CHAPTER 10 - Mr Jaffemrjaffe.net/GeomRef/unit10solutions.pdf · dicular to Perpendicular lines form right angles. Therefore ... Chapter 10 Worked-out ... 10 3 8 x 360 33 x 18 42 x

CHAPTER 10

204 GeometryChapter 10 Worked-out Solution Key

Copyright © McDougal Littell Inc. All rights reserved.

Think & Discuss (p. 593)

1. point B 2. one

3. The fireworks would not be so high above the Earth. Theships would be smaller and closer together.

Chapter 10 Study Guide (p. 594)

1. 2.

3. 4.

5.

6.

Solution:

7.

8.

a.

b.

� �3, �412�

midpoint of AB � ��3 � 92

, 0 � 9

2 � � 15

AB � ��3 � 9�2 � �0 � 9�2

A��3, 0�, B�9, �9� � 12.0

m�J � 48.4� m�L � 41.6� JL � �145

sin J �9

�145 sin L �

8�145

82 � 92 � �JL�2

�8, 10� y � 10

54 � 3y � 4y � 64

x � 8 3�18 � y� � 4y � 64

x � 18 � 10 3x � 4y � 64

x � 18 � y x � y � 18

x � 4 x � �16

x � 4 � 0 x � 16 � 0

�x � 16��x � 4� � 0

x2 � 12x � 64 � 0

64 � x2 � 12x

82 � x�x � 12� z � ±�6 y �

1175 or 232

5

z2 � 6 15y � 351

2z2 � 12 15y � 225 � 576

2z2 � 7 � 19 15�y � 15� � 242

x �52

x � 48 8x � 20

�96 � �2x 8x � 20 � 0

264 � 360 � 2x x2 � 8x � 16 � x2 � 36

132 �12��360 � x� � x �x � 4�2 � x2 � 62

c.

equation:

d. the segment with endpoints and

10.1 Guided Practice (p. 599)

1. 2. Both a chord and a secantintersect a circle at twopoints. A chord is a linesegment having its end-points on the circle, whilea secant is a line that pass-es through two points on acircle.

3. According to Thm. 10.1, is perpen-dicular to Perpendicular lines form right angles.Therefore

4. 6.5 cm

5. No; so by the Converse of the PythagoreanThm., is not a right so is not to If

were tangent to would be a right angle.Thus, is not tangent to

6. 4 7. 2 8. 5

10.1 Practice and Applications (pp. 599–602)

9. 10. 11.

12. 13. 14.

15. 16.

17. and are congruent because they have the sameradius, 22.5.

18. B 19. E 20. F 21. D 22. A 23. C 24. H

25. G 26. external 27. internal 28. internal

29. 30.

4 common tangents no common tangents(2 external, 2 internal)

�G�C

d � 8.8 cmd � 17.4 in.

d � 124 ftd � 52 in.r � 4 cm

r � 1.5 ftr � 3.35 in.r � 7.5 cm

�C.↔BD

�B�C,↔BD

AB.�BD�,�ABD52 � 52 � 72,

m�CPX � 90�.CP.

↔XYm�CPX � 90�;

chord

radiusdiameter

B� �5, �9�A� ��7, 0�

y � �34

x �94

�94

� b � �34

0 � ��34��3� � b m �

0 � 9�3 � 9

�9

�12

MCRBG-1001-SK.qxd 5-25-2001 11:38 AM Page 204

Geometry 205Chapter 10 Worked-out Solution Key


Chapter 10 continued

31. 32. center is radius is 2 units

2 common external tangents

33. center is ; radius is 2 units

34. The two circles intersect at one point,

35. The two circles have three common tangents, the lineswith equations and

36. No; so by the Converse of thePythagorean Thm., is not a right so is not

to Then is not tangent to

37. No; so by the Converse of thePythagorean Thm., is not a right so is not

to Then, is not tangent to

38. Yes; so by the Pythagorean Thm.

is a right so Therefore, is tan-gent to

39. Yes; so by the Pythagorean Thm.,

is a right so Therefore, is tan-gent to

40. 41.

42. 43.

44. Yes; is a chord. A diameter of a circle is the longestchord of the circle.

45. 46.

47. 48.

49. is tangent to at P, is tangent to at S,

is tangent to and is tangent to Then, and (2 tangent segments withthe same exterior endpoint are ) By the def. of congru-ence, and so

by the addition prop. of equality. Then, by theSegment Addition Post. and the Substitution prop.,

or

50. 51. QP > QRQR > QP

PS RT.PS � RT

QRPQ � QS � TQ �QS � QR,PQ � TQ

.QS QR.PQ TQ

�Y at R.↔RT�X at T,

↔RT�Y

↔PS�X

↔PS

x � ±2 x � ±1

x2 � 4 x2 � 1

3x2 � 12 � 0 5x2 � 5

2x � 5 � 3x2 � 2x � 7 5x2 � 9 � 14

x � 5

15 � 3x

2x � 7 � 5x � 8JK

HC

GD, HC, FA, or EB↔AF,

↔BE

r � 45 ft

16r � 720

16r � 720 � 0

� 53 ft 784 � r2 � r2 � 16r � 64

d � 45 � 8 282 � r2 � �r � 8�2

�C.

↔ABAB � AC.�,�ABC

202 � 212 � 292,

�C.

↔ABAB � AC.�,�ABC

162 � 122 � 202,

�C.↔ABAC.�

AB�,�ABC52 � 152 � 172,

�C.↔ABAC.�

AB�,�ABC52 � 142 � 152,

y � 4.y � 0,x � 4,

�4, 2�.�6, 2�

�2, 2� 52. The statements and cannot both betrue at once. Therefore, the assumption that and are not perpendicular must be false. Then

53. Assume that is not tangent to P, that is, there is anotherpoint X on that is also on X is on so

But the perpendicular segment from Q to is theshortest such segment, so cannot be bothequal to and greater than The assumption that such apoint X exists must be false. Then is tangent to P.

54.

55. Square; and are tangent to at A and B,respectively, so and are right angles. Then, bythe Interior Angles of a Quadrilateral Thm., is also aright angle. Then CABD is a rectangle. Opposite sides ofa rectangle are congruent, so and But and are radii, so and by theTransitive Prop. of Congruence, all 4 sides of CABD arecongruent. CABD is both a rectangle and a rhombus, so itis a square by the Square Corollary.

56. a.

b. The slope of j is 2. Since j is tangent to at P,The slopes of 2 perpendicular lines are nega-

tive reciprocals of each other. The slope of is so the slope of j is 2.

c.

d. Choose any point Q on the circle, determine the slopeof the radius to that point, and use its negativereciprocal along with the coordinates of Q to find the equation.

57.

—CONTINUED—

CQ,

y � 2x � 13

�13 � b

3 � 8�2� � b

�12,CP

CP � j.�C

m �5 � 34 � 8

� �12

CA CBCBCAAD CB.CA BD

�D�B�A

�CADBD

B

A

C

D

�QP.

QXQX > QP.�QP.

QX ��Q,�Q.��

� � QP.QP�

QP > QRQR > QP





10.1 Mixed Review (p. 602)

58. Let x represent the length of the third side.

59. Since the slope of slope of Sincethe slope of slope of Then,PQRS is a parallelogram by def.

60. so so

Then PQRS is a parallelogram by Thm. 6.6.

61. 62. 63.

64. 65. 66.

67. 68.

69.

70.

71.

Lesson 10.2


1. minor arc

2. no, are not arcsof the same nor of

3. 4. 5. 6. 7. 8. 40�220�100�180�300�60�

�s.�

MN�KL� and mMN� � 72�;mKL� � 72�;

m�A � 55.2� m�B � 34.8� CB � 11.5

cos A �8

14 sin B �814 82 � �CB�2 � 142

CB � 10.7 � 47� AB � 14.7

tan 43� �10CB

m�A � 90� � 43� sin 43� �10AB

� 15.2

m�C � 66.8� m�A � 23.2� AC � 2�58

tan C �146 tan A �

614 142 � 62 � �AC� 2

x � �9 9 � x

10x � 9x � 9 2x � 3x � 9

5

x � 1�

92x

2

x � 3�

3x

�8 � x x � 225

x � 77

3x � 4x � 8 10x � 24 18x � 1386

3

x � 2�

4x

103

�8x

33x

�1842

x � 28 x � 27 x � 635

3x � 84 2x � 54 5x � 33

x7

�123

x6

�92

x

11�

35

PS QR.PS � �41 � QRPQ SR.PQ � �125 � SR

PQ � SR.SR,PQ � �3 �PS � QR.QR,PS �

38 �

6 < x < 14

x > 614 > x

4 � x > 104 � 10 > x

9. is a diameter; a chord that is the perpendicular bisec-tor of another chord is a diameter.

10. Sample answer: the measure ofthe arc formed by two adjacent arcs is the sum of themeasures of the two arcs.

11. if a diameter of a circle isperpendicular to a chord, then the diameter bisects thechord and its arc.


12. minor arc 13. minor arc 14. semicircle

15. minor arc 16. major arc 17. semicircle

18. major arc 19. major arc 20.

21. 22. 23.

24. 25. 26.

27. 28. 29.

30. 31.

32. 33. 34.

35. and and are congruent(both have radius 4). By the Arc Add. Post.,

andsince

by theArc Add. Post. Since

36.

37.

38.

39. in a , 2 minor arcs are congruent if andonly if their corresponding chords are congruent.

40. in a , 2 minor arcs are congruent if and onlyif their corresponding chords are congruent.

41. in a circle, 2 chords are congruent if and onlyif they are equidistant from the center.AB AC;

�AB CD;

�AB� CB� ;

� 195� x � 15

� 13�15°� 24x � 360

mRST� � 6x° � 7x° 4x � 6x � 2�7x� � 360

x � 36

� 144� 5x � 180

mMB� � [4�36�]° 4x � x � 180

x � 70

� 110� 3x � 210

mBC� � [2�70� � 30]° x � 2x � 30 � 180

ABC� KML�.�D �N,130� � 85� � 215�mKML� � mKM� � mML� �

360� � 145� � 215�.mABC� � 360� � mAC� �AC� KL� ;�D �N,

mKL� � 145�70� � 75� � 145�.mAE� � mEC� �mAC� �

�N�DABC� KML�;AC� KL�145�145�145�

120�115�

65�60�65�

60�180�180�

305�300�55�

Q

J

N

M L

K60�55�

60�

BC CA, BD� DA� ;

mAB� � mAC� � mCB� ;

BC





42. in a circle, 2 chords are congruent if and onlyif they are equidistant from the center.

43. a diameter that is perpendicular to a chord bisectsthe chord and its arc.

44. 2 minor arcs are the same measure if their chords are the same measure ( so

).

45. in a circle, 2 chords are congruent if and only if theyare equidistant from the center.

46. 7; if a diameter of a circle is perpendicular to a chord,then the diameter bisects the chord and its arc.

47. Vertical Angles Thm., def. of minor arc

48. farther from the center

49. 50. 51. 3:00 A.M.

52. During step 1, the ski patrol marks off a chord of a circle.During steps 2 and 3, the patrol marks off a diameter ofthe circle, because this line is the perpendicular bisectorof the chord from step 1. The midpoint of the diameter isthe center of the circle.

53. This follows from the definition of the measure of aminor arc. (The measure of a minor arc is the measure ofits central angle.) If 2 minor arcs in the same circle orcongruent circles are congruent, then their central anglesare congruent. Conversely, if 2 central angles of the samecircle or congruent circles are congruent, then the mea-sures of the associated arcs are congruent.

54. Circles may vary; Sample answer: to find the center,draw a chord with the straightedge and construct its perpendicular bisector. Extend the perpendicular bisector so it touches 2 points on the circle. Construct the perpendicular bisector of the chord determined by the 2 points to find its midpoint, which will be the centerof the circle.

55. Yes; construct the perpendiculars from the center of thecircle to each chord. Use a compass to compare thelengths of the segments.

56. and are all radii, so they areall congruent. Therefore, by the SSSCongruence Post. Since the 2 triangles are congruent,

because corresponding parts of con-gruent triangles are congruent. By def. of congruent arcs,

57. Since by the def. ofcongruent arcs. and are all radii of so Then by theSAS Congruence Post., so corresponding sides and

are congruent.DCAB

�APB � �CPDPA � PB � PC � PD.�P,PDPA, PB, PC,

�APB � �CPDAB� � DC� ,

AB� � DC� .

�APB � �DPC

�APB � �DPCPDAP, PB, PCAB � DC;

90�15�

40�;

15;

mBED�� 110° � 60° � 170° � mCDE�BD � CE,170�;

40�;

ED � 10; 58. You would have to use the definition of � çs and theTransitive Prop. of Cong. to show that the appropriatesides and are �.

59. Draw radii and and sinceby the HL Congruence Thm.

Then, corresponding sides and are congruent asare corresponding angles GLJ and HLJ. By the def. ofcongruent arcs,

60. Assume center L is not on because both are radii. because is the perp. bisector of and by the Reflexive Prop. by the SSS Congruence Post.

because corresponding parts of congru-ent triangles are congruent. Since and forma linear pair and are congruent, both must be right angles.Therefore, Then and are both perpen-dicular to through J. This contradicts thePerpendicular Postulate. The assumption that center L isnot on must be incorrect, so L is on Therefore,

is a diameter of

61. Draw radii and and Also,since and and areright triangles and are congruent by the HL CongruenceThm. Corresponding sides and are congruent, so

and by the Multiplication prop. of equality,By Thm. 10.5 bisects and

bisects so and Then by theSubstitution Prop., or

62. Draw radii and andso by Thm. 10.5, bisects and

bisects By the definition of bisector, andSince then AB � DC and by the

Multiplication Prop. of equality, . Then bythe Substitution Prop., BE � CF, so and are right triangles and are congruent by theHL Congruence Thm. Therefore, because theyare corresponding sides of congruent triangles.

63.

64. 65. 66. 67.

68. a. Construct the perpendicular bisector of each chord.The point at which the bisectors intersect is the centerof the circle. Connect the center with any point on thecircle and measure the segment drawn.

—CONTINUED—

210�150�90�120�

60�

30�

0�

330�

300�

120�

150�

180�

210�

240�

90�

270�

A(2, 30°)

B(4, 120°)

C(4, 210°) D(4, 330°)

E(2, 150°)

2

PE � PF�PFC

�PEBBE � CF.

12AB �

12DC

AB � DC,12DC � CF.

12AB � BEDC.

PFABPEPF � DC,PE � ABPB � PC.PC.PB

AB � CD.AB � CDCD � 2CF.AB � 2BECD,

PFABPE2BE � 2CE.BE � CF

CFBE

�PFC�PEBPF � CD,PE � ABPE � PF.PB � PCPC.PB

�L.EFEF.EF

GHJLEFJL � GH.

�HJL�GJL�GJL � �HJL,

�GLJ � �HLJLJ � LJGH

EFGJ � HJGL � HLEF.

GE� � EH� .

JHGJ�LGJ � �LHJEF � GH,

LJ � LJ,LG � LH,LH.LG

�





68. —CONTINUED—

b. Construct s to the two tangents at the points of tangency. The intersection of the s is the center of the circle. Draw a segment from the center to anypoint on the circle. Measure the segment drawn.

c. The object would have to be small enough to be traced on a piece of paper and to allow you to do constructions. If it were too small, however, it wouldbe difficult to perform the constructions accurately.

69.


70. interior:

exterior:

71. lies in the interior

lies in the exterior

72. (1, �5) lies in the interior

(2, 1) lies in the exterior

73. lies in the interior

lies in the exterior

74. Rhombus; so PQRS is arhombus by the Rhombus Corollary.

75. Square; so PQRS is arhombus by the Rhombus Corollary; so

is a rectangle. (A parallelogram is a rectangle ifand only if its diagonals are congruent.) Then, PQRS is asquare by the Square Corollary.

76. 77.

x � 16 x � 12 x2 � 256 x2 � 144

8x

�x

32 9x

�x

16

PQRSPR � QS � 6,

PQ � QR � RS � PS � 3�2,

PQ � QR � RS � PS � �10,

�1, �2��0, 1�y

x

1

1

C(3, 2)(0, 1)

(1, �2)

B(0, 0)

A(�3, 2)

y

x

1

�1

A(�2, �3)B(0, �1)

C(2, �3)

(2, 1)

(1, �5)

��1, 2��1, 1�y

x

1

�1�1

B(0, 0)(1, 1)

C(4, �1)(1, �2)

A(�2, 3)

�2, 4��3, 1�y

x

1

�1

A(4, 2)B(0, 2)

C(3, 0)

(2, 4)

(3, 1)

x � 3�5

81 � 36 � x2

�

�78. 79.

Lesson 10.3

Activity 10.3: Investigating Inscribed Angles (p. 612)

Sample answers are given.Exploring the Concept

1–3. Circle 1

Investigate

1.

2.

Circle 2 Circle 3

Make a Conjecture

3. The measure of an inscribed angle is half the measure ofthe corresponding central angle.

Extension

About the star divides the into 7 arcs, so each has

measure Each inscribed corresp. to an arc with

measure Then


1. 2. No; according to Thm.10.11, since the oppositeangles are not supplemen-tary, the quadrilateral cannotbe inscribed in a circle.

intercepted arc

3. 4. 5.

6. 7. 8.

y � 100z � 75

x � 95y � 150x � 115

mLMK� � 210�mKML� � 180�mKL� � 40�

AC�

B A

C

540�

7� 77

17

� � 77�.

x �12

�1080�

7�3 �

360�

7�

1080�

7.

�360�

7.

� �77�;

U

R

S T

V

P

UR

S

T

V

P

TR

S

UV

P

x � 18 x � 14 x2 � 324 x2 � 196

9x

�x

36 4x

�x

49

Circle 1

Circle 2

Circle 3 28�28�28�56�

35�35�35�70�

45�45�45�90�

m�RVSm�RUSm�RTSm�RPS






9. 10. 11.

12. 13.

14.

15. inscribed angles and interceptthe same arc so their measures must be the same.

16. intercepts an arc whose chordis a diameter so the measure of the arc is and

can be found using the Triangle SumThm.

17. inscribed angles and intercept the same arc and inscribed angles and

intercept the same arc so and

18. 19. 20.

21.

22. Mult. by 3

Mult. by

23. Mult. by 7

Mult. by

24. Yes; every angle of a square measures so both pairsof opposite angles are always supplementary. Thereforethe square can always be inscribed in a circle, accordingto Thm. 10.11.

25. Yes; every angle of a rectangle is a right angle so bothpairs of opposite angles are always supplementary.Therefore, the rectangle can always be inscribed in a cir-cle, according to Thm. 10.11.

26. No; the opposite angles are not always supplementary, soaccording to Thm. 10.11, it cannot always be inscribed ina circle.

90�

x � 9m�C � 126�, m�D � 144�

y � 6m�A � 54�, m�B � 36�

150y � 900

�28x � 18y � �360→�2 14x � 9y � 180

28x � 168y � 1260→ 4x � 24y � 180

x � 25m�C � 50�, m�D � 105�

y � 5m�A � 130�, m�B � 75�

36y � 180

�6x � 42y � �360→�2 3x � 21y � 180

6x � 78y � 540→ 2x � 26y � 180

m�A � 60�, m�B � 60�, m�C � 60�

y � 20

92y � 90

x � 30 3y �32y � 90

x �32�20� 3y � x � 90

x �32y 12y � 4x � 360

2x �12�6y� 6y � 6y � 4x � 360

z � 180z � 160z � 112

y � 90y � 78y � 90

x � 65x � 80x � 90

m�QSR � 45�.m�PQS � 40��PRS

�PQS�QSR�QPRx � 45, y � 40;

m�LKMx � 90;180�

�MLKx � 90, y � 50;

�EDH�EGHx � 47;

m�ABC � 90�

m�ABC � 109�m�ABC � 55�

mBC� � 228�mBC� � 156�mCB� � 64�

27. No; the opposite angles of a kite are not always supple-mentary, so according to Thm. 10.11, it cannot always beinscribed in a circle.

28. No; the opposite angles are not always supplementary, soaccording to Thm. 10.11, it cannot always be inscribed ina circle.

29. Yes; both pairs of opposite angles of an isosceles trape-zoid are supplementary.

30. Construct a perpendicular toat A. According to Thm. 10.2,

a line that is perpendicular to aradius of a circle is tangent tothe circle.

31. is a diameter of

32. since it inter-cepts which is a diameter,the measure of the interceptedarc is is aninscribed angle so its measureis

33. a line perpendicular to a radius of a circle at its end-point is tangent to the circle.

34. Answers may vary. Sample answer: . When

As you drag C toward A, increases and decreases. As you drag Caway from A, decreases and increases.

35. isosceles; base angles; Exterior Angle;

36. Draw the diameter containing intersecting the circle at a point D. By the proof in Ex. 35,and By the Arc. Add. Post.,

By the Angle Add. Post.,By repeated applica-

tion of the Substitution prop.,

37. Draw the diameter containing intersecting the circleat point D. By the proof in Ex. 35,and By the Arc Add. Post.,

so by theSubtraction prop. of equality. By the Angle AdditionPost., so

by the Subtraction prop. of equality.Then, by repeated application of the Substitution prop.,m�ABC �

12mAC� .

m�CBDm�ABD �m�ABC �m�ABD � m�ABC � m�CBD,

mAC� � mAD� � mCD�mAC� � mCD� ,mAD� �m�DBC �

12mDC� .

m�ABD �12mAD�

QB,

m�ABC �12mAC� .

m�ABD � m�DBC � m�ABC.mAD� � mDC� � mAC� .

m�DBC �12mDC� .

m�ABD �12mAD�

QB,

2x�; 2x�; 2; 12mAC� ; 12mAC��A � �B;QB;

m�Bm�Am�Bm�A

↔CQ�

↔AB, m�A � m�B � 45�.

m�C � 90�

BA;

90�.

�CBA180�.

ACm�CBA � 90�;

A

M

B

C

�M.AC

A

MC

ACA

C





38. Given: with inscribed angles and

Prove:

Proof: Inscribed and both intercept

So m∠ADB � mand m∠ACB � m . By the Transitive Prop. of equality so

39. Given: with inscribed is a diameter of

Prove: is a right triangle.

Use the Arc Addition Post. to showthat and thus

Then use theMeasure of an Inscribed AngleThm. to show so that is a rt. and

is a rt.

Given: with inscribed is a right angle

Prove: is a diameter of

Use the Measure of an Inscribed Angle Thm., to show theinscribed right angle intercepts an arc with measure

Since intercepts an arc that is half themeasure of the circle, it must be a diameter.

40. Given: DEFG is inscribed in a circle.

Prove:

Proof:

so

By the Subst. Prop.,

Likewise and

so

By the Subst. Prop.,

41. Use the carpenter’s square to draw 2 diameters of the cir-cle. (Position the vertex of the tool on the circle and markthe 2 points where the sides intersect the Repeat, plac-ing the vertex at a different point on the circle. The centeris the point where the diameters intersect.)

42. m∠ADB � m∠ACB 43.

� (80°)

� 40°

B

C

44. a right triangle

45. GJ is the geometric mean of FJ and JH; Thm. 9.2 justi-fies this answer.

16 � x

64 � 4x

14x � 32 � 18x � 3212

2�7x � 16� � 18x � 3212

�.

m�D � m�F � 180�

12mGDE� � 180�.1

2mGFE� �

mGFE� � mGDE� � 360�;m�F �12mGDE�;

m�D �12mGFE�m�E � m�G � 180�.

12mDGF� �

12mDGF� � 180°.

mDGF� � mDEF� � 360�

m�G �12mDEF�;

m�E �12mDGF� ;

m�E � m�G � 180�m�D � m�F � 180�,

G

E

F

C

D

AC2�90�� 180�.

�O.AC

�B�ABC,�O

�.�ABC��Bm�B � 90�,

mABC� � 180�.mAEC� � mABC�

�ABC

�O.AC�ABC,

B

A

C

E

O

�O

�ADB � �ACB.m�ADB � m�ACB

AB�AB�AB�.

�ACB�ADB�

�ADB � �ACB

�ADB�ACB

B

A

X

D

C

�X 46. FJ � 6 in.; JH � 2 in.

GK � 2(GJ)

� 2(3.46)

� 6.92

47. Consider FLH. Suppose chord intersects at a point X. at X so because a diam-eter ⊥ to a chord bisects the chord. LX is the geometricmean of FX and XH, FX � 1, and XH � 7, so

LM � 2(LX)

� 2(2.65)

� 5.3 in.


48. 49.

50. 51.

52. 53.

54. 55.

56. 57. y

x�2

�2

P

P�

QR�

R

Q�

y

x�2

2�2

P

P�

Q

Q�

R

R�

y

x�2

2�2

P

P�

Q Q�

R

R�

y

x�2

2�2

P

P�

QQ�

R

R�

y � �45x � 16 y � �

12x

�16 � b 0 � b

�12 � ��45��5� � b 4 � ��1

2��8� y �

43x � 7 y � 3

7 � b 3 � b

7 �43�0� � b 3 � 3�0� � b

y � 2x � 9 y � �x � 8

�9 � b �8 � b

1 � 2�5� � b �6 � �1��2� � b

� 2.65 in.

LX � �7

7 � �LX�2

1

LX�

LX7

LX � MXLM � FHFHLM�

� 3.46 in.

GJ � 2�3

12 � �GJ�2

2

GJ�

GJ6

JHGJ

�GJFJ





58. 59. � 0.5 60. �

61. � 0.8660 62. � 1.7320

Quiz 1 (p. 620)

1. x � 90; Thm. 10.1 2. Thm. 10.3 3.

4. 5. 6. 7. 8. 9.

Lesson 10.4


1. The measure of each angle is equal to half the measure ofthe intercepted arc.

2.

3.

4.

5.

6.

7.


8.

9.

10.

11.

12.

13.

14. 15.

16.

17.

18.

19.

20.

21.

22. m�1 �12�142° � 52°� � 45�

m�1 �12�122° � 70°� � 26�

m�1 �12�105° � 51°� � 27�

m�1 � 180° � 77° � 103�

m�2 �12�32° � 122°� � 77�

m�1 �12�25° � 75°� � 50�

m�1 �12�130° � 95°� � 112.5�

x � 18

6x � 108

16x � 58 � 10x � 50

2�8x � 29� � 10x � 50

x � 25.4

127 � 5x

144 � 5x � 17� 2�84°� � 168�mAB�m�3 �

12�220°� � 110�

� 2�126°� � 252�mABC�� 2�36°� � 72�mDE�

m�2 �12�180°� � 90�

� 2�140°� � 280�mGHJ�m�1 �

12�220°� � 110�

m�1 �12�88° � 88°� � 88�

m�N �12�80° � 35°� � 22.5�

m�RQU �12�270° � 90°� � 90�

m�DBR �12�190° � 60°� � 65�

m�1 �12�55° � 65°� � 60�

� 2 � �105°� � 210�mSTU�

85.2�47�180�313�227�133�

47�x � 12;

�3�32

0.8660�32

12

13�623.

24.

25.

26. 27.

28.

29.

30.

31. 32. 33. 34.

35.

36. The measure of is equal to half the measure ofTheorem 10.12.

37. Diameter; a tangent line is to the radius drawn tothe point of tangency.

38. Draw intersecting the ç at P and let X be a point onthe upper semicircle. By Thm. 10.8, m∠PBC �

By the proof of Case 1 m∠PBA �

By the Arc Add. Post.,

By the Angle Add. Post., m∠ABP � m∠PBC �m∠ABC. By repeated application of the Substitution

Prop., 12mBPC�.m�ABC �

mBXP� � mPC� � mBPC� .

12mBXP�.1

2mPC�.

BQ

�90�;

AC� ;�BAC

� 0.7�

� 0.701�

� m�FCB � 0.128 � 0.573mSB�

m�TCF � tan�1�40.00054000 � � 0.573�

TF � 40.0005

�TF�2 � 1600.04

40002 � �TF�2 � 4000.22

m�TCE � tan�1�8.9444000 � � 0.128�

TE � 8.944

�TE�2 � 80.0001

40002 � �TE�2 � 4000.012

60�30�90�30�

60�

60�

a � 40

a � a � 70 � a � 30

a2

�12

�a � 70 � �a � 30�

a � 5 a � 15

15a � 75 8a � 120

15a �12�255 � 105� 8a � 10 � 130

m�1 �12�235° � 125°� � 55�

m�1 �12�235° � 125°� � 55�

m�1 �12�120° � 46°� � 37�





39. The proof would be similar, using the Angle Additionand Arc Addition Postulates, but you would be subtract-ing and instead of adding.

40. Reasons1. Given2. Through any 2 points there is exactly 1 line.3. Exterior Angle Thm.4. Measure of an Inscribed ∠ Thm.5. Measure of an Inscribed ∠ Thm.6. Substitution Property7. Distributive Property

41. Case 1: Draw Use theExterior Angle Thm. to showthat so that

Then use Thm. 10.12to show that and the Measure of anInscribed Angle Thm. to showthat Then

Case 2: Draw Use theExterior Angle Thm. to showthat sothat Then use Thm. 10.12 to showthat and

Then,

Case 3: Draw Use theExterior Angle Thm. to showthat so that

Then use the Measureof an Inscribed Angle Thm. to

show that and

Then,

42. E 43. C

44. is a right angle; circle P is inscribed in T, U,and V are points of tangency. By Thm. 10.1,and so and are right angles.Using the Interior Angles of a Quadrilateral Thm. is a right angle. By the Rectangle Corollary, quad. TPVRis a rectangle; so by the RhombusCorollary, quad. TPVR is a rhombus. By the SquareCorollary, quad TPVR is a square. According to Thm.10.3, and Using the SegmentAddition Postulate, and Since by Substitution Prop. and The perimeter of

and perim � so QR � QS � RS � QT � r � SV � r � QS. By theAddition and Subtraction Properties of Equality,

QR � RS � QS,r � SV � r � QS�QRS � QT �QT � r � QR.

SV � r � SRRV � r � TR,SV � VR � SR.QT � TR � QR

SU � SV.QT � QU

TP � PV � RV � TR,

�TPV�RVP�RTPPV � RS,

PT � QR�QRS;�R

12�mXY� � mWZ� �.m�3 �

�12mWZ� .m�WXZ

m�4 �12mXY�

m�WXZ.m�3 � m�4 �

m�WXZ,m�4 � m�3 �

XZ.

ZY

X

W

34

12�mPQR� � mPR� �.

m�2 �12mPR� .m�4 �

m�3 �12mPQR�

m�4.m�2 � m�3 �m�2 � m�4,m�3 �

PR.

R

P

2

3

4

Q

12�mBC� � mAC� �.m�1 �

m�ABC �12mAC� .

m�2 �12mBC�

m�ABC.m�1 � m�2 �

� m�ABC,m�2 � m�1

BC.

C

B

A

1

2

�ACB;

mPC�m�PBC

2r � QR � RS � QT � SV or 2r � QR � RS � (QT �SV). By substitution, 2r � QR � RS � (QU � US). QU � US � QS. Therefore r � (QR � RS � QS).

45.


46.

47.

48. 49.

50. 51.

Lesson 10.5

Activity 10.5 Investigating Segment Lengths (p. 628)

Investigate

1. yes; yes; yes

3. The products are equal.

4. The products remain equal.

Conjecture

5. The product of the segments of intersecting chords areequal.

Investigate

6. The products are equal.

Conjecture

7. The product of the length of the external segment of onesecant and the total length of the secant is equal to theproduct of the total length of the other secant and thelength of its external segment.

Extension

Tangent; the square of the length of thetangent segment is equal to the product of the lengths of thesecant segment and its external segment.

�EA�2 � EC � ED;

2 � x

8 � 4x x � 8

6x � 12 � 10x � 4 2x � 5 � x � 3

r � 19.2 ft

20r � 384

r2 � 20r � 100 � r2 � 484

x � 25 �r � 10�2 � r2 � 222

LP � 6

36 � �LP�2

9

LP�

LP4

LP � 20 LM � 16

LP2 � 400 9LM � 144

25LP

�LP16

912

�12LM

r � 1

r �12�3 � 4 � 5�

12

MCRBG-1005-SK.qxd 6-15-2001 3:37 PM Page 212





1. 2.

External segment is

3. 15; 18 4. 12; 15

5. 6. 2

7. 9 8.

Solution:

9. The segment from you to the center of the aviary is asecant segment that shares an endpoint with the segmentthat is tangent to the aviary. Let x be the length of theinternal secant segment (twice the radius of the aviary)and use Thm. 10.7. Since the radius isabout or 25 ft.


10. 9; 15 11. 45; 27 12. 16

13. 14.

15. 16.

17. 18.

6 � x

x � 4313 36 � x2

72x � 40�78� 72 � 2x2

x � 2

5x � 10 x � 8.5

15x � 10x � 10 24x � 204

15x � 10�x � 1� 24x � 12 � 17

13 � x

x � 28 195 � 15x

4x � 112 420 � 225 � 15x

4x � 16 � 7 12 � 35 � 15��5 � x� x � 12 x � 30 x � 20

x2 � 144 45x � 1350 9x � 180

502 ,

40�40 � x� � 602,

x � 1

x � 1x � �4

�x � 4��x � 1� � 0

x2 � 3x � 4 � 0x � 6

x2 � 3x � 4x2 � 36

x � �x � 3� � 22x2 � 4 � 9

x � 3; 22

6 � x4 � x

12 � 2x32 � 8x

16 � 4 � 2x96 � 8x � 64

42 � 2 � �2 � x�6 � 16 � 8 � �x � 8�

16; x � 8

x � 50x � 12

12x � 60015x � 180

12 � x � 15 � 40x � 15 � 10 � 18

AB

HF � HJ � HG � HKA

C

B

D

19. 20.

21.

Solution:

22.

Solution:

23.

24.

25.

Solution:

26.

27.

Solution: y � 3.81

y � 3.81 or y � �16.81

��13 ± 5�17

2 y �

�13 ± �132 � 4�1��64�2

0 � y2 � 13y � 64

64 � y2 � 13y

82 � y�y � 13� x � 7

3�14� � 6x

y � 2�13 x � 12

y2 � 2�26� 8�18� � 12�x�y � 10

y � 10y � �40 x � 42

�y � 40��y � 10� � 0 336 � 8x

y2 � 30y � 400 � 0 400 � 64 � 8x

400 � y�30 � y� 400 � 8�8 � x�Solution: x � 16.49

x � 16.49 or x � �45.49

��29 ± �3841

2 x �

�29 ± �841 � 4�1��750�2�1�

x2 � 29x � 750 � 0

x�x � 29� � 15�50�Solution: x � 7.38

x � 7.38 or x ��16.38

��9 ± �565

2 x �

�9 ± �81 � 4�1��121�2�1�

x2 � 9x � 121 � 0

121 � x�x � 9�x � 8

x � 8x � �18

�x � 18��x � 8� � 0

x2 � 10x � 144 � 0

144 � x�10 � x�x � 4

x � 4x � �16

�x � 16��x � 4� � 0

x2 � 12x � 64 � 0

64 � 12x � x2

64 � x�12 � x� x � 24 x � 82

3

x2 � 576 88 � 36 � 6x

x2 � 12�48� 4�22� � 6�6 � x�





28.

29. 4.875 ft; the diameter through A bisects the chord intotwo 4.5 ft segments. Use Thm. 10.15 to find the length ofthe part of the diameter containing A. Add this length to 3and divide by 2 to get the radius.

30.

31. Draw and . Then inscribed angles ∠EBC and∠EDA intercept the same arc, so ∠EBC � ∠EDA. ∠E � ∠E by the Reflexive Prop. of Congruence, so

by the AA Similarity Thm. Then, sincelengths of corresponding sides of similar triangles areproportional,

By the Cross Product Prop.,

32. is tangent to a circle; is a secant of the same cir-cle. Draw and ∠ADE is an inscribed ∠, som∠ADE � ∠CAE is formed by a tangent and achord so m∠CAE � Then ∠ADE � ∠CAE.Since ∠E � ∠E by the Reflexive Prop. of Cong.,

ACE � DAE by the AA Similarity Post. Then sincelengths of corresponding sides of � s are proportional,

By the Cross Products Property,

33. 34.

35. The distances over which most of the inhabitants of Earthare able to see are relatively short and the curvature ofEarth over such distances is so small as to be unnoticeable.

36. 37.

38. By applying the substitution property andusing the results from Exercises 36 and 37, the two areequal since and both AB and AE are positive.

39. Sample answer: Conjecture: If tangent segments to twointersecting çs share a common endpoint outside the 2çs, the tangent segments are �. The conjecture is nottrue in general, however. Let A be the common endpointof the tangent segments and C and D be the points ofintersection of the 2 çs. The conjecture is true if andonly if A, C, and D are collinear.


40. 41.

Midpoint Midpoint � �3, 0�� 12, 4�

� 10 � �29 � 5.39

AB � �36 � 64 AB � �25 � 4

�AB�2 � �AE�2

AB � AE;

�AE�2 � AC�AD��AB�2 � AC�AD�

x �1

8000 mi

18000 � 5280

1 � 121 � 8 in. 1�1� � 8000x

�EA�2 � EC � ED.

EAEC

�EDEA

.

�

��

12mAC� .

12mAC� .

AD.ACEDEA

EA � EB � EC � ED.

EAEC

�EDEB

.

�BCE ~ �DAE

BCAD

BC � BA � 16,008 mi

BA � 2500�41 � 16,008 mi

�BA�2 � 12,500�20,500�

CN � 18

15�12� � 10CN 42. 43.

Midpoint: Midpoint:

44. 45.

Midpoint: Midpoint:

46. 47.

48. 49.

50. 51.

52.

53.

C��4, 5�, D��9, 6�A��8, 11�, B��3, 7�,C��2, 2�, D��7, 3�A��6, 8�, B��1, 4�,

y

x�2

�2

A

D C

B

A�

D� C�

B�

C��5, 2�, D��0, 3�A��1, 8�, B��6, 4�,C��2, 2�, D��7, 3�A��6, 8�, B��1, 4�,

y

x�2

�2 2

A

D C

B

A�

D�

C�

B�

y �37x �

817 y � �5x � 51

817 � b �51 � b

9 �37��6� � b �1 � �5��10� � b

y � �13x �

103 y � x � 9

�103 � b 9 � b

�4 � �13�2� � b 9 � 1�0� � b

y � �32x � 17 y �

12x

17 � b 0 � b

8 � �32�6� � b �1 �

12��2� � b

��2, �2��4, �92�

� 14 � 15.26

AB � �196 � 0 AB � �64 � 169

��112 , 1��7

2, 32� � 15 � 17.49

AB � �81 � 144 AB � �81 � 225





54.

Quiz 2 (p. 635)

1. 2.

3. 4.

5.

Solution:

6.

Solution:

7. Solve (Thm. 10.17) or solve(the Pythagorean Thm.);

Lesson 10.6


1. , where (h, k) is the center ofthe ç and r is the radius.

2. The center is and the radius is units. Placethe point of a compass at (3, 4), set the radius at 3 units,and swing the compass to draw a full circle.

3. Center: radius: 2;

4. Center: radius:

5. Center: radius:

6.

10.6 Guided Practice and Applications (pp. 638–640)

7. center: radius: 4 8. center: radius: 5

9. center: radius: 2 10. center: radius: 6��2, 3�;�0, 0�;�5, 1�;�4, 3�;

x2 � y2 � 10

� �10

r � �1 � 9

2; �x � 2�2 � �y � 2�2 � 4��2, 2�;4; �x � 2�2 � y2 � 16�2, 0�;

x2 � y2 � 4�0, 0�;

�9 � 3�3, 4�

�x � h�2 � �y � k�2 � r2

50.025 ft.r2 � 492�r � 20�2 �20�2r � 20� � 492

x � 5

x � 5x � �20

�x � 20��x � 5� � 0

x2 � 15x � 100 � 0

100 � x�x � 15�x � 6

x � 6x � �24

�x � 24��x � 6� � 0

x2 � 18x � 144 � 0

8�18� � x�x � 18� x � 26

x � 6.25 56 � 82 � x

16x � 100 28 �12�82 � x�

� 139

x �12�110 � 168�202

A��6, 212�, B��1, �11

2�, C��2, �312�, D��7, �21

2�A��6, 8�, B��1, 4�, C��2, 2�, D��7, 3�

y

x

�6

2

A

DC

B

A�

D� C�

B�

11. center: radius: 1

12. center: radius:

13. center: radius: 2;



16. center: radius: 2.5;



19. 20.

21.

22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. exterior 34. interior 35. on the circle 36. exterior

37. interior 38. on the circle 39. exterior 40. interior

�x �12�2

� �y �12�2

�14�x � 5�2 � �y � 1�2 � 49

y

x

1

�1

�1

y

x

2

�2�2

�x � 3�2 � �y � 4�2 � 16�x � 3�2 � y2 � 9

y

x�2

2�2

y

x

2

�1

x2 � �y � 4�2 � 1x2 � y2 � 25

y

x

1

�1 1

y

x

2

�22�2

�x � 3�2 � �y � 2�2 � 4

� 2

�x � 5�2 � �y � 3�2 � 16 r � �4 � 0

�x � 1�2 � �y � 2�2 � 25x2 � y2 � 9

� 5 � 3

r � �9 � 16 r � �9

�x � 1�2 � �y � 3�2 � 36

�x � 3�2 � �y � 2�2 � 4

�x � 4�2 � y2 � 16x2 � y2 � 1

x2 � y2 � 36�0, 0�;�x � 2�2 � �y � 2�2 � 16�2, 2�;

�x � 0.5�2 � �y � 1.5�2 � 6.25�0.5, 1.5�;

�x � 3�2 � �y � 3�2 � 1�3, 3�;x2 � �y � 1�2 � 4�0, 1�;

�x � 3�2 � �y � 2�2 � 4��3, 2�;

12�1

2, �34�;

��5, �3�;





41.

42. J: A; K: B; L: B and C; none; N: C

43. 44.

45. Answers may vary.

46. Any way it is rolled, the width is the same.

47. image:

48. 49. D 50. C

51. Yes; let P be the point of intersection and r the radius ofçB. Since the çs are externally tangent, and are ⊥ to the tangent line at P. Thus point B lies on theline through A and P. Point B also lies on the circle withcenter P and radius r. Use these facts to write two equations in terms of the coordinates of point B and solve this system. Point B is the solution to this systemthat lies outside circle A. Use the coordinates of point Band the radius to write the equation for circle B.

52. 53.


54. kite, rhombus, rectangle, parallelogram

55. parallelogram, rectangle, rhombus, kite, isoscelestrapezoid

56. kite, rhombus, rectangle, parallelogram

57.

58.

59.

60.

61. No; P is not equidistant from the sides of

62. Yes; P is equidistant from the sides of �A.

�A.

�2, 13�; 13.2

�3, �11�; 11.4

��8, �2�; 8.2

��6, 7�; 9.2

b � �2

0 � �b � 2�2 b � 0

25 � 25 � �b � 2�2 9 � 9 � b2

5 � �25 � �b � 2�2 3 � ��3�2 � �b � 0�2

PBAP

�x � p�2 � �y � q�2 � q2

�x � 2�2 � �y � 4�2 � 16y

x

2

�22�2

�x � 3�2 � y2 � 49�x � 3�2 � y2 � 1

y

x

1

�11�1

A

B

C

C: �x � 2�2 � �y � 5�2 � 4B: �x � 5�2 � �y � 3�2 � 6.25,A: x2 � y2 � 9, Lesson 10.7

Activity 10.7 (p. 641)

Investigate1. Yes; m and k are � and the distance from m to k is AB, so

Y is AB units from k. çP has radius AB, so Y is AB unitsfrom P. This is also true of point Z.

2. If B is dragged to the right, m moves up and circle P getslarger. If B is dragged to the left, m moves down and circle P gets smaller; yes

3. parabola

Conjecture4. The set of points in a plane that are equidistant from a

line and a point in the plane is a parabola.

ExtensionThe locus of points equidistant from and the line

is the parabola with equation


1. exterior 2. The locus of pointsthat are equidistantfrom A and B formthe segment of theperpendicular bisector of thatlies on the paper.

3. B 4. A 5. D 6. C

7. The two points on the inter-section of the bisector of

and çA with radius 5

8. The intersection of with radius 3 and with radius5; 2, 1, or 0 points depending on whether the distancebetween C and D is less than, equal to, or greater than,3 � 5 � 8 units.


9. 10.

The points on with The 2 lines parallel to kradius 1 inch on opposite sides of k and

1 in. away

�P

k

n

m

1 in.

1 in.

P1 in.

�D�C

AB�

y

x

2

�22�2

A B

AB

A B

y � x2.y � �14

�0, 14�





11. 12.

The points on with The points on and beyondradius 1 inch and the 2 lines parallel to j on points in the interior of opposite sides of j and �C 1 in. from j

13. 14.

A line � to both j and k and The bisector of halfway between them

15. 16.

A � with center C and The bisectors of all 4radius half that of the formed by the intersection original � of r and s

17. 18.

All points except A on Line q, the ⊥ bisector of

two rays, and ,such that m∠MAN � 60°

and is the bisector of ∠MAN

19. 20.

21. 22.

midpoint �

,

so ⊥ bisector has slope � �1

23. y � 4, y � �2

y � �x � 6

6 � b

3 � �1�3� � b

m �5 � 15 � 1

� 1

�3, 3��x � 5�2 � �y � 5�2 � 9K�5, 5�M�1, 1�y � 3x � 3

→AB

AB→AN

→AM

BA

q

BA

N

M

30�

30�

�

s

r

C

�A

A

k

j

�C

j

n

m

1 in.

1 in.

C1 in.

24.

points on the x-axis with x-coordinate greater than �4and less than 4

25.

2 points, and the intersections of with and

26.

Let d be the distance from R to k; the locus of points is 4points if points if points if point if and 0 points if

27.

Let d be the distance from R tothe bisector of the locusof points is 2 points if 1 point if , and 0 points if d > 4.

d � 4d < 4,

PQ;�

y

x

2

�22�2

R

QP

d > 4

y

x

2

�22�2

R

QP

d � 4

y

x�2

2�2

R

QP

d < 4

d > 3.d � 3,1 < d < 3, 1d � 1, 2d < 1, 3

R

d < 1

kk

R

d � 1

k

R

1 < d < 3

k

R

d � 3

k

R

d > 3

y � 4y � 2y � x�4, 4�,�2, 2�

y

x

k

1

�11�1

D

C

B

A

y

x

1

�11�1





28. 29.

30.

No; your friend is more than 14 mi from the epicenter.

31. Let d be the distance from P to k. If the locusis 2 points. If the locus is 1 point. If thelocus is 0 points.

32. the with center at the midpoint of and radius

33. D 34. C 35.


36. 22 37. 69 38. 70 39.

40. 41.

42. 43.

44. 45.

�x � 4�2 � �y � 5�2 � 1x2 � �y � 7�2 � 100

y

x

1

1

y

x

4

�48�8

�x � 6�2 � �y � 4�2 � 9x2 � y2 � 81

y

x

2

�22�2

y

x

3

�33�3

x � 17

x � 24 270 � 100 � 10x

x2 � 16�36� 9�30� � 10�10 � x� x � 17.5

12x � 210

9 ft

5 ft

2 ft

6 ft

1 ft � 0.5 cm

12ABAB�

d > 4,d � 4,0 < d < 4,

� 26.2 mi

d � �9 � 676

d � ��0 � ��3��2 � ��6 � 20�2

�0, �6�

BC

A

y

x

12

12

Quiz 3 (p. 648)

1. 2.

3. 4.

5.

6. The points that are in both theexterior of the with center Pand radius 6 units and the interior of the circle with centerP and radius 9 units

7.

A set of points formed by 2 rays on opposite sides of each to and 4 cm from it, and a semicircle withcenter A and radius 4 cm

8.

The points that are on the field and on or outside the circle whose center is the center of the field and whoseradius is 10 yd

Math & History (p. 648)

1. June 21: 75 min

Dec. 21: 46 min

50 yd

10 yd

100 yd

→AB�

→AB,

4 cm

4 cmA B

�

6

9

P

�x � 2�2 � �y � 2�2 � 74

� �74

r � �25 � 49

�x � 4�2 � �y � 7�2 � 25�x � 1�2 � y2 � 36

y

x

2

2�2

y

x

2

�22�2

�x � 3�2 � �y � 3�2 � 49x2 � y2 � 100

y

x

2

�22�2

y

x

4

�44�4





Chapter 10 Review (pp. 650–652)

10.1 Tangents to Circles

1. 2. C, D, or R 3. 4. P 5.

6. 7. 8. R

9. Yes; is tangent to at point B so (Thm 10.1).

10. According to Thm. 10.3, since the segments from point Sare tangent to they are congruent. So and

SCD is isosceles.

10.2 Arcs and Chords

11. 12. 13. 14. 15.

16.

10.3 Inscribed Angles

17. True; the sides of the triangles opposite the inscribedangles are diameters, so the inscribed angles are rightangles.

18. False; in kite ABCD, the diagonals are so .Then m∠AED � 90°. If m∠ACD � m∠AED �

(90°) � 45°, then m∠DAE � 45° as well. ABCD is akite, so Then AED � AEB (HL Cong.Thm.) and m∠BAE � 45°. This would make ∠DAB aright angle and so make a diameter, which wouldmean that E and F would be the same point and the diagonals of ABCD would bisect each other. That contradicts the given that ABCD is a kite. So the assumption that m∠ACD � m∠AED must be false.

19. True; ABCD is inscribed in a circle, so opposite anglesare supplementary.

10.4 Other Angle Relationships in a Circle

20. 112 21. 55 22. 64 23. 94

10.5 Segment Lengths in Circles

24. 25.

26.

Solution: x � 10

x � �40, x � 10

�x � 40��x � 10� � 0

x2 � 30x � 400 � 0

400 � x�30 � x� x � 34.4

10x � 100 � 444 x � 5

10�x � 10� � 12�37� 16x � 80

12

DB

��AD � AB.

12

12

AC � DB�

324�

275�85�239�118�62�

�

SC � SD�Q,

↔BC � PB�P

↔BC

↔BF

↔BC

QEBF or BNBN

10.6 Equations of Circle

27.

28.

29.

10.7 Locus

30. The perpendicular bisectorof contains the pointsequidistant from R and S.

31. Parallel lines 4 in. from and every point between and each line

32. Two points where the cir-cles with center B andradius 4 cm and center Aand radius 3 cm intersect

Chapter 10 Test (p. 653)

1. Thm. 10.3; HL Congruence Thm.

BA

��

4 in.

4 in.

n

l

m

RS

T

S

R

y

x

1

�1�1

�x � 6�2 � y2 � 10

y

x

4

2

�x � 4�2 � �y � 1�2 � 16

y

x

2

�22�2

�x � 2�2 � �y � 5�2 � 81

MCRBG-100R-SK.qxd 5-25-2001 11:36 AM Page 219




2.

3. and are rt because tangents are toradii at the points of tangency. and are radii of

so they are by the Reflexive Prop. ofCongruence. So by the HL Cong. Thm.

because they are corresponding angles.By the defs. of congruence, minor arcs, and congruentarcs,

4. is a short leg: 4; long leg:hypotenuse: 8.

m∠MHP �

5. According to Thm. 10.5, since and

6. so by Thm. 10.4

7. and

8. 9.

10. 11.

12. a. supplementary

b. congruent

13. 14. 15.

x � 6 x � 4 x � 6

x2 � 3�12� 3�12� � x�9� 3x � 18

m�CBE � m�CAE

m�CDE � 2m�CAE

E

D

C

B

A

m�3 � 37�

m�2 � 66�m�2 � 75�

m�1 � 29�m�1 � 120�

m�3 � 45�

m�2 � 90�m�2 � 145�

m�1 � 90�m�1 � 72.5�

FB� � EC�FB � EC

FE� � BC� .FE � BC

� BA� .FA�FH � BHAD � FB,

m�PHJ � m�MHJ � m�MHP � 156.9� � mJN�tan�1 PM

HM � tan�1 34 � 36.9�

m�MHJ � 120� � mJM�m�MHK � 60� � m�JHK

4�3;30� � 60� � 90��,�HMK

LM� � JL� .

�KHM � �KHJ�KJH � �KMH

HK � HK� .�HHMHJ

��s�KMH�KJH

PK � 4�3 � 3

MK � 4�3MP � 3

�MK�2 � 48�MP�2 � 9

�MK�2 � 42 � 82�MP�2 � 42 � 52

JK � 4�3 � 6.9

42 � �JK�2 � 82 16. 17.

Two points, (0, 0) and (4, 4), where the line

and circle with center and radius 4 units intersect

18.

19.

Let A and B be the ends of the cable. The locus consists ofthe points on or inside a region bounded by two semicircleswith centers A and B and radius 3.5 ft and two segments onopp. sides of to and 3.5 ft from

Chapter 10 Standardized Test (pp. 654–655)

1. D 2. C 3. E 4. B 5. B 6. C 7. A

8. 9. D

A

10. 11.

12. 13. is inscribed

in the � and is

the hypotenuse. Thehypotenuse of aninscribed right triangle is a diameter.

14. 15.

16. the point �32, 2�

r �52

� 5

d � �9 � 16

�x �32�2

� �y � 2�2 �254P: �3

2, 2�

mEH� � 240�

BCmEF� � 40�

�ABCmFG� � 50�

m�H � 45� x � 5

m�G � 140� 7x2 � 175

m�F � 135� 7x2 � 5 � 180

m�E � 40� x2 � 15 � 6x2 � 10 � 180

�x � 1�2 � �y � 5�2 � 34

Midpoint � �1, 5� � �34

d � �25 � 9

AB.ABAB, both �

A B

3.5 ft

3.5 ft

�x � 30�2 � �y � 30�2 � 900

�4, 0�y � x

�x � 4�2 � �y � 6�2 � 64

y

x

1

1

y

x

2

�22





17. a. b. c.

d. e. f.

18. Their corresponding chords are congruent soby Thm. 10.4 the arcs are congruent.

19. 20.

21. 22.

all points on the perpen- all points on the lines

dicular bisector of parallel to and 2 in. from

and all points less

than 2 in. from

23. A 4 in. segment of the bisector of with midpointthe midpoint of

24. 100 25.

26. Farther away; this will increase the measure of thesmaller intercepted arc and thereby decrease

27. 28.

Use Pyth. Thm or Thm. 10.17.

r � 7.5 ft

d � 15 ft

25 � 5d � 100

5�5 � d� � 100

5 ft10 ft

m�B.

m�3 � 25�

m�1 � 80�; m�2 � 40�;

ABAB

�

B

2 in.

2 in.A

↔AB

↔AB

AB

2 in.

A B2 in.

BA

� 74�mHD�� 106�,mBH�x � 74

2x � 148

32 � 180 � 2x

16 �12�180 � x � x� 60� � m �AG

� x°mHD�240° � 180° � mAG�� 180� � x°mBH�120° �

12�180° � mAG� �

BC� � CD� ;

45�45�90�

x � 24 x � 3

180� x2 � 18�32� 10x � 30





Algebra Review (pp. 656–657)

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14. 15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

x � 15.38% x � 400%

8 � x�52� 200 � x�50� x � 22% x � 25%

11 � x�50� 34 � x�136� $12.50 � xx � 142

x � 0.50�25�x � 0.71�200�x � 51 milesx � 36

x � 0.15�340�x � 0.30�120� x � �6

�5x � 30 x � 3

x � 6x � 30 16x � 48

12x � 3�x � 5� 2x � 14x � 48

x � 13 x � 4

2x � 26 5x � 20

2x � 16 � 10 5x � 7 � 13

12x � 3x�x � 2� � 9x

3x � 62x � 14x2 � �25 � x

�6S

6� s

b � �c2 � a2

�S6

� s

92 � b2 � c2 S � 6s2

V

�r2 � h Vlw

� h

V � �r2h V � lwh

P2

� w � l

P � 2w

2� l 3�V � S

P � 2l � 2w V � S3

�A�

�� r

C

2�� r �A

�� r

C � 2�r A � �r2

2Ah

� b2 � b1 2Ab

� h

A �12

h�b1 � b2� A �12

bh

3�6�2V2�

� r

3�3V4�

� r Al

� w

V �43

�r3 A � lw

31. 32.

33. 34.

35. 36.

37.

38.

39. 40.

41.

42. 43.

44. 45.

46.

�m � 2m � 2

m2 � 4m � 4

m2 � 4�

�m � 2�2

�m � 2��m � 2�

�t � 1t � 1

t2 � 1

t2 � 2t � 1�

�t � 1��t � 1��t � 1��t � 1�

�5h � 1h � 1

�9s � 1s � 3

�y � 6

12 � y

36s2 � 4s4s2 � 12s

�4s�9s � 1�4s�s � 3�

2y � 1224 � 2y

�2�y � 6�

2�12 � y�

�7d � 13d � 4

14d2 � 2d6d2 � 8d

�2d�7d � 1�2d�3d � 4�

�x � 3

6x � 1 �

a � 2a � 8

5x2 � 15x30x2 � 5x

�5x�x � 3�5x�x � 1�

5a � 105a � 40

�5�a � 2�5�a � 8�

� 3

9w3 � 27w3w3 � 9w

�9w�w2 � 3�3w�w2 � 3�

� x

�5x2 � x�5x � 1

�x�5x � 1�

5x � 1

16a3

8a� 2a2

5x10x2 �

12x

x � 16.67%

3 � x�18�25.95�0.08� � $2.08

x � 25 meters x � 10

16 � 0.64x 3 � 0.30x


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CHAPTER 10 - Mr Jaffemrjaffe.net/GeomRef/unit10solutions.pdf · dicular to Perpendicular lines form right angles. Therefore ... Chapter 10 Worked-out ... 10 3 8 x 360 33 x 18 42 x