5 0
c52
x 5 }2 (212)
x 5 16
(3a 1 5)(a 2 2) 5 0
8.
1
80 5 96 2 x
3(3a 2 a 2 10) 5 0
2
5
a 5 2}3
1
22. x8 5 } (1528 1 608)
2
Chapter 10 Test (p. 712)
1.
AB 5 AD
5x 2 4 5 3x 1 6
2x 5 10
x55
2. AD2 1 DC 2 5 AC 2
122 1 r 2 5 (r 1 6)2
144 1 r 2 5 r 2 1 12r 1 36
108 5 12r
95r
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 10,
Chapter 10,
continued
AB 5 AD
3.
18.
2x 2 1 8x 2 17 5 8x 1 15
x 2 1 28x 5 1040
2x 2 5 32
x 2 1 28x 2 1040 5 0
}}
228 6 Ï282 2 4(1)(21040)
2(1)
x 5 16
2
x 5 }}}
x54
C
C
CB > C
A
CD
CG 5 mŽFHG 5 1368
5. mF
L 5 3608 2 2248 5 1368
mJC
CG > JCL
F
CN is not congruent to QCR because they are not part of
6. M
}
228 6 Ï4944
4. mCD 5 mŽCED 5 608 5 mAB
two congruent circles.
}
}
}
7. AB is perpendicular to and bisects CD, so AB is a
diameter.
}
}
}
8. AB is perpendicular to and bisects CD, so AB is a
diameter.
9. 202 1 ZY 2 5 252
ZY 2 5 225
}
ZY 5 15
}
}
AB is perpendicular to but does not bisect XY, so AB is
not a diameter.
C
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x(x 1 28) 5 20(20 1 32)
x 5 }}
2
x ø 21.2
19.
(x 1 2)2 1 (y 2 5)2 5 169
(x 2 (22))2 1 (y 2 5)2 5 132
The center is (22, 5) and the radius is 13.
Chapter 10 Algebra Review (p. 713)
1. 6x 2 1 18x 4 5 6x 2 (1 1 3x2)
2. 16a2 2 24b 5 8(2a2 2 3b)
3. 9r 2 2 15rs 5 3r(3r 2 5s)
4. 14x 5 1 27x 3 5 x 3(14x2 1 27)
5. 8t 4 1 6t 2 2 10t 5 2t(4t 3 1 3t 2 5)
6. 9z3 1 3z 1 21z2 5 3z(3z2 1 1 1 7z)
7. 5y 6 2 4y 5 1 2y 3 5 y 3(5y3 2 4y 2 1 2)
8. 30v 7 2 25v 5 2 10v 4 5 5v 4(6v 3 2 5v 2 2)
9. 6x3y 1 15x 2y 3 5 3x 2y(2x 1 5y2)
1
1
10. mŽABC 5 } mAC 5 }(1068) 5 538
2
2
C
12. mC
JG 5 2mŽJHG 5 2(438) 5 868
C
HJ 5 3608 2 mC
JG 5 3608 2 868 5 2748
mG
10. x 2 1 6x 1 8 5 (x 1 4)(x 1 2)
11. mFD 5 2mŽFED 5 2(828) 5 1648
11. y 2 2 y 2 6 5 (y 2 3)(y 1 2)
1
13. mŽ1 5 }(2388) 5 1198
2
14. 3s 2 1 2s 2 1 5 (3s 2 1)(s 1 1)
1
1
14. mŽ2 5 }(528 1 1128) 5 } (1648) 5 828
2
2
C C
1
CC )
428 5 2 (1688 2 mA
CC
848 5 1688 2 mA
CC 5 848
mA
1
15. mŽB 5 }(mAD 2 mAC )
2
}
16. 8 + x 5 14 + 4
8x 5 56
x57
17. 122 5 9(9 1 x)
144 5 81 1 9x
63 5 9x
75x
12. a2 2 64 5 (a 2 8)(a 1 8)
13. z2 2 8z 1 16 5 (z 2 4)(z 2 4) 5 (z 2 4)2
15. 5b2 2 16b 1 3 5 (5b 2 1)(b 2 3)
16. 4x 4 2 49 5 (2x 2 2 7)(2x 2 1 7)
17. 25r 2 2 81 5 (5r 2 9)(5r 1 9)
18. 4x 2 1 12x 1 9 5 (2x 1 3)(2x 1 3) 5 (2x 1 3)2
19. x2 1 10x 1 21 5 (x 1 3)(x 1 7)
20. z2 2 121 5 (z 2 11)(z 1 11)
21. y 2 1 y 2 6 5 ( y 1 3)(y 2 2)
22. z 2 1 12z 1 36 5 (z 1 6)(z 1 6) 5 (z 1 6)2
23. x 2 2 49 5 (x 1 7)(x 2 7)
24. 2x 2 2 12x 2 14 5 2(x 2 2 6x 2 7) 5 2(x 2 7)(x 1 1)
TAKS Practice (pp. 716–717)
1. C; The graph shows that as time increases, distance
increases linearly, then stays constant, then increases
linearly again. So, the graph represents a car that moves
at a constant speed, then stops for a period of time, then
moves at a constant speed again.
2. H; The volume depends on the side length of the base s
and the height h.
3. D; The vertex of the parabola is (0, 21), not (0, 1). The
minimum value is 21, not 0. The parabola does not have
a maximum value. The axis of symmetry is the y-axis.
Geometry
Worked-Out Solution Key
331
Chapter 10,
continued
4. J; 38 is contained in the corresponding set for the
dependent variable of f(x) 5 3x 2 2 10 because 4 is
contained in the replacement set for the independent
variable x, and f(4) 5 3(4)2 2 10 5 3(16) 2 10 5
48 2 10 5 38.
5. C; The vertex of the parabola is (21, 5), not (23, 0). The
axis of symmetry is x 5 21, not y 5 21. The parabola
does not have a minimum value. The maximum value is 5.
6. H; For each value of x shown in the table, the
corresponding value of f(x) is 3 less than the square of x.
So, the expression x 2 2 3 corresponds to the function f(x)
in the table.
13. C;
Given: A(1, 1), B(2, 3), C(3, 2)
Given: L(3, 0), M(5, 4)
y2 2 y1
321
}
2
Slope of AB: m 5 }
5}
5 }1 5 2
x 2x
221
2
420
4
}
5}
5 }2 5 2
Slope of LM: m 5 }
523
x 2x
2
y2 2 y1
223
}
21
Slope of BC: m 5 }
5}
5}
5 21
x 2x
322
1
}
221
1
5}
5 }2
Slope of AC: m 5 }
x 2x
321
2
3(x 2 1 3x 2 18) 5 0
or
x1650
x 5 26
The roots of the quadratic equation are 3 and 26.
8. H; The rate of change is the slope.
y2 2 y1
1 2 (22)
3
m5}
5 } 5 }5
x2 2 x1
2 2 (23)
y2 2 y1
y24
}
Slope of MN: m 5 }
5}
5 21
x25
x 2x
2
1
y 2 4 5 21(x 2 5)
y 2 4 5 2x 1 5
3
5
y 5 2x 1 9
The rate of change is }.
y2 2 y1
9. C;
Number Price
Price
Total dollar
Number
of hats + of 1 1 of T-shirts + of 1 5 amount
sold
hat
T-shirt
sold
sold
12
+ x
1
10
+
y
5
158
8
+ x
1
15
+
y
5
172
10. F;
6(x 2 2 2x 1 4) 1 3x 2 3x(7 1 2x)
5 6x 2 2 12x 1 24 1 3x 2 21x 2 6x 2
5 230x 1 24
5 24 2 30x
11. B;
In a right triangle, a 2 1 b2 5 c 2.
3.32 1 5.62 0 6.52
10.89 1 31.36 0 42.25
42.25 5 42.25 The side lengths 3.3, 5.6, and 6.5 can be the side lengths
of a right triangle.
12. J;
V 5 Bh
5.25 5 B(1.5)
5.25
1.5
}5B
3.5 5 B
The area of one of the bases of the box is
3.5 square inches.
332
Geometry
Worked-Out Solution Key
y20
y
1
}
5}
5}
5 }2
Slope of LN: m 5 }
x23
x2 2 x1
x23
2y 5 x 2 3
2(2x 1 9) 5 x 2 3
22x 1 18 5 x 2 3
21 5 3x
75x
y 5 2x 1 9 5 27 1 9 5 2
Vertex N should be placed at the coordinates (7, 2).
14. Of the 600 students surveyed, 16% or 0.16(600) 5 96
people said that sausage is their favorite topping.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x53
1
If n ABC and nLMN are similar and oriented the same
}
}
way, then the coresponding sides BC and MN have the
}
}
same slope and the corresponding sides AC and LN have
the same slope. If the coordinates of point N are (x, y):
3(x 2 3)(x 1 6) 5 0
or
1
y2 2 y1
3x 2 1 9x 2 54 5 0
x2350
1
}
}
The corresponding sides AB and LM have the
same slope.
2
7. B;
1
y2 2 y1