CHAPTER 9 - Mr Jaffemrjaffe.net/GeomRef/Ch9solns.pdf · Chapter 9continued 24. 25. 26. 27. 28. d 16...

CHAPTER 9

182 GeometryChapter 9 Worked-out Solution Key

Copyright McDougal Littell Inc. All rights reserved.

Lesson 9.1

Think and Discuss (p. 525)

1. and are both to in a plane, 2 lines to the same line are .

2. According to Theorem 4.8, if the hypotenuse and leg ofone right triangle are congruent to the hypotenuse andcorresponding leg of another right triangle, the twotriangles are congruent.

Skill Review (p. 526)

1.

is a right triangle.

2. 3.

4.

Developing Concepts Activity (p. 527)

3. All of the triangles are similar.

9.1 Guided Practice (p. 531)

1. geometric mean 2. KML; JMK 3.

4. JM 5. JK 6. LJ 7. KM 8. KM 9. LK

10.

DF 48.3

53.4FD

FD43.6

DC 53.4

9772

72DC

MK

92

x

9 2x

3x 9 5x

3x 3 5x

x 3

5

x3

y

x1

1

leg, altitude

leg, altitude

altitude

hypotenuse

K

J

L

JKL

mL 180 30 60 90

BD;EDAB

9.1 Practice and Applications (pp. 531534)

11.

12.

13. 14. 15.

16.

17.

18.

19.

20.

21.

22.

23.

x 4

x2 16

x4

4x

CBA ~ DBC ~ DCA

x 50

32x 1600

x

40

4032

RSQ ~ TRQ ~ TSR

x 68.3

15x 1024

x

32

3215

LKJ ~ MLJ ~ MKL

x 16

25x 400

x

20

2025

GEF ~ HGF ~ HEG

x 9

144 16x

1216

x

12

CBA ~ DBC ~ DCA

GFE ~ HFG ~ HGE; EH

SQR ~ TQS ~ TSR; RQ

ZYX ~ WYZ ~ WZX; ZW

x 15 x 6 x 33

13

15 x2 36 x2 12x 400

5x

x3

4x

x9

x

20

2012

SRQ ~ TSQ ~ TRS

S

Q

T

S

R T

Q

S R

5.

Two angles of are con-gruent to two angles of therefore bythe AA Sim. Post.

ABC JKLABC,

JKL

30

60

A

B

C

MCRBG-0901-SK.qxd 5-25-2001 11:17 AM Page 182

Geometry 183Chapter 9 Worked-out Solution Key


Chapter 9 continued

24.

25. 26. 27.

28.

d 16 c

29.

30.

Solution:

31. About 76 cm; and are congruent righttriangles by the SSS Congruence Postulate, so is aperpendicular bisector of . By Geometric MeanTheorem 9.3, the altitude from D to hypotenuse divides into segments of lengths about 23.7 cm and 61.1 cm. By Geometric Mean Theorem 9.2, the length of the altitude to the hypotenuse of each right triangle isabout 38 cm long, so the crossbar should be about

or 76 cm long.2 38,BD

ACAC

BDAC

ADCABC

x 3

x 3 x 21

x 3 0 x 21 0

x 21x 3 0

x2 18x 63 0

x2 18x 81 144

x 9

18

8x 9

z 5313

y 40 x 4223

24z 1280 y2 1600 24x 1024

32z

2440

66

23

y

y24

2432

32x

e 715

4

e2 49 154 4

334

1214

e

e33

4

16 1214

ce

ed

c 1214

16c 196

1614

14c

m 1047

7m 74 x 9.6 x 33

7m 49 25 20x 192 x2 27

57

m 7

5 x

12

1620

x3

9x

x 126 314 11.2

x2 126

x7

18x

HGE ~ FHE ~ FGH 32.

33.

34. Given is a right triangle and altitude is drawnto hypotenuse ; and are right trianglesby the definition of right triangles; because all right angles are congruent; byreflexive property for angles; therefore by the AA Similarity Postulate; becauseall right angles are congruent; by the reflexiveproperty for angles; therefore by the AASimilarity Postulate; by theAngle Addition Postulate; because the two acute angles in a right triangle arecomplementary; mACD mB by the Transitive andSubtraction Properties of Equality, so bythe def. of congruent ; because allright angles are congruent; so by theAA Similarity Postulate.

35. From Ex. 34, . Corresponding side

lengths are in proportion, so

36. From Ex. 34, and .Corresponding side lengths are in proportion, so

and

37. Values of the ratios will vary, but will not be equal. Thetheorem says they are equal.

38. The ratios are equal.

39. The ratios are equal when the triangle is a right trianglebut are not equal when the triangle is not a right triangle.

ABAC

ACAD

.ABBC

BCBD

ABC ~ ACDABC ~ CBD

BDCD

CDAD

.

CBD ~ ACD

DCA ~ DBCCDA CDB

ACD B

mDCB mB 90mACD mDCB 90

ACB ~ ADCA AADC ACB

ACB ~ CDBB B

CDB ACBDCADBCAB

CDABC

1.5 m2 Area of CAB 1221.5

0.96 m2 Area of DAC 121.61.2

0.54 m2 Area of DCB 121.20.9

DB 0.9 m AD 1.6 m CD 1.2 m

DB1.5

1.52.5

AC2

2

2.5 CD2

1.52.5

DCB ~ DAC ~ CAB

h 64.4 ft

5.5h 30.25 324

h 5.5

18

185.5

xywy

wyzy

18 ft

5 ft12

W Y

h

X

ZNot drawn to scale

xy h 5.5


Chapter 9 continued



40. Using the right triangle, calculate the values of

These proportions should be true:

and Now drag C to change the value of

(so mC 90) and recalculate and

The values of the ratios will vary but

and

41. D

42.

C

43. Method 1

Measure the distance from the ground to the persons eye level (DC) and the distance from the person to the

building (AC). Use the proportion and solve

for BC (the height of the building). One advantage of this method is you only need two measurements. Onedisadvantage is you need a friend to help.

Method 2

Measure the length of the buildings shadow (QS), theheight of the pole (NP) and the length of the poles

shadow (MP). Use the proportion and solve

for RS (the height of the building). One advantage is itcan be done by one person. One disadvantage is it mustbe done when the building and pole cast a shadow.

9.1 Mixed Review (p. 534)

44. 45. 46.

47. If the measure of one of the angles of a triangle is greaterthan then the triangle is obtuse; true.

48. If the corresponding angles of two triangles are congru-ent, then the two triangles are congruent; false.

49. 50.

51.

62.5 m2 A 1212 135

31.5 cm2 36 in.2 A 74.5 A 12612

90,

d 9 x 8

d2 81 x2 64 n 13

d2 18 99 14 x2 78 n2 169

MPQS

NPRS

BCAC

ACDC

DC 6

AD 18 DC

14424

AD 24 6 DC12

1224

ACAD

.ABAC

CBDB

ABCB

ACAD

.

ABAC

,CBDB

,ABCB

,mC

ABAC

ACAD

.

ABCB

CBDB

ACAD

.ABAC

,

CBDB

,ABCB

,Lesson 9.2


1. Sample answer: In a right triangle, the square of thelength of the hypotenuse is equal to the sum of thesquares of the lengths of the legs.

2. A, C

3. 4.

no yes

5. 6.

no 3.3 ft


7. 8.

yes no

9. 10.

yes yes

11. 12.

no no

13. 14.

no yes

15. 16.

no

x 83 142 x

x2 192 392 x2 64 x2 256 196 196 x2 82 x2 162 142 142 x2

x 21 x 83

x2 441 x2 192

400 x2 841 64 x2 256

202 x2 292 82 x2 162

13 x x 42

13 x2 x2 32

4 9 x249 x2 81

22 32 x2 72 x2 92

41 x x 80

1681 x2 x2 6400

81 1600 x21521 x2 7921

92 402 x2 392 x2 892

x 35 97 x

x2 45 9409 x236 x2 814225 5184 x2 62 x2 92 652 722 x2

d 11 x 43

d2 11 x2 48

52 d2 6242 x2 82

x 6 5 x

x2 36 5 x2x2 82 10222 12 x2




Chapter 9 continued

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27.

28.

21 m2A 1210.54base a b 3 7.5 10.5 m

b 7.5 m

b2 72.25 16 56.25

42 b2 8.52 a 3 m

a2 25 16 9

a2 42 52 25.2 cm2

A 1277.2 b 7.2

b2 51.75

12.25 b2 64

3.52 b2 82 32.7 m2 35.7 cm2

A 12319 5A 12937 b 319 b 37

b2 171 b2 63

25 b2 196 81 b2 144

52 b2 142 92 b2 122 r 468 s 12

r2 219,024 s2 144

354,025 r2 573,049 1225 s2 1369

5952 r2 7572 352 s2 372 r 99 s 24

r2 9801 s2 576

400 r2 10,201 324 s2 900

202 r2 1012 182 s2 302 15 t 20 t

225 t2 400 t2 81 144 t2 144 256 t2 92 122 t2 122 162 t2

58 x 213 x

58 x2 52 x2 9 49 x2 16 36 x2 32 72 x2 42 62 x211 4 712 8 4

b 4 b 8

b2 16 b2 64

9 b2 25 36 b2 100

32 b2 52 62 b2 102 29.

30.

32. The minimum distance of the base of the ladder from the wallis or 2.5 feet. The ladder, ifplaced 2.5 feet from the wall,will reach feet up the wall.

33.

34.

35.

48 in.

12 12 24

r 3 in.4 6 in.2 12 in.2 h 84.9 in.

h2 7201

36002 h2 36012 300 ft 1 in. 3601 in.

300 ft 3600 in.

39 in. 39 in. 16 in. 94 in.

39 in. c

1521 c2 1296 225 c2

362 152 c2 2 ft 6 in. 30 inches

3 ft 36 inches

100 6.25 9.7

104

10 ft

ladder wall

2.5 ft

9.7 ft

100 m2A 122010d2 5 5 10

d1 12 8 20

b 5

b2 25

144 b2 169

122 b2 132 104 cm2

A 12810 16 b 8

b2 100 36 64

62 b2 102

10 cm

10 cm

10 cm

6 cm

b

31.

Distance from pitchersplate to home is 50 feet.The distance from secondbase to home is about 91.9feet so the distance fromsecond to the pitchersplate is orabout 41.9 feet.

91.9 50

91.9 ft c

8450 c2 652 652 c2


Chapter 9 continued



36.

Method 2 uses less ribbon.

37. The area of the large square is Also, the area ofthe large square is the sum of the areas of the four con-gruent right triangles plus the area of the small square, or

Thus,and so Subtracting fromeach side gives

38. The area of the trapezoid is Also, the area ofthe trapezoid is equal to the sum of the areas of the twocongruent right triangles plus the area of the isosceles triangle or

Thus

and so

Subtracting from eachside gives

39. a. b.

yes No. The longest space in theroom is the diagonal of theroom which is only about17.5 ft long.

c.

The length of the diagonal of the base is The length of the diagonal of the box is

40. The length of one side of the rhombus is

or Multiplying the length of one side by 4gives the perimeter of the rhombus, which is

or

41.


42. 43. 44.

45. 46.

47. 48. 49 2 36549 2 1225413 2 20822 2 8

14 2 149 2 96 2 6

a 32 cm, b 0.7532 24 cm 32 x

40 1.25x

40 1.5625x2 40 x2 0.5625x2 80 2x2 0.75x2 P 2a2 b2; a x, b 0.75x

2a2 b2.412a2 b2

12a

2 b2.

12a2 12b2l 2 w 22 h2 l 2 w 2 h2.

l2 w2.

d l2 w2 h2

17.5 ft 9.8 ft

BD 208 100BD 80 16

413 45

AB 144 64AB 82 42a2 b2 c2.

2aba2 2ab b2 2ab c2.

12a b2 a b

12c

2,

12 a b 12 a b 12 c2.

12a b2.

a2 b2 c2.2aba2 2ab b2 2ab c2.

a b2 412 a b c2,412 a b c2.

a b2.

35.0 in. r

1224 r2 324 900 r2 182 302 r2 49. 50. no 51. no 52. no 53. no

54. yes

55. Sample answer: slope of slope

of Both pairs of opposite sidesare parallel, so PQRS is a parallelogram by the definitionof a parallelogram.

56. Sample answer: slope of slopeof Both pairs of opposites sidesare parallel, so PQRS is a parallelogram by the definitionof a parallelogram.

Lesson 9.3

Activity 9.3 Investigating Sides and Angles of Triangles (p. 542)

Construct

Constructions may vary.

Investigate

Values in tables may vary.

Conjecture

4. when

when

when


1. Sample answer: If the square of the length of the longestside of a triangle is equal to the sum of the squares of thelengths of the other two sides, then the triangle is a righttriangle.

2. acute:

right:

obtuse:

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

7. No; the sum of while Since the two numbers are not equal, the triangles formedby the crossbars and the sides are not right triangles sothe crossbars are not perpendicular.


8. 9.

right right

10. 11.

not right right

26 26529 < 542.89

262 ? 12 52232 ? 20.82 10.52

7921 79219409 9409

892 ? 802 392972 ? 652 722

452 2025.222 382 1928,

c > 30c2 > 242 182c 30c2 242 182c < 30c2 < 242 182

mC < 90AC2 BC2 > AB2mC > 90AC2 BC2 < AB2mC 90AC2 BC2 AB2

QR 38 slope of PS.PQ 3 slope of RS;

QR 54 slope of PS.

PQ 112 slope of RS;

73 2 147




Chapter 9 continued

12. 13.

; not right ; not right

14. 15.

; right ; right

16. 17. not a triangle

; obtuse

18. 19.

; acute ; right

20. 21.

; right ; acute

22. 23.

; acute ; right

24. 25.

; acute ; obtuse

26. Rectangle; the quadrilateral has two pairs of congruent opposite sides. Each triangle formed by either diagonal is a right triangle because in each case

Therefore, the quadrilateral has four right angles. The quadrilateral is a rectangle.

27. Square; the diagonals bisect each other, so the quadri-lateral is a parallelogram. The diagonals are congruent,so the parallelogram is a rectangle. so the diagonals intersect at right angles to form perpen-dicular lines; thus, the parallelogram is also a rhombus. A quadrilateral that is both a rectangle and a rhombus is a square.

28. Rhombus; the diagonals bisect each other so the quadri-lateral is a parallelogram. so the diagonalsintersect at right angles to form perpendicular lines so theparallelogram is a rhombus.

29. slope of

slope of

Since , so is a right

angle. Therefore, is a right triangle by the defini-tion of a right triangle.

ABC

ABCAC BC3443 1,

BC 7 33 0

43

;

AC 6 34 0

34

;

32 42 52,

12 12 2 2,

142 82 265 2; 260 260.

30 < 30.252501 > 2500

5 25 ? 30.25100 2401 ? 2500

5 52 ? 5.52102 492 ? 50221,025 21,02541 > 25

289 20,736 ? 21,02516 25 ? 25

172 1442 ? 145242 52 ? 52221 > 1961156 1156

100 121 ? 196256 900 ? 1156

102 112 ? 142162 302 ? 34249 4983 > 81

13 36 ? 4916 67 ? 81

132 62 ? 7242 67 2 ? 92389 < 676

100 289 ? 676

102 172 ? 2621225 122510,201 10,201

441 784 ? 1225400 9801 ? 10,201

212 282 ? 352202 992 ? 1012560 < 56927 < 29

435 2 ? 202 13233 2 ? 22 52 30.

therefore is a right triangle by theConverse of the Pythagorean Theorem.

31. Computing slopes is easier because it involves two calcu-lations, not three. Computing slopes also does not involvesquare roots.

32.

The triangle is a right triangle.

33.

The triangle is an acute triangle.

34. Since is obtuse and isobtuse. and are a linear pair and are thereforesupplementary. By the definition of supplementaryangles, Since is obtuse,

Therefore, is an acuteangle by definition of an acute angle.

35. Since is obtuse and is obtuse. By the Triangle Sum Theorem,

is obtuse, soIt follows that Vertical angles

are congruent, so By substitution,By the definition of an acute angle, is

acute.1m1 < 90.

mABC m1.mABC < 90.mC > 90.

CmA mABC mC 180.C

ABC10 2 22 < 42,

1m1 < 90.mABC > 90.ABCmABC m1 180.

ABC1ABCABC22 32 < 42,

27 > 26

10 2 17 2 ? 262 17

RQ 0 42 1 12 10

PR 1 02 2 12 26

PQ 1 42 2 12

x2

1

R(0, 1)

P (1, 2)Q(4, 1)

y

80 45 125

45 2 35 2 ? 552 55

RQ 6 52 2 02 35

PR 3 62 4 22 45

PQ 3 52 4 02y

x6

2

R(6, 2)

P (3, 4)

Q(5, 0)

ABC25 25 50;

52 52 52 2 5

distance from A to C 4 02 6 32 52

distance from B to A 3 42 7 62 5

distance from B to C 3 02 7 32


Chapter 9 continued



36. If a, b, and c are a Pythagorean triple, then Let k represent a positive integer. Multiplying both sides of the equation by gives the equation

or by theDistributive Property. So by aproperty of powers. Since k 0, ka, kb, and kc representthe side lengths of a right triangle by the Converse of thePythagorean Theorem.

37. A, C, D 38. rectangle

39.

40.

so the is acute.

Cincinnati is not directly north of Tallahassee. It is north-west of Tallahassee.

41. Reasons

1. Pythagorean Theorem

2. Given

3. Substitution property of equality

5. Converse of the Hinge Theorem

6. Given, def. of right angle, def. of acute angle, andsubstitution property of equality

7. Def. of acute triangle ( is the largest angle of.)

42. Given: In

Prove: is an obtuse triangle.

Plan for Proof: Draw right triangle PQR with side lengthsa, b, and hypotenuse x. Compare lengths c and x.

Statements Reasons

1. 1. Pythagorean Theorem

2. 2. Given

3. 3. Substitution prop. of equality

4. 4. A property of square roots

5. 5. Converse of Hinge Thm.

6. is an obtuse angle. 6. Given, def. of rt. angle, def.of obtuse angle, subst. prop.of equality

7. is an obtuse 7. Def. of obtuse triangle triangle. ( is the largest of

.)ABCC

ABC

C

mC > mP

c > x

c2 > x2c2 > a2 b2x2 a2 b2

b

a

x

P R

Q

b

a

c

C B

A

ABC

ABC, c2 > a2 b2ABC

C

509,796 < 521,210

7142 ? 5992 4032343,768,681 343,768,681

18,5412 ? 13,5002 12,709244,209,201 44,209,20128,561 28,561

66492 ? 48002 460121692 ? 1192 1202

ka2 kb2 kc2k2a2 k2b2 k2c2k2a2 b2 k2c2,k2

a2 b2 c2. 43.

Statements Reasons


2. 2. Given


4. 4. A property of square roots

5. 5. Converse of Hinge Thm.

6. is a right angle. 6. Given, def. of rt. angle, def.of obtuse angle, subst. prop.of equality

7. is a right 7. Def. of right triangletriangle. ( is the largest .)

44.

is obtuse is acute

is obtuse is acute

A

45.

B

46.

Statements Reasons

1. is an altitude. 1. Given

2. 2. Def. of altitude

3. MPN and QPN are 3. Def. of perpendicularright angles.

4. and are 4. Def. of right triangleright triangles.


6. 6. Addition and Substitution properties of equality

7. t is the geometric mean of 7. Givenr and s.

8. 8. Def. of geometric mean

9. 9. Cross product prop.


11. r s MQ 11. Given (diagram)

12. 12. Substitution prop. ofequality

13. is a right triangle. 13. Converse of thePythagorean Thm.

MQN

MN2 NQ2 MQ2

r s2s2 2rs r2 MN2 NQ2 t2 rs

rt

ts

s2 2t2 r2s2 t2 r2 t2 MN2 NQ2 NQ2 r2 t2MN2 s2 t2,

QPNMPN

NPMQ

NP

mD < 90, so mE mF > 90

mA > 90, so mB mC < 90

DA

DEFABC

8324 > 82816925 < 7225

6724 1600 ? 82815629 1296 ? 7225

822 402 ? 912772 362 ? 852N

LMN

N

mN mR

c x

c2 x2c2 a2 b2x2 a2 b2




Chapter 9 continued


47.

48.

49.

50.

51. 52.

53.

54.

55. an enlargement with center C and scale factor

56. reduction with center C and scale factor

57.

Quiz 1 (p. 549)

1. 2.

3. 4.

5. 6.

7.

8.

No; the square of the longest side is larger than the sumof the squares of the smaller sides.

Lesson 9.4

Activity 9.4 Developing Concepts (p. 550)

Exploring the Concept

1. Triangles may vary.

2. side length 3 cm:

side length 4 cm:

side length 5 cm:

c 52 cm

52 52 c2 c 42 cm

42 42 c2 c 32 cm

32 32 c2

47,961 > 47,824

2192 ? 1682 1402 x 122

x2 62 182 x 65 x 210

x2 122 182 32 x2 72 BD 12 AC 25

180 15BD 9AC 225

9

15

BD20

9

15

15AC

BDCDB ~BDA ~ CBA

x 9

y 11 4x 36

2y y 11 5x x 36

35

74

824

8

26

466

26

3

1218

12

32

422

22

45

45

53

11

31111

15 6 90 310

14 6 84 221

6 8 48 43

22 2 44 211

Conjecture

3. The length of the hypotenuse is the product of the lengthof one side and

Exploring the Concept

4. Triangles may vary.

6. triangle with side length 4 cm: side lengths: 2 cm, 4 cm,

triangle with side length 6 cm: side lengths: 3 cm, 6 cm,

triangle with side length 8 cm: side lengths: 4 cm, 8 cm,

Conjecture

7.

ratio of hypotenuse:longer leg:shorter leg


1. Right triangles with angle measures and

2. According to the AA Similarity Postulate, since twoangles of one triangle are congruent to two angles of theother triangle, the two triangles are similar.

3. true 4. false 5. false 6. true 7. true

8. true 9. 10.

11.


12. 13.

14.

15. 16.

17. 18.

19. 20.

h 163

3

f 83

3

n 6 f 3 8

m 12; p 63q 162; r 16

c 42

d 42

c 5; d 53d 2 8

e 22

a 123; b 24x 5; y 52

92

2 k h

92

k

9 2k

h k

a 2; b 23x 42

306090454590

2:3:1

longer legshorter leg

: 23

2 3;

333

3; 43

4 3

hypotenuseshorter leg

: 42

2; 63

2; 84

2

43 cm

33 cm

23 cm

2.


Chapter 9 continued



21.

22.

23.

24.

25.

26.

27.

28. 29.

30.

31.

I used the Pythagorean theorem in each right triangle,working from left to right.

32. Going from left to right: triangle 1

33. Going from left to right: triangle 3

34.

35. Let Then By the PythagoreanTheorem,

by a property of square roots.Thus the hypotenuse is times as long as a leg.

36. Construct on so that CD BC a. Then by the SAS Cong. Post.

and because they are correspondingparts of congruent triangles. Therefore and

is equiangular so it is also equi-lateral. Since it is equilateral, If and

, then The sidelengths are in the following ratio: hypotenuse:longer leg:shorter leg : Therefore, in a triangle, the hypotenuse is twice as long as the shorter legand the longer leg is times as long as the shorter leg.

37. C 38. A; 6 63 12 28.4 cm

3

3060903 a:a.2a

AC 2a2 a2 3 a.AB 2aBC aAB 2a.

BAD30 30 60.mBAD mBAC mCAD mCAD 30.

mD 60BAC CAD,

B DABCADCBC

CD

2DE 2x2 2 x

2x2 DE2;x2 x2 DE2;EF x.DF x.

n 1

w 7u 5s 3

v 6t 2r 2

y 23 3.5 cm; x 2 2 4 cm

x 1.42 2.0 cmx 3 cm 1.7 cmA 612423 41.6 ft2A 523 17.3 m2A 12663 31.2 ft2A 12438 27.7 ft2

x 18.4 in.

x2 338

2x2 67626 in.

x

x

x2 x2 262 12.7 in. x

162 x2 81 81 x2 92 92 x2 s 9 in.

9 in.

9 in.

9 in. x 9 in.

4s 36

x 4.3 cm

x2 18.75

6.25 x2 25

2.52 x2 52

5 cm 5 cmx

5 cm2.5 cm

39. Stage 1:

Stage 3:

40. The pattern of the lengths is where the numberof the stage.

41. Substitute 8 for n into the formula and

simplify.


42. Let x length of the third side; x; x ;

43. 44. 45.

46. 47. AA Similarity Postulate

48. SAS Similarity Theorem 49. SSS Similarity Theorem

Math & History

1. area of triangles:

area of square:

2.

Lesson 9.5


1.

2. The value of a trigonometric ratio depends only on themeasure of the acute angle, not on the particular right triangle used to compute the value.

3. 4. 5. 6. 7. 8.34

45

35

43

35

45

tan A BCAC

cos A ACAB

sin A BCAB

a2 b2 c2 2ab b2 2ab a2 c2

b a2 b2 2ab a2

412 ab 2ab

B0, 10A4, 5P8, 3Q1, 2

5 cm < x < 23 cm 9 > 1414 9 >

12n

128

1

16

n 1

2n,

x 1

22

x2 18

2x2 14

x2 x2 122

x 12

x2 12

2x2 1

x2 x2 12 Stage 2:

Stage 4:

x 14

x2 1

16

2x2 18

x2 x2 1222

x 12

x2 14

2x2 12

x2 x2 122




Chapter 9 continued

9.

d 16.6, or about 17 ft


10.

11.

12.

13.

14.

15.

16. 0.7431 17. 0.9744 18. 6.3138 19. 0.4540

20. 0.3420 21. 0.0349 22. 0.9781 23. 0.8090

24. 0.4245 25. 0.4540 26. 0.8290 27. 2.2460

28.

x 10.0 y 8.0

sin 37 6x

tan 37 6y

tan K 35

0.6cos K 5

34 0.8575

sin K 3

34 0.5145tan J

53

1.6667

cos J 3

34 0.5145sin J

534

0.8575

tan H 12

0.5cos H 25

0.8944

sin H 15

0.4472tan G 21

2

cos G 15

0.4472sin G 25

0.8944

tan F 247

3.4286cos F 7

25 0.28

sin F 2425

0.96tan D 7

24 0.2917

cos D 2425

0.96sin D 7

25 0.28

tan Y 23

0.6667cos Y 3

13 0.8321

sin Y 2

13 0.5547tan X

32

1.5

cos X 2

13 0.5547sin X

313

0.8321

tan A 86

1.3333cos A 6

10 0.6

sin A 8

10 0.8tan B

68

0.75

cos B 8

10 0.8sin B

610

0.6

tan S 2845

0.6222cos S 4553

0.8491

sin S 2853

0.5283tan R 4528

1.6071

cos R 2853

0.5283sin R 4553

0.8491

sin 25 7d

29.

30.

31.

32.

33.

34.

35.

36.

37. 38.

39. vertical drop,

40. 41.

42.

43.

44. The tangent of one acute angle of a right triangle is thereciprocal of the tangent of the other acute angle. Thesine of one acute angle of a right triangle is the same asthe cosine of the other acute angle and the cosine of oneacute angle of a right triangle is the same as the sine ofthe other acute angle.

tan B ba

cos B ac

sin B bc

tan A ab

cos A bc

sin A ac

x 2.9 ft

tan 20 x8

x 16.4 in.

x 30

sin 45 26 d 714.1 m

sin 45 30

26 x tan 55

d500

1409.3 ft

sin 20 482d

x 5500 5018 482 ft

d 36.0 m h 13.4 m

tan 42 d

40 tan 13

h58.2

A 12

116 3 34.9 m2

A 12

126.9 41.6 m2

A 12

22 22 4 cm2 y 14.9 x 16.0

tan 22 6y

sin 22 6x

w 3.3 v 9.6

tan 70 9w

sin 70 9v

u 3.4 t 7.3

cos 65 u8

sin 65 t8

s 2.9 r 4.9

tan 36 s4

cos 36 4r

s 31.3 t 13.3

cos 23 s

34 sin 23

t34


Chapter 9 continued



45. Procedures may vary. One method is to reason that sincethe tangent ratio is equal to the ratio of the lengths of thelegs, the tangent is equal to 1 when the legs are equal inlength, that is, when the triangle is a triangle. Tan A > 1 when and when

since increasing the measure of in-creases the length of the opposite leg and decreasing themeasure of decreases the length of the opposite leg.

46. is not a right triangle, so you cannot use thetrigonometric ratios.

47. Reasons

1. Given

2. Pythagorean Theorem

3. Division property of equality

5. Substitution property of equality

48.

49.

50.

51.

1

sin 132 cos 132 0.22502 0.97442cos 13 0.9744sin 13 0.2250

34

14

1

sin 602 cos 602 32 2

122

cos 60 12

sin 60 32

24

24

1

sin 452 cos 452 22 2

22 2

cos 45 22

sin 45 22

14

34

1

sin 302 cos 302 122

32 2

cos 30 32

sin 30 12

BC 11.0 x 9

sin 55 9

BC sin 30

x18

x

C

B

A30

18

55

ABC

A

AmA < 45,tan A < 1mA > 45,

454590

52. Statements Reasons

1. is a right 1. Giventriangle with sidelengths a, b, and hypotenuse c.

2. 2. Def. of tangent

3. 3. Def. of sine and cosine

4. 4. Subst. prop of equality anddividing and simplifyingfractions

5. 5. Transitive property ofequality

53. ; D 54. C

55.


56. enlargement;

scale factor

2

57.

58. 59.

yes no

x 569 x 168

x2 1725 x2 28,224

x2 2500 4225 x2 9025 37,249

x2 502 652 x2 952 1932 NP 7.73 QP 3.27

NP2 59.7429 49 15QP

18.27NP

NP3.27

715

QP7

MNP ~ NQP ~ MQN

PR 8

QR 10

63

5

10

R

R

Q

Q

P

P

4

8

6

3

h 46 ft

79.62 33.26 h

x y h

y 33.26 x 79.62

tan 29 y

60 tan 53

x60

sin 25 8

CD

tan A sin Acos A

sin Acos A

acbc

ab

cos A bc; sin A

ac

tan A ab

ABC




Chapter 9 continued

60.

no

Quiz 2 (p. 566)

1.

2.

3.

4.

5.

6.

7.

d 4887.3 ft

tan 11 950d

y 22.1 x 9.3

cos 25 20y

tan 25 x

20

y 15.9 x 8.5

sin 62 y

18 cos 62

x18

y 11.9 x 15.6

tan 40 10y

sin 40 10x

3.9 in.2

A 12

31.53h 1.53 in.

3 in. 3 in.

3 in.1.5 in.

h

5.7 in.

x 42 in.4 in.

4 in.

4 in. x 4 in.

3.5 m

x 23 m

4 m 4 m

4 m2 m

x

82.1 x

6740.41 x2 1840.41 4900 x2

42.92 702 x2 Lesson 9.6

Activity 9.6 Developing Concepts (p. 567)

1.

2.

3.

4. The values are approximately equal.


1. To solve a right triangle is to find the measures of allangles and the lengths of all sides of the triangle.

2. true 3. false 4. 5.

6. 7.

8.

9.

10. mX 30


11. 12.

13.

14. 15. 16.

17. 18. 19.

20. 21. mA 6.3mA 65.6

mA 50.2mA 81.4mA 20.5

mA 30mA 45mA 26.6

mS 41.1

sin S 4873

73 QS

mQ 48.9 5329 QS2 sin Q

5573

482 552 QS2

x 2 y 3.5

cos 60 x4

sin 60 y4

d 60

mE 56.6 mD 33.4d2 3600

sin E 91

109 sin D

60109

912 d2 1092

65 c

mB 30.5 mA 59.5 4225 c2

sin B 3365

sin A 5665

332 562 c2

mA 84.3mA 64.2

mA 79.5mA 35.0

tan1 0.75 36.9cos1 0.8 36.9sin1 0.6 36.9

tan A 34

0.75cos A 45

0.8sin A 35

0.6

C

B

A

5 cm

4 cm

3 cm


Chapter 9 continued



22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

mB 90 56 34

a 7.4 c 8.9

tan 56 a5

cos 56 5c

mY 38

mY 90 52

z 13.8 x 10.9

cos 52 8.5z

tan 52 x

8.5

mT 70

mT 90 20

t 11.3 s 4.1

cos 20 t

12 sin 20

s12

mP 64

mP 90 26

p 4.0 q 2.0

cos 26 p

4.5 sin 26

q4.5

TS 11.0

mR 61.3 mS 28.7 TS2 120.25

cos R 6

12.5 sin S

612.5

62 TS2 12.52

NQ 13.0 mP 72.9 mN 17.1 NQ2 168.96

cos P 4

13.6 sin N

413.6

42 NQ2 13.62

ML 4.5

mK 29.6 mL 60.4 ML2 20.64

cos K 8

9.2 sin L

89.2

82 ML2 9.22

6.3 GH

mH 18.4 mG 71.6 40 GH

sin H 2

40 sin G

640

22 62 GH2

9.9 DE

mD 45 mE 45 98 DE2

sin D 7

98 sin E

798

72 72 DE2

29 AB

mA 46.4 mB 43.6 841 AB2

sin A 2129

sin B 2029

202 212 AB2 32.

33.

34. 35.

36. 37.

38. 39.

x 239.4 in. or 19 ft 11 in.

40.

41. Answers may vary.

42. 43.

44. Sample answer: riser length: 6 in.; tread length: 12 in.

45. Sample answer: The riser-to-tread ratio affects the safetyof the stairway in several ways. First, the deeper the treadthe more of a persons foot can fit on the step. Thismakes a person less likely to fall. Also, the smaller theangle of inclination the less steep the stairway. Thismakes the stairs less tiring to climb, and therefore, safer.

x 26.6

tan x 6

12

x 42.5x 32.5

tan x 8.25

9 tan x

711

8 in.

54.6 in.

55.2 in.8.33

y 4.1

sin y 17

240

57,311 x2 x 15.4

2402 172 x2 tan x 4855

17,625

77.8 in. AB

sin A 0.4626 6057 AB2

sin A BCAB

36

6057 692 362 AB2

mB 62.4 1.9167

tan1B 6936

tan B 6936

mL 56

mL 90 34

5.9 m 7.2

tan 34 4

sin 34 4m

mF 39

mF 90 51

e 4.8 d 3.7

cos 51 3e

tan 51 d3




Chapter 9 continued

46. Draw an altitude, , from C to , and let CD h. In

rt. sin A . In rt. sin B . Thus,

and By the substitution prop. of equality, Dividing both

sides by gives or


47. 48. 49.

50. 51. 52.

53. 54.

55. 56.

57. 58.

59. not a triangle

60. 61.

acute right

62. 63.

obtuse right

64. not a triangle

Lesson 9.7


1. The magnitude of a vector is the distance from its initialpoint to its terminal point. The direction of a vector is theangle it makes with a horizontal line.

2.

3. is parallel to is parallel to

4.

5.

6.

7. MN\

3 12 4 12 29 5.42, 5PQ

\

5 12 4 22 25 4.54, 2AB

\

4 02 5 02 41 6.44, 52, 2

PQ\

AB\

MN\

;UV\

AB\

: 2, 2; PQ\

: 3, 3; MN\

: 0, 3; UV\

: 0, 2

12,769 12,76969,169 > 68,900

1132 ? 1122 1522632 ? 2502 802

72.25 72.2551,984 < 52,000

8.52 ? 7.72 3.622282 ? 2202 602

t 22

88 4t m 14

8t

4

11 m2

71

k 216 g 12.6

7k 1512 10g 126

7

18

84k

310

g

42

y 112 x 25

7y 784 6x 150

7

16

49y

x

30

56

3, 11, 21, 0

1, 32, 23, 2

asin A

b

sin B.

bsin B

a

sin A,sin A sin B

b sin A a sin B.h a sin B.h b sin A

ha

BCD,hb

ACD,

ABCD 8.

9.


10.

11.

12.

13.

14.

15.

16.

7.2

52

PQ\

4 22 3 726, 4y

x11

Q

P

10.8

116

PQ\

3 72 2 6210, 4y

x2

2

Q

P

5.8

34

PQ\

5 22 1 623, 5y

x1

1

Q

P

7.3

53

PQ\

2 02 7 022, 7

y

x1

1

Q

P

EF\

0 42 1 52 42 5.74, 4JK

\

2 12 2 42 35 6.73, 6RS

\

1 52 1 22 17 4.14, 1

4, 5 4, 2 0, 3

x 51.3 north of east

tan x 54


Chapter 9 continued



17.

18.

19.

20.

21.

22.

x 51.3 south of east

tan x 5040

64 miles per hour

AB\

40 02 50 020, 0 and 40, 50

x 9.5 north of east

tan x 1060

61 miles per hour

ST\

60 02 30 2020, 20 and 60, 30

3

9

PQ\

0 32 5 523, 0y

x1

1

QP

4.1

17

PQ\

6 52 0 421, 4y

x1

1

Q

P

8.2

68

PQ\

6 22 3 12

8, 2y

x1

1Q

P

7.2

52

PQ\

5 12 0 426, 4

y

x1

1

Q

P

23.

24.

25. 26.

27. 28.

29. yes; no

30. Round 2; the vectors have the same magnitude and opposite directions. In Round 1, team A won; since has a greater magnitude than represents a greaterforce applied.

31.

32.

33.

34.

u\

v\

3, 3

v\

: 1, 6

u\

: 2, 3y

x1

1

u

u v

v

u\

v\

5, 2

v\

: 3, 6

u\

: 2, 4y

x1

1

u

u v

v

u\

v\

1, 5

v\

: 5, 3

u\

: 6, 2y

x1

5

v

u

u v

u\

v\

6, 5

v\

: 2, 4

u\

: 4, 1y

x1

1

v

u

u v

CA\

CB\

,CA

\

GH\

and JK\

EF\

and CD\

EF\

and CD\

EF\

, CD\

, and AB\

x 38.7 south of west

tan x 4050

64 miles per hour

OP\

0 502 0 4020, 0 and 50, 40

x 45 north of west

tan x 4040

57 miles per hour

LM\

10 502 10 50210, 10 and 50, 50




Chapter 9 continued

35. 36. 37.

38. 39. 40.

41.

42.

43.

the speed at which the skydiver is falling, taking intoaccount the breeze.

44.

45. The new velocity is

46.

Sample answer:

The component form must give the same magnitude as

AB\

10.JK\

.

AB\

3, 1JK

\

10

30, 120.

10

10EW

DOWN

UP

x 71.6

tan x 12040

s\

20 602 140 202 126.5 mih

10

10EW

DOWN

UP

v

us

v\

: 40, 0

u\

: 0, 120

0, 04, 42, 3

10, 108, 74, 11 47. When the magnitude of is k times the magnitudeof and the directions are the same. When themagnitude of is times the magnitude of and thedirection of is opposite the direction of . Justificationsmay vary.

48. a.

b.

c. Answers may vary.

49.

50.

51.

Total distance

52. The answer to Ex. 50 is a vector which gives the finalposition of the bumper car, while the answer to Ex. 51 isa number which gives the total distance traveled by thebumper car.

53. Since and are right angles and all right anglesare congruent, . Since is equilateral,

so andby the Alternate Interior Angles

Theorem. An equilateral triangle is also equiangular, soBy the definition of con-

gruent angles and the substitution property of equality,. by the AAS

Congruence Theorem. Corresponding parts of congruenttriangles are congruent, so By the definition ofmidpoint, B is the midpoint of

54. 55. 56.

57.

58.

59.

60. 7 x2 49 14x x2x 112 x2 22x 121x 72 x2 14x 49x 12 x2 2x 1

y 60y 30y 90

x 30x 120x 45

DE.DB EB.

ADB CEBDBA EBC

mBAC mBCA 60.

EBC BCADBA BACDE AC, AB BC.

ABCD EED

59.1 50.9 62.6 172.6 ftCA

\

182 602 62.6 ftBC

\

362 362 50.9 ftAB

\

542 242 59.1 ft18, 60 18, 60 0, 0

CA\

: 18, 60

AB\

BC\

18, 60

BC\

: 36, 36

AB\

: 54, 24


tan x 2

10

s\

10 02 2 02 10.2 mih

2

2EW

S

N

vu

s

u\

v\

u\

kv\

k < 0,u\

v\

k > 0,


Chapter 9 continued



Quiz 3 (p. 580)

1.

2.

3.

4.

5.

6.

7.

8.

7.8

PQ\

2 42 2 32PQ

\

: 6, 5y

x2

2

Q

P

5.1

PQ\

3 22 4 32PQ

\

: 5, 1y

x1

1

Q

P

mL 90 14.0 76.0

12.0

2 144.76 mK 14.0

32 2 12.42 sin K 3

12.4

mF 90 52.1 37.9

f 4.7

f 2 21.76 mG 52.1

62 f 2 7.62 sin G 6

7.6

mQ 90 75 15

q 2.1 p 7.7

cos 75 q8

sin 75 p8

mN 90 40 50

q 20.9 m 13.4

cos 40 16q

tan 40 m16

mY 45

y 12 z 17.0

tan 45 12y

sin 45 12z

mA 90 25 65

a 41.7 b 19.4

cos 25 a

46 sin 25

b46

9.

10.

11.

north of east

12. 13. 14.

15. 16. 17.

Chapter 9 Review (pp. 582584)

9.1 Similar Right Triangles

1.

2.

3.

9.2 The Pythagorean Theorem

4. 5.

yes no

s 45 8.9 20 t

s2 80 400 t2 82 s2 122 122 162 t2

23.8

z 97 x 48

z2 567 y 21 27x 1296

z

27

21z

48 27 y 3627

x

36

y 12 x 15

y2 144 x2 225

y

16

9y

25x

x9

y 35 x 4

y2 45 36 9x

5y

y9

6x

96

0, 36, 132, 1

2, 82, 44, 2

x 69.4

tan x 83y

x

2

2

T

S

13.0

PQ\

2 52 6 52PQ

\

: 7, 11y

x2

4

P

Q

5.8

PQ\

3 02 4 12PQ

\

: 3, 5y

x1

1

Q

P




Chapter 9 continued

6. 7.

yes no

9.3 The Converse of the Pythagorean Theorem

8. 9.

obtuse right

10. not a triangle

11.

acute

9.4 Special Right Triangles

12.

13.

14. ;

15.

9.5 Trigonometric Ratios

16.

17.

tan N 1235

0.3429cos N 3527

0.9459

sin N 1237

0.3243tan P 3512

2.9167

cos P 3537

0.3243sin P 3537

0.9459

tan L 6011

5.4545cos L 1161

0.1803

sin L 6061

0.9836tan J 1160

0.1833

cos J 6061

0.9836sin J 1161

0.1803

A 12

1893 813 140.3 cm2altitude 93 cm

longer leg 63 in.shorter leg 12

12 6 in.

18 in.2 122 in.

A 32 2 P 432

leg 62

32

hypotenuse 232 6

81 < 89

92 ? 32 45 2

1681 1681100 > 85

412 ? 402 92102 ? 62 72

t 213 7.2 r 30

52 t2 r2 900

42 62 t2 r2 162 34218.

9.6 Solving Right Triangles

19.

20.

21.

9.7 Vectors

22.

23.

24.

PQ\

9 4 13 3.6PQ

\

: 3, 2y

x1

2 Q

P

PQ\

144 25 13PQ

\

: 12, 5y

x2

2

Q

P

PQ\

1 16 17 4.1PQ

\

: 1, 4y

x2

1

Q

P

17 s

mT 61.9 289 s2mR 28.1

sin T 1517

82 152 s2 tan R 8

15

40 f 12.9 d 15.3

mF 90 50 cos 50 f

20 sin 50

d20

8.9

x 45

mZ 41.8mX 48.2 x2 80

sin Z 8

12 cos X

812

82 x2 122

tan A 7

42 1.2374cos A

429

0.6285

sin A 79

0.7778tan B 42

7 0.8081

cos B 79

0.7778sin B 42

9 0.6285

MCRBG-090R-SK.qxd 5-25-2001 11:38 AM Page 199

Chapter 9 continued



25.

Chapter 9 Chapter Test (p. 585)

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

6.

7. is a kite. The diagonals are perpendicular and thequadrilateral has two pairs of consecutive congruentsides, but opposite sides are not congruent.

8. 9.

acute

10.

side length

11.

12.

13.

4.5

mR 41.8 mP 48.2 QR 25

sin R 46

cos P 46

42 QR2 62

mF 90 25 65

DE 25.7 DF 28.4

tan 25 12DE

sin 25 12DF

mK 90 30 60

JL 7.8 KL 4.5

cos 30 JL9

sin 30 KL9

6 in.

31.2 in.2 183

12

663

A 12

d1d2

6 in.

6 in.6 in.

30

3 3 in. 3 in.

60

6 in.

40 < 54

210 2 ? 29 2 52 b 112PR 4 36 210

b2 12,544QR 9 16 5

152 b2 1132PQ 25 4 29

WXYZ

DBA; DAC


tan x 9

14

u\

v\

196 81 277 16.6u\

v\

14, 9 14.

15.

16.

17. 18. 19.

Chapter 9 Standardized Test (pp. 586587)

1.

C

2. C

3. D

4. B 5. D

6.

D

7.

E

8. 9.

A B

mA 38.0x 53.1

sin A 8

13 tan x

129

y 20.5 x 18.8

cos 67 8y

tan 67 x8

x 128 82 in.

2x2 256

P 482 322 in. x2 x2 162

P 2125 214 50.4 in.A 1114 154 in.2x 11 x 9

x 11x 9 0

x2 2x 99 0

x 12 100

x 1

20

5x 1

2, 34, 12, 8

DE 32.7 BC 22.9

tan 35 22.9DE

sin 35 BC40

mBCA 90 35 55

AB 13.1 CD 6.4

sin 50 10AB

sin 40 CD10

x 36.9 south of east

tan x 34

LM\

16 9 5LM

\

: 4, 3y

x1

1

L

M




Chapter 9 continued

10.

B

11. D

12.

13.

14.

15.

16. ; ;

17.

18.

19. a. b.

c. d.

e.

20. As the sun rises, the value of b decreases.

21.

22. The value of the expression increases as the sunapproaches the horizon.

Chapter 9 Cumulative Practice (pp. 588589)

1. No; if two planes intersect, then their intersection is aline. The three points must be collinear, so they cannot bethe vertices of a triangle.

2. always 3. never 4. always

x 43.6

tan x 5

5.25

b 1.8 cm

tan 70 5b

b 2.9 cm b 4.2 cm

tan 60 5b

tan 50 5b

b 6.0 cm b 8.7 cm

tan 40 5b

tan 30 5b

A 12

3032.3 15 709.5 square units

AE 37.3 FE 18.6

cos 30 32.3AE

tan 30 FE

32.3

mBGF 135mFEA 60mABC 75

AF 103 15 32.3DC 30

GC 450 21.2FG 15

BC 302 42.4BD 103 3 30

A 25 12 12724 216 square units y 73.7 x 16.3

sin y 2425

sin x 7

25

p 7 24 25 56

x 7

242 x2 252AB

\

8 12 3 92 15

y 7 x 8

y 4 11 2 x 6 5. is the median from point B,and it is given that Thus, bythe SSS Congruence Postulate. Also,since corresponding parts of congruent triangles arecongruent. By the definition of an angle bisector,bisects

6. Sample answer: Given quadrilateral where and Since there are in a quadrilateral,

and are opposite angles, and and are opposite angles. and Since both pairs of opposite angles are congruent,quadrilateral is a parallelogram.

7. Yes; clockwise and counterclockwise rotational symmetryof

8.

9.

10.

11.

12.

scalene right triangle

13. A1, 2, B3, 5, C5, 6

125 55

BC 4 121

AC 36 64 10

AB 16 9 5

y 34

x 72

y 2 34

x 64

y 2 34

x 2

midpoint: 5 12 , 6 2

2 2, 2slope of perpendicular bisector

34

m 6 2

5 1

86

43

z 55; x 90; y 65

y 113 x 24

y 67 180 8x 192

y 324 5 180 8x 12 180

y 3x 5 180 3x 5 5x 7 180

y 16

x 544y 64

26 x 10 90 26 4y 90

120.

ABCD

B D.A CDBCA

mD 360 37 143 37 143.360

mC 37.mB 143,mA 37,ABCD

ABC.BD

ABD CBDABD CBDAB CB.

BD BD,AD CD,BD


Chapter 9 continued



14.

15.

16. 17. 18.

19. No; in ABCD, the ratio of the length to width is 8:6 or4:3. In APQD, the ratio of length to width is 4:6, or 2:3.Since these ratios are not equal, the rectangles are notsimilar.

20. is a midsegment ofBy the Midsegment

Theorem, Since the lines containing these segments are parallel,

and by the Corresponding Angles

Postulate. Since two angles of are congruent totwo angles of the two triangles are similar bythe AA Similarity Postulate.

21. Yes; the ratios all equal so the triangles aresimilar by the SSS Similarity Theorem.

22. segment 1 4.2 cm

segment 2 4.8 cm

23. The image with scale factor has endpoints

and its slope is The image with scale

factor has endpoints and its slope is

The two image segments are parallel.136

.

6, 4.5;3, 212

133

2

136

.4, 3;

2, 43

13

x 4.2

15x 63

8x 63 7x

x

9 x

78

23,

69,

812, and

1218

BAC,BDE

BED BCA,BDE BAC

DE AC.ABC.DE

A

B

D E

C

y 1012

2y 21 x 337

x 445

9y 7y 21 7x 24 24 5x

79

y

y 3 37

x8

12x

52

A3, 6, B7, 9, C9, 2

C6, 5B5, 3A2, 1y

x1

1

A

B

A

B

C

C 24.

25. 26.

27. 28.

acute

29. Let x the measure of the smaller acute angle.

30.

31.

south of east

32. Construct a circle inscribed in the triangle by bisectingtwo angles of the triangle. The point at which the bisectors intersect is the center of the circle. Construct a segment from the center of the circle perpendicular to a side of the triangle. The length of this segment is the radius of the desired circle.

33. 34.

35.

d 189.4 mi

d2 35,856.25 mi

d2 127.52 1402 11.25 in.

36 13.5 2

w 6.75 in.

16w 108 x 20 gallons

1636

3w

376x 7520

P 3 3 5 5 16 37616

470

x

x 7.1

tan x 2

16

u\

v\

4 256 16.1u\

v\

2, 16y

x2

2

v

u

u v

mS 90 57 33

TR 10.9 in. ST 16.8 in.

cos 57 TR20

sin 57 ST20

tan x 8

15

cos x 1517

sin x 8

17

26

3

361 < 369

8243

223

33

192 ? 152 122

26 XY 4 XY

676 XY2 12 3XY

102 242 XY2 23XY

3

23

4 ZX ZP 23

16 ZX2 ZP2 12

23 2 22 ZX2 ZP6

2

ZP




Chapter 9 continued

Project: Investigating Fractals (pp. 590591)

Investigation

1. stage 1: ; stage 2:

2. stage 3: ; stage 4: ; the length at stage 1 is timesthe length at stage 0, and similarly the length at stage 2 is

times the length at stage 1.

3. At each stage the length would be times the previousstage, so the length gets increasingly large.

4. Yes; the graph is a curve that increases sharply as n, thenumber of stages, increases

Stages of a Koch Snowflake

5.

6.

7. P 343n

Stage 0 Stage 1 Stage 2

43

43

43

25681

6427

169

43

Stage, n 0 1 2 3 4

Perimeter, P 3 4 25627649

163


CHAPTER 9 - Mr Jaffemrjaffe.net/GeomRef/Ch9solns.pdf · Chapter 9continued 24. 25. 26. 27. 28. d 16...

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Transcript of CHAPTER 9 - Mr Jaffemrjaffe.net/GeomRef/Ch9solns.pdf · Chapter 9continued 24. 25. 26. 27. 28. d 16...