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CHAPTER 11
Geometry, Concepts and Skills 185Chapter 11 Worked-Out Solution Key
FH**** is a diameter; FG**** is a radius (as is GH**** );
G is the center; B is a point of tangency.
2. (0, 4)
11.1 Guided Practice (p. 591)
1. Sample answer:
2. B 3. E 4. D 5. A 6. C 7. F
8. (3, 3) 9. (3, 0) and (3, 6)
10. (0, 3), (3, 0), or (6, 3) 11. (3, 0) and (0, 3)
12. (3, 3) and one of the following: (3, 0), (6, 3), (3, 6)
11.1 Practice and Applications (pp. 591–593)
13. r � �12
�d � �12
�(15) � 7.5 cm
14. r � �12
�d � �12
�(6.5) � 3.25 in.
15. r � �12
�d � �12
�(3) � 1.5 ft 16. r � �12
�d � �12
�(8) � 4 m
17. d � 2r � 2(26) � 52 in. 18. d � 2r � 2(62) � 124 ft
19. d � 2r � 2(8.7) � 17.4 m 20. d � 2r � 2(4.4) � 8.8 cm
21. chord 22. tangent 23. diameter 24. radius
25. point of tangency 26. center of the circle
tangent
diameter
radius
chord
27. chord 28. radius 29. diameter 30. secant
31. tangent 32. secant
33. EG**** is a chord (as is EF**** );
EG&̂( is a secant;
EF**** is a diameter;
CE**** is a radius (as are CF**** and CG**& );
D is a point of tangency.
34. MN**& is a chord (as is JL**** );
MN^&( is a secant;
JL**** is a diameter;
KR**** is a radius (as are KJ**** and KL**** );
U is a point of tangency.
35. LM**** is a chord (as is PN**** );
LM&̂( is a secant;
PN**** is a diameter;
QR**** is a radius (as are QP**** and QN**** );
K is a point of tangency.
36. FA&̂( and EB&̂( are secants.
37. Any two of GD**&, HC**** , FA**** , and EB**** are chords.
38. Yes; the diameter is the longest chord and must passthrough the center of the circle. Since HC**** does not passthrough the center, it is shorter than the diameter.
39. Yes, a line can be drawn through points K, G, and J thatwill be tangent to the circle.
40. The center of �A is (2, 2); the center of �B is (6, 2).
41. The length of the radius of �A is 2 units; the length ofthe radius of �B is 2 units.
So, TABC is not a right triangle and AB**** is not perpen-dicular to AC**** . Therefore, AB**** is not tangent to �C.
21. No; AB2 � AC2 � BC2
142 � 52 � 152
196 � 25 � 225
221 � 225
So, TABC is not a right triangle and AB**** is not perpen-dicular to AC**** . Therefore, AB**** is not tangent to �C.
22. AB**** � AD**** , BC**** � DC****
23. aABC � aADC, aBAC � aDAC, aDCA � aBCA
24. T ABC �T ADC
25.
26. Yes. Explanations may vary. Sample answer: JM**** � KM****since tangent segments from the same point are congruent.LJ**** � LK**** since all the radii of a circle are congruent.LM**** � LM**** by the Reflexive Property of Congruence.Then T JLM �T KLM by the SSS Congruence Postulate.
27. Let R represent the center of Earth.
BA2 � AR2 � BR2
BA2 � 39602 � (12,500 � 3,960)2
BA2 � 15,681,600 � 270,931,600
BA2 � 255,250,000
BA � 15,977 miles
BA � BC � 15,977 miles
28. BC2 � FB2 � CF2
39602 � FB2 � 3960.22
15,681,600 � FB2 � 15,683,184.04
FB2 � 1584.04
FB � 39.8
BE2 � CB2 � CE2
BE2 � 39602 � 3960.012
BE2 � 15,681,600 � 15,681,679.2
BE2 � 79.2
BE � 9
FE � BE � FB
� 9 � 39.8
� 49 miles
M
K
J
L
Geometry, Concepts and Skills 187Chapter 11 Worked-Out Solution Key
Chapter 11 continued
188 Geometry, Concepts and SkillsChapter 11 Worked-Out Solution Key
3. The measure of an arc is the degree measure of the related central angle (or 360� minus the measure of therelated central angle), while an arc length is a portion ofthe circumference of a circle.
42. If two cities differ by 180� on the wheel, then it is 3:00P.M. in one city when it is 3:00 A.M. in the other city.
43. No; the circles are not congruent.
44. Yes; aACD � aBCE since they are vertical angles;
mADs � mBEs, so ADs � BEs.
45. Yes; UWs and XZs are arcs of congruent circles with thesame measure.
46. No; F is not the center of the circle, so you cannot determine the measures of JKs and GHs.
47. Length of ABr� �34650�
�� � 2π(3) � �
34
�π � 2.36 cm
48. Length of ABr� �36600�
�� � 2π(7) � �
73
�π � 7.33 in.
49. Length of ABr� �13
26
00
�
�� � 2π(10) � �
230�π � 20.94 ft
50. Length of ABr� �33600�
�� � 2π(4) � �
23
�π � 2.09 cm
51. Length of ABr� �37650�
�� � 2π(12) � 5π � 15.71 m
52. Length of ABr� �13
56
00
�
�� � 2π(6) � 5π � 15.71 in.
53. No; they have the same arc length only if the two circlesare congruent circles.
54. Arc length � �03.6406�
�� � 2π(5588) � 45 cm
11.3 Standardized Test Practice (p. 607)
55. C; length of ACs � �180
3�
6�
0�
40��� 2π(8) � 19.6 ft
11.3 Mixed Review (p. 607)
56. sin 56� � �4x
� 57. sin 34� � �9x
�
4 sin 56� � x 9 sin 34� � x
3.3 � x; 5.0 � x;
cos 56� � �4y
� cos 34� � �9y
�
4 cos 56� � y 9 cos 34� � y
2.2 � y 7.5 � y
58. sin 48� � �1y4�
14 sin 48� � y
10.4 � y;
cos 48� � �1x4�
14 cos 48� � x
9.4 � x
11.3 Algebra Skills (p. 607)
59. �4
200
kmkm� � �
4200
�
�
22
� � �2
100�
60. �752
fint.
� � �5
7�
212
ini.n.
� � �67
02
iinn
.
.� � �
67
02
�
�
11
22
� � �56
�
61. �3
2y7ar
fdts
� � �32�
73ftft
� � �297
fftt
� � �297
�
�
99
� � �13
�
62. �84
lobzs
� � �8 �
41o6zoz
� � �1
428
ozoz
� � �1
428
�
�
44
� � �312�
Quiz 1 (p. 607)
1. tangent 2. secant 3. diameter 4. chord
5. radius 6. point of tangency
Geometry, Concepts and Skills 189Chapter 11 Worked-Out Solution Key
Chapter 11 continued
7. QP � RP
x � 15
8. QP � RP 9. PQ � PR
x � 1 � 2x � 7 3x � 8 � x
�x � �6 2x � 8
x � 6 x � 4
10. Length of ABr� �37600�
�� � 2π(3) � �
76
�π � 3.67 cm
11. Length of ABr� �13
56
00
�
�� � 2π(7) � �
365�π � 18.33 m
12. Length of ABr� �32650�
�� � 2π(10) � �
21
58�π � 4.36 ft
Lesson 11.4
11.4 Checkpoint (pp. 608–610)
1. JM � HM � 12
2. SR � SN � NR 3. ED � AB
� 15 � 15 x � 4
� 30
4. HG � FG 5. maZWY � maVWU
x � 2 � 5 (x � 10)� � 40�
x � 3 x � 30
11.4 Guided Practice (p. 610)
1. BE**** is a diameter.
2. PQs and RSr are congruent arcs.
3. PT � RT 4. maMHN � maJHK
x � 1 � 6 (x � 5)� � 50�
x � 7 x � 45
5. DE � AB
x � 8
11.4 Practice and Applications (pp. 611–612)
6. No; AB**** is not perpendicular to CD**** , so AB**** is not a diameter of the circle.
7. No; AB**** does not bisect CD**** , so AB**** is not a diameter ofthe circle.
8. Yes; AB**** is a perpendicular bisector of CD**** , so AB**** is adiameter of the circle.
9. AB � DE 10. maACB � maDCE
x � 7 35� � (x � 6)�
29 � x
11. DB � AB
3x � 1 � x � 7
2x � 6
x � 3
12. AB**** � BC**** (given) and ABs � BCs (If two chords are congruent, then their corresponding minor arcs are congruent.)
13. ABs � CDs (given) and AB**** � CD**** (If two minor arcs are congruent, then their corresponding chords are congruent.)
14. AB**** � CD**** (given) and ABs � CDs (If two chords are congruent, then their corresponding minor arcs are congruent.) aAQB � aCQD by the definition of themeasure of a minor arc.
15. DF � FE � 10 16. mDCs � mBDs � 40�
17. mDCs � mEBs � 110� 18. mBAr� mEDs
mEDCt � mEDs � mDCs (2x � 10)� � (x � 30)�
� 60� � 110� x � 40
� 170�
19. AF � FB 20. maACB � maDCE
x � 6 � 3x � 8 (x � 45)� � 4x�
14 � 2x 45 � 3x
7 � x 15 � x
21. Answers may vary, but the re-creation should be based onthe method shown in Example 2 on page 609.
22. The searcher is constructing a chord of the beacon’s circle and the perpendicular bisector of the chord, whichis a diameter of the circle. By locating the midpoint ofthe diameter, the searcher locates the center of the circle,which is the location of the beacon.
11.4 Standardized Test Practice (p. 612)
23. a. In a circle, if two chords are congruent, then their corresponding minor arcs are congruent.
b. mADs � mEBs
(15x � 40)� � (10x � 10)�
5x � 50
x � 10
c. mADs � (15x � 40)�
� (15 � 10 � 40)�
� (150 � 40)�
� 110�;
mBEs � mADs � 110�
d. mBDs � 360� � mADs � mBEs � mAEs
� 360�� � 110� � 110� � 40�
� 100�
11.4 Mixed Review (p. 612)
24. mDEs � mABs � 40�
25. mBCs � 180� � mABs � mCDs
� 180� � 40� � 65�
� 75�
190 Geometry, Concepts and SkillsChapter 11 Worked-Out Solution Key
Step 3. Angle measures will vary, but the measures of allthree angles will be equal.
1, 2. Answers will vary, but in each case,
maRTS � maRUS � maRVS � �12
�maRPS.
3. The measure of an inscribed angle is half the measure ofthe corresponding central angle.
11.5 Checkpoint (pp. 615–616)
1. maBAC � �12
�mBCs � �12
�(90�) � 45�
2. maDEF � �12
�mDFs � �12
�(160�) � 80�
3. maKMP � �12
�mKNPt
120� � �12
�mKNPt
240� � mKNPt
4. By Theorem 11.8, the triangle is a right triangle. So, x � 90 and y � 90 � 35 � 55.
5. By Theorem 11.8, the triangle is a right triangle.So, y � 90 and 2x � 90 so x � 45.
6. By Theorem 11.8, the triangle is a right triangle.So, y � 90 and x � 90 � 60 � 30.
7. x� � 95� � 180� 8. x� � 90� � 180�
x � 85; x � 90;
y� � 100� � 180� y� � 90� � 180�
y � 80 y � 90
9. x� � 50� � 180�
x � 130;
y� � 80� � 180�
y � 100
11.5 Guided Practice (p. 617)
1. aA, aB, aC, aD 2. aA and aC, aB and aD
3. maLJK � �12
�mKLs 4. maKJL � �12
�mKMLt
20� � �12
�mKLs 90� � �12
�mKMLt
40� � mKLs 180� � mKMLt
5. maLJK � �12
�mLMKt 6. x� � �12
�(230�)
105� � �12
�mLMKt x � 115
210� � mLMKt
7. 75� � �12
�y� 8. x� � 85� � 180�
150 � y; x � 95;
z� � �12
�(150�) y� � 80� � 180�
z � 75 y � 10
11.5 Practice and Applications (pp. 617–619)
9. maABC � �12
�mACs 10. maABC � �12
�mACs
� �12
�(110�) � �12
�(218�)
� 55� � 109�
11. maABC � �12
�mACs 12. maLMN � �12
�mLNs
� �12
�(180�) � �12
�(68�)
� 90� � 34�
13. maPQR � �12
�mPRs 14. maUTS � �12
�mUSs
� �12
�(134�) � �12
�(238�)
� 67� � 119�
15. maBAC � �12
�mBCs 16. maBAC � �12
�mBCs
32� � �12
�mBCs 78� � �12
�mBCs
64� � mBCs 156� � mBCs
17. maBAC � �12
�mBCs 18. maRST � �12
�mRUTt
114� � �12
�mBCs 120� � �12
�mRUTt
228� � mBCs 240� � mRUTt
Geometry, Concepts and Skills 191Chapter 11 Worked-Out Solution Key
Chapter 11 continued
19. maPQN � �12
�mPNs 20. maXYZ � �12
�mXWZt
50� � �12
�mPNs 103� � �12
�mXWZt
100� � mPNs 206� � mXWZt
21. maEAB � �12
�mBEs 22. maBDE � �12
�mBEs
47� � �12
�mBEs � �12
�(94�)
94� � mBEs � 47�
23. maAED � maDCE � maBDE � 180�
maAED � 80� � 47� � 180�
maAED � 53�
24. maABD � �12
�mADs
(180� � 47� � 80�) � �12
�mADs
53� � �12
�mADs
106� � mADs
25. maABD � �12
�mADs
� �12
�(106�)
� 53�
26. mDEs � mDAr� mABs � mEBs � 360�
mDEs � 106� � 100� � 94� � 360�
mDEs � 60�
27. Yes; Sample answer: maBAC � 47� � maCDE (fromEx. 22) and maDCE � maACB (vertical angles), so T ABC �T DEC by the AA Similarity Postulate.
28. T ABC is an inscribed triangle and AC**** is a diameter,so T ABC is a right triangle with diameter AC**** ; x � 90;y � 90 � 30 � 60.
29. T KLM is an inscribed triangle and KM**& is a diameter,so T KLM is a right triangle with diameter KM**&; x � 90;y � 90 � 40 � 50.
30. T PQR is an inscribed triangle and PR**** is a diameter,so T PQR is a right triangle with diameter PR**** ; y � 90;x � 90 � 58 � 32.
31. Sample Answer: Position the vertex of the tool on the circle and mark the two points at which the sides intersectthe circle; draw a segment to connect the two points,forming a diameter of the circle. Repeat these steps,placing the vertex at a different point on the circle. Thecenter is the point at which the two diameters intersect.
32. x� � 114� � 180� 33. x� � 100� � 180�
x � 66; x � 80;
y� � 92� � 180� y� � 102� � 180�
y � 88 y � 78
34. x� � 115� � 180�
x � 65;
y� � y� � 180�
2y � 180
y � 90
35. Yes; both pairs of opposite angles are right angles, whichare supplementary angles.
36. Yes; both pairs of opposite angles of an isosceles trapezoidare supplementary.
37. No; if a rhombus is not a square, then the opposite anglesare not supplementary.
38. Yes; both pairs of opposite angles are right angles, whichare supplementary angles.
Step 3. The measure of aAEB is 50� for every student.
Step 4. Answers will vary, but the measure of aAEB will
always be �12
��mABr� mCDs �.
Step 5. The measure of an angle formed by intersectingchords of a circle is equal to half the sum of the measures of the intercepted arcs.
11.6 Checkpoint (pp. 621–622)
1. x� � �12
��mABr� mDCs � 2. x� � �12
��mABr� mCDs �
x� � �12
�(190� � 70�) x� � �12
�(66� � 70�)
x � �12
�(260) x � �12
�(136)
x � 130 x � 68
3. 72� � �12
��mABr� mCDs �
72� � �12
�(x� � 99�)
72 � �12
�x � 49.5
22.5 � �12
�x
45 � x
4. PN � PL � PM � PK 5. AB � BC � DB � BE
12 � 6 � 12 � x 6 � 6 � x � 9
72 � 12x 36 � 9x
6 � x 4 � x
6. RV � VT � SV � VU 7. FH � HJ � EH � HG
6 � 4 � x � 8 x � 8 � 4 � 10
24 � 8x 8x � 40
3 � x x � 5
8. CG � GE � DG � GF 9. KN � NM � LN � NJ
5 � 5 � 5 � x 16 � 24 � 12 � x
25 � 5x 384 � 12x
5 � x 32 � x
11.6 Guided Practice (p. 623)
1. Points B and E lie inside the circle.
2. ma1 � �12
��mADs � mBCs � 3. ma1 � �12
��mADs � mBCs �
� �12
�(55� � 65�) � �12
�(88� � 88�)
� �12
�(120�) � �12
�(176�)
� 60� � 88�
4. ma1 � �12
��mBCs � mADs �
� �12
�(110� � 168�)
� �12
�(278�)
� 139�
5. x � 4 � 2 � 6 6. x � 7 � 14 � 8 7. x � 15 � 10 � 18
4x � 12 7x � 112 15x � 180
x � 3 x � 16 x � 12
Geometry, Concepts and Skills 193Chapter 11 Worked-Out Solution Key
Chapter 11 continued
11.6 Practice and Applications (pp. 623–625)
8. C 9. B 10. D 11. A
12. x� � �12
��mBCs � mADs � 13. x� � �12
��mCDs � mABs �
x� � �12
�(134� � 162�) x� � �12
�(75� � 25�)
x � �12
�(296) x � �12
�(100)
x � 148 x � 50
14. x� � �12
��mABs � mDCs �
x� � �12
�(130� � 96�)
x � �12
�(226)
x � 113
15. 55� � �12
��mCDs � mBAr � 16. 59� � �12
�� mABs � mDCs �
55� � �12
�(89� � x�) 59� � �12
�(70� � x�)
55 � 44.5 � �12
�x 59 � 35 � �12
�x
10.5 � �12
�x 24 � �12
�x
21 � x 48 � x
17. 129� � �12
�� mADs � mBCs �
129� � �12
�(x� � 72�)
129 � �12
�x � 36
93 � �12
�x
186 � x
18. Yes; Sample answer: If two chords intersect and themeasure of each angle formed is the same as the measureof the arc intercepted by the angle, then the circle isdivided into two pairs of congruent arcs since verticalangles are congruent. Suppose one pair of arcs has measure x� and an angle intercepting the arc is a1.
Then ma1 � �12
�(x� � x�) � �12
�(2x�) � x�.
The angle formed by the chords has the same measure asits intercepted arc, so it is a central angle.
19. CN � 10 � 15 � 12 20. x � 9 � 12 � 15
10CN � 180 9x � 180
CN � 18 x � 20
21. x � 4 � 6 � 8 22. x � 10 � 14 � 5
4x � 48 10x � 70
x � 12 x � 7
23. 180� � x� � �12
��mBCs � mADs �
180� � x� � �12
�(122� � 32�)
180 � x � �12
�(154)
180 � x � 77
103 � x
24. 180� � x� � �12
��mABs � mDCs �
180� � x� � �12
�(128� � 100�)
180 � x � �12
�(228)
180 � x � 114
66 � x
25. 180� � x� � �12
��mBCs � mADs �
180� � x� � �12
�(90� � 150�)
180 � x � �12
�(240)
180 � x � 120
60 � x
26. EA � EB � EC � ED; If two chords intersect inside a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.
11.6 Standardized Test Practice (p. 625)
27. a. BC � 9 � 12 � 3
9BC � 36
BC � 4
b. maACB � �12
��mABs � mDEs�
� �12
�(100� � 80�)
� �12
�(180�)
� 90�
c. mAEs � mABs � mBDs � mDEs � 360�
mAEs � 100� � 36� � 80� � 360�
mAEs � 216� � 360�
mAEs � 144�
d. Yes; Sample answer: by the Vertical Angles Theorem,
aACB � aECD. Also, �AC
CE� � �
192� � �
43
� � �CC
DB�
so the two sides that include the congruent angles areproportional. Thus, T ACB �T ECD by the SASSimilarity Theorem.
194 Geometry, Concepts and SkillsChapter 11 Worked-Out Solution Key
4. The measure of an inscribed angle is half the measure ofits intercepted arc;
maABC � �12
�mACs
51� � �12
�x�
102 � x
5. x� � maB � �12
�mADs � maC � 58�;
y� � maD � �12
�mBCs � maA � 41�
6. The opposite angles of an inscribed quadrilateral are supplementary;
maB � maD � 180�
x� � 105� � 180�
x � 75;
maA � maC � 180�
98� � y� � 180�
y � 82
7. x� � �12
��mABs � mDCs �
x� � �12
�(80� � 44�)
x � �12
�(124)
x � 62
8. 135� � �12
��mABs � mDCs �
135� � �12
�(x� � 107�)
135� � �12
�x � 53.5
81.5 � �12
�x
163 � x
9. x � 14 � 7 � 12
14x � 84
x � 6
11.6 Technology Activity (p. 626)
1. Product will vary, but in each case, JH � JG � JK � JL.
2. The relationship JH � JG � JK � JL is true.
3. no
4. If JL&̂( and JG&̂( are secants of a circle intersecting at a pointJ outside the circle and points K, L, G, and H are points on the circle as shown in the diagram on page 626, thenJH � JG � JK � JL.
5. Measures will vary, but in each case,
maKJH � �12
��mLGs� mKHs �. If 2 secants intersect outside
a circle, then the measure of the angle formed is half thepositive difference of the measures of the intercepted arcs.
Lesson 11.7
11.7 Checkpoint (pp. 627–629)
1. x2 � y2 � r22. x2 � y2 � r2
x2 � y2 � 22 x2 � y2 � 32
x2 � y2 � 4 x2 � y2 � 9
3. (x � h)2 � (y � k)2 � r2
(x � (�4))2 � (y � (�6))2 � 52
(x � 4)2 � (y � 6)2 � 25
4. y
x
1
1
Geometry, Concepts and Skills 195Chapter 11 Worked-Out Solution Key
Chapter 11 continued
5.
11.7 Guided Practice (p. 629)
1. C
2. radius 2, center (0, 0); 3. radius 4, center (2, 0);
(x � h)2 � (y � k)2 � r2 (x � h)2 � (y � k)2 � r2
(x � 0)2 � (y � 0)2 � 22 (x � 2)2 � (y � 0)2 � 42
x2 � y2 � 4 (x � 2)2 � y2 � 16
4. radius 2, center (�2, 2);
(x � h)2 � (y � k)2 � r2
(x � (�2))2 � (y � 2)2 � 22
(x � 2)2 � (y � 2)2 � 4
11.7 Practice and Applications (pp. 630–632)
5. B; radius 2, center (3, 0); 6. A; radius 2, center (0, 0);
(x � 3)2 � y2 � 4 x2 � y2 � 4
7. C; radius 2, center (�3, 0);
(x � 3)2 � y2 � 4
8. radius 6, center (0, 0); 9. radius 1, center (0, 0);
10. radius 7, center (2, 6); 11. radius 4, center (4, 3);
12. radius 5, center (5, 1); 13. radius 6, center (�2, 3);
y
x
2
�2
y
x
2
2
y
x
1
1
y
x
2
�2
y
x
2
2
y
x
2
2
y
x
1
1
14. radius 2, center (2, �5); 15. radius 8, center (0, 5);
16. radius 2, center (�3, 2);
(x � h)2 � (y � k)2 � r2
(x � (�3))2 � (y � 2)2 � 22
(x � 3)2 � (y � 2)2 � 4
17. radius 2, center (0, 1); 18. radius 1, center (3, 3);
(x � h)2 � (y � k)2 � r2 (x � h)2 � (y � k)2 � r2
(x � 0)2 � (y � 1)2 � 22 (x � 3)2 � (y � 3)2 � 1
x2 � (y � 1)2 � 4
19. radius 2.5, center (0.5, 1.5);
(x � h)2 � (y � k)2 � r2
(x � 0.5)2 � (y � 1.5)2 � 2.52
(x � 0.5)2 � (y � 1.5)2 � 6.25
20. radius 4, center (2, 2); 21. radius 6, center (0, 0);
(x � h)2 � (y � k)2 � r2 x2 � y2 � r2
(x � 2)2 � (y � 2)2 � 42 x2 � y2 � 62
(x � 2)2 � (y � 2)2 � 16 x2 � y2 � 36
22. (x � h)2 � (y � k)2 � r2
(x � 0)2 � (y � 0)2 � 102
x2 � y2 � 100
23. (x � h)2 � (y � k)2 � r2
(x � 4)2 � (y � 0)2 � 42
(x � 4)2 � y2 � 16
24. (x � h)2 � (y � k)2 � r2
(x � 3)2 � (y � (�2))2 � 22
(x � 3)2 � (y � 2)2 � 4
25. (x � h)2 � (y � k)2 � r2
(x � (�1))2 � (y � (�3))2 � 62
(x � 1)2 � (y � 3)2 � 36
26. (x � h)2 � (y � k)2 � r2
(x � (�3))2 � (y � 5)2 � 32
(x � 3)2 � (y � 5)2 � 9
27. (x � h)2 � (y � k)2 � r2
(x � 1)2 � (y � 0)2 � 72
(x � 1)2 � y2 � 49
y
x
2
2
y
x�1
1
196 Geometry, Concepts and SkillsChapter 11 Worked-Out Solution Key
38. Sample answer: The student did not subtract the coordinates of the center from x and y in the equation and did not square the radius. The equation should be (x � 1)2 � (y � 2)2 � 4.
The coordinates of the image of the point (x, y) after a90� clockwise rotation about the origin are (y, �x).
28.
D(1, �4) → D�(4, 1)
E(2, 0) → E�(0, 2)
F(5, �2) → F�(2, 5);
The coordinates of the image of the point (x, y) after a90� clockwise rotation about the origin are (�y, x).
x
y
�2E D�
D
E�
F �6
F
x
y
6J
J�
K�
K LM
M� L �
6
Y
W�
Z
P
Z�
Y�
W
XX�
T
SPR�
T�
R S�
B
B�
A
A�
C
C�
D
D�
P
YX X�
Y�
Z �
W
W�
Z
P
S
S�
R
P
R�
T
T�
�
A
B
P
CA�
C�
B�
29.
A(�1, 1) → A�(�1, �1)
B(2, 4) → B�(�4, 2)
C(5, 2) → C�(�2, 5);
The coordinates of the image of the point (x, y) after a90� clockwise rotation about the origin are (�y, x).
30.
X(�2, �3) → X�(2, 3)
O(0, 0) → O�(0, 0)
Z(3, �4) → Z�(�3, 4);
The coordinates of the image of the point (x, y) after a180� rotation about the origin are (�x, �y).
31. The design has rotational symmetry about its center; itcan be mapped onto itself by a clockwise or counter-clockwise rotation of 180�.
32. The design has rotational symmetry about its center; itcan be mapped onto itself by a clockwise or counter-clockwise rotation of 90� or 180�.
33. Yes. The image can be mapped onto itself by a clockwiseor counterclockwise rotation of 180� about its center.
34. Yes; the answer would change to a clockwise or counter-clockwise rotation of 90� or 180� about its center. If youdisregard the colors of the figures, then, for example, thegreen fish in the middle map onto the orange fish and theorange fish map onto the green fish.
35. The center of rotation is the center of the circle.
36. Yes; this piece could be hung upside down because theimage can be mapped onto itself by a clockwise or counterclockwise rotation of 180� about its center.
11.8 Standardized Test Practice (p. 638)
37. D; in a 90� clockwise rotation,(x, y) → (y, �x).
38. H
11.8 Mixed Review (p. 639)
39. A � bh � 7 � 13 � 91 ft2
40. A � bh � 9 � 8 � 72 cm2
x
y
4
Z �
Z
X �
O�
6
OX
x
y
2
B �
B
A�
C� 6
CA
Geometry, Concepts and Skills 199Chapter 11 Worked-Out Solution Key
Chapter 11 continued
41. A � �12
�h(b1 � b2)
� �12
�(10)(6 � 10)
� �12
�(10)(16)
� 80 m2
11.8 Algebra Skills (p. 639)
42. �4�2� � 6.5 43. �9�0� � 3�1�0� � 9.5
44. �2�5�6� � 16 45. �0� � 0
Quiz 3 (p. 639)
1. center (�1, 6), radius 5
2. (x � h)2 � (y � k)2 � r2
(x � 0)2 � (y � (�4))2 � 32
x2 � (y � 4)2 � 9
3. 4.
5. 6.
7. Yes; it can be mapped onto itself by a clockwise or counterclockwise rotation of 90� or 180� about its center.
8. Yes; it can be mapped onto itself by a clockwise or counterclockwise rotation of 180� about its center.
9. no
10.
A(2, 4) → A�(�2, �4)
B(1, 1) → B�(�1, �1)
C(�1, 3) → C�(1, �3);
The coordinates of the image of the point (x, y) after a180� rotation about the origin are (�x, �y).
x
y
4B �
B
A �
C �
6
CA
y
x
1
1
y
x�1
1
y
x
1
1
y
x
2
2
11.
A(1, 4) → A�(�4, 1)
B(4, 4) → B�(�4, 4)
C(4, 1) → C�(�1, 4)
D(2, 1) → D�(�1, 2);
The coordinates of the image of the point (x, y) after a 90� counterclockwise rotation about the origin are (�y, x).
12.
A(2, 0) → A�(0, �2)
B(4, �1) → B�(�1, �4)
C(3, �3) → C�(�3, �3);
The coordinates of the image of the point (x, y) after a90� clockwise rotation about the origin are (y, �x).
11.8 Technology Activity (p. 640)
1. T A B C is a rotation of T ABC about point P.
2. AP � A P
3. The measure of aAPA is twice the measure of the acuteangle formed at the intersection of lines m and k.
4. The measure of aAPA changes so it is still twice themeasure of the acute angle formed at the intersection oflines m and k.
5. yes
Chapter 11 Summary and Review (pp. 641–645)
1. A secant is a line that intersects a circle in two points.
2. A polygon is inscribed in a circle if all of its vertices lieon the circle.
3. A line in the plane of a circle that intersects the circle inexactly one point is called a tangent.
4. If the endpoints of an arc are the endpoints of a diameter,then the arc is a semicircle.
5. An inscribed angle is an angle whose vertex is on a circleand whose sides contain chords of the circle.
6. A chord is a segment whose endpoints are points on a circle.
x
y
�6 BA �
C�
2
C
A
B �
x
y
2
B � B
A �
C �
D �
8
CD
A
200 Geometry, Concepts and SkillsChapter 11 Worked-Out Solution Key
AB&̂( and CD&̂( do not intersect, and they are not parallel.The two lines are skew lines.
13. Since alternate exterior angles are congruent, you can usethe Alternate Exterior Angles Converse to show l m.
14. Since alternate interior angles are congruent, you can usethe Alternate Interior Angles Converse to show l m.
15. Since same-side interior angles are supplementary, youcan use the Same-side Interior Angles Converse to showl m.
A
B
C D
Geometry, Concepts and Skills 203Chapter 11 Worked-Out Solution Key
Chapter 11 continued
16. In a plane, if two lines are perpendicular to the same line,then they are parallel to each other.
17. In an isosceles triangle, the base angles are congruent,and by the Triangle Sum Theorem, the sum of the anglesof a triangle is 180�.
180� � 124� � 2x�
56 � 2x
x � 28
So, the measure of each base angle is 28�.
18. AB2 � BC2 � AC2
82 � 62 � 122
64 � 36 � 144
100 144
obtuse
19. BC**** is the shortest side, so aA is the smallest angle;AC**** is the longest side, so aB is the largest angle.
20. Yes; aA � aJ, aC � aL and the included sides, AC****and JL**** are congruent. So, T ABC �T JKL by the ASACongruence Postulate.
21. No; you only know that GF**** � DE**** and GE**** � GE**** bythe Reflexive Property of Congruence.
22. Yes; Sample answer: PQ**** � RQ**** and PS**** � RS**** andtheir included angles, aP and aR, are congruent. So,T PQS �T RQS by the SAS Congruence Postulate.
23. Yes; hypotenuses XU**& and VU**** are congruent and legs XY**** and VW**** are congruent in right triangles T XYU andT VWU. So, T XYU �T VWU by the HL CongruenceTheorem.
24. AB � AD 25. x� � 63� � 180�
2x � x � 8 x � 117;
x � 8; y� � x�
y� � 90� y � 117
y � 90
26. maK � maL � maM � maN � 360�
85� � 120� � x� � 88� � 360�
x � 293 � 360
x � 67
27. x� � 70� � 180�
x � 110;
y� � 110� � 180�
y � 70;
z� � 70� � 180�
z � 110
28.
2(2x) � 2(3x) � 60
4x � 6x � 60
10x � 60
x � 6
length � 3x � 3(6) � 18 cm
width � 2x � 2(6) � 12 cm
29. Sample answer: aCBD � aCAE and aBCD � aACE(Reflexive Property of Congruence). Therefore,T BCD �T ACE by the AA Similarity Postulate.
30. BD � �12
�AE
5 � �12
�x
10 � x
31. A � bh � bh
� (16)(18) � (20 � 16)(6)
� (16)(18) � (4)(6)
� 288 � 24
� 312 m2
32. A � �12
�h(b1 � b2)
� �12
�(8)(12 � 19)
� �12
�(8)(31)
� 124 ft2
33. A � �12
�d1 d2
� �12
�(16)(10)
� 80 cm2
34. A � πr235. A � bh
� π(18)2 84 � 14h
� 324π 6 m � h
� 1018 yd2
2x
3x
204 Geometry, Concepts and SkillsChapter 11 Worked-Out Solution Key