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Page 182
CHAPTER 9
Lesson 9.1
9.1 Practice and Applications (pp. 531–534)
Think and Discuss (p. 525)
11. Q
Q
S
1. AB and ED are both to BD; in a plane, 2 lines to the same line are .
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.
13.
1. mL 180 30 60 90
JKL is a right triangle.
3.
leg, altitude
x
T
R
14.
4
x
x
9
36 x2
1
3
x6
15.
x
5
x
3
15 x2
x 15
17. SQR ~ TQS ~ TSR; RQ
18. GFE ~ HFG ~ HGE; EH
1
19. CBA ~ DBC ~ DCA
L
x
12
16 12
x
x3
5
3
4.
S
16. ZYX ~ WYZ ~ WZX; ZW
altitude
leg, altitude
T
12x 400
y
1
K
x
20
20 12
x 33
J
hypotenuse
R
12. SRQ ~ TSQ ~ TRS
Skill Review (p. 526)
2.
S
5.
B
3x 3 5x
x9
60°
3x 9 5x
A
9 2x
20. GEF ~ HGF ~ HEG
30°
C
Two angles of JKL are congruent to two angles of ABC,
therefore ABC JKL by
the AA Sim. Post.
9
x
2
144 16x
20
x
20 25
25x 400
x 16
21. LKJ ~ MLJ ~ MKL
x
32
32 15
Developing Concepts Activity (p. 527)
15x 1024
3. All of the triangles are similar.
x 68.3
9.1 Guided Practice (p. 531)
1. geometric mean
4. JM
10.
5. JK
2. KML; JMK
6. LJ
7. KM
22. RSQ ~ TRQ ~ TSR
3. MK
8. KM
9. LK
72
97
72 DC
DC 53.4
FD
53.4
FD
43.6
DF 48.3
40
x
40 32
32x 1600
x 50
23. CBA ~ DBC ~ DCA
4
x
4
x
x2 16
x4
182
Geometry
Chapter 9 Worked-out Solution Key
Copyright © McDougal Littell Inc.
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11:17 AM
Page 183
Chapter 9 continued
24. HGE ~ FHE ~ FGH
xy h 5.5
32.
18
h 5.5
18
5.5
x2 126
x 126 314 11.2
25.
x
9
3
x
26.
x2 27
x
16
12 20
5 m7
7
5
27.
20x 192
7m 49 25
x 9.6
7m 74
x 33
m 10
28.
4
7
Z
h 64.4 ft
Not drawn to scale
CD 1.5
2
2.5
CD 1.2 m
AC
2
2
2.5
DB 1.5
1.5
2.5
AD 1.6 m
DB 0.9 m
1
2 1.20.9
Area of DAC 121.61.2
0.96 m2
Area of CAB 1221.5
e
c
e d
d 16 c
1
16 12
4
1
124
e
3
3
4
34. Given ABC is a right triangle and altitude CD is drawn
e
33
e2 49 15
44
e
715
4
663
y
y
24
32 24
z
40
y2 1600
2
3
1.5 m2
4
2
24 32
32
x
24x 1024
30.
Y
1
5 2 ft
0.54 m2
1
4
x 42
W
5.5h 30.25 324
Area of DCB 16 14
14
c
c 12
h
18 ft
33. DCB ~ DAC ~ CAB
16c 196
29.
X
wy
xy
wy
zy
x
18
7
x
24z 1280
z 53
y 40
1
3
8
x9
18
x9
x2 18x 81 144
35. From Ex. 34, CBD ~ ACD. Corresponding side
x2 18x 63 0
lengths are in proportion, so
x 21x 3 0
x 21 0
x 21
x30
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 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 is
about 38 cm long, so the crossbar BD should be about
2 38, or 76 cm long.
Copyright © McDougal Littell Inc.
All rights reserved.
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; ADC 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; m∠ACD m∠B by the Transitive and
Subtraction Properties of Equality, so ACD B by
the def. of congruent ; CDA CDB because all
right angles are congruent; so DCA ~ DBC by the
AA Similarity Postulate.
CD
BD
.
CD
AD
36. From Ex. 34, ABC ~ CBD and ABC ~ ACD.
Corresponding side lengths are in proportion, so
AB
BC
AB
AC
.
and
BC BD
AC AD
37. Values of the ratios will vary, but will not be equal. The
theorem says they are equal.
38. The ratios are 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.
Geometry
Chapter 9 Worked-out Solution Key
183
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Page 184
Chapter 9 continued
AB CB
,
,
CB DB
AB AC
CB
AB
,
. These proportions should be true:
AC AD
CB
DB
AC
AB
and
. Now drag C to change the value of
AC AD
AB CB AB
mC (so m∠C 90) and recalculate
,
,
, and
CB DB AC
40. Using the right triangle, calculate the values of
CB
AC
AB
. The values of the ratios will vary but
AD
CB
DB
AC
AB
and
.
AC
AD
Lesson 9.2
9.2 Guided Practice (p. 538)
1. Sample answer: 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
5
4. x2 82 102
x2 36
x2
5 x
41. D
x6
no
DC 12
42.
12
24
AD 24 6
AD 18
144
DC 24
yes
5. 42 x2 82
6. 52 d2 62
x2 48
d2 11
C
x 43
3.3 ft
no
DC 6
43. Method 1
9.2 Practice and Applications (pp. 538–541)
Measure the distance from the ground to the person’s
eye level (DC) and the distance from the person to the
BC
AC
building (AC). Use the proportion
and solve
AC DC
for BC (the height of the building). One advantage of
this method is you only need two measurements. One
disadvantage is you need a friend to help.
Measure the length of the building’s shadow (QS), the
height of the pole (NP) and the length of the pole’s
MP NP
shadow (MP). Use the proportion
and solve
QS
RS
for RS (the height of the building). One advantage is it
can be done by one person. One disadvantage is it must
be done when the building and pole cast a shadow.
652 722 x2
8. 62 x2 92
4225 5184 x2
36 x2 81
7.
9409 45. 14 x2 78
46. d2 18 99
n ± 13
x2 64
d2 81
d ±9
x ±8
47. If the measure of one of the angles of a triangle is greater
1521 1
51. A in.2
1
2 12
50.
A 74.5
31.5
cm2
10.
92 402 x2
7921
81 1600 x2
x2 6400
1681 x2
x2
x 80
yes
41 x
yes
11. 72 x2 92
12. 22 32 x2
49 x2 81
4 9 x2
x2 32
13 x2
x 42
no
13 x
no
2
13. 8 x 16
14. 202 x2 292
64 x2 256
400 x2 841
x2 192
x2 441
2
2
x 83
48. If the corresponding angles of two triangles are congru49. A 2 612
no
392 x2 892
than 90, then the triangle is obtuse; true.
ent, then the two triangles are congruent; false.
x 35
yes
9.1 Mixed Review (p. 534)
44. n2 169
x2 45
x2
97 x
9.
Method 2
36
d 11
no
x 21
yes
2
15. 14 14 x
16. 82 x2 162
196 196 x2
64 x2 256
2
135
2
392 62.5 m2
x2
142 x
x2 192
x 83
no
184
Geometry
Chapter 9 Worked-out Solution Key
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5-25-2001
11:17 AM
Page 185
Chapter 9 continued
17. 62 b2 102
36 b2
18. 32 b2 52
9
100
b2
b2 64
b4
42 62 x2
32 72 x2
16 36 x2
9 49 x2
52 x2
58 x2
21.
b8
A
144 256 t2
81 144 t2
400 t2
225 t2
20 t
15 t
s2
324 s2
900
182
302
20.
22.
122
400 r2
10,201
r2
9801
s 576
1225 s2 1369
26. 52 b2 142
81 b2 144
25 b2 196
b2 63
27.
b5
d1 12 8 20
d2 5 5 10
A 122010
100 m2
b 319
A
cm2
1
2
319 5
33.
42 b2 8.52
b2 72.25 16 56.25
b 7.5 m
base a b 3 7.5 10.5 m
9.7 ft
2.5 ft
3 ft 36 inches
1521 c2
39 in. c
b 7.2
a3m
10 ft
1296 225 c2
b2 51.75
a2 25 16 9
wall
362 152 c2
12.25 b2 64
28. a2 42 52
ladder
2 ft 6 in. 30 inches
3.52 b2 82
25.2 cm2
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.
base of the ladder from the wall
10
is 4 or 2.5 feet. The ladder, if
placed 2.5 feet from the wall,
will reach 100 6.25 9.7
feet up the wall.
32.7 m2
A 1277.2
91.9 ft c
32. The minimum distance of the
b2 171
b 37
35.7
b2 25
r 468
25. 92 b2 122
37 8450 c2
1012
r2 219,024
s 12
A
144 b2 169
354,025 r2 573,049
s2 144
1
2 9
31. 652 652 c2
5952 r2 7572
24.
16
30. 122 b2 132
r 99
s 24
352 s2 372
1
2 810
104 cm2
t2
r2
202
2
23.
b2 100 36 64
58 x
92
10 cm
62 b2 102
t2
162
b
6 cm
11 4 7
213 x
10 cm
10 cm
b2 16
b8
19.
25
12 8 4
122
29.
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.
A 1210.54
21 m2
Copyright © McDougal Littell Inc.
All rights reserved.
Geometry
Chapter 9 Worked-out Solution Key
185
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11:17 AM
Page 186
Chapter 9 continued
36.
182 302 r2
49. 73 147
324 900 54. yes
2
r2
1224 r2
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 congruent 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 2 a b2. Also, the area of
1
the trapezoid is equal to the sum of the areas of the two
congruent right triangles plus the area of the isosceles
triangle or
12 a b 12 a b 12 c2. Thus
b2 a b 12c2, and so
2 2ab b2 2ab c2. Subtracting
a
2ab from each
side gives a2 b2 c2.
1
2 a
39. a. AB 82 42
b. AB 144 64
45
BD 208 100
9.8 ft
No. The longest space in the
room is the diagonal of the
room which is only about
17.5 ft long.
h2
The length of the diagonal of the base is l2 w2.
The length of the diagonal of the box is
40.
l 2 w 22 h2 l 2 w 2 h2.
2
2
The length of one side of the rhombus is 12a 12b
or 12a2 b2. Multiplying the length of one side by 4
gives the perimeter of the rhombus, which is
412a2 b2 or 2a2 b2.
41.
56. Sample answer: slope of PQ 3 slope of RS; slope
3
of QR 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
(p. 542)
Construct
Constructions may vary.
Investigate
Values in tables may vary.
Conjecture
AC2 BC2 < AB2 when mC > 90
9.3 Guided Practice (p. 545)
1. Sample answer: 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
P 2a2 b2; a x, b 0.75x
4. D
5. D
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.
40 x2 0.5625x2
40 1.5625x2
40 1.25x
9.3 Practice and Applications (pp. 546–548)
32 x
8. 972 ? 652 722
a 32 cm, b 0.7532 24 cm
2
45. 22 8
2
2
44. 14 14
2
46. 413 208
47. 549 1225
7921 7921
right
43. 9 9
2
9. 892 ? 802 392
9409 9409
9.2 Mixed Review (p. 541)
42. 6 6
6. A
7. No; the sum of 222 382 1928, while 452 2025.
80 2x2 0.75x2
186
5
of QR 4 slope of PS. Both pairs of opposite sides
are parallel, so PQRS is a parallelogram by the definition
of a parallelogram.
AC2 BC2 > AB2 when mC < 90
17.5 ft
yes
53. no
4. AC2 BC2 AB2 when mC 90
413
BD 80 16
c. d 52. no
11
Method 2 uses less ribbon.
w2
51. no
55. Sample answer: slope of PQ 2 slope of RS; slope
35.0 in. r
l2
50. no
2
48. 49 36
2
Geometry
Chapter 9 Worked-out Solution Key
2
10. 23
right
? 20.8 10.5
2
2
11. 262 ? 12 52
529 < 542.89
26 26
not right
right
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11:16 AM
Page 187
Chapter 9 continued
12. 33 2 ? 22 52
13. 435 2 ? 202 132
27 < 29; not right
14.
202
992
?
1012
5
560 < 569; not right
15.
212
28
2
distance from B to A 3 42 7 62
2
? 35
400 9801 ? 10,201
441 784 ? 1225
10,201 10,201; right
1225 1225; right
16. 102 172 ? 262
30. distance from B to C 3 02 7 32
52
distance from A to C 4 02 6 32
5
17. not a triangle
52 52 52 2
100 289 ? 676
25 25 50; therefore ABC is a right triangle by the
Converse of the Pythagorean Theorem.
389 < 676; obtuse
18.
42
67 2
?
92
24.
2
2
? 7
31. Computing slopes is easier because it involves two calcu-
16 67 ? 81
13 36 ? 49
83 > 81; acute
49 49; right
20. 162 302 ? 342
22.
19. 13 6
2
21. 102 112 ? 142
256 900 ? 1156
100 121 ? 196
1156 1156; right
221 > 196; acute
42
52
?
52
lations, not three. Computing slopes also does not involve
square roots.
23.
172
144
2
2
? 145
41 > 25; acute
21,025 21,025; right
?
502
25. 5 52
?
PR 3 62 4 22
Q(5, 0)
2
289 20,736 ? 21,025
492
45
P(3, 4)
16 25 ? 25
102
PQ 3 52 4 02
y
32.
35
6 x
R(6, 2)
RQ 6 52 2 02
55
45 2 35 2
5.52
?
55 2
100 2401 ? 2500
5 25 ? 30.25
80 45 125
2501 > 2500; acute
30 < 30.25; obtuse
The triangle is a right triangle.
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
142 82 265 2; 260 260. Therefore, the
quadrilateral has four right angles. The quadrilateral
is a rectangle.
PQ 1 42 2 12
y
33.
26
P(1, 2)
Q(4, 1)
PR 1 02 2 12
1
10
x
2
R(0, 1)
RQ 0 42 1 12
27. Square; the diagonals bisect each other, so the quadri-
lateral 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 is
a square.
28. Rhombus; the diagonals bisect each other so the quadri-
lateral is a parallelogram. so the diagonals
intersect at right angles to form perpendicular lines so the
parallelogram is a rhombus.
32
42
52,
63 3
;
40 4
73
4
slope of BC ;
30
3
3
4
Since
1, AC BC , so ABC is a right
4
3
angle. Therefore, ABC is a right triangle by the definition of a right triangle.
29. slope of AC Copyright © McDougal Littell Inc.
All rights reserved.
17
10 2
17 2
?
26 2
27 > 26
The triangle is an acute triangle.
34. Since 22 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, m1 < 90. 1 is an acute
angle by definition of an acute angle.
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.
Geometry
Chapter 9 Worked-out Solution Key
187
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11:16 AM
Page 188
Chapter 9 continued
36. If a, b, and c are a Pythagorean triple, then a2 b2 c2.
Let k represent a positive integer. Multiplying both
sides of the equation by k2 gives the equation
k2a2 b2 k2c2, or k2a2 k2b2 k2c2 by the
Distributive Property. So ka2 kb2 kc2 by a
property of powers. Since k 0, ka, kb, and kc represent
the side lengths of a right triangle by the Converse of the
Pythagorean Theorem.
37. A, C, D
38. rectangle
1192 1202
39. 1692 ?
66492 ? 48002 46012
✓
28,561 28,561
44,209,201 44,209,201
Statements
x2
1.
a2
Reasons
b2
1. Pythagorean Theorem
2. c2 a2 b2
2. Given
3. c2 x2
3. Substitution prop. of equality
4. c x
4. A property of square roots
5. mN mR
5. Converse of Hinge Thm.
6. N is a right angle.
6. Given, def. of rt. angle, def.
of obtuse angle, subst. prop.
of equality
✓
18,5412 ? 13,5002 12,7092
7. LMN is a right
✓
343,768,681 343,768,681
43.
7. Def. of right triangle
(N is the largest .)
triangle.
2
2
2
40. 714 ? 599 403
44. 77 36
2
2
822 402 ? 912
2
? 85
509,796 < 521,210 so the is acute.
5629 1296 ? 7225
6724 1600 ? 8281
Cincinnati is not directly north of Tallahassee. It is northwest of Tallahassee.
6925 < 7225
8324 > 8281
ABC is obtuse
DEF is acute
A is obtuse
D is acute
41. Reasons
1. Pythagorean Theorem
A
2. Given
45. mA > 90, so mB mC < 90
3. Substitution property of equality
mD < 90, so mE mF > 90
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 hypotenuse x. Compare lengths c and x.
Q
C
a
B
P
a
1. Given
2. NPMQ
2. Def. of altitude
3. ∠MPN and ∠QPN are
3. Def. of perpendicular
right angles.
4. MPN and QPN are
4. Def. of right triangle
5. MN2 s2 t2,
5. Pythagorean Theorem
6. MN2 NQ2 6. Addition and
s2 t2 r2 t2 s2 2t2 r2
R
Reasons
1. x2 a2 b2
1. Pythagorean Theorem
2. c2 > a2 b2
2. Given
3. c2 > x2
3. Substitution prop. of equality
4. c > x
4. A property of square roots
5. mC > mP
5. Converse of Hinge Thm.
6. C is an obtuse angle.
6. Given, def. of rt. angle, def.
of obtuse angle, subst. prop.
of equality
triangle.
Reasons
1. NP is an altitude.
NQ2 r2 t2
Statements
7. ABC is an obtuse
Statements
x
b
b
46.
right triangles.
A
c
B
7. Def. of obtuse triangle
(C is the largest of
ABC .)
Substitution properties
of equality
7. t is the geometric mean of
7. Given
r and s.
8.
r
t
t
s
8. Def. of geometric mean
9. t2 rs
9. Cross product prop.
10. MN NQ 2
2
10. Substitution prop. of
s2 2rs r2 r s2
11. r s MQ
equality
11. Given (diagram)
12. MN NQ MQ
2
2
2
12. Substitution prop. of
equality
13. MQN is a right triangle.
13. Converse of the
Pythagorean Thm.
188
Geometry
Chapter 9 Worked-out Solution Key
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11:16 AM
Page 189
Chapter 9 continued
9.3 Mixed Review (p. 549)
47. 22
48. 6
Conjecture
2 44 211
3. The length of the hypotenuse is the product of the length
of one side and 2.
8 48 43
6 84 221
15 6 90 310
49. 14
50.
3
311
51.
11
11
12
53.
18
54.
24
8
Exploring the Concept
4. Triangles may vary.
4
45
52.
5
5
6. triangle with side length 4 cm: side lengths: 2 cm, 4 cm,
23 cm
12
42
22
32
2
triangle with side length 6 cm: side lengths: 3 cm, 6 cm,
33 cm
8
46 26
26
6
3
triangle with side length 8 cm: side lengths: 4 cm, 8 cm,
43 cm
55. an enlargement with center C and scale factor
3
5
56. reduction with center C and scale factor
57. 5x x 36
7
4
2y y 11
4x 36
Conjecture
7.
y 11
33
43
longer leg 23
:
3;
3;
3
shorter leg 2
3
4
x9
ratio of hypotenuse:longer leg:shorter leg 2:3:1
Quiz 1 (p. 549)
1. CDB ~BDA ~ CBA
3.
9
15
15 AC
9.4 Guided Practice (p. 554)
BD
9
15
20
1. Right triangles with angle measures 454590 and
306090
180 15BD
AC 25
2. According to the AA Similarity Postulate, since two
BD 12
5. 3 x 7
2
2. BD
4.
9AC 225
2
hypotenuse 4
6
8
: 2; 2; 2
shorter leg 2
3
4
angles of one triangle are congruent to two angles of the
other triangle, the two triangles are similar.
6. x 12 18
2
2
x 210
2
2
x 65
7. x2 62 182
3. true
4. false
8. true
9. x 42
7. true
10. a 2; b 23
9 2k
8. 2192 ? 1682 1402
9
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
2
k
92
kh
2
9.4 Practice and Applications (pp. 554–556)
Activity 9.4 Developing Concepts (p. 550)
12. x 5; y 52
13. a 123; b 24
14. e 22
Exploring the Concept
15. d
1. Triangles may vary.
2. side length 3 cm:
6. true
hk
11.
x 122
5. false
2 8
d 42
32 32 c2
c 32 cm
side length 4 cm: 42 42 c2
c 42 cm
c 42
17. q 162; r 16
19. f
3 8
f
side length 5 cm: 52 52 c2
c 52 cm
h
Copyright © McDougal Littell Inc.
All rights reserved.
16. c 5; d 53
18. m 12; p 63
20. n 6
83
3
163
3
Geometry
Chapter 9 Worked-out Solution Key
189
MCRBG-0903-SK.qxd
5-25-2001
11:16 AM
Page 190
Chapter 9 continued
2.52 x2 52
21.
39. Stage 1: x2 x2 12
6.25 x2 25
5 cm
5 cm
2x2 1
x2 18.75
x
1
x 2
2
x 4.3 cm
2.5 cm
5 cm
92 92 x2
x
9 in.
Stage 3: x2 x2 9 in.
81 81 x2
162 x2
1
4
x
1
2
2x2 676
x
x2 338
x 18.4 in.
Stage 4: x2 x2 27.7 ft2
1
8
x2 1
16
x
1
4
of the stage.
31.2
41.
ft2
26. A 5 23 17.3 m2
1
1
16
simplify.
28
22
x2 1
22
1
2n
2
1
1
8
40. The pattern of the lengths is
x
1
2 43 8
1
2 6 63
2
2x2 x
26 in.
12
1
4
23. x2 x2 262
25. A x2 2x2 9 in.
12.7 in. x
1
2
9 in.
s 9 in.
12
2x2 1
x
2
22. 4s 36
24. A Stage 2: x2 x2 , where n the number
Substitute 8 for n into the formula
1
2n
and
27. A 6 2 4 23 41.6 ft2
9.4 Mixed Review (p. 557)
28. x 3 cm 1.7 cm
42. Let x length of the third side; 14 9 > x; x 9 > 14;
1
30. y 23 3.5 cm; x 2
31. r 2
s 3
29. x 1.42 2.0 cm
2 4 cm
5 cm < x < 23 cm
t2
v 6
43. Q1, 2
u 5
w 7
46. B0, 10
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
Theorem, x2 x2 DE2; 2x2 DE2;
DE 2x2 2 x by a property of square roots.
Thus the hypotenuse is 2 times as long as a leg.
→
36. Construct CD on BC so that CD BC a. Then
ADC ABC by the SAS Cong. Post. B D
and BAC CAD, because they are corresponding
parts of congruent triangles. Therefore mD 60 and
mCAD 30. mBAD mBAC mCAD 30 30 60. BAD is equiangular so it is also equilateral. Since it is equilateral, AB 2a. If BC a and
AB 2a, then AC 2a2 a2 3 a. The side
lengths are in the following ratio: hypotenuse:longer leg:
shorter leg 2a:3 a:a. Therefore, in a 306090
triangle, the hypotenuse is twice as long as the shorter leg
and the longer leg is 3 times as long as the shorter leg.
37. C
190
44. P8, 3
45. A4, 5
47. AA Similarity Postulate
48. SAS Similarity Theorem
49. SSS Similarity Theorem
Math & History
1. area of triangles: 4
34. n 1
35. Let DF x. Then EF x. By the Pythagorean
2
12 ab 2ab
area of square: b a2 b2 2ab a2
2. 2ab b2 2ab a2 c2
a2 b2 c2
Lesson 9.5
9.5 Guided Practice (p. 562)
1. sin A BC
AB
cos A AC
AB
tan A BC
AC
2. The value of a trigonometric ratio depends only on the
measure of the acute angle, not on the particular right
triangle used to compute the value.
3.
4
5
4.
3
5
5.
4
3
6.
3
5
7.
4
5
8.
3
4
38. A; 6 63 12 28.4 cm
Geometry
Chapter 9 Worked-out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.
MCRBG-0904-SK.qxd
5-25-2001
11:38 AM
Page 191
Chapter 9 continued
9. sin 25 7
d
29. sin 23 30. cos 36 9.5 Practice and Applications (pp. 562–565)
45
0.8491
53
28
0.5283
53
cos R 45
tan R 1.6071
28
28
sin S 0.5283
53
45
cos S 0.8491
53
28
tan S 0.6222
45
6
0.6
10
11. sin B 6
tan B 0.75
8
6
0.6
10
cos A 3
12. sin X 13
0.8321
3
1.5
2
tan X tan D tan J sin Y 13
2
0.5547
0.5547
13
24
0.96
25
7
0.2917
24
sin F 24
0.96
25
2
0.8944
2
2
1
cos H 15. sin J 2
cos X cos D 5
tan G 8
1.3333
6
tan A 7
0.28
25
0.8321
7
cos F 0.28
25
14. sin G 8
sin A 0.8
10
2
0.6667
3
13
13. sin D 8
0.8
10
cos B tan Y 3
cos Y 2
5
5
34
cos K 24
tan F 3.4286
7
1
cos G tan H 0.8575
cos J sin K 0.4472
5
1
sin H 0.8944
5
1.6667
3
5
0.4472
1
0.5
2
3
34
0.5145
3
0.5145
34
3
5
0.8575 tan K 0.6
34
5
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. tan 37 6
y
y 8.0
Copyright © McDougal Littell Inc.
All rights reserved.
cos 23 t 13.3
d 16.6, or about 17 ft
10. sin R t
34
sin 37 6
x
s 31.3
4
r
tan 36 r 4.9
31. sin 65 s
4
s 2.9
t
8
cos 65 t 7.3
32. sin 70 s
34
u
8
u 3.4
9
v
tan 70 v 9.6
9
w
w 3.3
6
33. sin 22 x
tan 22 6
y
y 14.9
x 16.0
1
2
34. A 2 22 22 4 cm
35. A 1
126.9 41.6 m2
2
1
2
36. A 116
37. tan 13 3 34.9 m2
h
58.2
38. tan 42 h 13.4 m
d
40
d 36.0 m
39. vertical drop, x 5500 5018 482 ft
sin 20 482
d
1409.3 ft
40. tan 55 30
26 x
30
x
26
sin 45
x 16.4 in.
d
500
41. sin 45 d 714.1 m
42. tan 20 x
8
x 2.9 ft
43. sin A a
c
cos A b
c
tan A a
b
sin B b
c
cos B a
c
tan B b
a
44. The tangent of one acute angle of a right triangle is the
reciprocal of the tangent of the other acute angle. The
sine of one acute angle of a right triangle is the same as
the cosine of the other acute angle and the cosine of one
acute angle of a right triangle is the same as the sine of
the other acute angle.
x 10.0
Geometry
Chapter 9 Worked-out Solution Key
191
MCRBG-0904-SK.qxd
5-25-2001
11:38 AM
Page 192
Chapter 9 continued
45. Procedures may vary. One method is to reason that since
52. Statements
the tangent ratio is equal to the ratio of the lengths of the
legs, the tangent is equal to 1 when the legs are equal in
length, that is, when the triangle is a 454590
triangle. Tan A > 1 when mA > 45, and tan A < 1 when
mA < 45, since increasing the measure of A increases the length of the opposite leg and decreasing the
measure of A decreases the length of the opposite leg.
1. ABC is a right
b
c
x
55
x
sin 30 18
9
sin 55 BC
x9
47. Reasons
55. tan 53 x
60
2
9.5 Mixed Review (p. 566)
2
2 cos 45 sin 45°2 cos 45°2 50. sin 60 3
2
3
2
enlargement;
56.
Q
51. sin 13 0.2250
2
Q
2
2
3
2
10
P
5
P 4
8
6
3
2
QR 10
PR 8
R
R
2 2
1
4 4
57. MNP ~ NQP ~ MQN
1
2
2
2 3
1
2
2
3 1
1
4 4
cos 13 0.9744
sin 13°2 cos 13°2 0.22502 0.97442
7
QP
15
7
18.27
NP
NP
3.27
49 15QP
NP2 59.7429
QP 3.27
58.
x2 952 1932
Geometry
x2 502 652
x2 2500 4225
28,224
x2 1725
x2
x 168
yes
Chapter 9 Worked-out Solution Key
NP 7.73
59.
x2 9025 37,249
1
192
scale factor 6
22 22
cos 60 sin 60°2 cos 60°2 y 33.26
h 46 ft
1 3
1
4 4
2
y
60
xyh
3
1
sin 30° cos 30° 2
49. sin 45 tan 29 79.62 33.26 h
5. Substitution property of equality
2
54. C
x 79.62
3. Division property of equality
2
equality
8
;D
CD
2. Pythagorean Theorem
2
dividing and simplifying
fractions
5. Transitive property of
53. sin 25 1. Given
4. Subst. prop of equality and
sin A
cos A
5. tan A C
BC 11.0
cos 30 a
3. Def. of sine and cosine
c
a
sin A
c
a
4.
cos A
b
b
c
30
1
2
2. Def. of tangent
3. cos A ; sin A B
48. sin 30 a
b
2. tan A trigonometric ratios.
A
1. Given
triangle with side
lengths a, b, and
hypotenuse c.
46. ABC is not a right triangle, so you cannot use the
18
Reasons
x 569
no
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11:38 AM
Page 193
Chapter 9 continued
42.92 702 x2
60.
Lesson 9.6
1840.41 4900 x2
Activity 9.6 Developing Concepts (p. 567)
6740.41 x2
1.
B
82.1 x
no
5 cm
3 cm
Quiz 2 (p. 566)
x 23 m
1.
4m
A
3.5 m
4m
cos A x 42 in.
4 in.
1. To solve a right triangle is to find the measures of all
angles and the lengths of all sides of the triangle.
4 in.
2. true
4. mA 35.0
3. false
6. mA 64.2
4 in.
1
h 1.53 in. A 31.53
2
3.
3 in.
3.9 in.2
3 in.
h
sin A 10
x
x 15.6
x
18
x 8.5
x
20
x 9.3
7. tan 11 950
d
d 4887.3 ft
Copyright © McDougal Littell Inc.
All rights reserved.
56
65
sin B mA 59.5
4225 c2
33
65
mB 30.5
65 c
sin D 9. 912 d2 1092
60
109
sin E mD 33.4
d2 3600
1.5 in.
3 in.
5. mA 79.5
7. mA 84.3
8. 332 562 c2
6. tan 25 3
0.75
4
9.6 Guided Practice (p. 570)
5.7 in.
x
5. cos 62 tan A 4. The values are approximately equal.
4m
4. sin 40 4
0.8
5
3. sin1 0.6 36.9 cos1 0.8 36.9 tan1 0.75 36.9
2m
4 in.
3
0.6
5
2. sin A x
2.
C
4 cm
91
109
mE 56.6
d 60
tan 40 10
y
y 11.9
sin 62 y
18
y 15.9
cos 25 20
y
y 22.1
10. sin 60 y
4
cos 60 y 3.5
x
4
m∠ X 30°
x2
9.6 Practice and Applications (pp. 570–572)
55
73
mQ 48.9
12. sin Q 11. 482 552 QS2
5329 QS2
73 QS
13. sin S 48
73
mS 41.1
14. mA 26.6
15. mA 45
17. mA 20.5
18. mA 81.4
20. mA 65.6
21. mA 6.3
16. mA 30
19. mA 50.2
Geometry
Chapter 9 Worked-out Solution Key
193
MCRBG-0904-SK.qxd
5-25-2001
11:38 AM
Page 194
Chapter 9 continued
22. 202 212 AB2
841 AB2
sin B 20
29
sin A 21
29
mB 43.6
mA 46.4
7
sin E 98
7
sin D 98
mE 45
mD 45
32. tan 51 98 DE2
40 GH
2
sin H 40
mG 71.6
mH 18.4
mF 39
33. sin 34 4
m
tan 34 8
9.2
sin L ML2 20.64
mL 60.4
cos K 5.9
mL 56
8
9.2
mK 29.6
69
36
35. tan1B 1.9167
36. 692 362 AB2
ML 4.5
26. 42 NQ2 13.62
NQ2 168.96
4
cos P 13.6
mN 17.1
mP 72.9
37.
38. tan x 4855
17,625
x 15.4
sin S TS2 120.25
6
12.5
mS 28.7
cos R 6
12.5
39.
2402 172 x2
57,311 x2
sin y TS 11.0
p
4.5
p 4.0
sin A 0.4626
x 239.4 in. or
19 ft 11 in.
mR 61.3
cos 26 36
BC
AB
6057
sin A 77.8 in. AB
NQ 13.0
27. 62 TS2 12.52
69
36
mB 62.4
6057 AB2
4
sin N 13.6
4
mL 90 34
34. tan B 25. 82 ML2 9.22
q
4.5
q 2.0
e 4.8
m 7.2
6
sin G 40
6.3 GH
28. sin 26 3
e
mF 90 51
9.9 DE
24. 22 62 GH2
cos 51 d 3.7
29 AB
23. 72 72 DE2
d
3
17
240
y 4.1
40.
55.2 in.
8.33°
54.6 in.
8 in.
mP 90 26
41. Answers may vary.
mP 64
29. sin 20 s
12
cos 20 t
12
t 11.3
s 4.1
cos 52 8.5
z
z 13.8
x 10.9
mY 90 52
mY 38
31. cos 56 5
c
x 32.5°
tan x mT 70
x
8.5
7
11
43. tan x 8.25
9
x 42.5°
44. Sample answer: riser length: 6 in.; tread length: 12 in.
mT 90 20
30. tan 52 42. tan x tan 56 c 8.9
a
5
6
12
x 26.6°
45. Sample answer: The riser-to-tread ratio affects the safety
of the stairway in several ways. First, the deeper the tread
the more of a person’s foot can fit on the step. This
makes a person less likely to fall. Also, the smaller the
angle of inclination the less steep the stairway. This
makes the stairs less tiring to climb, and therefore, safer.
a 7.4
mB 90 56 34
194
Geometry
Chapter 9 Worked-out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.
MCRBG-0905-SK.qxd
5-25-2001
11:38 AM
Page 195
Chapter 9 continued
46. Draw an altitude, CD , from C to AB , and let CD h. In
h
h
. In rt. BCD, sin B . Thus,
b
a
h b sin A and h a sin B. By the substitution
prop. of equality, b sin A a sin B. Dividing both
a
b
sides by sin A sin B gives
, or
sin B sin A
a
b
.
sin A sin B
8. tan x rt. ACD, sin A x 51.3 north of east
9. 4, 5 4, 2 0, 3
9.7 Practice and Applications (pp. 576–579)
\
\
\
12.
47. 3, 2
48. 2, 2
49. 1, 3
50. 1, 0
51. 1, 2
52. 3, 1
y
2, 7
Q
PQ 2 02 7 02
\
53
7.3
6x 150
7y 784
x 25
y 112
3
g
10 42
56.
1
14.
y
x
1
P
84
7
18
k
10g 126
3, 5
Q
PQ 5 22 1 62
\
7k 1512
g 12.6
57.
13.
7
49
54.
16
y
x
5
53.
30 6
55.
3, 6 JK 2 12 2 42 35 6.7
4, 4 EF 0 42 1 52 42 5.7
10. 4, 1 RS 1 52 1 22 17 4.1
11.
9.6 Mixed Review (p. 572)
5
4
34
k 216
m 7
2
1
4
8
t
11
58.
x
1
88 4t
m 14
5.8
P
1
t 22
59. not a triangle
15.
60. 2282 ? 2202 602
61. 8.52 ? 7.72 3.62
51,984 < 52,000
72.25 72.25
acute
62.
2632
?
802
63.
1132
?
1122
69,169 > 68,900
12,769 12,769
obtuse
right
PQ 3 72 2 62
\
Q
right
2502
10, 4
y
P
152
64. not a triangle
116
10.8
2
x
2
16.
6, 4
y
1
Lesson 9.7
1
x
PQ 4 22 3 72
\
52
P
7.2
9.7 Guided Practice (p. 576)
1. The magnitude of a vector is the distance from its initial
point to its terminal point. The direction of a vector is the
angle it makes with a horizontal line.
\
\
\
Q
\
2. AB : 2, 2; PQ : 3, 3; MN : 0, 3; UV : 0, 2
\
\
\
\
3. UV is parallel to MN ; AB is parallel to PQ
4. 2, 2
4, 2 PQ 5 12 4 22 25 4.5
2, 5 MN 3 12 4 12 29 5.4
\
5. 4, 5 AB 4 02 5 02 41 6.4
\
6.
\
7.
Copyright © McDougal Littell Inc.
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Geometry
Chapter 9 Worked-out Solution Key
195
MCRBG-0905-SK.qxd
5-25-2001
11:38 AM
Page 196
Chapter 9 continued
17.
23. 10, 10 and 50, 50
6, 4
y
1
LM 10 502 10 502
\
P
x
1
PQ 5 12 0 42
\
52
57 miles per hour
7.2
tan x Q
40
40
x 45 north of west
24. 0, 0 and 50, 40
OP 0 502 0 402
\
18.
8, 2
y
PQ 6 2
\
P
Q
2
64 miles per hour
3 1
2
tan x 68
1
x 38.7 south of west
8.2
x
1
40
50
\
\
\
\
25. EF , CD , and AB
\
\
26. EF and CD
\
\
27. EF and CD
\
28. GH and JK
29. yes; no
19.
1, 4
y
P
1
1
x
30. Round 2; the vectors have the same magnitude and
PQ 6 52 0 42
\
17
4.1
Q
\
opposite directions. In Round 1, team A won; since CA
has a greater magnitude than CB , CA represents a greater
force applied.
\
\
\
31.
u : 4, 1
y
\
v : 2, 4
\
v
20.
u
1
3, 0
y
\
u v 6, 5
uv
x
1
\
P
PQ 0 32 5 52
Q
9
\
32.
u : 6, 2
y
\
3
v : 5, 3
5
1
\
\
u v 1, 5
v
x
1
uv
u
1
x
21. 0, 20 and 60, 30
ST 60 02 30 202
\
\
33.
61 miles per hour
u : 2, 4
y
uv
1
\
v : 3, 6
x
1
10
tan x 60
\
\
u v 5, 2
v
u
x 9.5 north of east
22. 0, 0 and 40, 50
AB 40 02 50 02
\
\
34.
u : 2, 3
y
64 miles per hour
tan x 50
40
x 51.3 south of east
196
Geometry
Chapter 9 Worked-out Solution Key
\
v : 1, 6
u
\
1
\
u v 3, 3
v
uv
1
x
Copyright © McDougal Littell Inc.
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MCRBG-0905-SK.qxd
5-25-2001
11:38 AM
Page 197
Chapter 9 continued
35. 4, 11
36. 8, 7
38. 2, 3
37. 10, 10
39. 4, 4
40. 0, 0
\
41. u : 0, 120
\
of u and the directions are the same. When k < 0, the
magnitude of v is k times the magnitude of u and the
direction of v is opposite the direction of u . Justifications
may vary.
\
\
\
v : 40, 0
42.
\
47. When k > 0, the magnitude of v is k times the magnitude
48. a.
UP
\
\
N
s
2
W
u
2
v
u
E
s
S
\
b. s 10 02 2 02 10.2 mih
tan x v
10
W
E
10
2
10
x 11.3 north of east
c. Answers may vary.
DOWN
\
49. AB : 54, 24
\
\
43. s 20 602 140 202 126.5 mih
the speed at which the skydiver is falling, taking into
account the breeze.
44. tan x \
\
AB BC 18, 60
\
50. CA : 18, 60
18, 60 18, 60 0, 0
120
40
BC 362 362 50.9 ft
CA 182 602 62.6 ft
\
51. AB 542 242 59.1 ft
x 71.6
45.
BC : 36, 36
\
The new velocity is
30, 120.
UP
\
Total distance 59.1 50.9 62.6 172.6 ft
52. The answer to Ex. 50 is a vector which gives the final
position of the bumper car, while the answer to Ex. 51 is
a number which gives the total distance traveled by the
bumper car.
53. Since D and E are right angles and all right angles
10
W
E
10
DOWN
\
46. JK 10
\
Sample answer: AB 3, 1
The component form must give the same magnitude as
JK . AB 10.
\
\
are congruent, D E . Since ABC is equilateral,
AB BC. DE AC, so DBA BAC and
EBC BCA by the Alternate Interior Angles
Theorem. An equilateral triangle is also equiangular, so
mBAC mBCA 60. By the definition of congruent angles and the substitution property of equality,
DBA EBC . ADB CEB by the AAS
Congruence Theorem. Corresponding parts of congruent
triangles are congruent, so DB EB. By the definition of
midpoint, B is the midpoint of DE.
54. x 45
55. x 120
56. x 30
y 90
y 30
2
2
x
1
x
2x
1
57.
2
2
58. x 7 x 14x 49
y 60
2
2
59. x 11 x 22x 121
60. 7 x2 49 14x x2
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Geometry
Chapter 9 Worked-out Solution Key
197
MCRBG-0905-SK.qxd
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11:38 AM
Page 198
Chapter 9 continued
Quiz 3 (p. 580)
1. sin 25 \
9.
b
46
cos 25 b 19.4
Q
5.8
a 41.7
12
z
1
12
y
\
10.
y 12
z 17.0
x
1
P
tan 45 y
mY 45
3. tan 40 PQ 3 02 4 12
\
a
46
mA 90 25 65
2. sin 45 PQ : 3, 5
y
PQ : 7, 11
P
PQ 2 52 6 52
\
4
m
16
cos 40 m 13.4
13.0
16
q
2
q 20.9
x
Q
mN 90 40 50
4. sin 75 p
8
cos 75 p 7.7
q
8
11.
T
6
7.6
2
f 7.6
62
2
2
2
S
f 4.7
mF 90 52.1 37.9
3
12.4
32 2 12.42
mK 14.0
12. 4, 2
13. 2, 4
15. 2, 1
16. 6, 13
144.76
y
5
y
9
6 9
6
1. x
\
PQ : 5, 1
PQ 3 2
\
P
2
36 9x
4 3
2
5.1
Q
2.
1
x
1
\
8.
PQ : 6, 5
y
PQ 2 4
\
P
2
2
3.
2 3
7.8
2
17. 0, 3
9.1 Similar Right Triangles
mL 90 14.0 76.0
y
14. 2, 8
Chapter 9 Review (pp. 582–584)
2
12.0
7.
x
f 2 21.76
mG 52.1
6. sin K 8
3
x 69.4 north of east
q 2.1
mQ 90 75 15
5. sin G tan x y
x
2
y2 45
x4
y 35
25
x
x
9
x2 225
y
9
16
y
y2 144
x 15
y 12
36
x
27 36
27x 1296
48 27 y
y 21
z
21
27
z
z2 567
z 97
x 48
23.8
Q
9.2 The Pythagorean Theorem
4. 122 162 t2
5. 82 s2 122
400 t2
s2 80
s 45 8.9
20 t
yes
198
Geometry
Chapter 9 Worked-out Solution Key
no
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MCRBG-090R-SK.qxd
5-25-2001
11:38 AM
Page 199
Chapter 9 continued
6. r2 162 342
7. 42 62 t2
r2 900
52 t2
r 30
18. sin B 42
0.6285
9
cos B 7
0.7778
9
tan B 42
0.8081
7
sin A 7
0.7778
9
cos A 42
0.6285
9
tan A 7
1.2374
42
t 213 7.2
yes
no
9.3 The Converse of the Pythagorean Theorem
8. 102 ? 62 72
9. 412 ? 402 92
100 > 85
1681 1681
obtuse
right
9.6 Solving Right Triangles
10. not a triangle
11. 9
2
x2 80
? 3 45 2
2
sin Z mX 48.2
8
12
mZ 41.8
8.9
acute
d
20
20. sin 50 9.4 Special Right Triangles
21.
6
32
13. leg 2
P 432 14. shorter leg 8
15
tan R 15
17
mT 61.9
\
22.
PQ : 1, 4
y
PQ 1 16 17 4.1
\
P
1
x
2
Q
9.5 Trigonometric Ratios
cos J 60
0.9836
61
\
23.
11
0.1803
61
tan L 60
5.4545
11
17. sin P 35
0.9459
37
cos P 35
0.3243
37
35
tan P 2.9167
12
12
0.3243
sin N 37
35
cos N 0.9459
27
12
tan N 0.3429
35
PQ : 12, 5
y
PQ 144 25 13
\
60
0.9836
sin L 61
cos L Copyright © McDougal Littell Inc.
All rights reserved.
sin T 9.7 Vectors
1
18 93 813 140.3 cm2
2
11
tan J 0.1833
60
289 s2
40
17 s
15. altitude 93 cm
11
0.1803
61
mF 90 50
82 152 s2
18 in.2
1
12 6 in.; longer leg 63 in.
2
f
20
f 12.9
mR 28.1
A 32 2
122 in.
cos 50 d 15.3
12. hypotenuse 232 6
16. sin J 8
12
x 45
81 < 89
A
cos X 19. 82 x2 122
P
2
2
x
Q
\
24.
PQ : 3, 2
y
PQ 9 4 13 3.6
\
2
P
Q
1
x
Geometry
Chapter 9 Worked-out Solution Key
199
MCRBG-090R-SK.qxd
5-25-2001
11:38 AM
Page 200
Chapter 9 continued
\
\
\
25. u v 14, 9
14.
y
u v 196 81 277 16.6
\
LM 16 9 5
\
tan x \
9
14
M
x 32.7 north of east
1. E
2. A
3. C
4. D
x
5. B
15. sin 40 quadrilateral has two pairs of consecutive congruent
sides, but opposite sides are not congruent.
8. PQ 25 4 29
?
sin 35 b2 12,544
PR 4 36 210
b 112
29 2 52
sin 50 10
AB
AB 13.1
16. mBCA 90 35 55
9. 152 b2 1132
QR 9 16 5
CD
10
CD 6.4
7. WXYZ is a kite. The diagonals are perpendicular and the
3
4
x 36.9 south of east
1
6. DBA; DAC
210 2
tan x 1
Chapter 9 Chapter Test (p. 585)
BC
40
tan 35 BC 22.9
17. 2, 8
22.9
DE
DE 32.7
18. 4, 1
19. 2, 3
Chapter 9 Standardized Test (pp. 586–587)
40 < 54
acute
5
x1
20
x1
1.
10.
1
A d1d2
2
6 in.
3 in.
3 3 in.
x 12 100
x2 2x 99 0
1
663 2
6 in.
6 in.
x 11x 9 0
183
30
60
6 in.
x 11
31.2 in.2
KL
11. sin 30 9
JL
cos 30 9
KL 4.5
JL 7.8
12
DF
tan 25 DF 28.4
4. B
QR 25
4
6
mP 48.2
C
D
5. D
P 482 322 in.
2x 256
2
x 128 82 in.
12
DE
7. tan 67 DE 25.7
cos P C
6. x2 x2 162
x
8
D
cos 67 x 18.8
mF 90 25 65
13. 42 QR2 62
154 in.2
3. P 2125 214 50.4 in.
mK 90 30 60
12. sin 25 x9
2. A 1114 side length 6 in.
sin R 4
6
Geometry
Chapter 9 Worked-out Solution Key
8
y
y 20.5
E
8. tan x mR 41.8
4.5
200
LM : 4, 3
L
12
9
x 53.1
A
9.
sin A 8
13
mA 38.0
B
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5-25-2001
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Page 201
Chapter 9 continued
10. 2 x 6
y 4 11
x8
5. BD is the median from point B, AD CD, BD BD,
y7
B
\
11. AB 8 12 3 92 15
12. 24 2
x2
D
252
6. Sample answer: Given quadrilateral ABCD
x7
p 7 24 25 56
13. sin x 7
25
sin y x 16.3
14. A 25
and it is given that AB CB. Thus, ABD CBD by
the SSS Congruence Postulate. Also, ABD CBD
since corresponding parts of congruent triangles are
congruent. By the definition of an angle bisector, BD
bisects ABC.
24
25
y 73.7
12 2724
216 square units
1
15. BD 103 3 30
BC 302 42.4
FG 15
GC 450 21.2
DC 30
AF 103 15 32.3
where mA 37, mB 143, and mC 37.
Since there are 360 in a quadrilateral,
mD 360° 37° 143° 37° 143.
A and C are opposite angles, and B and D
are opposite angles. A C and B D.
Since both pairs of opposite angles are congruent,
quadrilateral ABCD is a parallelogram.
7. Yes; clockwise and counterclockwise rotational symmetry
of 120.
8. 26 4y 90
26 x 10 90
4y 64
x 54
16. mABC 75; mFEA 60; mBGF 135
FE
17. tan 30 32.3
32.3
cos 30 AE
FE 18.6
AE 37.3
1
18. A 3032.3 15 709.5 square units
2
5
b
19. a. tan 30 b. tan 40 b 8.7 cm
b 6.0 cm
5
d. tan 60 b
5
c. tan 50 b
b 2.9 cm
b 4.2 cm
e. tan 70 5
b
5
b
20. As the sun rises, the value of b decreases.
5
5.25
x 43.6
y 324 5 180
8x 192
y 67 180
x 24
y 113
10. z 55; x 90; y 65
11. m 8
4
62
5 1
6
3
slope of perpendicular bisector midpoint:
3
4
52 1, 6 2 2
2, 2
3
6
y2 x
4
4
12. AB 16 9 5
approaches the horizon.
AC 36 64 10
BC 4 121
Chapter 9 Cumulative Practice (pp. 588–589)
1. No; if two planes intersect, then their intersection is a
line. The three points must be collinear, so they cannot be
the vertices of a triangle.
3. never
y 3x 5 180
8x 12 180
3
7
y x
4
2
22. The value of the expression increases as the sun
2. always
9. 3x 5 5x 7 180
3
y 2 x 2
4
b 1.8 cm
21. tan x y 16
125 55
scalene right triangle
13. A1, 2, B3, 5, C5, 6
4. always
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Geometry
Chapter 9 Worked-out Solution Key
201
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Page 202
Chapter 9 continued
14.
A2, 1
y
C
24.
B5, 3
C6, 5
25.
x
1
23 2 22 ZX2
ZP2 12
16 ZX2
ZP 23
A
1
2
ZP
6
ZP
23
XY
A
B
C
B
4 ZX
3
23
26. 102 242 XY2
12 3XY
676 XY2
4 XY
26 XY
27. 192 ? 152 122
15. A3, 6, B7, 9, C9, 2
12 5
16.
x
2
7x 24
4
x4
5
3
x3
7
26
3
29. Let x the measure of the smaller acute angle.
2y 21
19. No; in ABCD, the ratio of the length to width is 8:6 or
sin x 8
17
tan x 8
15
30. sin 57 4:3. In APQD, the ratio of length to width is 4:6, or 2:3.
Since these ratios are not equal, the rectangles are not
similar.
cos x ST
20
8
12
2
21. Yes; the ratios 9 , 12 , and 18 all equal 3 , so the triangles are
ST 16.8 in.
x
7
9x 8
31.
TR 10.9 in.
8x 63 7x
u v 4 256 16.1
v
tan x \
\
2
16
u
x
2
32. Construct a circle inscribed in the triangle by bisecting
two 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.
1
4
has endpoints 2, 3
3
13
13
and 4, 3; its slope is 3 . The image with scale
2
6
1
factor has endpoints 3, 2 and 6, 4.5; its slope is
2
13
. The two image segments are parallel.
6
23. The image with scale factor
uv
x 7.1 south of east
segment 2 4.8 cm
x 4.2
\
u v 2, 16
y
2
segment 1 4.2 cm
15x 63
TR
20
mS 90 57 33
similar by the SSS Similarity Theorem.
22.
15
17
cos 57 \
DE is a midsegment of
ABC. By the Midsegment
Theorem, DE AC. Since
D
E
the lines containing these
segments are parallel,
A
C
BDE BAC and
BED BCA, by the Corresponding Angles
Postulate. Since two angles of BDE are congruent to
two angles of BAC, the two triangles are similar by
the AA Similarity Postulate.
B
6
3
3
acute
1
y 10
2
20.
82 22
3
43
361 < 369
7
y
18.
9 y3
9y 7y 21
3
x
17.
7 8
24 5x
28.
376 470
16
x
34. P 3 3 5 5 16
3
16
36 w
376x 7520
x 20 gallons
16w 108
w 6.75 in.
36 13.5 2
11.25 in.
35. d 127.5 140
2
2
2
d2 35,856.25 mi
d 189.4 mi
202
Geometry
Chapter 9 Worked-out Solution Key
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MCRBG-090R-SK.qxd
5-25-2001
11:38 AM
Page 203
Chapter 9 continued
Project: Investigating Fractals (pp. 590–591)
Investigation
1. stage 1:
4
3
2. stage 3:
64
27
; stage 2:
16
9
4
256
; stage 4: 81 ; the length at stage 1 is 3 times
the length at stage 0, and similarly the length at stage 2 is
4
3 times the length at stage 1.
3. At each stage the length would be
4
3
times the previous
stage, so the length gets increasingly large.
4. Yes; the graph is a curve that increases sharply as n, the
number of stages, increases
Stages of a Koch Snowflake
5.
Stage 0
6.
Stage 1
Stage, n
Perimeter, P
Stage 2
0
3
1
2
3
4
4
16
3
64
9
256
27
7. P 3 3
4 n
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Geometry
Chapter 9 Worked-out Solution Key
203