hscc geom wsk 10 - schoolwires

Copyright © Big Ideas Learning, LLC Geometry 341All rights reserved. Worked-Out Solutions

Chapter 10

Chapter 10 Maintaining Mathematical Profi ciency (p. 527) 1. (x + 7)(x + 4) = x ⋅ x + x ⋅ 4 + x ⋅ 7 + 7 ⋅ 4

= x2 + 4x + 7x + 28

= x2 + 11x + 28

The product is x2 + 11x + 28.

2. (a + 1 )(a − 5) = a ⋅ a − a ⋅ 5 + 1 ⋅ a − 1 ⋅ 5

= a2 − 5a + a − 5

= a2 − 4a − 5

The product is a2 − 4a − 5.

3. (q − 9)(3q − 4) = q ⋅ 3q − q ⋅ 4 − 9 ⋅ 3q + 9 ⋅ 4

= 3q2 − 4q − 27q + 36

= 3q2 − 31q + 36

The product is 3q2 − 31q + 36.

4. (2v − 7)(5v + 1) = 2v ⋅ 5v + 2v ⋅ 1 − 7 ⋅ 5v − 7 ⋅ 1

= 10v2 + 2v − 35v − 7

= 10v2 − 33v − 7

The product is 10v2 − 33v − 7.

5. (4h + 3)(2 + h) = (4h + 3)(h + 2)

= 4h ⋅ h + 4h ⋅ 2 + 3 ⋅ h + 3 ⋅ 2

= 4h2 + 8h + 3h + 6

= 4h2 + 11h + 6

The product is 4h2 + 11h + 6.

6. (8 − 6b)(5 − 3b) = 8 ⋅ 5 − 8 ⋅ 3b − 6b ⋅ 5 − 6b ⋅ (−3b)

= 40 − 24b − 30b + 18b2

= 40 − 54b + 18b2

The product is 18b2 − 54b + 40.

7. x2 − 2x = 5

x2 − 2x + (−1)2 = 5 + (−1)2

(x − 1)2 = 6

x − 1 = ± √—

6

x = 1 ± √—

6

The solutions are x ≈ 1 + √—

6 = 3.45 and x = 1 − √

— 6 ≈ −1.45.

8. r 2 + 10r = −7

r 2 + 10r + 52 = −7 + 52

(r + 5)2 = 18

r + 5 = ± √—

18

r = −5 ± 3 √—

2

The solutions are r = −5 + 3 √—

2 ≈ − 0.76 and r = −5 − 3 √

— 2 ≈ −9.24.

9. w2 − 8w = 9

w2 − 8w + (−4)2 = 9 + (−4)2

(w − 4)2 = 25

w − 4 = ± √—

25

w = 4 ± 5

The solutions are w = 4 + 5 = 9 and w = 4 − 5 = −1.

10. p 2 + 10p − 4 = 0

p2 + 10p = 4

p 2 + 10p + 52 = 4 + 52

(p + 5)2 = 29

p + 5 = ± √—

29

p = −5 ± √—

29

The solutions are p = −5 + √—

29 ≈ 0.39 and p = −5 − √

— 29 ≈ −10.39.

11. k2 − 4k − 7 = 0

k2 − 4k = 7

k2 − 4k + (−2)2 = 7 + (−2)2

(k − 2)2 = 11

k − 2 = ± √—

11

k = 2 ± √—

11

The solutions are k = 2 + √—

11 ≈ 5.32 and k = 2 − √

— 11 ≈ −1.32.

12. −z2 + 2z = 1

z2 − 2z = −1

z2 − 2z + (−1)2 = −1 + (−1)2

(z − 1)2 = 0

z − 1 = 0

z = 1

The solution is z = 1.

13. Sample answer: (2n + 1)(2n + 3); 2n + 1 is positive and odd when n is a nonnegative integer. The next positive, odd integer is 2n + 3.

Chapter 10 Mathematical Practices (p. 528) 1. Sketch circles A and B with radius 3 units and circle C so

that it passes through the centers of circles A and B. C must be 3 units from A and B, so C must lie on an intersection of circles A and B.

A B

C

342 Geometry Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 10

2. The distance from the center of circle D to any point on circle D is greater than 6. Circles A, B, and C each have diameters of 6 units, and because circle D is tangent to the circles on the outside and whether circle A, circle B, or circle C intersect or contain point D, the radius of circle D must be greater than 6.

A B

C

D

10.1 Explorations (p. 529) 1. Chord: a segment whose endpoints lie on the circle

Secant: a line that intersects a circle at two points

Tangent: a line in the plane that intersects the circle at exactly one point

Radius: a segment whose endpoints are the center and any point on a circle

Diameter: a chord that contains the center of the circle

2. a. Check students’ work.

b. Check students’ work.

c. The distance between the pencil points are the radius and half the diameter.

3. A chord is a segment with endpoints that lie on a circle. A diameter is a chord that passes through the center of a circle. A radius is a segment with one endpoint on the center of a circle and one endpoint on the circle. A secant line is a line that passes through two points on a circle. A tangent line is a line that passes through only one point on a circle.

4. Diameters are a subset of chords. A diameter is a chord that passes through the center of a circle.

5. To draw a circle with a diameter of 8 inches, use two pencils tied together with a string that is 4 inches long.

10.1 Monitoring Progress (pp. 530 –533) 1. — AG is a chord and — CB is a radius.

2. ⃖ &&⃗ DE is a tangent line and — DE or — DB is a tangent segment.

3. The circles have 4 common tangents, 2 external tangents, and 2 internal tangents.

4. The circles have 1 common external tangent.

5. There are no common tangents.

6. Use the Converse of the Pythagorean Theorem (Thm. 9.2).

52 32 + 42

25 9 + 16

25 = 25

△CDE is a right triangle with the right angle at ∠D.

Therefore, — DE is tangent to ⊙C at point D.

7. (r + 18)2 = 242 + r2

r2 + 36r + 324 = 576 + r2

36r + 324 = 576

36r = 252

r = 7

The radius of ⊙Q is 7.

8. By the External Tangent Congruence Theorem (Thm. 10.2), x2 = 9. By taking the square root of each side, x = ±3.

10.1 Exercises (pp. 534 –536)

Vocabulary and Core Concept Check 1. Chords and secants both intersect the circle in two points. A

chord is a segment and a secant is a line.

2. When the context is measure, it refers to length.

3. Coplanar circles that have a common center are called concentric circles.

4. Tangent is the segment that does not belong, because a tangent segment is on the outside of the circle. The other three, chord, radius, and diameter, are segments that are on the interior of the circle.

Monitoring Progress and Modeling with Mathematics 5. The name of the circle is C.

6. Two radii are — CA and — CD .

7. Two chords are — BH and — AD .

8. — AD is a diameter.


Chapter 10

9. ⃖ &&⃗ KG is a secant.

10. A tangent line is ⃖ &&⃗ EG , and a point of tangency is F.

11. There are 4 common tangents.

12. There are no common tangents.

13. There are 2 common tangents.

14. There is 1 common tangent.

15. The common tangent is an external tangent because it does not intersect the segment that joins the centers of the two circles.

16. The common tangent is an internal tangent because it intersects the segment that joins the centers of the two circles.

17. The common tangent is an internal tangent because it intersects the segment that joins the centers of the two circles.

18. The common tangent is an external tangent because it does not intersect the segment that joins the centers of the two circles.


52 32 + 42

25 9 + 16

25 = 25

△CAB is a right triangle with the right angle at ∠A.

Therefore, — AB is tangent to ⊙C at point A.


182 92 + 152

324 81 + 225

324 ≠ 306

△ ABC is not a right triangle. Therefore, — AB is not a tangent segment.


602 402 + 482

3600 1600 + 2304

3600 ≠ 3904

△DAB is not a right triangle. Therefore, — AB is not a tangent segment.


202 ____ 122 + 162

400 ____ 144 + 256

400 = 400

△ ABC is a right triangle, with the right angle at ∠A.

Therefore, — AB is tangent to ⊙C at point A.

23. (r + 16)2 = 242 + r 2

r 2 + 32r + 256 = 576 + r 2

32r + 256 = 576

32r = 320

r = 10

The radius of ⊙C is 10.

24. (r + 6)2 = 92 + r 2

r 2 + 12r + 36 = 81 + r 2

12r + 36 = 81

12r = 45

r = 3.75

The radius of ⊙C is 3.75.

25. (r + 7)2 = 142 + r 2

r 2 + 14r + 49 = 196 + r 2

14r + 49 = 196

14r = 147

r = 10.5

The radius of ⊙C is 10.5.

26. (r + 18)2 = 302 + r 2

r 2 + 36r + 324 = 900 + r 2

36r + 324 = 900

36r = 576

r = 16

The radius of ⊙C is 16.


Chapter 10

27. Sample answer: 28. Sample answer:

C

2 in.

A

C

4.5 cm

A

29. 2x + 7 = 5x − 8 30. 3x + 10 = 7x − 6

7 = 3x − 8 10 = 4x − 6

15 = 3x 16 = 4x

5 = x 4 = x

31. 2x 2 + 4 = 22 32. 3x 2 + 2x − 7 = 2x + 5

2x 2 = 18 3x 2 − 7 = 5

x 2 = 9 3x 2 = 12

x = ±3 x 2 = 4

x = ±2

33. ∠Z is a right angle, not ∠YXZ. — XY is not tangent to ⊙Z.

34. The diameter is 15 and the radius is 15 — 2 = 7.5.

35. There are two possible points of tangency from a point outside the circle, one from a point on the circle, and none from a point inside the circle.

D A

CB

36. When the tangent lines are parallel.

Sample answer: The tangents in the fi gure are perpendicular to the same diameter and, therefore, are parallel.

AC

B

37. The perimeter of the quadrilateral is

6.4 + 4.5 + 6.4 + 8.3 = 25.6 units.

V

Y

T

W

X

8.3

4.5

6.4

6.4

3.3

3.3

3.13.1

1.2 1.2

5.2

5.2

38. a.

CA

B D

b. ∠ACB is a right angle because it was given that the radii were perpendicular. Because — CA and — CB are radii of the same circle, — CA ≅ — CB . The tangent lines are perpendicular to the radii by the Tangent Line to Circle Theorem (Thm. 10.1). So, ∠A and ∠B are right angles. Because ACBD is a quadrilateral, ∠D is also a right angle. — AD ≅ — DB by the External Tangent Congruence Theorem (Thm. 10.2). Therefore, quadrilateral ADBC is a square.

39. By the Tangent Line to a Circle Theorem (Thm. 10.1), ∠PEB and ∠PMB are right angles. By the External Tangent Congruence Theorem (Thm. 10.2), BE = BM and because all radii in a circle are equal, PE = PM. By the SAS Congruence Theorem (Thm. 5.5), △PEB ≅ △PMB. Because corresponding angles of congruent triangles are congruent, ∠PBE ≅ ∠PBM. Hence, the path of the bike trail will bisect the angle formed by the nature trails.

40. To fi nd the distance between the centers of the gears represented by points L and P, divide the region into a rectangle and a triangle.

NM

P

L

1.8 in. 1.8 in.

2.5 in.

17.6 in.

17.6 in.4.3 in.

Apply the Pythagorean Theorem to the right triangle to fi nd LP.

LP 2 = 2.52 + 17.62

LP 2 = 6.25 + 309.76

LP 2 = 316.01

LP = √—

316.01 ≈ 17.8 inches

The distance between the centers of the gears is about 17.8 inches.

41. Sample answer: Every point is the same distance from the center, so the farthest two points can be from each other is opposite sides of the center.

42. — PA ≅ — PB ≅ — PC by the External Tangent Congruence Theorem (Thm. 10.2).


Chapter 10

43. Given — RS is a common internal tangent in ⊙A and ⊙B.

Prove AC

— BC

= RC —

SC

AB

C

R

S

STATEMENTS REASONS

1. — RS is a common internal tangent in ⊙A and ⊙B.

1. Given

2. — RS ⊥ — AR and — RS ⊥ — BS 2. Tangent Line to Circle Theorem (Thm. 10.1)

3. ∠CRA and ∠CSB are right angles.

3. Defi nition of perpendicular

4. ∠RCA ≅ ∠SCB 4. Vertical Angles Congruence Theorem (Thm. 2.6)

5. △RAC ∼ △SBC 5. AA Similarity Triangle Theorem (Thm. 8.3)

6. AC — BC

= RC — SC

6. Corresponding sides of similar fi gures are proportional.

44. Yes, it is true.

— AZ ≅ — AW , — BW ≅ — BX , — CX ≅ — CY , and — DY ≅ — DZ . So, (AW + WB) + (CY + YD) = (AZ + DZ ) + (BX + CX ).

45. 2x − 5 = x + 8

x = 13

2x − 5 + 4y − 1 = x + 8 + x + 6

2x + 4y − 6 = 2x + 14

4y − 6 = 14

4y = 20

y = 5

46. Given — SR and — ST are tangent to ⊙P.

Prove — SR ≅ — ST

R

ST

P

By the Tangent Line to Circles Theorem (Thm. 10.1), m∠PRS = m∠PTS = 90°. — PT ≅ — PR and — PS ≅ — PS , so

△PTS ≅ △PRS by the HL Congruence Theorem (Thm. 5.9). So, — SR ≅ — ST .

47. a. Given Line m is tangent to ⊙Q at point P.

Prove m ⊥ — QP

P R

Q

m

Assume that — mP is not perpendicular to line — QP . Then the perpendicular segment from Q to m intersects m at some other point R. Construct — QR . The perpendicular segment is the shortest distance from Q to m. So, QR < QP, R must be inside ⊙Q, and m must be a secant line. This is a contradiction, so m must be tangent to ⊙Q.

b. Given m ⊥ — QP

Prove Line m is tangent to ⊙Q.

Assume m is not tangent to ⊙Q. Then m must intersect

⊙Q at a second point R. — QP and — QR are both radii of ⊙Q,

so — QP ≅ — QR . Because m ⊥ — QP , QP< QR. This is a contradiction, so m must be tangent to ⊙Q.

48. Use the Pythagorean Theorem to fi nd the length of — AE .

AE 2 + 42 = 122

AE 2 = 144 − 16 AE = √

— 144 − 16

= √—

128

= √—

64 ⋅ 2

= 8 √—

2

P A

FC

E

BD

12

12

8

Because — AB , — AC , and — BC are all tangents to circle O, and — PD ,

— PF , and — PE are radii and perpendicular to the tangents at the

points of tangency, — PD is an altitude to △APB, — PE is an

altitude to △BPC, and — PF is an altitude to △CPA.

Area of △APB = 1 — 2 ⋅ r ⋅ 12 = 6r

Area of △BPC = 1 — 2 ⋅ r ⋅ 8 = 4r

Area of △CPA = 1 — 2 ⋅ r ⋅ 12 = 6r

The total area is 6r + 4r + 6r = 16r.

Area of △ ABC = 1 — 2 ⋅ 8 √

— 2 ⋅ 8 = 32 √

— 2

16r = 32 √—

2

r = 32 √—

2 —

16 = 2 √

— 2 ≈ 2.83

The radius of circle P is 2 √—

2 ≈ 2.83.


Chapter 10

Maintaining Mathematical Profi ciency 49. m∠JKM = m∠JKL + m∠LKM 50. AC = AB + BC

m∠JKM = 15° + 28° 10 = AB + 7

m∠JKM = 43° 3 = AB

10.2 Explorations (p. 537) 1. a. m . BC ≈ 36.9° Use the Law of Cosines to verify.

BC = √—— (4 − 5)2 + (3 − 0)2 = √

— 1 + 9 = √

— 10

a2 = b2 + c2 − 2bc cos A

( √— 10 ) 2 = 52 + 52 − 2(5)(5) cos A

10 = 50 − 50 cos A

−40 = −50 cos A

cos A = 40 — 50 = 4 — 5

m∠A = cos−1 ( 4 — 5 ) ≈ 36.9°

b. m . BC ≈ 53.1° Use the Law of Cosines to verify.

BC = √—— (3 − 5)2 + (4 − 0)2 = √

— 4 + 16 = √

— 20 = 5 √

— 2

a2 = b2 + c2 − 2bc cos A

( √— 20 ) 2 = 52 + 52 − 2(5)(5) cos A

20 = 50 − 50 cos A

−30 = −50 cos A

cos A = 30 — 50 = 3 — 5

m∠A = cos−1 ( 3 — 5 ) ≈ 53.1°

c. m . BC ≈ 16.26° Use the Law of Cosines to verify.

BC = √—— (4 − 3)2 + (3 − 4)2 = √

— 1 + 1 = √

— 2

a2 = b2 + c2 − 2bc cos A

( √— 2 ) 2 = 52 + 52 − 2(5)(5) cos A

2 = 50 − 50 cos A

−48 = −50 cos A

cos A = 48 — 50 = 24

— 25

m∠A = cos−1 ( 24 — 25 ) ≈ 16.26°

d. m . BC ≈ 106.26° Use the Law of Cosines to verify.

BC = √—— (4 − (−4))2 + (3 − 3)2 = √

— (4 + 4)2 = √

— 82 = 8

a2 = b2 + c2 − 2bc cos A

(8)2 = 52 + 52 − 2(5)(5) cos A

64 = 50 − 50 cos A

14 = −50 cos A

cos A = − 14 — 50 = − 7 — 25

m∠A = cos−1 ( − 7 — 25 ) ≈ 106.26°

2. Circular arcs are measured in degrees or radians and are based on the measure of the central angle.

3. a.

−2−3−4−5−6−7 −1 0

0

−2

−3

−4

−5

−1

1

2

3

4

5

1 2 3 4 5 6 7

A B

C

b.

−2−3−4−5−6−7 −1 0

0

−2

−3

−4

−5

−1

1

2

3

4

5

1 2 3 4 5 6 7

A B

C

c.

−2−3−4−5−6−7 −1 0

0

−2

−3

−4

−5

−1

1

2

3

4

5

1 2 3 4 5 6 7

A B

C

d.

−2−3−4−5−6−7 −1 0

0

−2

−3

−4

−5

−1

1

2

3

4

5

1 2 3 4 5 6 7

A B

C

10.2 Monitoring Progress (pp. 539 –541) 1. . TQ is a minor arc, so m . TQ = 120°.

2. . QRT is a major arc, so m . QRT = m . QR + m . RS + m . ST

= 60° + 100° + 80° = 240°.

3. . TQR is a semicircle, so m . TQR = m . TQ + m . QR

= 120° + 60° = 180°.

4. . QS is a minor arc, so m . QS = m . QR + m . RS

= 60° + 100° = 160°.

5. . TS is a minor arc, so m . TS = 80°.


Chapter 10

6. . RST is a semicircle, so m . RST = m . RS + m . RS T

= 100° + 80° = 180°.

7. . AB ≅ . CD , because the circles are congruent and m . AB = m . CD .

8. Because the two circles have different radii, the circles are not congruent and, therefore, . MN is not congruent to . PQ .

10.2 Exercises (pp. 542 –544)

Vocabulary and Core Concept Check 1. If ∠ACB and ∠DCE are congruent central angles of ⊙C,

then . AB and . DE are congruent arcs.

2. The circle with a diameter of 6 inches does not belong. The others have a diameter of 12 inches.

Monitoring Progress and Modeling with Mathematics 3. The minor arc is . AB and it has a measure of 135°. The major

arc is . ADB and its measure is 360° − 135° = 225°.

4. The minor arc is . EF and it has a measure of 68°. The major arc is . FGE and its measure is 360° − 68° = 292°.

5. The minor arc is . JL and it has a measure of 120°. The major arc is . JKL and its measure is 360° − 120° = 240°.

6. The minor arc is . MN and it has a measure of 170°. The major arc is . NPM and its measure is 360° − 170° = 190°.

7. . BC is a minor arc and it has a measure of 70°.

8. . DC is a minor arc and it has a measure of 65°.

9. . ED is a minor arc and it has a measure of 45°.

10. . AE is a minor arc and it has a measure of 70°.

11. . EAB is a semicircle and it has a measure of 180°.

12. . ABC is a semicircle and it has a measure of 180°.

13. . BAC is a major arc and it has a measure of 290°.

14. . EBD is a major arc and it has a measure of 315°.

15. a. m . JL = m . JK + m . KL = 53° + 79° = 132° b. m . KM = m . KL + m . LM = 79° + 68° = 147° c. m . JLM = m . JK + m . KL + m . LM = 53° + 79° + 68° = 200° d. m . JM = 360° − 200° = 160°

16. a. m . RS = m . QRS − m . QR = 180° − 42° = 138° b. m . QRS = 180° c. m . QST = m . QR + m . RS + m . ST = 42° + 138° + 42° = 222° d. m . QT = m . QTS − m . TS = 180° − 42° = 138°

17. a. m . AE = m . AH + m . HG + m . GF + m . FE

= 17° + 26° + 28° + 32° = 103° b. m . ACE = m . AB + m . BC + m . CD + m . DE

= 66° + 55° + 89° + 47° = 257° c. m . GDC = m . GF + m . FE + m . ED + m . DC

= 28° + 32° + 47° + 89° = 196° d. m . BHC = 360° − m . BC = 360° − 55° = 305° e. m . FD = m . FE + m . ED = 32° + 47° = 79° f. m . FBD = 360° − (m . FE + m . ED )

= 360° − (32° + 47°) = 281°

18. Football: m . VW = 20% ⋅ 360° = 72° Soccer: m . WX = 30% ⋅ 360° = 108° Volleyball: m . XY = 15% ⋅ 360° = 54° Cross-Country: m . YZ = 20% ⋅ 360° = 72° None: m . ZV = 15% ⋅ 360° = 54°

19. . AB ≅ . CD because they are in the same circle and m . AB = m . CD .

20. . LP is not congruent to . MN because the circles are not congruent.

21. . VW ≅ . XY because the circles are congruent and m . VW = m . XY .

22. . QRS is not congruent to . FGH because the circles are not congruent.

23. (2x − 30)° + x° = 180° 3x − 30 = 180

3x = 210

x = 70

m . AB = 2 ⋅ 70° − 30° = 110°

24. 6x° + 7x° + 7x° + 4x° = 360° 24x = 360

x = 15

m . RST = 13 ⋅ 15 = 195°

25. Your friend is correct. The arcs must be in the same circle or congruent circles.

26. Your friend is not correct. . AMB is a semicircle, so x° + 4x° = 180°.

27. . AD is the minor arc. The red arc is the major arc, . ABD .

28. . JK and . NP are not in the same circle. . JK ≅ . RQ or . LM ≅ NP.

29. m . ACD = 360° − 20° = 340°

PC

BA

D20°

m . AC = 180° − 20° = 160°


Chapter 10

30. Sample answer:

R

C B

A

E

D20°

25°70°

60°

m . AE = 60° + 25° + 70° − 20° = 135°

31. The measure of each arc is 360° — 20

= 18°.

32. a. Each arc measure is 360° — 20

= 15°.

b. The measure of the minor arc from the Tokyo time zone to the Anchorage time zone is 6 ⋅ 15° = 90°.

c. If two locations differ by 180° on the wheel, then it is 3 p.m. at one location when it is 3 a.m. at the other location.

33. Given ⊙O with center O(0, 0) and radius r, ⊙A with center A(a, 0) and radius s

Prove ⊙O ∼ ⊙A

x

y

Or

As

Translate ⊙A left a units so that point A maps to point O. The image of ⊙A is ⊙A′ with center O, so ⊙A′ and ⊙O are concentric circles. Dilate ⊙A′ using center of dilation O and

scale factor r — s , which maps the points s units from point O to

the points r — s (s) = r units from point O. So, this dilation maps

⊙A′ to ⊙O. Because a similarity transformation maps ⊙A to ⊙O, ⊙O ∼ ⊙A.

34. yes; — CD is the radius of ⊙C and — DC is the radius of ⊙D. Because the radii are the same, the circles are congruent.

35.

A

C

B

D

a. Given — AC ≅ — BD

Prove ⊙A ≅ ⊙B

Translate ⊙B so that point B maps to point A. The image of ⊙B is ⊙B′ with center A. Because — AC ≅ — BD , this

translation maps ⊙B′ to ⊙A. A rigid motion maps ⊙B to ⊙A, so ⊙A ≅ ⊙B.

b. Given ⊙A ≅ ⊙B

Prove — AC ≅ — BD

Because ⊙A ≅ ⊙B, the distance from the center of the circle to a point on the circle is the same for each circle.

So, — AC ≅ — BD .

36. The circles are similar because they are concentric, with different radii.

37.

A

C

B

D

E

a. Given ∠BAC ≅ ∠DAE

Prove . BC ≅ . DE

m . BC = m∠BAC, m . DE = m∠DAE and m∠BAC = m∠DAE, so m . BC = m . DE . Because . BC and . DE are in the same circle, . BC ≅ . DE .

b. Given . BC ≅ . DE

Prove ∠BAC ≅ ∠DAE

m . BC = m∠BAC and m . DE = m∠DAE. Because . BC ≅ . DE , ∠BAC ≅ ∠DAE.

38. The circumference of a circle is 2πr. So, the arc length is

s = 2πr ⋅ θ — 360° = πrθ —

180° .

Maintaining Mathematical Profi ciency 39. c2 = a2 + b2

172 = x2 + 82

289 = x2+ 64

225 = x2

x = √—

225 = 15

Because the side lengths 8, 15, and 17 are integers that satisfy the equation c2 = a2 + b2, they form a Pythagorean triple.

40. hypotenuse = leg ⋅ √—

2

x = 13 √—

2 ≈ 18.38

Because 13 √—

2 is not an integer, the side lengths do not form a Pythagorean triple.

41. c2 = a2 + b2

x2 = 72 + 112

x2 = 49 + 121

x = √—

170 ≈ 13.04

Because √—


42. c 2 = a2 + b2

142 = x2 + 102

196 = x2 + 100

x = √—

96 ≈ 9.80

Because √—



Chapter 10

10.3 Explorations (p. 545) 1. a. The diameter with an endpoint of (4, 3) will pass through

the center and the other endpoint of the diameter will be (−4, −3).

−2−3−4−5−6−7 −1 0

0

−2

−3

−4

−5

−1

1

2

3

4

5

1 2 3 4 5 6 7

b. The diameter with an endpoint of (0, 5) will pass through the center and the other endpoint of the diameter will be (0, −5).

−2−3−4−5−6−7 −1 0

0

−2

−3

−4

−5

−1

1

2

3

4

5

1 2 3 4 5 6 7

c. The diameter with an endpoint of (−3, 4) will pass through the center and the other endpoint of the diameter will be (3, −4).

−2−3−4−5−6−7 −1 0

0

−2

−3

−4

−5

−1

1

2

3

4

5

1 2 3 4 5 6 7

d. The diameter with an endpoint of (−5, 0) will pass through the center and the other endpoint of the diameter will be (5, 0).

−2−3−4−5−6−7 −1 0

0

−2

−3

−4

−5

−1

1

2

3

4

5

1 2 3 4 5 6 7

2. Check students’ work. The perpendicular bisector is a diameter. The results are the same. A perpendicular bisector of a chord is a diameter of the circle.

3. Check students’ work.

DF = EF

The results are the same. If a chord is perpendicular to a diameter of a circle, then the diameter is a perpendicular bisector of the chord.

4. A chord is a diameter of a circle when it is a perpendicular bisector of a chord or it passes through the center of the circle.

10.3 Monitoring Progress (pp. 547 –548) 1. Because — AB ≅ — BC , then by the Congruent Corresponding

Chords Theorem (Thm. 10.6), . AB ≅ . BC . Therefore, m . BC = 110°.

2. m . AC + m . CB + m . AB = 360° 150° + 2m . AB = 360° 2m . AB = 210° m . AB = 105°

3. CE = 2 ⋅ 5 = 10

4. m . CD = m . DE

9x° = (80 − x)° 10x = 80

x = 8

m . CE = 9x° + (80 − x)° = 8x + 80 = 8 ⋅ 8 + 80 = 64 + 80 = 144°

5. PN = NO

3x = 7x − 12

−4x = −12

x = 3

(3x)2 + 122 = (NK)2

(3 ⋅ 3)2 + 122 = (NK)2

81 + 144 = (NK)2

NK = √—

225 = 15

The radius of circle N is 15 units.

10.3 Exercises (pp. 549 –550)

Vocabulary and Core Concept Check 1. To bisect a chord is to divide the chord into two segments of

equal length.

2. no; One chord does not necessarily bisect another chord.

Monitoring Progress and Modeling with Mathematics 3. Because AB = DE and are in the same circle, then the

corresponding minors arcs are equal. Therefore, mAB = 75°.


Chapter 10

4. m . TU = m . UV , so TU = UV = 5.

5. By segment addition, WY = WC + CY and ZX = ZC + CX. Because WC = CX and ZC = YC, WY = ZX. By arc addition, m . XYZ = m . XY + m . YZ , m . WZY = m . WZ + m . ZY , and m . XYZ = m . XWZ . By substitution, m . XY + m . YZ = m . WZ + m . ZY , m . XY + 60° = 110° + 60°, m . XY + 60° = 170°, m . XY = 110°. Therefore, m . XYZ = 110° + 60° = 170°.

6. Because ⊙C ≅ ⊙P and QR = LM, LM = 11. Therefore, QR = 11.

7. By the Perpendicular Chord Bisector Theorem (Thm. 10.7), JE = EG. Therefore, x = 8.

8. By the Perpendicular Chord Bisector Theorem (Thm. 10.7), m . RS = m . ST . Therefore, m . RS = 40°.

9. By the Perpendicular Chord Bisector Theorem (Thm. 10.7):

5x − 6 = 2x + 9

3x − 6 = 9

3x = 15

x = 5

10. By the Perpendicular Chord Bisector Theorem (Thm. 10.7):

5x + 2 = 7x − 12

−2x + 2 = −12

−2x = −14

x = 7

11. — AC and — DB are not perpendicular. . BC is not congruent to . CD .

12. Sample answer: Draw the perpendicular bisector of the control panels and fi nd the midpoint; Because the control panels are parallel and the congruent chords of the cross section, they are equal distances from the center, and their perpendicular bisectors form a diameter.

13. — AC ≅ — AD and ∠CAB ≅ ∠DAB. By the Refl exive Property of Congruence, — AB ≅ — AB . So, by the SAS Congruence Theorem, △CAB ≅ △DAB. Because the triangles are congruent, — BC ≅ — BD and therefore — AB is a perpendicular bisector of — CD . So, — AB is a diameter.

14. (AE)2 + (ED)2 = (AD)2

32 + (ED)2 = 52

9 + (ED)2 = 25

(ED)2 = 16

ED = 4

Because CE ≠ ED, — AB is not a diameter.

15. AQ = QB

4x + 3 = 7x − 6

−3x + 3 = −6

−3x = −9

x = 3

QG 2 = QB 2 + BG2

QG 2 = (7x − 6)2 + 82

QG 2 = (7 ⋅ 3 − 6)2 + 82

QG 2 = 152 + 82

QG 2 = 225 + 64

QG = √—

289 = 17

The radius of circle Q is 17 units.

16. AD = BC

4x + 4 = 6x − 6

−2x + 4 = −6

−2x = −10

x = 5

QD2 = (2x + 2)2 + 52

QD2 = (2 ⋅ 5 + 2)2 + 52

QD2 = 122 + 52

QD2 =144 + 25

QD = √—

169 = 13

The radius of circle Q is 13 units.

17. The perpendicular bisectors intersect at the center, so the right triangle with legs of 6 inches and 3.5 inches have a hypotenuse equal to the length of the radius.

r 2 = 62 + 3.52

r 2 = 36 + 12.25

r 2 = 48.25

r 2 = √—

48.25 ≈ 6.95

The radius is about 6.95 inches. So, the diameter is about 2 ⋅ 6.95 = 13.9 inches.

18. a. — AB is a diameter by the Perpendicular Chord Bisector Converse (Thm. 10.8).

b. — AB ≅ — CD by the Congruent Corresponding Chords Theorem (Thm. 10.6).

c. — JH bisects — FG and . FG by the Perpendicular Chord Bisector Theorem (Thm. 10.7).

d. — NP ≅ — LM by the Equidistant Chords Theorem (Thm. 10.9).


Chapter 10

19. a.

A

BC

D P

Given — AB and — CD are congruent chords.

Prove . AB ≅ . CD

Because PA = PB = PC = PD, △PDC ≅ △PAB by the SSS Congruence Theorem (Thm. 5.8). So, ∠DPC ≅ ∠APB and . AB ≅ . CD .

b. Given . AB ≅ . CD

Prove — AB ≅ — CD

PA = PB = PC = PD, and because . AB ≅ . CD , ∠DPC ≅ ∠APB. By the SAS Congruence Theorem (Thm. 5.5), △PDC ≅ △PAB, so — AB ≅ — CD .

20. Sample answer: — AC ≅ — RC , so . AC ≅ . RC and m . AR = 360 − 2m . AC . m . AC is an integer, so 2m . AC is even and 360 − 2m . AC is even.

21.

A

B

C8

6

10

D

P

Using △PDB, m∠DPB = sin−1 ( 6 — 10 ) ≈ 36.87°. Using △PCA, m∠CPA = sin−1 ( 8

— 10 ) ≈ 53.13°. m∠BPA = m∠CPA − m∠DBP ≈ 53.13° − 36.87° = 16.26° Therefore, m . AB ≈ 16.26°.

22. Given — EG is a diameter of ⊙L, — EG ⊥ — DF .

Prove — DC ≅ — FC , . DG ≅ . FG

E

F

G

D

CL

— LD ≅ — LF and — LC ≅ — LC , so △LDC ≅ △LFC by the HL Congruence Theorem (Thm. 5.9). Then — DC ≅ — FC and ∠DLC ≅ ∠FLC. So, . DG ≅ . FG .

23. Given — QS is a perpendicular bisector of — RT .

Prove — QS is a diameter of the circle L.

P

T

R

QS

L

— TP ≅ — PR , — LP ≅ — LP , and — LT ≅ — LR , so △LPR ≅ △LPT by the SSS Congruence Theorem (Thm. 5.8). Then ∠LPT ≅ ∠LPR, so m∠LPT = m∠LPR = 90°. By defi nition, — LP is a perpendicular bisector of — RT , so L lies on — QS . Because — QS contains the center, — QS is a diameter of ⊙L.

24. ∠A ≅ ∠C because they intercept the same arc, . BD . ∠APB ≅ ∠CPD, therefore, △APB ∼ △CPD by the

AA Similarity Theorem. So, AP —

BP = CP

— DP

.

25.

ADE

G

F

C

B

If — AB ≅ — CD , then — GC ≅ — FA . Because — EC ≅ — EA , △ECG ≅ △EAF by the HL Congruence Theorem (Thm. 5.9), so — EF = — EG . If — EF = — EG , then because — EC ≅ — ED ≅ — EA ≅ — EB , △AEF ≅ △BEF ≅ △DEG ≅ △CEG by the HL Congruence Theorem (Thm. 5.9). Then — AF ≅ — BF ≅ — DG ≅ — CG , so — AB ≅ — CD .

26. yes; The diameter of the tire that is perpendicular to the ground is also perpendicular to — AB , so it bisects . AB .

Maintaining Mathematical Profi ciency 27. (n − 2) ⋅ 180° = (4 − 2) ⋅ 180° = 360° 360° − (32° + 25° + 44°) = 360° − 101° = 259° So, m∠M = 259°.

28. (n − 2) ⋅ 180° = (54 − 2) ⋅ 180° = 540° 540° − (85° + 134° + 97° + 102°) = 540° − 418° = 122° So, m∠T = 122°.

10.1–10.3 What Did You Learn? (p. 551) 1. Sample answer: The External Tangent Congruency Theorem

(Thm. 10.2) could be used.

2. None of the arcs overlapped at more than one point, and point E was on . CD . There are other cases, such as point C could be on . AB .

3. A general conclusion was reached from several specifi c cases.


Chapter 10

10.1–10.3 Quiz (p. 552) 1. The circle is P. 2. — PN and — PK are radii.

3. — KN is a diameter. 4. — JL is a chord.

5. ⃖ &&⃗ SQ is a secant. 6. ⃖ &&⃗ QR is a tangent line.

7. (x + 9)2 = 152 + x2 8. 6x − 3 = 3x + 18

x2 + 18x + 81 = 225 + x2 3x − 3 = 18

18x + 81 = 225 3x = 21

18x = 144 x = 7

x = 8

9. . AE is a minor arc with a measure of 180° − 36° = 144°.

10. . BC is a minor arc with a measure of 180° − (67° + 70°) = 43°.

11. . AC is a minor arc with a measure of 67° + 43° = 110°.

12. . ACD is semicircle with a measure of 180°.

13. . ACE is a major arc with a measure of 180° + 36° = 216°.

14. . BEC is a major arc with a measure of 360° − 43° = 317°.

15. . JM ≅ . KL because they are in the same circle and m . JM = m . KL .

16. . PQ and . SR are not congruent because the circles are not congruent.

17. ⊙P ≅ ⊙Q and BD = EG. Therefore, . BAD ≅ . EFG and m . EFG = 30° + 70° = 100°.

18. PG = PJ

x + 5 = 3x − 1

−2x + 5 = −1

−2x = −6

x = 3

Apply the Pythagorean Theorem.

c2 = a2 + b2

PD2 = PJ 2 + JD2

PD2 = 82 + 152

PD2 = 64 + 225

PD2 = 289

PD = √—

289 = 17

The radius of ⊙P is 17 units.

19. a. Each arc measure is 360° — 12

= 30°.

b. The measure of the minor arc formed by the hour hand and minute hand at 7:00 is 360° − 210° = 150°.

c. Sample answer: The minor arc formed by the hour and minute hands when the time is 5:00 measures 150°.

10.4 Explorations (p. 553) 1. a. Check students’ work.

b. m∠CAD = 90° and m∠CBD = 45°. So, m∠CBD = 1 — 2 ⋅ m∠CAD or m∠CAD = 2 ⋅ m∠CBD.

c. Check students’ work. The measure of an inscribed angle is equal to half of the measure of the intercepted arc.


b. Sample answer: The angle measures of the quadrilateral sum to 360°. Opposite angles sum to 180°.

c. Check students’ work. The sum of the opposite angles in an inscribed quadrilateral is 180°.

3. Inscribed angles are one-half of the intercepted arcs. Opposite angles of an inscribed quadrilateral are supplementary.

4. m∠G + m∠E = 180° m∠G + 80° = 180° m∠G = 100° ∠E and ∠G are supplementary.

10.4 Monitoring Progress (pp. 555–557)

1. m∠HGF = 1 — 2 m . HF 2. m . TV = 2∠U

m∠HGF = 1 — 2 ⋅ 90° = 45° m . TV = 2 ⋅ 38° = 76°

3. m∠Y = m∠X by the Inscribed Angles of a Circle Theorem (Thm. 10.11). Therefore, m∠X = 72°.

4. By the Inscribed Right Triangle Theorem (Thm. 10.12), x = 90°. m . LK = 2 ⋅ 40° m . LM = 180° − 80° = 100° m∠LKM = 1 — 2 m . LM = 1 — 2 ⋅ 100° = 50° So, y = 50.

5. By the Inscribed Quadrilateral Theorem (Thm. 10.13):

m∠ABC + m∠CDA = 180° x° + 82° = 180° x = 98

m∠BAD + m∠DCB = 180° y° + 68° = 180° y = 112

6. By the Inscribed Quadrilateral Theorem (Thm. 10.13): m∠VST + m∠TUV = 180° c° + (2c − 6)° = 180° 3c − 6 = 180 3c = 186 c = 62 m∠SVU + m∠UTS = 180° 8x° + 10x° = 180° 18x = 180 x = 10


Chapter 10

7. Draw the circle that has the diagonal from the back left corner to the front right corner as a diameter.

10.4 Exercises (pp. 558 –560)

Vocabulary and Core Concept Check 1. If a circle is circumscribed about a polygon, then the polygon

is an inscribed polygon.

2. “Find m∠AGC ” is different. m∠AGC = 180° − (25° + 25°) = 180° − 50° = 130°. The other three are right angles and measure 90° by the Inscribed Right Triangle Theorem (Thm. 10.12).

Monitoring Progress and Modeling with Mathematics 3. m∠ABC = 1 — 2 m . BC

m∠ABC = 1 — 2 ⋅ 84° = 42°

4. m . DF + m . FG + m . GD = 360° m . DF + 70° + 120° = 360° m . DF + 190° = 360° m . DF = 170° m∠G = 1 — 2 m . DF

m∠G = 1 — 2 ⋅ 170° = 85°

5. m . LM + m . MN = 180° m . LM + 160° = 180° m . LM = 20° m∠N = 1 — 2 m . LM

m∠N = 1 — 2 ⋅ 20° = 10°

6. m . RS = 2 ⋅ m∠RQS 7. m . TV = 2 ⋅ m∠TUV

m . RS = 2 ⋅ 67° m . TV = 2 ⋅ 30° m . RS = 134° m . TV = 60° m . VU = 180° − 60° = 120°

8. m . WY = 2 ⋅ m∠WXY

m . WY = 2 ⋅ 75° m . WY = 150° m . WY + m . YX + m . XW = 360° 150° + 110° + m . XW = 360° 260° + m . XW = 360° m . XW = 100°

9. ∠DBC ≅ ∠DAC and ∠BCA ≅ ∠BDA by the Measure of an Inscribed Angle Theorem (Thm. 10.10).

10. ∠XWY ≅ ∠YZX and ∠WXZ ≅ ∠WYZ by the Measure of an Inscribed Angle Theorem (Thm. 10.10).

11. m∠EHF = m∠EGF by 12. m . PS = 2 ⋅ m∠PRSthe Inscribed Angles of a m . RS = 2 ⋅ 40°Circle Theorem (Thm. 10.11). m . RS = 80°Therefore, m∠EHF = 51°.


m∠QRS + m∠STO = 180° x° + 80° = 180° x = 100

m∠RST + m∠TQR = 180° y° + 95° = 180° x = 85


m∠DEF + m∠FGD = 180° m° + 60° = 180° m = 120

m∠EFG + m∠GDE = 180° 2k° + 60° = 180° 2k = 120

k = 60

15. m . KL + m . LM + m . MJ + m . JK = 360° 110° + 130° + 54° + m . JK = 360° 294° + m . JK = 360° m . JK = 66° m∠KLM = 1 — 2 ( m . MJ + m . JK ) 3a° = 1 — 2 (54° + 66°) 3a = 60

a = 20

m∠JML = 1 — 2 ( m . JK + m . KL ) 4b° = 1 — 2 (66° + 110°) 4b = 88

b = 22

16. m∠XYZ = 1 — 2 ( m . XZ ) 3x° = 1 — 2 (180°) 3x = 90

x = 30

m . YZ = 2 ⋅ 34° = 68° m . YZ + m . YX = 180° 68° + m . YX = 180° m . YX = 112° m∠YZX = 1 — 2 ( m . XY ) 2y° = 1 — 2 (112°) 2y = 56

y = 28

17. The inscribed angle was not doubled.

m∠BAC = 2(53)° = 106°

18. Place the right angle of the carpenter’s square on the edge of the circle and connect the points where the sides intersect the edge of the circle.


Chapter 10

19. m∠DAB + m∠BCD = 180° 26y° + 2x° = 180° 2x = 180 − 26y x = 90 − 13y m∠ABC + m∠ADC = 180° 3x° + 21y° = 180° 3(90 − 13y) + 21y = 180 270 − 39y + 21y = 180 270 − 18y = 180 −18y = −90 y = 5 x = 90 − 13 ⋅ 5 = 25 m∠DAB = 26 ⋅ 5 = 130° m∠ABC = 3x = 3 ⋅ 25 = 75° m∠BCD = 2 ⋅ 25 = 50° m∠ADC = 21y = 21 ⋅ 5 = 105°

20. m∠ABC + m∠ADC = 180° 4x° + 24y° = 180° x + 6y = 45 x = 45 − 6y m∠DAB + m∠BCD = 180° 9y° + 14x° = 180° 9y + 14(45 − 6y) = 180 9y + 630 − 84y = 180 630 − 75y = 180 −75y = −450 y = 6 x = 45 − 6 ⋅ 6 = 9 m∠DAB = 9y = 9 ⋅ 6 = 54° m∠ABC = 4x = 4 ⋅ 9 = 36° m∠BCD = 14x = 14 ⋅ 9 = 126° m∠ADC = 24y = 24 ⋅ 6 = 144°

21. m∠BAC = 1 — 2 m . BC

2x° = 1 — 2 ⋅ 6y° x = 3 — 2 y

m . AB + m . BC + m . CA = 360° 6y° + 6y° + 4 ( 3 — 2 y ) ° = 360° 12y + 6y = 360 18y = 360 y = 20 x = 30 m . AB = 6y = 6 ⋅ 20 = 120° m . BC = 6y = 6 ⋅ 20 = 120° m . CA = 4x = 4 ⋅ 30 = 120° m∠BAC = 1 — 2 m . BC = 1 — 2 ⋅ 120 = 60° m∠ABC = 1 — 2 m . AC = 1 — 2 ⋅ 120 = 60° m∠BCA = 1 — 2 m . AB = 1 — 2 ⋅ 120 = 60°

22. yes; ∠PTQ, ∠PSQ, and ∠PRQ are all congruent because they all intercept . PQ .

23. Construct an equilateral triangle by drawing a segment and rotating one image of the segment 60° clockwise about one endpoint of the original segment and the other image of the segment 60° counterclockwise about the other endpoint of the original segment. The endpoints of the images should meet to form the third vertex of the triangle. Then bisect two ofthe angles. Where the angle bisectors intersect is the center of the circle.

24. Construct a regular hexagon by drawing a segment and rotating one image of the segment 120° clockwise about one endpoint of the original segment and then rotate that image of the segment 120° clockwise about the endpoint of the image. Continue this pattern until the endpoints of the images meet to form a hexagon. Then bisect two of the angles. The point where the angle bisectors intersect is the center of the circle.

25. yes; Opposite angles are always supplementary.


27. no; Opposite angles are not always supplementary.




31. Moon A and moon C are 220,000 kilometers apart.

B

O

A

C

100,000 km

10,000 km

10,000 km100,000 km

32. You can sit at any point on the circle circumscribed about the triangle that seat F7 makes with the movie screen because the angle intercepts the same arc.


Chapter 10

33. The length of the hypotenuse of an inscribed right triangle in a circle is always two times the radius, which is the diameter.

34. ∠AYB is always a right angle for any non-endpoint location on semicircle . AYB .

35. The diagonals of a rectangle inscribed in a circle are diameters of the circle because each diagonal splits the rectangle into two right triangles.

36. yes; Every triangle has a circumcenter.

37. a. Case 1:

C

A

BQ

x °

Given ∠ABC is inscribed in circle Q. Let m∠ABQ = x°. Center Q lies on — BC .

Prove m∠ABC = 1 — 2 m . AC

— QB ≅ — QA , so △ABC is isosceles. By the Base Angles Theorem (Thm. 5.6), ∠QBA ≅ ∠QAB, so m∠BAQ = x°. By the Exterior Angles Theorem (Thm. 5.2), m∠AQC = 2x°.

Then m . AC = 2x°, so m∠B = x° = 1 — 2 (2x)° = 1 — 2 m . AC .

b. Case 2:

C

A

BDQ

Given ∠ABC is inscribed in circle Q. — DB is a diameter.


By Case 1, proved in part (a), m∠ABD = 1 — 2 m . AD and

m∠CBD = 1 — 2 m . CD . By the Arc Addition Postulate (Post. 10.1),

m . AD + m . CD = m . AC . By the Angle Addition Postulate

(Post 1.4), m∠ABD + m∠CBD = m∠ABC.

Then m∠ABC = 1 — 2 m . AD + 1 — 2 m . CD

= 1 — 2 ( m . AD + m . CD ) = 1 — 2 m . AC .

c. Case 3:

C

A

BDQ

Given ∠ABC is inscribed in circle Q. — DB is a diameter.


By Case 1, proved in part (a), m∠DBA = 1 — 2 m . AD and m∠DBC = 1 — 2 m . CD . By the Arc Addition Postulate

(Post. 10.1), m . AC + m . CD = m . AD , so m . AC = m . AD − m . CD . By the Angle Addition Postulate (Post. 1.4), m∠DBC + m∠ABC = m∠DBA, so m∠ABC = m∠DBA − m∠DBC. Then m∠ABC = 1 — 2 m . AD − 1 — 2 m . CD

= 1 — 2 (m . AD − m . CD )

= 1 — 2 m . AC .

38. Given ∠ADC and ∠ABC are inscribed angles intercepting . AC .

Prove ∠ADC ≅ ∠ABC

A

C

D

B

By the Measure of an Inscribed Angle Theorem (Thm. 10.10), m∠ABC = 1 — 2 m . AC and m∠ADC = 1 — 2 m . AC . By the Transitive Property of Equality, m∠ABC = m∠ADC.

39. To prove the conditional, fi nd the measure of the intercepted arc of the right angle and the defi nition of a semicircle to show the hypotenuse of the right triangle must be the diameter of the circle. To prove the converse, use the defi nition of a semicircle to fi nd the measure of the angle opposite the diameter.

40. By the Arc Addition Postulate (Postulate 10.1), m . EFG + m . EDG = 360° and m . FGD + m . DEF = 360°. Using the Measure of an Inscribed Angle Theorem, m . EDG = 2m∠F, m . EFG = 2m∠D, m . DEF = 2m∠G, and m . FGD = 2m∠E. By the Substitution Property of Equality, 2m∠D + 2m∠F = 360°, so m∠D + m∠F = 180°. Similarly, m∠E + m∠G = 180°.

41.

B5

43

A

J

K

C

Let x be a segment that extends from C to — AB . x is a diameter of the smallest circle that connects C and — AB . Because inscribed ∠C is a right angle, — JK is diameter. By the Right Triangle Similarity Theorem (Thm. 9.6), x — 3 = 4 — 5 .

x — 3 = 4 — 5

5x = 3 ⋅ 4

5x = 12

x = 12 — 5 = 2.4 units


Chapter 10

42. a. Because — FH is diameter, ∠FGH is a right angle. △FJG, △GJH, and △FGH are similar by the AA Similarity

Theorem (Thm. 8.3). Therefore, GJ —

JH = FJ

— JG

= FG —

GH because

corresponding parts of similar triangles are proportional.

b. By the board count, FJ = 6, so JH = 2.

FJ —

JG = GJ

— JH

6 — x = x —

2

x2 = 12

x = √—

12 = 2 √—

3

Therefore, JG = 2 √—

3 .

GK = 2 ⋅ 2 √—

3 = 4 √—

3

Maintaining Mathematical Profi ciency 43. 3x = 145 44. 1 —

2 x = 63

x = 145 — 3 x = 126

45. 240 = 2x 46. 75 = 1 — 2 (x − 30)

x = 120 150 = x − 30

180 = x


b. Sample answer: 49° and 131° c. Sample answer: 98° and 262° d. Check students’ work.

The measure of each angle between a chord and a tangent is half of its intercepted arc.


b. Sample answer: 60° c. Sample answer: 40° and 80°; The angle measure is half of

the sum of the measures of the intercepted arcs.

d. Check students’ work.

The measure of an angle between two chords inside is half of the sum of the measure of the arcs intercepted by the angle and its vertical angle.

3. When a chord intersects a tangent line, the angle formed is half of the measure of the intercepted arc. When a chord intersects another chord, the measure of the angle is half of the sum of the measures of the arcs intercepted by the angle and its vertical angle.

4. m∠1 = 1 — 2 ⋅ 148° = 74°

5. The sum of the measures of the arcs intercepted by the two angles is 55 ⋅ 2 = 110°.

10.5 Monitoring Progress (pp. 562 –565) 1. m∠1 = 1 — 2 ⋅ 210° = 105° 2. m . RST = 2 ⋅ 98° = 196°

3. m . XY = 2 ⋅ 80° = 160°

4. 180° − 102° = 78° 78° = 1 — 2 ⋅ (95° + y°) 156 = 95 + y

61 = y

5. m∠J = 1 — 2 ⋅ ( m . FG − m . KH ) 30° = 1 — 2 ⋅ (a° − 44°) 60 = a − 44

104 = a

6. m∠N = 180° − m∠KML

x° = 180° − 120° x = 60

7. m . PQ = 2x° m . PTQ + m . PQ = 360° m . PTQ + 2x° = 360° m . PTQ = 360° − 2x° m∠S = 1 — 2 ( m . PTQ − m . PQ ) 50° = 1 — 2 (360° − 2x° − 2x°)

50 = 1 — 2 (360 − 4x)

50 = 180 − 2x

−130 = −2x

65 = x

8. m∠ABD = sin−1 ( 4000 — 4002.73

) ≈ 87.8836 ≈ 88°

m∠CBD ≈ 2 ⋅ 88° = 176° m∠CBD = 180° − m . CD

176° = 180° − m . CD

−4° = −m . CD

m . CD = 4°

10.5 Exercises (pp. 566 –568)

Vocabulary and Core Concept Check 1. Points A, B, C, and D are on a circle, and ⃖ &&⃗ AB intersects ⃖ &&⃗ CD

at point P. If m∠APC = 1 — 2 ( m . BD − m . AC ) , then point P is outside the circle.

2. To fi nd the measure of a circumscribed angle, fi nd the measure of the central angle that intercepts the same arc and subtract it from 180°.

Monitoring Progress and Modeling with Mathematics 3. m . AB = 2 ⋅ 65° = 130°

4. m . DEF = 2 ⋅ 117° = 234°

5. m∠1 = 1 — 2 ⋅ 260° = 130°

6. m∠3 = 1 — 2 ⋅ (360° − 140°) = 1 — 2 ⋅ 220° = 110°


Chapter 10

7. x° = 1 — 2 ⋅ ( m . CD + m . BA ) x° = 1 — 2 ⋅ (145° + 85°)

x = 1 — 2 ⋅ 230 = 115

8. 180° − x° = 1 — 2 ⋅ ( m . KL + m . JM ) 180° − x° = 1 — 2 ⋅ (2x° − 30° + 30°)

180 − x = 1 — 2 ⋅ (2x)

180 − x = x

180 = 2x

90 = x

9. m∠E = 1 — 2 ( m . GD − m . DF ) 29° = 1 — 2 (114° − x°) 58 = 114 − x

−56 = −x

56 = x

10. m∠S = 1 — 2 ( m . UV − m . TW ) 34° = 1 — 2 [ (3x − 2)° − (x + 6)° ] 68 = 3x − 2 − x − 6

68 = 2x − 8

76 = 2x

38 = x

11. m∠T = 1 — 2 ( m . PQ − m . RS )

( x — 2 ) ° = 1 —

2 [ (x + 70)° − (x + 30)° ]

x = x + 70 − x − 30

x = 40

12. m∠X = 1 — 2 [ (360° − m . YZ ) − m . YZ ] (6x − 11)° = 1 — 2 [ (360° − 125°) − 125° ]

6x − 11 = 1 — 2 (360 − 250)

6x − 11 = 1 — 2 ⋅ 110

6x − 11 = 55

6x = 66

x = 11

13. m . MP = 2 ⋅ 73° = 146° m . MLP + m . MP = 360° m . MLP + 146° = 360° m . MLP = 360 − 146 = 214°

m∠N = 1 — 2 ( m . MLP − m . MP ) x° = 1 — 2 (214° − 146°)

x = 1 — 2 ⋅ 68

x = 34

14. m∠G = 1 — 2 [ 17x° − (360° − 17x°) ] 75° = 1 — 2 (17x° − 360° + 17x°)

75 = 1 — 2 (34x − 360)

75 = 17x − 180

255 = 17x

x = 15

15. ∠SUT is not a central angle.

m∠SUT = 1 — 2 ( m . ST + m . QR ) m∠SUT = 1 — 2 (46° + 37°)

= 1 — 2 ⋅ 83°

= 41.5°

16. The 1 — 2 was left out of the equation.

m∠1 = 1 — 2 (122° − 70°) = 1 — 2 ⋅ 52° = 26°

17. When a tangent and a secant intersect outside the circle, the measure of the angle is half of the difference of the measures of the intercepted arcs.

m∠1 = 1 — 2 (180° − 60°) = 1 — 2 (120°) = 60°

18. The measure of an intercepted angle is half the measure of its intercepted arc.

m∠2 = 1 — 2 (120°) = 60°

19. Because the sum of the angles of a triangle always equals 180°, solve the equation 960° + 90° + m∠3 = 180°.

60° + 90° + m∠3 = 180° 150° + m∠3 = 180° m∠3 = 30°

20. Because ∠2 and ∠5 sum to 90° and form a straight line with ∠4, the angles are supplementary.

90° + m∠4 = 180° m∠4 = 180°

21. ∠5 is complementary to ∠2, which measures 60°. m∠5 + 60° = 90° m∠5 = 30°

22. The triangle is equiangular, so m∠1 = m∠6 = 60°.

23. m∠ZWX = sin−1 ( 4000 — 4001.2

) ≈ 88.5967 ≈ 88.6°

m∠ZWX ≈ 2 ⋅ 88.6° = 177.2° m∠ZWX = 180° − m . CD

177.2° = 180° − m . CD

−2.8 = −m . CD

m . CD = 2.8°


Chapter 10

24. m∠FCT = cos−1 ( 4000 — 4000.2

) ≈ 0.5729

m∠ECT = cos−1 ( 4000 — 4000.01

) ≈ 0.1281

m∠FCE ≈ m∠FCT + m∠ECT = 0.5729 + 0.1281 = 0.7010

m∠FCE = m . SB ≈ 0.7

25. 40° = 1 — 2 (7x° − 3x°)

80 = 4x

20 = x

7x = 140

3x = 60

x = 20

m . CD = 360° − 10x° m . CD = 360° − 10 ⋅ 20° m . CD = 160°

26. Angle = 1 — 2 (b − a)

Angle = 1 — 2 (c)

1 — 2 (b − a) = 1 — 2 (c)

2 ⋅ 1 — 2 (b − a) = 2 ⋅ 1 — 2 (c)

b − a = c

c = b − a

27. m∠LPJ = 1 — 2 ( m . LJ − m . LK ) m . LJ + m . LK = 180° m . LJ = 180° − m . LK

m∠LPJ = 1 — 2 ( 180° − m . LK − m . LK ) m∠LPJ = 1 — 2 ( 180° − 2m . LK ) m∠LPJ = 90° − m . LK

∠LPJ is less than 90° and greater than 0°.

28. — AB is not a diameter, so m . AB ≠ 180°. So, x° < 180° and x° ≠ 90°.

29. Given ⃖ &&⃗ JL and ⃖ &&⃗ NL are secant lines that intersect at point L.

Prove m∠JPN > m∠JLN

J

K

M

N

LP

By the Angles Inside a Circle Theorem (Thm. 10.15), m∠JPN = 1 — 2 ( m . JN + m . KM ) . By the Angles Outside the

Circle Theorem (Thm. 10.16), m∠JLN = 1 — 2 ( m . JN − m . KM ) . Because the angle measures are positive,

1 — 2 ( m . JN + m . KM ) > 1 — 2 m . JN > 1 — 2 ( m . JN − m . KM ) ,

so, m∠JPN > m∠JLN.

30. yes; When the circumscribed angle is 90°, the central angle is 90°.

31. a. A C

B

AC

B

t

t

b. m∠BAC = 1 — 2 m . AB

2m∠BAC = m . AB

180° − m∠BAC = 1 — 2 m . AB

360° − 2m∠BAC = m . AB

c. If — AB is a diameter, m∠BAC = 90° and m . BA = 2m∠BAC = 360° − 2m∠BAC.

32. △ABC is an equilateral and equiangular triangle.

A

X

Y

Z

P

C

B

m∠YXZ = 60°, therefore m . YZ = 2 ⋅ 60° = 120°. m∠XZY = 60°, therefore m . XY = 2 ⋅ 60° = 120°. m∠XYZ = 60°, therefore m . XZ = 2 ⋅ 60° = 120°. m∠A = 1 — 2 ⋅ ( m . XZY − m . XY ) , therefore

m∠A = 1 — 2 ⋅ (240° − 120°) = 1 — 2 (120)° = 60°. m∠B = 1 — 2 ⋅ ( m . XYZ − m . XZ ) , therefore

m∠A = 1 — 2 ⋅ (240° − 120°) = 1 — 2 (120)° = 60°. m∠C = 1 — 2 ⋅ ( m . YXZ − m . YZ ) , therefore

m∠A = 1 — 2 ⋅ (240° − 120°) = 1 — 2 (120)° = 60°. Therefore △ABC is an equiangular and an equilateral triangle.

33. a. By the Tangent Line to Circle Theorem (Thm. 10.1), m∠BAC is 90°, which is half the measure of the semicircular arc.

b.

AC

BD

By the Tangent Line to Circle Theorem (Thm. 10.1), m∠CAD = 90°. m∠DAB = 1 — 2 m . DB and by part (a), m∠CAD = 1 — 2 mAD. By the Angle Addition Postulate (Post. 1.4), m∠BAC = m∠BAD + m∠CAD. So, m∠BAC = 1 — 2 m . DB + 1 — 2 mAD = 1 — 2 ( m . DB + m . AD ) . By the Arc Addition Postulate (Post. 10.1), m . DB + m . AD = m . ADB , so m∠BAC = 1 — 2 ( m . ADB ) .


Chapter 10

c.

AC

BD

By the Tangent Line to Circle Theorem (Thm. 10.1), m∠CAD = 90°. m∠DAB = 1 — 2 m . DB and by part (a), m∠DAC = 1 — 2 m . ABD . By the Angle Addition Postulate (Post. 1.4), m∠BAC = m∠DAC − m∠DAB. So, m∠BAC = 1 — 2 m . ABD − 1 — 2 m . DB = 1 — 2 ( m . ABD − m . DB ) . By the Arc Addition Postulate (Post. 10.1), m . ABD − m . DB = m . AB , so m∠BAC = 1 — 2 ( m . AB ) .

34. closer; The smaller arc needs to be 20°.

30° =?

1 — 2 (80° − 20°)

30° =?

1 — 2 (60°)

30° = 30°

35. Given Chords — AC and — BD intersect inside a circle.

Prove m∠1 = 1 — 2 ( m . DC + m . AB )

B

A

C

1

D

STATEMENTS REASONS

1. Chords — AC and — BD intersect.

1. Given

2. m∠ACB = 1 — 2 m . AB and

m∠DBC = 1 — 2 m . DC

2. Measure of an Inscribed Angle Theorem (Thm. 10.10)

3. m∠1 = m∠DBC + m∠ACB 3. Exterior Angle Theorem (Thm. 5.2)

4. m∠1 = 1 — 2 m . DC + 1 — 2 m . AB 4. Substitution Property of Equality

5. m∠1 = 1 — 2 ( m . DC + m . AB ) 5. Distributive Property

36.

AB

PC

∠PCB, ∠PBC, and ∠CAB intercept . BC .

37. Given a secant and a tangent line

Prove m∠1= 1 — 2 ( m . BC − m . AC )

A

B

C

12

By the Exterior Angle Theorem (Thm. 5.2), m∠2 = m∠1 + m∠ABC, so m∠1 = m∠2 − m∠ABC. By the Tangent and Intersected Chord Theorem (Thm. 10.14), m∠2 = 1 — 2 m . BC and by the Measure of an Inscribed Angle Theorem (Thm. 10.10), m∠ABC = 1 — 2 m . AC . By the

Substitution Property, m∠1 = 1 — 2 m . BC − 1 — 2 m . AC = 1 — 2 ( m . BC − m . AC ) ; Given two tangent lines

Prove m∠2 = 1 — 2 ( m . PQR − m . PR )

12

3

R

PQ

By the Exterior Angle Theorem (Thm. 5.2), m∠1 = m∠2 + m∠3, so m∠2 = m∠1 − m∠3. By the

Tangent and Intersected Chord Theorem (Thm. 10.14), m∠1 = 1 — 2 m . PQR and m∠3 = 1 — 2 m . PR . By the Substitution Property, m∠2 = 1 — 2 m . PQR − 1 — 2 m . PR = 1 — 2 ( m . PQR − m . PR ) ; Given two secant lines

Prove m∠3 = 1 — 2 ( m . XY − m . WZ )

12

3

Y

WX

By the Exterior Angle Theorem (Thm. 5.2), m∠1 = m∠3 + m∠WXZ, so m∠3 = m∠1 − m∠WXZ. By the Measure of an Inscribed Angle Theorem (Thm. 10.10), m∠1 = 1 — 2 m . XY and m∠WXZ = 1 — 2 m . WZ . By the Substitution Property, m∠3 = 1 — 2 m . XY − 1 — 2 m . WZ = 1 — 2 ( m . XY − m . WZ ) .

38.

B

A

CD

By the Angles Outside the Circle Theorem (Thm. 10.16),

m∠ADB = 1 — 2 ( ( 360° − m . AB ) − m . AB ) = 1 — 2 ( 360° − 2m . AB ) = 180° − m . AB

= 180° − m∠ACB.


Chapter 10

39. m∠P = 1 — 2 ( m . WZ − m . XY ) m . WZ + m . ZY = 200° m . WX = m . ZY

m . WZ + m . WX = 200° m . WX = 200° − m . WZ

m . WX + m . XY = 160° m . WX = 160° − m . XY

200° − m . WZ = 160° − m . XY

40° − m . WZ = −m . XY

40° = m . WZ − m . XY

m∠P = 1 — 2 ( m . WZ − m . XY ) m∠P = 1 — 2 (40°) = 20°

40. m∠DHC = 180° − 115° = 65° m∠DHC = 1 — 2 ( m . DC + m . EA ) 65° = 1 — 2 ( 85° + m . EA ) 130° = 85° + m . EA

m . EA = 45° m . EF = 20° m . FA = 45° − 20° = 25°

m∠DGB = 1 — 2 ( m . DCB + m . FA ) 90° = 1 — 2 ( m . DCB + 25° ) 180° = m . DCB + 25° m . DCB = 155°

m∠JAD = 60° m . AED = 2m∠JAD = 2 ⋅ 60° = 120° m . ED = m . AED − ( m . AF + m . FE ) m . ED = 120° −(25° + 20°) = 75°

m∠EHD = 1 — 2 ( m . ED + m . ABC ) 115° = 1 — 2 ( 75° + m . ABC ) 230° = 75° + m . ABC

m . ABC = 155° m . AB = 155° − 70° = 85° So, m . AB = 85° and m . ED = 75°.

Maintaining Mathematical Profi ciency 41. x2 + x − 12 = 0

(x − 3)(x + 4) = 0

x − 3 = 0 or x + 4 = 0

x = 3 x = −4

The solutions are x = 3 and x = −4.

42. x2 = 12x + 35

x2 − 12x − 35 = 0

Use completing the square.

x2 − 12x = 35

x2 − 12x + 62 = 35 + 62

(x − 6)2 = 71

x − 6 = ± √—

71

x = 6 ± √—

71

The solutions are x = 6 + √—

71 and x = 6 − √—

71 .

43. −3 = x2 + 4x

0 = x2 + 4x + 3

(x + 1)(x + 3) = 0

x + 1 = 0 or x + 3 = 0

x = −1 x = −3

The solutions are x = −1 and x = −3.


A

C

E

DF

B

b. Sample answer:

BF CF BF ⋅ CF

7.5 13.2 99

DF EF DF ⋅ EF

6.6 15 99

The products are equal.

c. Check students’ work. If two chords intersect inside a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

2. a. Check students' work.

b. Sample answer:

BE BC BE ⋅ BC

17.6 35.9 631.84

BF BD BF ⋅ BD

22 28.72 631.84

The products are equal.


Chapter 10

c. If two secants intersect outside a circle with a common endpoint, then the product of the lengths of the segments of one secant is equal to the product of the lengths of the segments of the other secant.

3. The products of the lengths of the segments of one chord or secant is equal to the product of the lengths of the segments of the other chord or secant.

4. AE ⋅ AF = AC ⋅ AD

18 ⋅ AF = 8 ⋅ 9

AF = 72 — 18 = 4

10.6 Monitoring Progress (pp. 570 –572) 1. x ⋅ 3 = 4 ⋅ 6 2. 2(x + 1) = 4 ⋅ 3

3x = 24 2x + 2 = 12

x = 8 2x = 10

x = 5

3. 6(6 + 9) = 5(5 + x)

6(15) = 25 + 5x

90 = 25 + 5x

65 = 5x

x = 13

4. 3(3 + x + 2) = (x + 1)(x + 1 + x − 1)

3(5 + x) = (x + 1)(2x)

15 + 3x = 2x2 + 2x

0 = 2x2 − x − 15

0 = (2x + 5)(x − 3)

0 = 2x + 5 or 0 = x − 3

x = − 5 — 2 x = 3

A negative value for x does not make sense. So, the only solution is x = 3.

5. x2 = 1 ⋅ 4

x = √—

4

x = ±2


6. 72 = 5 ⋅ (5 + x)

49 = 25 + 5x

24 = 5x

x = 24 — 5 = 4.8

7. 122 = x(x + 10)

144 = x2 + 10x

0 = x2 + 10x − 144

0 = (x − 8)(x + 18)

0 = x − 8 or 0 = x + 18

x = 8 x = −18


8. CB2 = CE ⋅ CD

352 = 14 ⋅ (2r + 14)

1225 = 28r + 196

1029 = 28r

1029 — 28 = r

The radius is approximately 1029 — 28 ≈ 36.75 feet.

10.6 Exercises (pp. 573 –574)

Vocabulary and Core Concept Check 1. The part of the secant segment that is outside the circle is

called an external segment.

2. A tangent intersects a circle at one point and a secant segment intersects a circle at two points.

Monitoring Progress and Modeling with Mathematics 3. x ⋅ 12 = 10 ⋅ 6 4. 9(x − 3) = 10 ⋅ 18

12x = 60 9x − 27 = 180

x = 5 9x = 207

x = 23

5. x (x + 8) = 8 ⋅ 6

x2 + 8x = 48

x2 + 8x − 48 = 0

(x − 4)(x + 12) = 0

x − 4 = 0 or x + 12 = 0

x = 4 x = −12


6. 2x ⋅ 12 = 15(x + 3) 7. 6 ⋅ 16 = 8(8 + x)

24x = 15x + 45 96 = 64 + 8x

9x = 45 32 = 8x

x = 5 x = 4

8. 5 ⋅ 12 = x(x + 4)

60 = x2 + 4x

0 = x2 + 4x − 60

(x + 10)(x − 6) = 0

x + 10 = 0 or x − 6 = 0

x = −10 x = 6



Chapter 10

9. 4 ⋅ 9 = (x − 2)(x − 2 + x + 4)

36 = (x − 2)(2x + 2)

36 = 2(x − 2)(x + 1)

18 = (x − 2)(x + 1)

18 = x2 − x − 2

0 = x2 − x − 20

(x + 4)(x − 5) = 0

x + 4 = 0 or x − 5 = 0

x = −4 x = 5


10. x ⋅ 45 = 27 ⋅ 50

45x = 1350

x = 30

11. x2 = 9 ⋅ 16

x = √—

144 = ±12


12. 242 = 12(12 + x)

576 = 144 + 12x

432 = 12x

36 = x

13. (x + 4)2 = x(x + 12)

x2 + 8x + 16 = x2 + 12x

8x + 16 = 12x

16 = 4x

x = 4

14. ( √— 3 ) 2 = x(x + 2)

3 = x2 + 2x

0 = x2 + 2x − 3

0 = (x + 3)(x − 1)

x + 3 = 0 or x − 1 = 0

x = −3 x = 1


15. The chords were used instead of the secant segments.

CF ⋅ DF = BF ⋅ AF

(CD + 4) ⋅ 4 = 8 ⋅ 3

4CD ⋅16 = 24

4 ⋅ CD = 8

CD = 2

16. 203,0002 = 83,000(DB)

(2.03 ⋅ 105)2 = (8.3 ⋅ 104)(DB)

4.1209 ⋅ 1010 = (8.3 ⋅ 104)(DB)

4.1209 ⋅ 1010 ——

8.3 ⋅ 104 = DB

4.9649 ⋅ 105 = DB

The distance from Cassini to Tethys is approximately 496,494 kilometers.

17. x ⋅ x = 188 ⋅ 62

x2 = 11,656

x ≈ ±107.96

A negative distance does not make sense. So, the distance from the end of the passage to either side of the mound is approximately 108 feet.

18. 4 ⋅ CN = 6 ⋅ 8

4CN = 48

CN = 12

CN is 12 centimeters. If the sparkle moves from C to D at a rate of 2 centimeters per second, it will take the sparkle 3 seconds to reach the outer circle. The sparkle will have to move at 4 centimeters per second from C to N in order to reach the outer circle at the same time.

19. Prove EA ⋅ EB = EC ⋅ ED

DB

C

EA

STATEMENTS REASONS

1. — AB and — CD are chords intersecting in the interior of the circle.

1. Given

2. ∠AEC ≅ ∠DEB 2. Vertical Angles Congruence Theorem (Thm. 2.6)

3. ∠CAB ≅ ∠CDB 3. Inscribed Angles of a Circle Theorem (Thm. 10.11)

4. △EAC ∼ △EDB 4. AA Similarity Theorem (Thm. 8.3)

5. EA —

ED = EC

— EB

5. Corresponding sidelengths of similar triangles are proportional.

6. EA • EB = EC • ED 6. Cross Products Property


Chapter 10

20. Prove EA ⋅ EB = EC ⋅ ED

D

BA

C

E

STATEMENTS REASONS

1. — EB and — ED are secant segments.

1. Given

2. ∠ABC ≅ ∠ADC 2. Inscribed Angles of a Circle Theorem (Thm. 10.11)

3. ∠E ≅ ∠E 3. Refl exive Property of Congruence (Thm. 2.2)

4. △EAD ∼ △ECB 4. AA Similarity Theorem (Thm. 8.3)

5. EA —

EC = ED

— EB

5. Corresponding side lengths of similar triangles are proportional.

6. EA • EB = EC • ED 6. Cross Products Property

21.

D

A

CO rr

x

yrE

By the Tangent Line to Circle Theorem (Thm. 10.1), ∠EAO is a right angle, which makes △AEO a right triangle. By the Pythagorean Theorem (Thm. 9.1), (r + y)2 = r 2 + x2. So, r 2 + 2yr + y2 = r 2 + x2. By the Subtraction Property of Equality, 2yr + y2 = x2. Then y(2r + y) = x2, so EC ⋅ ED = EA2.

22.

D

A

CE

STATEMENTS REASONS

1. — EA is a tangent segment and — ED is a secant segment.

1. Given

2. ∠E ≅ ∠E 2. Refl exive Property of Congruence (Thm. 2.2)

3. m∠EAC = 1 — 2 m . AC 3. Tangent and Intersected Chord Theorem (Thm. 10.14)

4. m∠ADC = 1 — 2 m . AC 4. Measure of an Inscribed Angle Theorem (Thm. 10.10)

5. m∠EAC = m∠ADC 5. Transitive Property of Equality

6. ∠EAC ≅ ∠ADC 6. Defi nition of congruence

7. △EAC ∼ △EDA 7. AA Similarity Theorem (Thm. 8.3)

8. EA —

ED = EC

— EA

8. Corresponding side lengths of

similar triangles are proportional.

9. EA2 = EC • ED 9. Cross Products Property

23. AB ⋅ AC = AD ⋅ AE

AB(AB + BC) = AD(AD + DE)

AB + BC = AD(AD + DE) ——

AB

BC = AD2 + (AD)(DE) ——

AB − AB

BC = AD2 + (AD)(DE) − AB2 ——

AB

24. Sample answer: Use Segments of Secants and Tangents Theorem (Thm. 10.20) to fi nd QR: QR2 = RS ⋅ RP. Then use the Pythagorean Theorem (Thm. 9.1) to fi nd PQ.

25. AB2 = BC ⋅ BE

122 = 8 ⋅ (8 + EC)

144 = 64 + 8EC

80 = 8EC

EC = 10

△CDP ~ △CPE by the AA Similarity Theorem (Thm. 8.3).

PD —

EP = PC

— EC

4 — r = r —

10

40 = r 2

r = √—

40 = 2 √—

10 ≈ 6.3

The radius of circle P is 2 √—

10 ≈ 6.3 units.

26. no; The side lengths are not proportional.

Maintaining Mathematical Profi ciency 27. x2 + 4x = 45

x2 + 4x + 22 = 45 + 22

(x + 2)2 = 49

x + 2 = √—

49

x = −2 ± √—

49

x = −2 + 7 = 5

x = −2 − 7 = −9



Chapter 10

28. x2 − 2x = 9

x2 + 2x + 12 = 9 + 12

(x + 1)2 = 10

x + 1 = √—

10

x = −1 ± √—

10

The solutions are x = −1 + √—

10 and x = −1 − √—

10 .

29. 2x2 + 12x + 20 = 34

2x2 + 12x = 14

x2 + 6x = 7

x2 + 6x + 33 = 7 + 32

(x + 3)2 = 16

x + 3 = √—

16

x = −3 ± 4

x = −3 + 4 = 1

x = −3 − 4 = −7


30. −4x2 + 8x + 44 = 16

x2 − 2x − 11 = −4

x2 − 2x = 7

x2 − 2x + 13 = 7 + 12

(x − 1)2 = 8

x − 1 = √—

8

x = 1 ± 2 √—

2

The solutions are x = 1 + 2 √—

2 and x = 1 − 2 √—

2 .

10.7 Explorations (p. 575) 1. a. Sample answer:

Radius Equation of circle

1 x2 + y2 = 1

2 x2 + y2 = 4

3 x2 + y2 = 9

4 x2 + y2 = 16

5 x2 + y2 = 25

6 x2 + y2 = 36

b. An equation of a circle with center (0, 0) and radius r is x2 + y2 = r2.

2. a. Sample answer:

Center Equation of circle

(0,0) x2 + y2 = 4

(2,0) (x − 2)2 + y2 = 4

(0,2) x2 + (y − 2)2 = 4

(−1,2) (x + 1)2 + (y − 2)2 = 4

(2,1) (x − 2)2 + (y − 1)2 = 4

(−3,−2) (x + 3)2 + (y + 2)2 = 4

b. An equation of a circle with center (h, k) and radius 2 is (x − h)2 + (y − k)2 = 4.

c. An equation of a circle with center (h, k) and radius r is (x − h)2 + (y − k)2 = r 2.

3. d = √——

(x − h)2 + (y − k)2

d 2 = ( √——

(x − h)2 + (y − k)2 ) 2 d 2 = (x − h)2 + (y − k)2

The equation is the same. The distance from the center to any point on the circle is the radius; therefore, d would represent the radius r.

4. The equation of a circle with center (h, k) and radius r in the coordinate plane is (x − h)2 + (y − k)2 = r 2.

5. Center: (4, −1), r = 3

(x − h)2 + (y − k)2 = r 2

(x − 4)2 + ( y − (−1) ) 2 = 32

(x − 4)2 + (y + 1)2 = 9

10.7 Monitoring Progress (pp. 576 −578) 1. (x − h)2 + (y − k)2 = r2

(x − 0)2 + (y − 0)2 = 2.52

x2 + y2 = 6.25

The standard equation of the circle is x 2 + y 2 = 6.25.

2. (x − h)2 + (y − k)2 = r 2

( x − (−2) ) 2 + ( y − 5)2 = 72

(x + 2)2 + (y − 5)2 = 49

The standard equation of the circle is (x + 2)2 + (y − 5)2 = 49.

3. r = √——

(3 − 1)2 + (4 − 4)2

= √—

(2)2 + (0)2

= √—

4

= 2

(x − h)2 + (y − k)2 = r 2

(x − 1)2 + (y − 4)2 = 4

The standard equation of the circle is (x − 1)2 + (y − 4)2 = 4.


Chapter 10

4. x2 + y2 − 8x + 6y + 9 = 0

x2 − 8x + y2 + 6y = −9

(x2 − 8x + 42) + (y2 + 6y + 32) = −9 + 42 +32

(x − 4)2 + (y + 3)2 = −9 + 16 + 9

(x − 4)2 + (y + 3)2 = 16

(x − 4)2 + ( y − (−3) ) 2 = 42

The center is (4, −3) and the radius is 4.

y

2 4 6 8

−4

−6

−8

−2

x

(4, −3)

5. r = √——

(x − h)2 + (y − k)2

= √——

(0 − 0)2 + (1 − 0)2

= √—

1

= 1

The radius of the circle is 1.

d = √——

(1 − 0)2 + ( √— 5 − 0 ) 2

d = √—

1 + 5 = √—

6

Because the radius is 1 and the distance between the center and the point ( 1, √

— 5 ) is greater than the radius, the point

( 1, √—

5 ) does not lie on the circle.

6. Three seismographs are needed to locate an earthquake's epicenter because two circles can intersect at more than one point.

10.7 Exercises (pp. 579−580)

Vocabulary and Core Concept Check 1. The standard equation of a circle with center (h, k) and

radius r is r 2 = (x − h)2 + (y − k)2.

2. The radius can be determined by using the distance formula to fi nd the distance between the center and a point on the circle.

Monitoring Progress and Modeling with Mathematics 3. The center is (0, 0) and the radius is 2.

(x − h)2 + (y − k)2 = r 2

(x − 0)2 + (y − 0)2= 22

x2 + y2 = 4

4. Using the endpoints of the diameter, (−12, −3) and (6, −3), the center is found by determining the midpoint.

( −12 + 6 — 2 , −3 + (−3) —

2 ) = ( −6 —

2 , −6 —

2 ) = (−3, −3).

Use the center and an endpoint of the diameter to fi nd the radius.

r = √——

(x − h)2 + (y − k)2

= √———

( 6 − (−3) ) 2 + ( −3 − (−3) ) 2

= √—

92 + 02

= √—

81

= 9

(x − h)2 + (y − k)2 = r2

( x − (−3) ) 2 + ( y − (−3) ) 2 = 92

(x + 3)2 + (y + 3)2 = 81

5. The center is (0, 0) and the radius is 7.

(x − h)2 + (y − k)2 = r2

(x − 0)2 + (y − 0)2 = 72

x2 + y2 = 49

6. The center is (4, 1) and the radius is 5.

(x − h)2 + (y − k)2 = r2

(x − 4)2 + (y − 1)2 = 52

(x − 4)2 + (y − 1)2 = 25

7. The center is (−3, 4) and the radius is 1.

(x − h)2 + (y − k)2 = r 2

( x − (−3) ) 2 + (y − 4)2 = 12

(x + 3)2 + (y − 4)2 = 1

8. The center is (3, −5) and the radius is 7.

(x − h)2 + (y − k)2 = r 2

(x − 3)2 + ( y − (−5) ) 2 = 72

(x − 3)2 + (y + 5)2 = 49

9. r = √——

(x − h)2 + (y − k)2

= √——

(0 − 0)2 + (6 − 0)2

= √—

62 = √—

36 = 6

(x − h)2 + (y − k)2 = r 2

(x − 0)2 + (y − 0)2 = 62

x2 + y2 = 36

10. r = √——

(x − h)2 + (y − k)2

= √——

(4 − 1)2 + (2 − 2)2

= √—

32 = √—

9 = 3

(x − h)2 + (y − k)2 = r2

(x − 1)2 + (y − 2)2 = 32

(x − 1)2 + (y − 2)2 = 9


Chapter 10

11. r = √——

(x − h)2 + (y − k)2

= √——

(3 − 0)2 + (−7 − 0)2

= √—

32 + (−7)2

= √—

9 + 49 = √—

58

(x − h)2 + (y − k)2 = r2

(x − 0)2 + (y − 0)2 = ( √— 58 ) 2

x2 + y2 = 58

12. The coordinates of the center should be subtracted.

( x − (−3) ) 2 + ( y − (−5) ) 2 = 9

(x + 3)2 + (y + 5)2 = 9

13. For the equation x2 + y2 = 49, the center is (0, 0) and the radius is 7.

y

84−4−8

−8

−4

4

8

x(0, 0)

14. For the equation (x + 5)2 + (y − 3)2 = 9, the center is (−5, 3) and the radius is 3.

y

−2−4−6−8

2

4

6

8

x

(−5, 3)

15. x2 − 6x + y2 = 7

x2 − 6x + (−3)2 + y2 = 7 + (−3)2

(x − 3)2 + y2 = 16

The center is (3, 0) and the radius is 4.

y

4 62

−4

4

x(3, 0)

16. x2 + y2 + 4y + 22 = 32 + 22

x2 + (y + 2)2 = 36


y

84−4−8

−4

8

x(0, −2)

17. x2 − 8x + y2 − 2y = −16

( x2 − 8x + (−4)2 ) + ( y2 − 2y + (−1)2 ) = −16 + (−4)2 + (−1)2

( x2 − 8x + (−4)2 ) + ( y2 − 2y + (−1)2 ) = 1

(x − 4)2 + (y − 1)2 = 1

The center is (4, 1) and the radius is 1.

y

4 62

4

6

2

x

(4, 1)

18. x2 + 4x + y2 + 12y = −15

( x2 + 4x + (2)2 ) + ( y2 + 12y + (6)2 ) = −15 + (2)2 + (6)2

( x2 + 4x + (2)2 ) + ( y2 + 12y + (6)2 ) = 25

(x + 2)2 + (y + 6)2 = 25

The center is (−2, −6) and the radius is 5.

y

42−2−4−6−8

−4

−6

−8

−2

x

−10

(−2, −6)

19. r 2 = (x − h)2 + (y − k)2

82 (2 − 0)2 + (3 − 0)2

64 4 + 9

64 ≠ 13

Because the radius is 8 and the distance between the center and the point (2, 3) is less than the radius, the point (2, 3) does not lie on the circle.


Chapter 10

20. r 2 = (x − h)2 + (y − k)2

32 (4 − 0)2 + ( √— 5 − 0 ) 2

9 16 + 5

9 ≠ 21

Because the radius is 3 and the distance between the center

and the point ( 4, √—

5 ) is greater than the radius, the point

( 4, √—

5 ) does not lie on the circle.

21. r 2 = (x − h)2 + (y − k)2

r 2 = ( √— 6 − 0 ) 2 + (2 − 0)2

r 2 = 6 + 4

r 2 = 10

r = √—

10

( √— 10 ) 2 (3 − 0)2 + (−1 − 0)2

10 9 + 1

10 = 10

Because the radius is √—

10 and the distance between the center and the point (3, −1) is equal to the radius, the point (3, −1) lies on the circle.

22. r2 = (x − h)2 + (y − k)2

r2 = (5 − 0)2 + (2 − 0)2

r2 = 25 + 4

r2 = 29

r = √—

29

( √— 29 ) 2 ( √—

7 − 0 ) 2 + (5 − 0)2

29 7 + 25

29 ≠ 32

Because the radius is √—

29 and the distance between the

center and the point ( √— 7 , 5 ) is greater than the radius, the

point ( √— 7 , 5 ) does not lie on the circle.

23. a. y

84−4−8

−8

−4

4

8

xZone 1

Zone 2

Zone 3

b. d 2 = 32 + 42 d 2 = 62 + 52

d 2 = 9 + 16 d 2 = 36 + 25

d 2 = 25 d 2 = 61

d = 5 d = √—

61 ≈ 7.8

Zone 2 Zone 3

d 2 = 12 + 22 d 2 = 02 + 32

d 2 = 5 d 2 = 9

d = √—

5 ≈ 2.2 d = √—

9 = 3

Zone 1 Zone 1

d 2 = 12 + 62

d 2 = 1 + 36

d 2 = 37

d = √—

37 ≈ 6.1

Zone 2

24. a. y

4 6 82−2

−2

4

6

x(0, 0)

(0, 5)

(6, 3)

Two of the circles overlap, so there are locations that may receive calls from more than one tower.

B

x

y

4

6

−2

4 62−2

A

8

b. City B has complete coverage. It’s entire radius is within the circle with center (6, 3). Parts of City A do not have coverage.


Chapter 10

25. y

2−2−6

−2

−6

2

x(0, 0)

(−2, −4)

The center of the image is (−2, 4) and the radius is 4.

(x − h)2 + (y − k)2 = r2

( x − (−2) ) 2 + (y − 4)2 = 42

(x + 2)2 + (y − 4)2 = 16

The equation of the image is (x + 2)2 + (y − 4)2 = 16. The equation of the image of a circle after a translation m units to the left and n units down is (x + m)2 + (y + n)2 = r2.

26. a. C; This circle has a center of (−3, 0) and a radius of 2.

b. D; This circle has a center of (0, 3) and a radius of 2.

c. B; This circle has a center of (3, 0) and a radius of 2.

d. A; This circle has a radius of (0, −3) and a radius of 2.

27. Determine the midpoint of two of the sides:

The midpoint of X(4, 5) and Y(4, 13) is

( 4 + 4 — 2 , 5 +13 —

2 ) = (4, 9).

The equation of the perpendicular bisector to — XY is y = 9. The midpoint of Y(4, 13) and Z(8, 9) is

( 4 + 8 — 2 , 13 + 9 —

2 ) = (6, 11).

Determine the equation of the line that is the perpendicular bisector of — YZ .

Slope of — YZ = ( 13 − 9 — 4 − 8

) = 4 — −4 = −1

The slope of the perpendicular bisector is 1.

y = mx + b

11 = 6 + b

5 = b

The equation is y = x + 5.

Where y = 9 and y = x + 5

x

y

4

6

8

10

12

2

4 6 82

X

Y

Z

intersect is the center of the circle, which is (4, 9).

Determine the radius:

(x − 4)2 + (y − 9)2 = r2

(4 − 4)2 + (5 − 9)2 = r2

(−4)2 = r2

16 = r2

The equation of the circle is (x − 4)2 + (y − 9)2 = 16.

28. The slope of the radius from (h, k) to (a, b) is k − b — h − a

. Because

the tangent line is perpendicular to the radius, the slope of

the tangent line is − ( h − a —

k − b ) . The equation of the tangent

line, where d is the y-intercept is:

y = − ( h − a — k − b

) x + d

Substitute (a, b) into this equation.

b = − ( h − a —

k − b ) a + d

Subtract these two equations.

y − b = − ( h − a — k − b

) x + ( h − a — k − b

) a + d − d

y − b = ( h − a — b − k

) x − ( h − a — b − k

) a y − b = h − a —

b − k (x − a)

29. (x − 4)2 + (y − 3)2 = 9, y = 6

(x − 4)2 + (6 − 3)2 = 9

(x − 4)2 + (3)2 = 9

(x − 4)2 + 9 = 9

(x − 4)2 = 0

√—

(x − 4)2 = √—

0

x = 4

The line is a tangent because the system has one solution, (4, 6).

30. (x + 2)2 + (y − 2)2 = 16, y = 2x − 4

(x + 2)2 + (2x − 4 − 2)2 = 16

(x + 2)2 + (2x − 6)2 = 16

(x + 2)2 + ( 2(x − 3) ) 2 = 16

(x2 + 4x + 4) + 4(x2 − 6x + 9) = 16

(x2 + 4x + 4) + (4x2 − 24x + 36) = 16

5x2 − 20x + 40 = 16

5(x2 − 4x + 8) = 16

(x2 − 4x + 8) = 16 — 5

x2 − 4x + (−2)2 = 16 — 5 − 8 + (−2)2

(x − 2)2 = 16 — 5 − 8 + 4

(x − 2)2 = 16 — 5 − 4

(x − 2)2 = 16 — 5 − 20

— 5

(x − 2)2 = − 4 — 5

Because the square root of a negative number does not yield a real number, the system has no solution. So, the line is none of these.


Chapter 10

31. (x − 5)2 + (y + 1)2 = 4, y = 1 — 5 x − 3

(x − 5)2 + ( 1 — 5 x − 3 + 1 ) 2 = 4

(x − 5)2 + ( 1 — 5 x − 2 ) 2 = 4

(x 2 − 10x + 25) + ( 1 — 25

x 2 − 4 — 5 x + 4 ) = 4

26 — 25

x 2 − 54 —

5 x + 29 = 4

26 — 25

x 2 − 54 —

5 x + 25 = 0

25 ⋅ 26 —

25 x 2 − 25 ⋅ 54

— 5 x + 25 ⋅ 25 = 0

26x 2 − 270x + 625 = 0

Use the Quadratic Formula; x = −b ± √—

b2 − 4ac ——

2a , where

a = 26, b = −270, and c = 625.

x = −(−270) ± √———

(−270)2 − 4 ⋅ 26 ⋅ 625 ————

2 ⋅ 26

x = 270 ± √——

72,900 − 65,000 ———

52

x = 270 ± √—

7900 ——

52

x = 270 + √—

7900 ——

52

= 6.9, y = 0.2 ⋅ 6.9 − 3 = −1.6 ⇒ (6.9,−1.6)

x = 270 − √—

7900 ——

52

= 3.5, y = 0.2 ⋅ 3.5 − 3 = −2.3 ⇒ (3.5,−2.3)

d = √———

(6.9 − 3.5)2 + (−1.6 − (−2.3))2

d = √——

(3.4)2 + (0.7)2 = √——

11.56 + 0.49 = √—

12.05

= 3.5 ≠ 8 (the diameter)

The system has two solutions, and (5, −1) is not on the line. So, the line is a secant line.

32. (x + 3)2 + (y − 6)2 = 25, y = − 4 — 3 x + 2

(x + 3)2 + ( − 4 — 3 x + 2 − 6 ) 2 = 25

(x + 3)2 + ( − 4 — 3 x − 4 ) 2 = 25

(x + 3)2 + (−1)2 ( 4 — 3 x + 4 ) 2 = 25

(x2 + 6x + 9) + ( 16 — 9 x

2 + 32 — 3 x + 16 ) = 25

25 — 9 x

2 + 50 — 3 x + 25 = 25

25 — 9 x

2 + 50 — 3 x = 0

9 ⋅ 25 — 9 x

2 + 9 ⋅ 50 — 3 x = 9 ⋅ 0

25x2 + 150x = 0

25x(x + 6) = 0

25x = 0 ⇒ x = 0

y = − 4 — 3 ⋅ 0 + 2 = 2 ⇒ (0, 2)

x + 6 = 0 ⇒ x = −6

y = − 4 — 3 ⋅ (−6) + 2 = 8 + 2 = 10 ⇒ (6, 10)

Because the equation y = − 4 — 3 x + 2 contains the center of the circle (−3, 6), and the system has two solutions, the line is a

secant line.

33. The center is (0, k) and passes through the points (−1, 0) and (1, 0). Find the radius.

(x − 0)2 + (y − k)2 = r2

(1 − 0)2 + (0 − k)2 = r2

1 + k2 = r2

r = √—

1 + k2

(x − 0)2 + (y − k)2 = ( √— 1 + k2 ) 2

x2 + y2 − 2yk + k2 = 1 + k2

x2 + y2 − 2yk − k2 + k2 = 1

x2 + y2 − 2yk = 1

Yes, the equation of the circle with center (0, k) and passing through the points (−1,0) and (1,0) is x2 + y2 − 2yk = 1.


Chapter 10

34. r 2 = (x − h)2 + y2

(4r)2 = (x −15r)2 + y2

16r 2 = (63 − 15r)2 + 162

16r 2 = 3969 − 1890r + 225r 2 + 256

−209 ⋅ r 2 = −1890r + 4225

0 = 209 ⋅ r 2 − 1890r + 4225

Use the Quadratic Formula; x = −b ± √—

b2 − 4ac ——

2a , where

a = 209, b = −1890, and c = 4225.

x = −(−1890) ± √———

(−1890)2 − 4(209)(4225) ————

2 ⋅ 209

x = 1890 ± √—

40,000 ——

418 = 1890 ± 200

— 418

x = 5 and x ≈ 4.04

Because the specifi cation was that all circles have integer radii, the value of x is 5.

The center of circle A is (15, 0) and the radius is 10. Therefore, the equation is (x − 15)2 + y2 = 102, or (x − 15)2 + y2 = 100.

Maintaining Mathematical Profi ciency 35. . RS is a minor arc with measure 53°.

36. . PR is a minor arc with measure m . PR = m . PQ + m . QR = 65° + 25° = 90°.

37. . PRT is a major arc with measure m . PRT = 53° + 25° + 65° + 90° = 270°.

38. . ST is a minor arc with measure 180° − 53° = 127°.

39. . RST is a semicircle with measure 180°.

40. . QS is a minor arc with measure 25° + 53° = 78°.

10.4 –10.7 What Did You Learn? (p. 581) 1. A ruler and a right angle used for mechanical drawing could

be used together to mark right angles. A ruler and a compass will also work.

2. ∠BAC could be acute or obtuse.

Chapter 10 Review (pp. 582–586) 1. — PK is a radius. 2. — NM is a chord.

3. &&&⃗ JL is a tangent. 4. — KN is a diameter.

5. ⃖ &&⃗ NL is a secant. 6. — PN is a radius.

7. The common tangent is an internal tangent.

8. The common tangent is an external tangent.

9. XY = XZ

9a2 − 30 = 3a

9a2 − 3a − 30 = 0

3a2 − a − 10 = 0

(3a + 5)(a − 2) = 0

3a + 5 = 0

a = −5 —

3

a − 2 = 0

a = 2

10. XY = XZ

2c2 + 9c + 6 = 9c + 14

2c2 = 8

c2 = 4

c = ± √—

4 = ±2

A negative value does not make sense. So, the only solution is c = 2.

11. (r + 3)2 = r 2 + 92

r 2 + 6r + 9 = r 2 + 81

6r + 9 = 81

6r = 72

r = 12

12. If 10 is the radius, then the diameter is 20.

522 482 + 202

2704 2304 + 400

2704 = 2704

— AB is tangent to circle C.

13. . KL is a minor arc, so m . KL = m∠KPL = 100°.

14. . LM is a minor arc, so m . LM = m∠LPM = 60°.

15. . KM is a minor arc, so m . KM = m∠KPM = 100° + 60° = 160°.

16. . KN is a minor arc, so m . KN = m∠KPN = 80°.

17. The red arcs are not congruent because the two circles are not congruent.

18. The red arcs are congruent because the circles are congruent and m . AB = m . EF .

19. — AB ≅ — ED by the defi nition of measure of minor arcs. So, by the Congruent Corresponding Chords Theorem (Thm. 10.6), m . ED = m . AB = 61°.

20. You are given — BE ≅ — ED and — CA is a perpendicular bisector. Therefore, by the Congruent Corresponding Chords Theorem (Thm. 10.6), m . AD = m . AB = 65°.

21. By the Equidistant Chords Theorem (Thm. 10.9), — AB ≅ — ED . Therefore, m . AB = m . ED = 91°.


Chapter 10

22. Using △KNQ: r 2 = (2x)2 + 102

Using △PLQ: r 2 = (3x − 12)2 + 102

QN = QP and by the Equidistant Chords Theorem (Thm. 10.9),

4x = 6x − 24

2x = 24

x = 12

r 2 = (2 ⋅ 12)2 + 102

r 2 = (24)2 + 102

r 2 = 676

r = √—

676 = 26

The radius is 26.

23. By the Measure of an Inscribed Angle Theorem (Thm. 10.10), x° = 2 ⋅ 40° = 80°.


q° + 80° = 180° q = 100

4r ° + 100° = 180° 4r = 80

r = 20

25. By the Inscribed Angles of a Circle Theorem (Thm. 10.11), m∠KJL = m∠KML.

14d ° = 70° d = 5

26. By the Inscribed Right Triangle Theorem (Thm. 10.12), △NPQ is a right triangle:

m∠NPQ = 90° 3y° = 90° y = 30

m∠PNQ + m∠PQN = 90° 50° + 4z° = 90° 4z = 40

z = 10

27. By the Inscribed Angles of a Circle Theorem (Thm. 10.11), m∠VRT = m∠VST and m∠RTS = m∠RVS.

m∠VRT = m∠VST

m° = 44° m∠RTS = m∠RVS

n° = 39° 28. By the Measure of an Inscribed Angle Theorem (Thm. 10.10),

c° = 1 — 2 ⋅ 56° = 28°.

29. By the Angles Outside the Circle Theorem:

x = 1 — 2 (250 − 110); x = 70

The value of x is 70.

30. By the Angles Inside the Circle Theorem (Thm. 10.15),

x = 1 — 2 (152 + 60)

x = 1 — 2 (212)

x = 106

31. By the Angles Outside the Circle Theorem (Thm. 10.16),

40 = 1 — 2 (96 − x)

80 = 96 − x

−16 = −x

x = 16

32. By the Tangent and Intersected Chord Theorem (Thm. 10.14), m . XYZ = 2 ⋅ 120° = 240°.

33. JK ⋅ JL = JM ⋅ JN

4 ⋅ (4 + 6) = x ⋅ (x + 3)

40 = x2 + 3x

0 = x2 + 3x − 40

(x + 8)(x − 5) = 0

x + 8 = 0 or

x − 5 = 0

x = −8 x = 5


34. TP ⋅ PR = QP ⋅ PS

(x + 3) ⋅ x = (6 − x) ⋅ 2x

x2 + 3x = 12x − 2x2

3x2 − 9x = 0

3x(x − 3) = 0

3x = 0 or

x − 3 = 0

x = 0 x = 3

Because 0 does not make sense, the only solution is x = 3.

35. WX ⋅ WX = XY ⋅ XZ

12 ⋅ 12 = 8 ⋅ (8 + x)

144 = 64 + 8x

80 = 8x

x = 10

36. AB ⋅ AB = AC ⋅ AD

20 ⋅ 20 = 12 ⋅ (12 + 2r)

400 = 144 + 24r

256 = 24r

r ≈ 10.7

The radius of the rink is approximately 10.7 feet.

37. The center of the circle is (4, −1) and the radius is 3.

(x − h)2 + (y − k)2 = r 2

(x − 4)2 + ( y − (−1) ) 2 = 32

(x − 4)2 + (y + 1)2 = 9


Chapter 10

38. The center of the circle is (8, 6) and the radius is 6.

(x − h)2 + (y − k)2 = r 2

(x − 8)2 + (y − 6)2 = 62

(x − 8)2 + (y − 6)2 = 36


(x − h)2 + (y − k)2 = r 2

(x − 0)2 + (y − 0)2 = 42

x 2 + y2 = 16


(x − h)2 + (y − k)2 = r 2

(x − 0)2 + (y − 0)2 = 92

x2 + y2 = 81

41. The center of the circle is (−5, 2) and the radius is 1.3.

(x − h)2 + (y − k)2 = r 2

( x − (−5) ) 2 + (y − 2)2 = 1.32

(x + 5)2 + (y − 2)2 = 1.69


(x − h)2 + (y − k)2 = r 2

(x − 6)2 + (y − 21)2 = 42

(x − 6)2 + (y − 21)2 = 16

43. The center of the circle is (−3, 2) and the radius is 16.

(x − h)2 + (y − k)2 = r2

( x − (−3) ) 2 + (y − 2)2 = 162

(x + 3)2 + (y − 2)2 = 256

44. The center of the circle is (10, 7) and the radius is 3.5.

(x − h)2 + (y − k)2 = r 2

(x − 10)2 + (y − 7)2 = 3.52

(x − 10)2 + (y − 7)2 = 12.25

45. The center of the circle is (0, 0) and the radius is 5.2.

(x − h)2 + (y − k)2 = r 2

(x − 0)2 + (y − 0)2 = 5.22

x2 + y2 = 27.04

46. r = √——

(x − h)2 + (y − k)2

= √——

(−7 + 7)2 + (1 − 6)2

= √—

02 + (−5)2

= √—

25 = 5

(x − h)2 + (y − k)2 = r2

( x − (−7) ) 2 + (y − 6)2 = 52

(x + 7)2 + (y − 6)2 = 25

47. x2 + y2 − 12x + 8y + 48 = 0

x2 − 12x + y2 + 8y = −48

x2 − 12x + (−6)2 + y2 + 8y + 42 = −48 + (−6)2 + 42

(x − 6)2 + (y + 4)2 = 4


y

2 4 6 8

−2

−4

−6

2

x

(6, −4)

48. r = √——

(x − h)2 + (y − k)2

= √——

(−5 − 0)2 + (0 − 0)2

= √—

(−5)2 + 02

= √—

25 = 5

52 (4 − 0)2 + (−3 − 0)2

25 16 + 9

25 = 25

Because the radius is 5 and the distance between the center and the point (4, −3) is equal to the radius, the point (4, −3) lies on the circle.

Chapter 10 Test (p. 587) 1. ∠1 is an inscribed angle, so m∠1 = 1 — 2 ⋅ 145 = 72.5°. ∠2 is a central angle, so m∠2 = 145°.

2. ∠1 and ∠2 intercept a circle, so m∠1 = 1 — 2 ⋅ 180 = 90° and

m∠2 = 1 — 2 ⋅ 180° = 90°.

3. ∠1 is outside the circle, so m∠1 = 1 — 2 ⋅ (96° − 38°) = 1 — 2 ⋅ 58° = 29°.

∠2 is inside the circle, so m∠2 = 1 — 2 (96° + 36°) = 66°. ∠3 is an inscribed angle, so

m∠3 = 180° − (180° − 66°) − 29° = 37°.

4. ∠1 and ∠2 are outside the circle. m∠1 = 1 — 2 ⋅ (77° − 48°) = 1 — 2 ⋅ 29° = 14.5°

360° − (215° + 48°) = 97° m∠2 = 1 — 2 (263° − 97°) = 1 — 2 ⋅ 166° = 83°

5. AG ⋅ GD = BG ⋅ GF 6. CE ⋅ CD = CB ⋅ CF

2 ⋅ 9 = 3 ⋅ GF CE ⋅ 9 = 3 ⋅ 12

18 = 3 ⋅ GF 9 ⋅ CE = 36

GF = 6 CE = 4


Chapter 10

7. CA ⋅ CA = CB ⋅ CF

CA2 = 3 ⋅ 12

CA2 = 36

CA = √—

36

CA = 6

8. Sample answer:

A B

C

D

E

a. m∠CDE = 1 — 2 ⋅ m . CAE

m∠CAE = 1 — 2 ⋅ m . CDE

The sum of the two arcs is 360°. So, m . CAE + m . CDE = 360°.

m∠CDE + m∠CAE = 1 — 2 ⋅ m . CAE + 1 — 2 ⋅ m . CDE

m∠CDE + m∠CAE = 1 — 2 ( m . CAE + m . CDE ) m∠CDE + m∠CAE = 1 — 2 ⋅ 360° m∠CDE + m∠CAE = 180° Therefore, ∠CDE and ∠CAE are supplementary.

b. m∠CBE = 1 — 2 ⋅ m . CDE and m∠CAE = 1 — 2 ⋅ m . CDE

m∠CBE = m∠CAE and both angles intercept the same arc. Therefore, ∠CBE ≅ ∠CAE.

9. 5x − 4 = 3x + 6 10. (6 + r)2 = 122 + r 2

2x = 10 36 + 12r + r 2 = 144 + r 2

x = 5 36 + 12r = 144

12r = 108

r = 9

11. r = √——

(x − h)2 + (y − k)2

r = √—— (−1 − 0)2 + (4 − 2)2

r = √—

1 + 4 = √—

5

x2 + (y − 2)2 = √— (5)2

x2 + (y − 2)2 = 5

A circle with center (0, 2) and point (−1, 4) on the circle has a radius of √

— 5 and has an equation of x2 + (y − 2)2 = 5.

( 2 √—

2 ) 2 + (−1 −2)2 5

8 + 9 5

17 > 5

The point, ( 2 √—

2 , −1 ) does not lie on the circle.

12. By the Equidistant Chords Theorem (Thm. 10.19), — ST ≅ — RQ . By the Congruent Corresponding Chords Theorem

(Thm. 10.6), . ST ≅ . RQ .

13. — CJ ≅ — CL and — CK ≅ — CK , so △CKJ ≅ △CKL by the HL Congruence Theorem (Thm. 5.9). Then ∠LCK ≅ ∠JCK, and by the Congruent Central Angles Theorem (Thm. 10.4), . JM ≅ . LM .

14. — GE is a perpendicular bisector of — DF , so DG = FG by the Perpendicular Bisector Theorem (Thm. 6.1). By the Congruent Corresponding Chords Theorem (Thm. 10.6), . DG ≅ . FG .

15. y

8 12 164

8

4

x

12

16

(13, 4)

Actor A (11, 4):

(11 − 13)2 + (4 − 4)2 42

(−2)2 16

4 < 16

yes; Actor A is inside the area that is illuminated by the light.

Actor B (8, 5):

(8 − 13)2 + (5 − 4)2 42

(−5)2 + 1 16

25 + 1 16

26 > 16

no; Actor B is outside the area that is illuminated by the light.

Actor C (15, 5):

(15 − 13)2 + (5 − 4)2 42

(2)2 + 1 16

5 < 16

yes; Actor C is inside the area that is illuminated by the light.

16. a. r 2 = 1602 + (r − 80)2

r 2 = 25,600 + r 2 − 160r + 6400

0 = 32,000 − 160r

160r = 32,000

r = 200

The radius is 200 feet.

b. S = 3.87 √— fr

= 3.87 √—

0.7 ⋅ 200

≈ 45.79

The car’s speed is about 45.79 miles per hour.

Chapter 10 Standards Assessment (pp. 588 –589) 1. a. — BG is a chord. b. — CD is a radius.

c. — AD is a diameter. d. — FE is a chord.


Chapter 10

2. Map Circle C to Circle C′ by using the translation (x, y) → (x − 2, y + 2) so that Circle C ′and Circle D have the same center at (0, 3). Dilate Circle C′ using a center of dilation (0, 3) and a scale factor of 4. Because there is a similarity transformation that maps Circle C to Circle D, Circle C is similar to Circle D.

3. Given △JPL ≅ △NPL

— PK is an altitude of △JPL.

— PM is an altitude of △NPL.

Prove △PKL ∼ △NMP

K

NML

P

J

m∠JPL = 90°, m∠LPN = 90°, and — PM and — PK are altitudes, so △JPL ∼ △PKL and △NPL ∼ △NMP by the Right Triangle Similarity Theorem (Thm. 9.6). By the Transitive Property, △PKL ∼ △NMP.

4. B;

x2 + y2 + 14x − 16y + 77 = 0

x2 + 14x + y2 − 16y = −77

(x2 + 14x + 72) + ( y2 − 16y + (−8)2 ) = −77 + 72 + (−8)2

(x2 + 14x + 72) + ( y2 − 16y + (−8)2 ) = −77 + 49 + 64

(x + 7)2 + (y − 8)2 = 36

The center is (−7, 8) and the radius is 6.

5. Slope of — WX = −2 − (−6) —

1 − (−7) = −2 + 6

— 1 + 7

= 4 — 8 = 1 —

2

Slope of — XY = −6 − (−2) —

3 − 1 = −6 + 2

— 2 = −4

— 2 = −2

Slope of — YZ = −10 − (−6) —— −5 − 3

= −10 + 6 — −8

= −4 — −8 = 1 —

2

Slope of — ZW = −10 − (−6) —— −5 − (−7) = −10 + 6

— −5 + 7 = −4

— 2 = −2

— WX ⊥ — XY , — WX ⊥ — WZ , — XY ⊥ — YZ , — YZ ⊥ — ZW

There are right angles at W, X, Y, Z. By defi nition, if a quadrilateral has four right angles it is a rectangle.

6. m∠ACB = 96° m∠JGK = 1 — 2 (60° + 132°) = 1 — 2 (192°) = 96° m∠JGK = m∠ACB

m∠LGM = 1 — 2 (60° + 132°) = 1 — 2 (192°) = 96° m∠LGM = m∠ACB

m∠STV = 1 — 2 (276° − 84°) = 1 — 2 (192°) = 96° m∠STV = m∠ACB

m∠VWU = 96° m∠VWU = m∠ACB

7. a. Inverse

b. Converse

c. Biconditional

d. Contrapositive

8. No, by the Inscribed Quadrilateral Theorem, A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary. The sum of the measures of the opposite angles are 70° + 70° = 140° and 110° + 110° = 220°, not 180°. Therefore, this quadrilateral cannot be inscribed in a circle.

hscc geom wsk 10 - schoolwires

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