CHAPTER 9

[Pages:22]CHAPTER 9

Lesson 9.1

Think and Discuss (p. 525)

1. Since angle B and angle D are both right angles, the skyscrapers must be parallel.

2. According to Theorem 4.8, if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and corresponding leg of another right triangle, the two triangles are congruent.

Skill Review (p. 526)

1. mL 180 30 60 90

JKL is a right triangle.

2.

J

3. y

1

hypotenuse

leg, altitude

x 1

altitude

K leg, altitude L

4. x 3 x 53

3x 3 5x 3x 9 5x 9 2x 9x 2

5.

B

60?

30?

A

C

Two angles of JKL are congruent to two angles of ABC, therefore the two triangles are similar.

Developing Concepts Activity (p. 527) 3. All of the triangles are similar.

9.1 Guided Pratice (p. 531) 1. geometric mean 2. KML; JMK 3. MK 4. JM 5. JK 6. LJ 7. KM 8. KM 9. LK

10. 97 72 72 DC DC 53.4

53.4 FD FD 43.6 DF 48.3

9.1 Practice and Applications (pp. 531?534)

11. Q

Q

S

S

RT

ST

R

12. SRQ ~ TSQ ~ TRS

13. x 20 20 12

14. 4 x x9

12x 400

36 x2

x 33.3

x6

16. ZYX ~ WYZ ~ WZX; ZW

17. SQR ~ TQS ~ TSR; RQ

18. GFE ~ HFG ~ HGE; EH

19. CBA ~ DBC ~ DCA

12 x 16 12

144 16x

x9

20. GEF ~ HGF ~ HEG

x 20 20 25

25x 400

x 16

21. LKJ ~ MLJ ~ MKL

x 32 32 15

15x 1024

x 68.27

22. RSQ ~ TRQ ~ TSR

x 40

40 32

32x 1600

x 50

23. CBA ~ DBC ~ DCA

x4 4x

x2 16

x4

15. 5 x x3 15 x2 x 15

182

Geometry Chapter 9 Worked-out Solution Key

Copyright ? McDougal Littell Inc. All rights reserved.

Chapter 9 continued

24. HGE ~ FHE ~ FGH

x 7

18 x

x2 126

x 126

25. x 9 3x

26. x 12 27. 20 16

5m7 75

x2 27

16x 240

7m 49 25

x 33

x 15

7m 74

m

4 10

7

28. 142 b2 162 b2 60 b 60

14 c 16 14

16c 196

c

1 12

4

1214 14

e 60

14e 94.88

e 6.78

14 60 6.78 d 14d 52.52

d 3.75

29. 32 24 x 32

66.67 y y 24

32 24 z 40

24x 1024

y2 1600

24z 1280

x 42.67

y 40

z 53.3

30.

x

18

9

x

8

9

x2 18x 81 144

x2 18x 63 0

x 21x 3 0

x 21 0

x30

x 21

x3

Solution: x 3

31. About 76 cm; ABC and ADC are congruent right triangles by the SSS Congruence Postulate, so AC is a perpendicular bisector of BD. By Geometric Mean Theorem 9.3, the altitude from D to hypotenuse AC divides AC into segments of lengths 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 is about 38 cm long, so the crossbar BD should be about

2 38, or 76 cm long.

32.

xy h 5.5

X

xy wy

wy zy

h 5.5 18 18 5.5

5.5h 30.25 324 h 64.4 ft

33. DCB ~ DAC ~ CAB

h

W 18 ft Y 512 ft Z

Not drawn to scale

CD 1.5 2 2.5

AC 2 2 2.5

DB 1.5 1.5 2.5

CD 1.2 m

AD 1.6 m

Area of DCB 121.20.9

0.54 m2

Area of DAC 121.61.2

0.96 m2

Area of CAB 1221.5

1.5 m2

DB 0.9 m

34. Given ABC is a right triangle and altitude CD is drawn to hypotenuse AB; DBC and DCA are right triangles by the definition of right triangles; CDB ACB because all right angles are congruent; B B by reflexive property for angles; therefore ACB ~ CDB by the AA Similarity Postulate; CDA ACB because all right angles are congruent; A A by the reflexive property for angles; therefore ACB ~ ADC by the AA Similarity Postulate; mACD mDCB 90 by the Angle Addition Postulate; mDCB mB 90

because the two acute angles in a right triangle are complementary; ACD B by the Addition Property; CDA CDB because all right angles are congruent; so DCA ~ DBC by the AA Similarity Postulate.

35. From Ex. 34, CBD ~ ACD. Corresponding side

lengths

are

in

proportion,

so

BD CD

CD .

AD

36. From Ex. 34, ABC ~ CBD and ABC ~ ACD. Corresponding side lengths are in proportion, so

AB BC and AB AC. BC BD AC AD

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

38. They become equal.

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

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Geometry Chapter 9 Worked-out Solution Key

183

Chapter 9 continued

AB CB 40. Using the right triangle, calculate the values of , ,

CB DB

AB ,

AC .

These

proportions

should

be

true:

AB

CB

AC AD

CB DB

and

AB AC

AC AD.

Now

change

the

value

of

C

and

recalculate CABB, DCBB, AACB, AADC. The values of the ratios

will

vary

but

AB CB

DCBB,

AB AC

AADC.

41. D

42. DC 12 12 24

DC

144 24

AD 24 6

AD 18

C

DC 6

43. Method 1

Measure the distance from the ground to the person's eye level and the distance from the person to the building. Use the proportion BC AC and solve for BC. One

AC DC advantage of this method is you only need two measurements. One disadvantage is you need a friend to help.

Method 2

Measure the length of the building's shadow, the height of the pole and the length of the pole's shadow. Use the proportion MP NP and solve for RS. One advantage

QS RS is it can be done by one person. One disadvantage is it must be done when the building and pole cast a shadow.

9.1 Mixed Review (p. 534)

44. n2 169

45. 14 x2 78 46. d2 18 99

n ? 13

x2 64

d2 81

x ?8

d ?9

47. If the measure of one of the angles of a triangle is greater than 90, 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. A 12612

50. A 4.5

36 in.2

31.5 cm2

51. A 1212 135

62.5 m2

Lesson 9.2

9.2 Guided Practice (p. 538)

1. In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.

2. A, C

3. 22 12 x2

4. x2 82 102

5 x2

x2 36

5 x

x6

no

yes

5. 42 x2 82

6. 52 d2 62

x2 48

d2 11

x 43

d 11 ft

no

9.2 Practice and Applications (pp. 538?541)

7. 652 722 x2 4225 5184 x2 9409 x2 97 x yes

9. 392 x2 892 1521 x2 7921 x2 6400 x 80 yes

11. 72 x2 92 49 x2 81 x2 32 x 42 no

13. 82 x2 162 64 x2 256 x2 192 x 83 no

15. 142 142 x2 196 196 x2 392 x2 142 x

8. 62 x2 92 36 x2 81 x2 45 x 35 no

10. 92 402 x2 81 1600 x2 1681 x2 41 x yes

12. 22 32 x2 4 9 x2 13 x2 13 x

no 14. 202 x2 292

400 x2 841 x2 441 x 21

yes 16. 82 x2 162

64 x2 256 x2 192 x 83

no

184

Geometry Chapter 9 Worked-out Solution Key

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Chapter 9 continued

17. 62 b2 102

18. 32 b2 52

36 b2 100

9 b2 25

b2 64

b2 16

b8

b4

12 8 4

11 4 7

42 62 x2

32 72 x2

16 36 x2

9 49 x2

52 x2

58 x2

213 x

58 x

19. 122 162 t2

20. 92 122 t2

144 256 t2

81 144 t2

400 t2

225 t2

20 t

15 t

21. 182 s2 302

22. 202 r2 1012

324 s2 900

400 r2 10,201

s2 576

r2 9801

s 24

r 99

23. 352 s2 372

24. 5952 r2 7572

1225 s2 1369

354,025 r2 573,049

s2 144

r2 219,024

s 12

r 468

25. 92 b2 122

26. 52 b2 142

81 b2 144

25 b2 196

b2 63

b2 171

b 37

A 12937

35.7 cm2

b 319

A 12319 5

32.7 m2

27. 3.52 b2 82

12.25 b2 64

b2 51.75

b 7.2 A 1277.2

25.2 cm2

28. a2 42 52

a2 25 16 9

a3m

42 b2 8.52

b2 72.25 16 56.25

b 7.5 m

base a b 3 7.5 10.5 m A 1210.54

21 m2

29.

10 cm

10 cm b

6 cm 10 cm

62 b2 102

b2 100 36 64

b8 A 12810 16

104 cm2

30. 122 b2 132

31. 652 652 c2

144 b2 169

8450 c2

b2 25

91.9 ft c

b5

d1 12 8 20 d2 5 5 10 A 122010

100 m2

Distance from pitcher's plate to home is 50 feet. The distance from second base to home is about 91.9 feet so the distance from second to the pitcher's plate is 91.9 50 or about 41.9 feet.

32. The minimum distance of the

base of the ladder from the wall

is

10 4

or

2.5

feet.

The

ladder,

if

ladder

wall

placed 2.5 feet from the wall,

will reach 100 6.25 9.7

10 ft 9.7 ft

feet up the wall.

2.5 ft

33.

3 ft 36 inches

2 ft 6 in. 30 inches

362 152 c2

1296 225 c2

1521 c2

39 in. c

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

34.

300 ft 3600 in.

300 ft 1 in. 3601 in.

36002 h2 36012

h2 7201

h 84.9 in.

35. r 3 in.4 6 in.2 12 in.2

12 12 24

48 in.

Copyright ? McDougal Littell Inc. All rights reserved.

Geometry Chapter 9 Worked-out Solution Key

185

Chapter 9 continued

36. 202 302 r2

400 900 r2

1300 r2

36.1 in. r

Method 2 uses less ribbon.

37. The area of the large square is a b2. Also, the area of

the large square is the sum of the areas of the four con-

gruent right triangles plus the area of the small square, or

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

and so a2 2ab b2 2ab c2. Subtracting 2ab from

each side gives a2 b2 c2.

38. The area of the trapezoid is 12a b2. Also, the area of the trapezoid is equal to the sum of the areas of the two

congruent right triangles plus the area of the isosceles

triangle or

1 2

a

b

1 2

a

b

1 2

c2.

Thus

12a b2 a b 12c2, and so

a2 2ab b2 2ab c2. Subtracting 2ab from each

side gives a2 b2 c2.

39. a. AB 82 42

b. AB 144 64

45

413

BD 80 16

BD 208 100

9.8 ft

17.5 ft

yes c. d l2 w2 h2

No. The longest space in the room is the diagonal of the room which is only approximately 17.5 ft.

The length of the diagonal of the base is l2 w2. The length of the diagonal of the box is

l 2 w 22 h2 l 2 w 2 h2.

40. The length of one side of the rhombus is 12a2 12b2

or 12a2 b2. Multiplying the length of one side by 4 gives the perimeter of the rhombus, which is

4 12a2 b2 or 2a2 b2.

41. P 2a2 b2; a x, b 0.75x

80 2x2 0.75x2

40 x2 0.5625x2

40 1.5625x2

40 1.25x

32 x

a 32, b 0.7532 24

9.2 Mixed Review (p. 541)

42. 6 2 6 43. 9 2 9 44. 14 2 14 45. 22 2 8 46. 413 2 208 47. 549 2 1225 48. 49 2 36

49. 73 2 147 50. no 51. no 52. yes 53. no

54. yes

55.

Slope

of

PQ

11 2

slope of RS; slope of

QR

5 4

slope

of

PS.

Both

pairs

of

opposite

sides

are

parallel, so PQRS is a parallelogram by the definition of

a parallelogram.

56. slope of PQ 3 slope of RS; slope of

QR

3 8

slope

of

PS.

Both

pairs

of

opposites

sides

are

parallel, so PQRS is a parallelogram by the definition of a

parallelogram.

Lesson 9.3

Activity 9.3 Investigating Sides and Angles of Triangles

Construct Constructions may vary. Investigate Values in tables may vary. Conjecture 4. AC2 BC2 AB2 when mC 90

AC2 BC2 < AB2 when mC > 90 AC2 BC2 > AB2 when mC < 90

9.3 Guided Practice (p. 545)

1. If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle.

2. acute: c2 < 242 182 c < 30

right: c2 242 182 c 30

obtuse: c2 > 242 182 c > 30

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

7. No; the sum of 222 382 1928, while 452 2025. Since the two numbers are not equal, the triangles formed by the crossbars and the sides are not right triangles so the crossbars are not perpendicular.

9.3 Practice and Applications (pp. 546?548)

8. 972 ? 652 722

9. 892 ? 802 392

9409 9409

7921 7921

right

right

10. 232 ? 20.82 10.52 11. 262 ? 12 52

529 < 542.89

26 26

not right

right

186

Geometry Chapter 9 Worked-out Solution Key

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Chapter 9 continued

12. 33 2 ? 22 52

13. 435 2 ? 202 132

27 < 29

560 < 569

not right

not right

14. 202 992 ? 1012

15. 212 282 ? 352

400 9801 ? 10,201

441 784 ? 1225

10,201 10,201

1225 1225

right

right

16. 102 172 ? 262

17. not a triangle

100 289 ? 676

389 < 676

obtuse

18. 42 67 2 ? 92

19. 132 62 ? 72

16 67 ? 81

13 36 ? 49

83 > 81

49 49

acute

right

20. 162 302 ? 342

21. 102 112 ? 142

256 900 ? 1156

100 121 ? 196

1156 1156

221 > 196

right

acute

22. 42 52 ? 52

23. 172 1442 ? 1452

16 25 ? 25

289 20,736 ? 21,025

41 > 25

21,025 21,025

acute 24. 102 492 ? 502

right

25. 5 52 ? 5.52

100 2401 ? 2500

5 25 ? 30.25

2501 > 2500

30 < 30.25

acute

obtuse

26. The quadrilateral has two pairs of congruent opposite sides. The triangle formed by the diagonal is a right

triangle because 142 82 265 2; 260 260.

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

27. Square; the diagonals bisect each other, so the quadrilateral is a parallelogram; the diagonals are congruent,

so the parallelogram is a rectangle. 12 12 2 2,

so the diagonals intersect at right angles to form perpendicular lines; thus, the parallelogram is also a rhombus. A quadrilateral that is both a rectangle and a rhombus must be a square.

28. Rhombus; the diagonals bisect each other so the quadrilateral is a parallelogram. 32 42 52, so the diagonals intersect at right angles to form perpendicular lines so the parallelogram is a rhombus.

29.

slope of

AC

6 4

3 0

43;

slope

of

BC

7 3

3 0

43;

Since

3 4

4 3

1, AC BC, so ABC is a right

angle. Therefore, ABC is a right triangle by the definition of a right triangle.

30. distance from B to C 3 02 7 32

5

distance from B to A 3 42 7 62

52

distance from A to C 4 02 6 32

5

52 52 52 2

25 25 50; therefore ABC is a right triangle.

31. Computing slopes is easier because it involves two calculations, not three. Computing slopes also does not involve square roots.

32.

y

PQ 3 52 4 02

P(3, 4) 2

R(6, 2)

Q(5, 0) 6x

45 PR 3 62 4 22

35 RQ 6 52 2 02

55

45 2 35 2 ? 552

80 45 125

The triangle is an right triangle.

33.

y

PQ 1 42 2 12

P(1, 2) 1

R(0, 1)

Q(4, 1)

2

x

26 PR 1 02 2 12

10 RQ 0 42 1 12

17

10 2 17 2 ? 262

27 > 26

The triangle is an acute triangle.

34. Since x2 32 < 42, ABC is obtuse and ABC is obtuse. 1 and ABC are a linear pair and are therefore supplementary. By the definition of supplementary angles, mABC m1 180. Since ABC is obtuse, mABC > 90. Therefore, m 1 < 90. 1 is an acute angle by definition of an acute angle.

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Geometry Chapter 9 Worked-out Solution Key

187

Chapter 9 continued

35. Since 10 2 22 < 42, ABC is obtuse and

C is obtuse. By the Triangle Sum Theorem, mA mABC mC 180. C is obtuse, so mC > 90. It follows that mABC < 90. Vertical angles are congruent, so mABC m1. By substitution, m1 < 90. By the definition of an acute angle, 1 is

acute.

36. If a, b, and c are a Pythagorean triple, then a2 b2 c2. Let k represent a positive integer. Multiplying a, b, and c by k gives the equation k2a2 k2b2 k2c2. Dividing both sides of the equation by k2 yields a2 b2 c2. By the Converse of the Pythagorean Theorem, ABC is a right triangle and a, b, and c represent the side lengths of ABC.

37. A, C, D 38. rectangle

39. 1692 ? 1192 1202 66492 ? 48002 46012

28,561 28,561

44,209,201 44,209,201

18,5412 ? 13,5002 12,7092

343,768,681 343,768,681

40. 7142 ? 5992 4032 509,796 < 521,210 Cincinnati is not directly north of Tallahassee. It is northeast 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, and substitution property of equality 7. Def. of acute triangle (C is the largest angle of ABC.)

42. Given: In ABC, c2 > a2 b2 Prove: ABC is an obtuse triangle. Plan for Proof: Draw right triangle PQR with side lengths a, b, and x. Compare lengths c and x.

A

c b

Q x

b

C

a

BP

a

R

Statements 1. x2 a2 b2 2. c2 > a2 b2 3. c2 > x2

4. c > x 5. m C > m P 6. C is an obtuse angle

7. ABC is an obtuse triangle

43. Statements 1. x2 a2 b2 2. c2 a2 b2 3. c2 x2

4. c x 5. m N m R 6. N is an right angle

7. LMN is a right triangle

Reasons 1. Pythagorean Theorem 2. Given 3. Substitution prop. of

equality 4. A property of square roots 5. Converse of Hinge Thm. 6. Given, def. of rt. angle, def.

of obtuse angle, sub. prop. of equality 7. Def. of obtuse triangle (C is the largest of ABC).

Reasons

1. Pythagorean Theorem 2. Given 3. Substitution prop. of

equality 4. A property of square roots 5. Converse of Hinge Thm. 6. Given, def. of rt. angle, def.

of obtuse angle, sub. prop. of equality 7. Def. of right triangle (N is the largest ).

44. 772 362 ? 852

822 402 ? 912

5629 1296 ? 7225 6724 1600 ? 8281

6925 < 7225

8324 > 8281

ABC is obtuse

DEF is acute

A is obtuse

D is acute

A

45. mA > 90, mB mC < 90

mD < 90, mE mF > 90

B

46.

Statements

Reasons

1. NP is an altitude

1. Given

2. mNPM mNPQ 90 2. Def. of an altitude

3. r t ts

3. Given

4. MQN is a right triangle

4. Theorem 9.2

188

Geometry Chapter 9 Worked-out Solution Key

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Chapter 9 continued

9.3 Mixed Review

47. 22 2 44 211

48. 6 8 48 43 49. 14 6 84 221 50. 15 6 90 310

51.

3 11

311 11

52.

4 5

45 5

53. 12 12 42 22 18 32 2

54.

8 24

8 26

46 6

26 3

55.

an

enlargement

with center

C

and

scale

factor

7 4

56.

reduction

with

center

C

and

scale

factor

5 3

57. 5x x 36

2y y 11

4x 36

y 11

x9

Quiz 1 (p. 549)

1. CDB ~ BDA ~ CBA 2. BD

3. 9 15 15 AC

4. 9 x 15 20

9AC 225

180 15x

AC 25

x 12

5. 32 x2 72

6. x2 122 182

x 210 6.3

x 65 13.4

7. x2 62 182

x 122 17.0

8. 2192 ? 1682 1402

47,961 > 47,824

No; the square of the longest side is larger than the sum of 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: 32 32 c2

c 32 4.2 cm side length 4 cm: 42 42 c2

c 42 5.7 cm side length 5 cm: 52 52 c2

c 52 7.1 cm

Copyright ? McDougal Littell Inc. All rights reserved.

Conjecture

3. The length of the hypotenuse is the product of the length of one side and 2.

Exploring the Concept

4. Triangles may vary.

6. triangle with side length 4 cm: side lengths: 2 cm, 4 cm, 23 cm

triangle with side length 6 cm: side lengths: 3 cm, 6 cm, 33 cm

triangle with side length 8 cm: side lengths: 4 cm, 8 cm, 43 cm

Conjecture

7.

hypotenuse :

4

2;

6

2;

8

2

shorter leg 2 3 4

longer leg :

shorter leg

23 2

3;

33 3

3;

43 4

3

ratio of hypotenuse: longer leg:shorter leg:2:3:1

9.4 Guided Practice (p. 554)

1. Right triangles with angle measures 454590 and 306090.

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

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

8. true 9. x 42 10. a 2; b 23

11. h k

9 2k

9 2

k

92 k h 2

9.4 Practice and Applications (pp. 554?556)

12. x 5; y 52 14. e 22

15. d 2 8

d 42 c 42 17. q 162; r 16

19. f 3 8

f 83 3

h 163 3

13. a 123; b 24 16. c 5; d 53

18. m 12; p 63 20. n 6

Geometry Chapter 9 Worked-out Solution Key

189

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