Solution Manual for Basic Technical Mathematics 10th Edition by Washington

ISBN 0133083500 9780133083507

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0133083500-9780133083507/

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0133083500-9780133083507/

Chapter 2

GEOMETRY
° ° ° °
2.1 Lines and Angles 15. ∠ BOD = 180 − 90 − 50 =40

1. ∠ ABE = 90° 16. ∠ 3 = 180° − 35° =145°

° ° °
2. ∠ QOR = 90° − 32° = 58° 17. ∠1 = 180 −145 = 35 = ∠ 2 = ∠ 4

° ° °
3. 4; ∠ BOC and ∠ COA, 18. ∠ 5 = ∠ 3 =145 19. ∠ 1 = 62
∠ COA and ∠ AOD,

° °
20. ∠ 2 + ∠ 1 = 180 ⇒ ∠ 2 = 180 − ∠ 1
∠ AOD and ∠ DOB, ° ° °

= 180 − 62 = 118
and, ∠ DOB and ∠ BOC.
° ° °
21. ∠ 3 = 90 −62 = 28
7.75 x
4. (a) =
5.65 6.50 ° °
22. ∠ 1 + ∠ 3 = 90 ∠ 3 + ∠ 4 = 180
7.75( 6.50) ° ° °

x= ∠ 3 = 90 −62 ∠ 4 = 180
5.65 ° °
∠ 3 =28 = 180
x = 8.92 ft
°
(b) More vertical ∠ 4 =152

° ° °
5. ∠ EBD and ∠ DBC are acute angles. 23. ∠ BDE = 90 − 44 = 46
° ° °
∠ BDF = 180 − 46 = 134
6. ∠ ABE and ∠ CBE are right angles.
° ° °
24. ∠ BDE = 90 − 44 = 46
° ° °
7. ∠ ABC is a straight angle. ∠ ABE = 90 + 46 = 136

°
8. ∠ ABD is an obtuse angle. 25. ∠ DEB = 44
 ° °
9. The complement of ∠ CBD = 65 is 25 . 26. ∠ DBE = 46°
 ° 27. (DFE = 90 ° − (FDE
10. The supplement of ∠ CBD = 65 is

= 90 ° − 44°
180° −65° = 115°.
°
= 46
11. Sides BD and BC are adjacent to < DBC
°
28. (ADE = (ADB + 90
= (90 − 44 ) + 90°
° °
12. The angle adjacent to ∠ DBC is ∠ DBE
°
= 136
° ° °
13. ∠ AOB = 90 + 50 = 140
29. a 3.05 3.05
° ° ° = ⇒ a = 4.75⋅ = 4.53 m
14. ∠ AOC = 90 − 50 =40 4.75 3.20 3.20
3Section 2.1 Lines and Angles Chapter 2 GEOMETR4Y5

3.20 6.25 2 3
30. = 43. = ⇒ AB = 3.225

3.05 b 2.15 AB

b = 5.96 m AC = 2.15 + 3.23 = 5.38 cm

c 5.50 5.50 x 590

31. = = 44. =

3.05 a 4.53 860 550

4.53c = 15.4025 x = 920 m
c = 3.40 m °

45. (1 + (2 + (3 = 180 ,
32.
4.75
=
6.25 ( (1, (2, and (3 form a straight line )
5.05 d

d = 6.64 m °
46. (4 + (2 + (5 = 180 , ( (1 = (4, (3 = (5 )
°
33. (BHC = (CGD = 25
°
47. The sum of the angles of ABD is 180 .
°
34. (AHC = (CGE = 45
° 48.
35. (BCH = (CHG = (HGC = (GCD = 65
B C

° 1 3 7
2

36. (HAB = (JHA = (FGE = (DEG = 70

6
4 5 7
°
(GHA = (HGE = 110 A D
37.

° °
38. (CGF = (CHJ = 115 (1 + (2 + (3 = 180 , (1 = (4
°
(4 + (2 + (3 = 180
°
° ° (5 + (6 + (7 = 180
39. (A = ( x +10 ) , (B = ( 4x − 5 ) ° °

(a) x +10 = 4x − 5 (4 + ( (2 + (3 ) + ( (5 + (6 ) + (7 = 180 + 180

°

x=5
° = 360
°
(b) x+10 +4x−5 =180 The sum of the angles of ABCD = 360
°
x = 35

° °
40. (A = ( x + 20 ) , (B = ( 3x − 2 )
°
(a) x + 20 + 3x − 2 = 90
°
x = 18
(b) x + 20 = 3x − 2
°
x = 11
° °
41. ∠ BCD = 180 − 47
°
= 133
4Section 2.1 Lines and Angles Chapter 2 GEOMETR4Y5

° °
42. ? = 90 − 28
5Section 2.1 Lines and Angles Chapter 2 GEOMETR4Y5

°
= 62
6Section 2.1 Lines and Angles Chapter 2 GEOMETR4Y5

12. p = 0.862 + 0.235 + 0.684 = 1.781 in.

2.2 Triangles 1.781
s= = 0.8905
° ° 2
1. ∠ 5 = 45 ⇒ ∠ 3 = 45
° ° ° ° A = s ( s −a )( s −b )( s −c )
∠ 2 =180 −70 −45 = 65

= .8905(.8905 − .862)(.8905 − .235)(.8905 − .684)

1 1
2. A = bh = (61.2 )(5.75 ) A = 0.586 in.2
2 2

A = 176 in.2 1 1
13. A = bh = (3.46)(2.55) = 4.41 ft2

2 2
3. AC 2 = AB2 + BC2
= 6.252 + 3.22 1 1
14. A = bh = (234)(342) = 40,000 mm2

AC = 6.252 + 3.22 2 2

AC = 7.02 m
15. Area

h 24 = 1.428(1.428 − 0.9860)(1.428 − 0.986)

4. =
3.0 4.0
h = 18 ft (1.428 − 0.884)

= 0.390 m2
° ° ° °
5. ∠ A =180 −84 − 40 = 56
3(320)
s= = 480
° ° °
16. 2
6. (A = 90 − 48 = 42

3
A = s ( s − a)
7. This is an isosceles triangle, so the base angles
3
are equal. ∠ A = 180 − (66 + 66
° ° °
) = 48° = 480 ( 480 − 320 )

= 44,000 yd2

1
( ° ° °

8. ∠ A =
2
180 −110 ) = 35 17. p = 205+ 322 + 415
p = 942 cm
1 1
9.
A= bh = (7.6)(2.2) = 8.4 ft2
18. p = 23.5 + 86.2 + 68.4

2 2
p = 178 in.
1 1
10. A = bh = (16.0 )(7.62 ) = 61.0 mm2
7Section 2.1 Lines and Angles Chapter 2 GEOMETR4Y5

2 2 19. 3(21.5) =64.5cm

11. Area = 471 (471−205 )(471−415 )(471−322 ) 20. Perimeter = 2 ( 2.45 ) + 3.22 = 8.12 in.

= 32,300 cm2
21. c = 13.82 + 22.72 = 26.6 ft

22. c2 = a2 + b2
8Section 2.1 Lines and Angles Chapter 2 GEOMETR4Y5

= 2.482 +1.452 c = 2.87 m

23. b = 5512 −1752 = 522 cm
S7ection
4 2.2 Triangles Chapter 2 GEOMETR4Y7

24. c2 = a 2 + b 2 31. An equilateral triangle.

0.8362 = a2 + 0.4742
32. Yes, if one of the angles of the triangle is obtuse.
a = 0.689 in.

25. ∠ B = 90° −23° = 67° 33.

B
D
2 2 2
26. c = 90.5 + 38.4
c = 98.3 cm 1
27. Perimeter = 98.3+ 90.5 +38.4 = 227.2 cm A 2
C
1
°
28. A = (90.5)(38.4) = 1740 cm2 (A + (B = 90
2 (1 + (B = 90
°

⇒ (A = (1
29.
B
redraw ΔBDC as
B/2
B
D 2
A' A/2 C'
C 1 D
A/2 C °
A (1 + (2 = 90
°
(1 + (B = 90
ΔADC ∼ ΔA ' DC ' ⇒ (DA 'C ' = A / 2 ⇒ (2 = (B
( between bisectors = (BA ' D
and ΔADC as
B °
ΔBA 'C ', + ( (BA ' D + A / 2 ) = 90 C
2 2

°
−⎛ A B⎞
from which (BA ' D = 90 ⎜ +
⎝ 2 2⎠
1 D
D A
⎛ A +B ⎞ 90
D
°
or (BA ' D = 90 − ⎜ ⎟ = 90 − = 45D ΔBDC and ΔADC are similar.
2
⎝ 2 ⎠

30. 34. Comparing the original triangle

F B
2

B C

1
A C
A D
E
(A = (D since ΔAFD is isosceles. Since AF = FD to the two smaller triangles shows that all three
(ΔAFD is isosceles) and since B and C are mid- are similar.
points, AB = CD which means ΔBAE and ΔCED
35. ∠ LMK and ∠ OMN are vertical angles and thus
are the same size and shape. Therefore, BE = EC
equal ⇒ ∠ KLM = ∠ MON . The corresponding
from which it follows that the inner ΔBCE is
angles are equal and the triangles are similar.
isosceles.
S8ection
4 2.2 Triangles Chapter 2 GEOMETR4Y7

°
36. ∠ ACB = ∠ ADC = 90 ; ∠ A = ∠A; ∠ DCA = ∠ CBA, (18.0
2
− y ) = y2 + 8.0 2
therefore +ACB ∼+ADC
18.02 − 2(18.0) y + y2 = y2 + 8.02

37. Since +MKL ∼+MNO; KN = KM − MN; 15 −9 = 6 y = 7.2 ft (two significant digits)

KM LM 6 LM
= KM; = ; = ; 9LM = 72; LM = 8 3(1600)
MN MO 9 12 44. = 2400
2
3
AB 12 A = 2400 ( 2400 −1600 )
38. =

12 9 A = 1,100, 000 km2

AB =16
1 1 2
39. p = 6 + 25+ 29 = 60 45. A = bh = (8.0)(15) = 60 ft
2 2
A = 30(30 − 6)(30 − 25)(30 − 29 ) = 60

Yes, the triangle is perfect. 46. d = 7502 + 5502 = 930 m

sidewalk
80 .0 in. ramp

40. 4.0 in. curb 47.

80 .0 in.
2.50 ft
street 20.0 L
= ⇒street = 4.0 (20.0 )
4.0 1 x

ramp = ( 4.0 ( 20.0 ) ) + 4.0 2 = 80.09993758, 10.0 ft

calculator 6.00 ft

ramp = 80 in. (two significant digits)

10.02 = x2 + 6.002
180° − 50° = 65° x = 8.00
41. ∠=
2 2
L= (8.00 + 2.50 ) + 6.00 2
° ° °
42. angle between tower and wire = 90 −52 = 38 L = 12.09338662, calculator
L = 12.1 ft (three significant digits)
48. Taking the triangles in clockwise order and using
Pythagorean Theorem together with side opposite
°
30 angle is half the hypotenuse gives side opposite
°
30 angle and third side, respectively.
43. tree

49. d = 182 + 122 + 82 = 23 ft

18.0 ft

x 1
50. = , x = 38 m
break 45.6 1.12
18.0 - y
S0ection
5 2.2 Triangles Chapter 2 GEOMETR5Y0

8.0 ft
4S9ection 2.3 Quadrilaterals Chapter 2 GEOMETR4Y9

51. 55. Redraw ΔBCP as

B

6.00

P C
12.0 - PD

4.5 ΔAPD is
= 5.4
A
z 1.2 + z
z = 6.0 m
x2 = z2 + 4.52 10.0
x = 7.5 m
2 P D
y 2 = (1.2 + 6 ) + 5.4 2

y = 9.0 m 6.00 10.0

from which ΔBCP ∼ ΔADP, so =
12.0 − PD PD
52. ⇒ PD = 7.50 and PC = 12.0 − PD = 4.50

l = PB + PA = 4.502 + 6.002 + 7.502 + 10.02

2.5 H d l = 20.0 mi

H 1 1
56. wd + 160 = w( d + 16)
1.25 1.25 2 2
1 1
d 2 = 1.252 + 5.02 wd + 160 = wd + 8w
2 2
d = 5.6 ft
8w = 160
w = 20 cm
53. d = w −12 = 8 cm

d
4.0
2.3 Quadrilaterals
6.0 4.0

d 4.0 4.0 (8.0 ) 1.

= ⇒d =
8.0 6.0 6.0 2
⎛ 4.0 (8.0) ⎞
l 2 = 8.02 + d 2 = 8.02 + ⎜ ⎟
⎝ 6.0 ⎠
l = 9.6 ft

2. L = 4s + 2w + 2l
ED 312
54. = = 4 (21) + 2 (21) + 2 (36)
80 50
ED =499 ft = 198 in.
5S9ection 2.3 Quadrilaterals Chapter 2 GEOMETR4Y9

1 1 1
3. A = bh = ( 72 )(55 ) = 2000 ft2 20. A = (392 + 672 )(201 ) = 107,000 cm2
1
2 2 2
A2 = bh = 72(55) = 4000 ft2
1 1 21. p = 2b +4a
A3 = h ( b1 + b2 ) = ( 55)( 72 + 35)
2 2
= 2900 ft2 22. p = a +b+b+a+ ( b −a) + ( b −a) = 4b A =
The total lawn area is about 8900 ft2 .
2 2
23. b × h + a = bh + a
4. 2 ( w + 3.0) + 2w = 26.4
2
2w+6.0+2w = 26.4 24. A = ab + a(b − a) = 2ab − a
4w = 20.4
w = 5.1 mm 25. The parallelogram is a rectangle.
w+ 3.0 = 8.1 mm
26. The triangles are congruent. Corresponding sides
5. p = 4s = 4(65) = 260 m and angles are equal.

6. p = 4(2.46) = 9.84 ft
27.
7. p = 2 (0.920 ) + 2 (0.742 ) =3.324 in.

s
8. p = 2(142) + 2(126) = 536 cm

9. p = 2l + 2w = 2 (3.7 ) + 2 ( 2.7 ) =12.8 m s

10. p = 2 ( 27.3 ) + 2 (14.2) = 83.0 in.
s2 + s2 = 24.02
11. p = 0.362 + 0.730 + 0.440 + 0.612 = 2.144 ft 2s 2 = 24.02
24.02
s2=
12. p = 272 + 392 + 223 + 672 = 1559 cm 2

2 2 2 A = s2 = 288 cm2
13. A = s = 2.7 = 7.3 mm

2 2
14. A = 15.6 = 243 ft 28.
2 B A
A

15. A = 0.920 (0.742 ) = 0.683 in.2

C

B
16. A = 142 (126) =17,900 cm2 2 A B

2 °
17. A = bh = 3.7 ( 2.5 ) = 9.3 m At top 2(B + 2(A = 180
°
(B + (A = 90
18. 19. A = 27.3 (12.6 ) =344 in.2
 S1ection
5 2.3 Quadrilaterals Chapter 2 GEOMETR5Y1

In triangle (A + (B
A = (1
+ (C =
/ 2) °
180
(29.8) 90
°

(61.2 +
+73.0) (C
= =
2000 °
180
ft2
(
C
=
9
0
°
S2ection
5 2.3 Quadrilaterals Chapter 2 GEOMETR5Y1

29.
23 4 36.
l
1

6 5 w =l - 18

sum of interior angles

p = 2l + 2(l −18) = 180 from which
= (1+ (2 + (3 + (4 + (5 + (6
°
= 180 + 180
° l = 54 in.
° w = l −18 = 54 −18
= 360
w = 36 in.

30. S = 180 (n − 2)
S 37.
(a) n = + 2
180
°
3600
(b) n = °
+2
180
n = 22
w+ 2.5 = 4w −4.7
A = area of left rectangle + area of right rectangle w =2.4 ft
31.
A = ab + ac 4w = 9.6 ft
A = area of entire rectangle
2
A = a ( b + c) which illustrates the distributive 38. A = 1.80 × 3.50 = 6.30 ft
property.

39. A = 2(area of trapezoid − area of window )

⎛1 ⎞
2 = 2 ⎜ ( 28 + 16 ) ⋅ 8 −12 ( 3.5 ) ⎟
32. A = ( a + b )( a + b ) = ( a + b ) ⎝ 2 ⎠

A = ab + ab + a 2 + b 2 = 268 ft2

A = a 2 + 2ab + b 2 which illustrates that the square 1 gal

=
x
ht of trapezoid

of the sum is the square of the first term plus twice 320 ft2 268 ft 2
⎛ 28 −16 ⎞
2
the product of the two terms plus the square of the
x = 0.84 gal = 102 − ⎜ ⎟
second term. ⎝ 2 ⎠

= 8.0 ft
33. The diagonal always divides the rhombus into two
congruent triangles. All outer sides are always l
equal. 40,
w = 70 yd

34. 16 2 +122 = 400 = 20

p 320 outer edge of the walkway:

35. (a) For the courtyard: s = = = 80. For the
4 4 l=
 S3ection
5 2.3 Quadrilaterals Chapter 2 GEOMETR5Y1

1302 − 702
S4ection
5 2.3 Quadrilaterals Chapter 2 GEOMETR5Y1

p = 4( 80 + 6) = 344 m. p = 2l + 2w
(b) A = 862 − 802 = 996 p = 2 1302 − 702 + 2(70)
A = 1000 m2 (2 significant digits) p = 360 yd
S5ection
5 2.3 Quadrilaterals Chapter 2 GEOMETR5Y1

41. 1
A=
2
( 2.27)( 1.86) + s ( s −1.46)( s − d )( s −1.74)
2

h A = 3.04 km
44.
w
w
= 1.60 ⇒ w = 1.60h
h
2
43.32 = h 2 + w2 = h 2 + (1.6h)
h = 22.9 cm
w = 1.60h = 36.7 cm 50 ( 2w ) + 5 ( 2w ) + 5w + 5w = 13, 200
w =110 m
l = 2w = 220 m
42.
30.0
°
45. 360 . A diagonal divides a quadrilateral into two
triangles, and the sum of the interior angles of
30.0 h °
each triangle is 180 .

15.0 60.0
1 ⎛ d2 ⎞ 1 ⎛ d2 ⎞
46. A = d + d
2 1 ⎜2 ⎟ 2 1 ⎜2 ⎟
⎝ ⎠ ⎝ ⎠
h = 30.02 +15.02 1
A= dd
1 2
2
1
A = 6⋅ (30.0 + 60.0) 30.02 +15.02
⋅
2
A = 9060 in.2
2.4 Circles
°
1. ∠ OAB + OBA + ∠ AOB = 180
43.
° ° °
1.74 ∠ OAB + 90 + 72 = 180
°
∠ OAB = 18
1.46 1.86
d
2
2. A = π r 2 = π ( 2.4 )
A = 18 km2
2.27
2πs πs
3. p = 2s + = 2s +
d = 2.272 +1.86 2 4 2
1 ( )
For right triangle, A = (2.27 )(1.86 ) p = 2 (3.25 ) +
2
1.46 + 1.74 +d p =11.6 in. π 3.25
For obtuse triangle, s = 2

π s2 2
π (3.25)
A= =

and A = s (s −1.46)(s − d )(s −1.74) 4 4

2
S6ection
5 2.3 Quadrilaterals Chapter 2 GEOMETR5Y1

A = 8.30 in.
A of quadrilateral = Sumof areas of two triangles,
S3ection
5 2.4 Circles Chapter 2 GEOMETR5Y3

°
4. AC = 2⋅ ∠ ABC ∠ ABT = 90
° ° °
= 2(25
°
) ∠ CBT = ∠ ABT − ∠ ABC = 90 − 65 = 25 ;
°
= 50
° ∠ CAB = 25

20. ∠ BTC = 65° ; ∠ CBT = 35° since it is

5. (a) AD is a secant line.
°

(b) AF is a tangent line. complementary to ∠ ABC = 65 .

21. ARC BC = 2 (60 ) = 120°

°
6. (a) EC and BC are chords.
(b) ∠ ECO is an inscribed angle.
p
BC = 2 ( 60 ) = 120
° °
22.
7. (a) AF ⊥ OE. p ° ° °
AB + 80 + 120 = 360
(b) +OCE is isosceles.
p
AB = 160
°

8. (a) EC and p
EC enclose a segment.

(b) radii OE and OB enclose a sector ∠ ABC = (1/ 2 ) ( 80 ) = 40 since the measure of an
23.

° °
with an acute central angle. inscribed angle is one-half its intercepted arc.

1
9. c = 2π r = 2π (275) = 1730 ft
24.
∠ ACB =
2
(160° ) = 80°

10. c = 2π r = 2π (0.563) = 3.54 m ° ⎛ π ⎞

25. 022.5 ⎜ ⎟ = 0.393 rad
⎝ 18⎠0°
11. d = 2r; c = π d = π (23.1 ) = 72.6 mm
° ° π rad
26. 60.0 = 60.0 ⋅ = 1.05 rad
12. c = π d = π (8.2 ) = 26 in. 180
°

13. A = π r 2 = π (0.0952 2 ) = 0.0285 yd2

° °
27. 125.2 = 125.2π rad/180 = 2.185 rad

π rad
2
14. A = π r 2 = π ( 45.8 ) = 6590 cm2 28. 3230 ° = 3230 ° ⋅ = 56.4 rad
180

2 2
15. A = π ( d / 2 ) = π ( 2.33 / 2 ) = 4.26 m2 1 πr
29. P = ( 2π r + 2r = 2r

4 2
1 1 2 2
16. A = π d = π (1256 ) = 1, 239, 000 ft
2

4 4 30. Perimeter = a + b + ⋅ 2π r + r
4
17. ∠ CBT = 90° − ∠ ABC = 90° − 65° = 25°
 1
S4ection
5 2.4 Circles Chapter 2 GEOMETR5Y3

1 1
31. Area = πr2− r2

18. ∠B
CT
 S5ection
5 2.4 Circles Chapter 2 GEOMETR5Y3

= 90° , 4
any 2
angle 32.
such as
Ar
∠ BCA
inscrib ea
ed in a =
semicir 1
cle is a
right (a
angle r)
and
+
∠ BCT 1
is
π
2
supple r
2
mentar
4
y to
∠ BCA.

19. A tangent to a circle is perpendicular to the 33. All are on the same diameter.
radius drawn to the point of contact. Therefore,
S6ection
5 2.4 Circles Chapter 2 GEOMETR5Y3

34.
A 39. A

s
45 0
r B r
°
AB = 45
°
s 45
=
2π r 360° O B
r
π
s= ⋅ r

4 Asegment = area of quarter circle − area of triangle

1 1
Asegment = π r 2 − ⋅ r ⋅ r
35.

⋅ 224 2 2
r (π − 2 )
Asegment =
6.00
4

40. A
6.00

A of sector = A of quarter circle − A of triangle 2r h

1 2 1
A = ⋅ π ( 6.00 ) − ( 6.00 ) ( 6.00 ) 2
4 2
A = 10.3 in.2

B
36. ∠ ACB = ∠ DCE (verticalangles) 2 2 2

∠ BAC = ∠ DEC and AB = r + r

∠ ABC = ∠ CDE (alternate interior angles) AB = 2r

Therefore, the triangles are similar since 1 2r
AC = AB = in right triangle OAC
corresponding angles are equal. 2 2
2

2 ⎛ 2r ⎞ 2
r= ⎜ ⎟ + ( r − h) from which
c d
37. c = 2π r ⇒ π = ; d = 2r ⇒ r = from which ⎝2⎠

2r 2
c ( )
π= d h=

2⋅ 2 38.
c
π = ⋅ π is the ratio of the circumference to
d
S7ection
5 2.4 Circles Chapter 2 GEOMETR5Y3

dia
me 2−2r
2

ter.

5
d 5 1 5
S8ection
5 2.4 Circles Chapter 2 GEOMETR5Y3

4= 41. h
= =
=⋅ 11
.5
k
m
d
c 4 4 4 16
r = 6378 - 11.5 km
16 16
c= d ⇒π = = 3.2 5
5
r = 6378 km
which is incorrect since π = 3.14159⋅ ⋅ ⋅
S9ection
5 2.4 Circles Chapter 2 GEOMETR5Y3

( 6378 + 11.5)
2
= d 2 + 63782 πd 2 ,
using A =
d = 383 km 4
2
π (12.0 + 2 ( 0.60 ) ) π (12.0 )
2

A= −
42. 4 4

0.346 km A = 23.8 m2
d
45. C = 2π r = 2π ( 3960) = 24,900 mi
6378km
46. 11( 2π r ) = 109
6378 km
r = 1.58 mm

2 ⎛12.0 ⎞
2
(6378+ 0.346) =d 2 +63782
A π⎜ 2 ⎟ 4
basketball
= ⎝ ⎠ =
47.
d = 66.4 km Ahoop 2 9
π ⎛ 18.0 ⎞
⎜ ⎟
⎝ 2⎠
43. d =
682 + 582
volume π r 2L
1
d = 89> 85 48. flow rate = =
time t
68 km π2 2πr 2
⋅ r
2 1
2 flowrate = =
t t
r22 = 2⋅ 1r 2
r2 = 2r1
58 km
d

49. c = 112; c = π d ; d = c / π = 112 / π = 35.7 in.

Signal cannot be received. π

50. A =
2
( 90 2
− 452 )

π d2 , A = 9500 cm2
44. Using A=
4
2
π (12.0 + 2( 0.60 ) ) 51. Let D = diameter of large conduit, then
π (12.0 )
2

A= − D = 3d where d = diameter of smaller conduit

4 4 π π
A = 23.8 m 2 F= D2 = 7⋅ ⋅ d2
4 4
2
7d 7d 2 7d 2
0.60 m F= = =
D2 ( 3d )2 9d 2
7
F=
9
S10ection
5 2.4 Chapter 2 GEOMETR5Y3
Circles
7
The smaller conduits occupy of the larger
d = 12.0 m 9
 S11ection
5 2.4 Chapter 2 GEOMETR5Y3
Circles

conduits.
S12ection
5 2.4 Chapter 2 GEOMETR5Y3
Circles

3
52. A of room = A of rectangle + A of circle
4
3 2
2.5 Measurement of Irregular Areas
A = 24 ( 35 ) + π ( 9.0 ) = 1030.85174
4 1. The use of smaller intervals improves the approxi-
A = 1000 ft2 , two significant digits mation since the total omitted area or the total
extra area is smaller.
3
53. Length = ( 2 ) ( 2π )( 5.5 ) + ( 4 )( 5.5 ) = 73.8 in.

4 2. Using data from the south end gives five intervals.

Therefore, the trapezoidal rulemust be used since
54. d = 4⋅ 12.52 −2.252 Simpson's rule cannot be used for an odd number
2π

d = 309 km of intervals.
3. Simpson's rule should be more accurate in that it
r plane accounts better for the arcs between points on the
upper curve.
12.5 4. The calculated area would be too high since each
2.25
trapezoid would include more area than that under
ITC the curve.
AP
2.0
5. A = ⎡ 0.0 + 2( 6.4 ) + 2 ( 7.4) + 2 ( 7.0 ) + 2(6.1)
trap
2 ⎣
55. Horizontally and opposite to original direction
⎡ ⎣ +2 ( 5.2 ) + 2( 5.0 ) + 2 ( 5.1) + 0.0⎤ ⎦
2
56. Let A be the left end point at which the dashed Atrap = 84.4 = 84 m to two significant digits
lines intersect and C be the center of the gear.
°
Draw a line from C bisecting the 20 angle. Call 6. Asimp

the intersection of this line and the extension of the h

= ( y + 4y + 2y + 4y +⋅ ⋅ ⋅ + 2y + 4y + y)
0 1 2 3 n −2 n −1 n
upper dashed line B, then 3

° ° 20
= (0 + 4 6.4 + 2 7.4 + 4 7.0 + 2 6.1 + 4 5.2 )

360 15 ° ( ) ( ) ( ) ( ) (
= ⇒ ∠ ACB =7.5 3
24 teeth tooth
° + 2( 5.0)+ 4 (5.1 )+ 0) = 88 m 2
° 20 °
∠ ABC = 180 − = 170
2
1.00
A = (0 + 4 0.52 + 2 0.75 + 4 1.05)

1
∠ x + ∠ ABC + ∠ ACB =180
° 7. simp ( ) ( ) (
2 3

1 + 2(1.15) + 4 (1.00 ) + 0.62) = 4.9 ft2

∠ x +170 ° + 7.5° =180°

S13ection
5 2.4 Chapter 2 GEOMETR5Y3
Circles

2 1 °
∠ x = 2.5
S0ection
6 2.4 Circles Chapter 2 GEOMETR6Y0
2
x=5
° 1
8. Atrap = (0 + 2( 0.52) + 2( 0.75) + 2(1.05) + 2(1.15)
2
+ 2(1.00) + 0.62) = 4.8 ft2

0.5
9. Atrap = ⎡ 0.6 + 2( 2.2 ) + 2 ( 4.7 ) + 2(3.1) + 2 ( 3.6 )
⎣
2
⎡ ⎣ +2 (1.6 ) + 2 ( 2.2 ) + 2 (1.5 ) + 0.8⎤ ⎦
Atrap = 9.8 m2
S7ection
5 2.5 Measurement of Irregular Areas Chapter 2 GEOMETR5Y7

0.5 17. Atrap

10. A = (0.6+ 4( 2.2 ) + 2( 4.7 ) + 4(3.1) + 2 (3.6 )
simp
3 0.500
= ⎡ ⎣ 0.0 + 2(1.732) + 2(2.000) + 2(1.732)
+ 4( 1.6) + 2 ( 2.2 ) + 4( 1.5) + 0.8) 2

= 9.3 mi2 + 0.0] = 2.73 in.2

This value is less than 3.14 in.2 because all of the

10 trapezoids are inscribed.
11. A = (38+ 2 ( 24 ) + 2 ( 25 ) + 2(17) +2 ( 34 )
2

( ) ( )
+ 2 ( 29 ) + 2( 36) + 2 ( 34) + 30) 0.250
18. Atrap = (0.000 + 2 1.323 + 2 1.732
2 2
2 ⎛ 23 km ⎞ 2
A =2330 mm ⎜ 2 2 ⎟
⎝ 10 mm ⎠ + 2(1.936) + 2(2.000) + 2(1.936)
A = 12, 000 km2 + 2(1.732) + 2(1.323) + 0.000)
= 3.00 in.2
4.0 The trapezoids are small so they can get closer

12. A= 2⋅ ⎡ 2 (55.0 ) +2 ( 2 (54.8 )) +2 ( 2 (54.0 ))

2⎣
to the boundary.
+ 2 ( 2 ( 53.6 )) + 2 ( 2 (51.2 )) + 2 ( 2 ( 49.0 ) )
+ 2 ( 2 ( 45.8 ) ) + 2 ( 2 ( 42.0 ) ) + 2 ( 2 ( 37.2 ) ) 0.500
19. A = (0.000 + 4(1.732) + 2 (2.000 )
simp
3
+ 2 ( 2 (31.1)) + 2 ( 2 ( 21.7 ) ) + 2 ( 0.0 )
+ 4(1.732) + 0.000)
A = 7500 m2 2
= 2.98 in.
45 The ends of the areas are curved so they can get

13. Atrap =
2
[170 + 2( 360 ) + 2( 420 ) + 2( 410 ) + 2( 390) closer to the boundary.
+ 2(350) + 2(330) + 2(290) + 230]

A = 120, 000 ft2 0.250

20. A = (0.000 + 4 1.323 + 2 1.732
simp
trap
( ) ( )
45 3
14. (230 + 4 ( 290 ) + 2( 330 ) + 4 ( 340 ) + 4(1.936) + 2(2.000) + 4(1.936)
Asimp = 3
+ 2(1.732) + 4(1.323) + 0.000)
+ 2 ( 390 ) + 4( 410 ) + 2 ( 420 ) + 4 ( 360 ) +170) = 3.08 in.2
= 120, 000 ft2 The areas are smaller so they can get closer
to the boundary.
50
15. Asimp = (5 + 4(12) + 2( 17) + 4( 21) + 2 ( 22 )
3

+ 4(25) + 2(26) + 4(16) + 2(10 )

+ 4(8) + 0)
= 8100 ft2
2.0
S8ection
5 2.5 Measurement of Irregular Areas Chapter 2 GEOMETR5Y7

16. Atrap = (3.5 + 2(6.0) + 2(7.6) + 2(10.8) + 2(16.2)

2
+ 2(18.2) + 2(19.0) + 2(17.8) + 2(12.5)
+ 8.2)
= 229 in.2
S9ection
5 2.5 Measurement of Irregular Areas Chapter 2 GEOMETR5Y7

Acircle = π r 2 = π ( d / 2 ) = π ( 2.5 / 2 ) = 4.9in.2

2 2

Atotal = 229 − 2(4.9) = 219 in. 2

S10ection
5 2.5 Measurement of Irregular Chapter 2 GEOMETR5Y7
Areas

1 1
15. S = ps = (3×1.092 )(1.025 ) = 3.358 m2

2.6 Solid Geometric Figures 2 2

1. V1 = lwh1, V2 = ( 2l )( w )( 2h ) = 4lwh = 4V1 16. S = 2π rh = 2π ( d / 2 ) h = 2π (250 / 2 )( 347 )

The volume is four times as much. = 273,000 ft2

2. s 2 = r2 + h 2 1⎛4 3 ⎞ 2 ⎛ 0.83 ⎞
3
3

2 2 2 17. V = ⎜ π r ⎟ = π⎜ ⎟ = 0.15 yd
17.5 = 11.9 + h 3
23 2
⎝ ⎠ ⎝ ⎠
h = 12.8 cm

22.4 2 2 2 2
2 18. b = =11.2; h = s−b = 14.2 −11.2
1 1 ⎛ 11.9 ⎞ 2
3. V = π r 2 h = π ( 2 (10.4))
⎜ ⎟ = 8.73 m
3 3 2
⎝

V = 771 cm3 1 1 2
V = Bh = ( 22.4 ) ( 8.73 ) = 1460 m3
2 ⎛ 122 ⎞ 2 3
3 3
4. V = π ( 40.0 ) + π ( 40.0 )
⎜2 ⎟3 19. s = = = 3.40 cm
⎝ ⎠ h2 + r 2 0.2742 + 3.39
2

V = 441,000 ft3 A = π r 2 + π rs = π ( 3.39 ) + π ( 3.39 )( 3.40 )

2

= 72.3 cm2
5. V = e3 = 7.153 = 366 ft3

2
20. There are four triangles in this shape.
6. V = π r 2 h = π ( 23.5 ) ( 48.4 ) = 84,000 cm3 2

2 ⎛ 3.67 1
s = 3.67 − = 3.18, A = ps
⎜
2
⎝2 2
7. A = 2π r 2 + 2π rh = 2π ( 689 ) + 2π ( 689 )( 233) 1

= 3,990,000 mm2
=
2
(4 × 3.67)(3.18) = 23.3 in.2

2 2 2 4 3 4 4 d
8. A = 4π r = 4π ( 0.067 ) = 0.056 in. 21. V = π r = ⎛d
3
= π
3
⎞
π ⎜⎟
3 3 ⎝2⎠ 3 8
4 4 1
9. V = π r3 = π 0(.877 = )2.83 yd
3 3
V= πd3

3 3 6

22. A = A + A
1 1 2
10. V = π r 2 h = π ( 25.1) ( 5.66 ) = 3730 m3 flat curved
1
3 3 =πr + 2
⋅ 4π r 2
2
S11ection
5 2.5 Measurement of Irregular Chapter 2 GEOMETR5Y7
Areas

11. S = π rs = π ( 78.0)( 83.8) = 20,500 cm

2 = 3πr 3

1 1 23.
cylinder
=
( ) 2=
12. S = ps = (345 )(272 ) = 46,900 ft2
S0ection
6 2.5 Measurement of Irregular Areas Chapter 2 GEOMETR6Y0
2
V π 2r h
6

2 2 Vcone 1
3 π r 2h 1

1 1 2 3
13. V = Bh = (0.76 ()1.30) = 0.25 in.
3 3

2
14. V = Bh = ( 29.0 ) (11.2 ) = 9420 cm3
5S9ection 2.6 Solid Geometric Figures Chapter 2 GEOMETR5Y9

1
24. π r 2 = (π r 2 + π rs) 2 2
4 34. s = h + r = 3.502 +1.802 = 3.94 in.
4r = r2 + rs
2
2

S = π rs = π (1.80)(3.94) = 22.3 in.

3r 2 =rs

r 1 4 4
= 3 3

s 3 35. V = π r = π (d / 2 )
3 3
2

final surface area 4π(2r ) 4 4 3

25. = = = π (165 / 2)

original surface area 4π r 2 1 3

= 2.35 ×106 ft3
2

lb 5280 ft ft
26. 62.4 ⋅ 1 mi2 ⋅ ⋅
⋅ 1 in.

ft3 mi2 12 in. 4

36. V = π r3 + π r 2 h
ton 3
=1.45×108 lb ⋅ 4 3 2
2000 lb = π ( 2.00 ) + π ( 2.00 ) ( 6.5 )
= 72,500 ton 3
= 115 ft3
27. A = 2lh + 2lw + 2wh
1 1
= 2( 12.0)( 8.75) + 2( 12.0)( 9.50 ) + 2 ( 9.50 )(8.75 ) 37. A = l 2 + ps = 162 + (4)(16) 82
2 2 + 402
A = 604 in.2
= 1560 mm2

28. V1 = 50.0× 78.0×3.50 =13,650 ft 38.

1 0.06
V2= × 78.0 × 5.00 × 50.0 = 9750 ft3
2

V = V + V = 13, 650 + 9750 = 23,400 ft3

1 2

0.96
2 2
29. V = π r 2 h = π ( d / 2 ) h = π ( 4.0 / 2 ) ( 3, 960, 000 )
= 5.0×107 ft3 or 0.00034 mi3 2

1 0.06 ⎞
30. V = h ( a + ab + b ) N⋅ π ⎛
2 2
(0.96) = 76

3 ⎟
2 ⎠
⎜
⎝
1 N = 280 revolutions
= ( 0.750 ) (2.502 + 2.50( 3.25 ) + 3.252 )
3
= 6.23 m3 29.8
39. c = 2π r = 29.8 ⇒ r =
2π
6S9ection 2.6 Solid Geometric Figures Chapter 2 GEOMETR5Y9

31. ( ) 4 4 ⎛ 29.8 ⎞
V = π r =3π
V = 1.80 3.93 −1.80 1.50 = 9.43 ft
2 2 3
3

⎜ ⎟
3 3 ⎝2π
7S9ection 2.6 Solid Geometric Figures Chapter 2 GEOMETR5Y9

32. There are three rectangles and two triangles in V = 447 in.3
this shape.
8S9ection 2.6 Solid Geometric Figures Chapter 2 GEOMETR5Y9

⎛1 40. ( )( )
A = 2 ⎞ ( 3.00 )( 4.00 ) + 3.00 ( 8.50) + 4.00 ( 8.50 )
2
⎜ Area = π ⋅ 3 + 0.25 4.25 = 41 in. 2
⎝
⎟ 2
⎠ 3.002 + 4.002 =114 cm
+ 8.50

1 1
33. V = BH = 2(502 (1)60) = 3,300,000 yd 3

3 3
S1ection
6 2.6 Solid Geometric Figures Chapter 2 GEOMETR6Y1

41. V = cylinder + cone (top of rivet)

1 Chapter 2 Review Exercises
= π r 2h + π r2h
3
° ° °
1 2 1. ∠ CGE = 180 −148 = 32
2
= π ( 0.625 / 2 ) ( 2.75 ) + π (1.25 / 2 ) ( 0.625 )
3
° ° ° °

= 1.10 in.3 2. ∠ EGF = 180 −148 − 90 = 58

° ° °
42. p = 18 = 2r + π r 3. ∠ DGH = 180 −148 = 32
r

18 ° ° °
=
4. ∠ EGI = 180 −148 + 90 = 122
π+2

1 ⎛ 18 ⎞2 (0.075)
V=
⋅ π

⎜ ⎟ 5. c = 9 2 + 40 2 = 41
2 ⎝ π +2

V = 1.4 m2
6. c2 = a2 + b2 = 142 + 482 ⇒ c = 50
⎛4 ⎞
4
43. V = π r 3 = 0.92V = 0.92 π r3

2 2 1 ⎜ 1 ⎟ 7. c2 = a 2 + b 2 = 4002 + 5802 ⇒ c = 700

3 3
2
⎝ ⎠
r = 0.97r
1
2 2 2 2 2 2

8. c = a + b ⇒ 6500 = a + 5600 ⇒ a = 3300

radius decreased by 3%

44. 9. a = 0.7362 − 0.3802 = 0.630

10. a = 1282 − 25.12 = 126

2 2 2 2 2 2
11. c = a + b ⇒ 36.1 = a + 29.3 ⇒ a = 21.1
2 2 2 2 2 2
12. c = a + b ⇒ 0.885 = 0.782 + b ⇒ b = 0.414
9 y
=
12 12 − x 13. P = 3s =3 ( 8.5 ) = 25.5 mm
3
y = (12 − x)
4
14. p = 45 = 4 (15.2) =60.8 in.
1
⋅ π ⋅ 9 22 ⋅ 12 1

3
= ⋅ π ⋅ ⎜ (12 − x ) ⎟ ⋅ (12 −
x)
S1ection
6 2.6 Solid Geometric Figures Chapter 2 GEOMETR6Y1

2 3 ⎝4 1 1 2
⎠ A = bh = (0.125 0.188
)(
= 0.0118 ft
)
15.
x = 12 − 3 864 16.
x = 2.50 cm
 S2ection
6 2.6 Solid Geometric Figures Chapter 2 GEOMETR6Y1

2 2

1 1
s = (a + b + c) = (175 + 138 +
2 2
119) = 216
S3ection
6 2.6 Solid Geometric Figures Chapter 2 GEOMETR6Y1

A = s ( s −a )( s −b )( s −c )
= 216(216 −175)(216 −138)(216 −119)
A = 8190 ft2
17. C = π d = π (98.4) = 309 mm

18. p = 2l + 2w = 2 (2980 ) + 2 (1860 ) = 9680 yd

 1 1 2 2
19. A = h (b +b ) = ( 34.2 )( 67.2 +126.7 ) = 3320 in. 34. AD = 6 2 + ( 4 + 4) = 10
1 2
2 2
2 BE
⎛ 32.8 35. = ⇒ BE = 2.4
20. A = π r = π
2
= 845 m2

⎜ ⎟ 4 10
2
⎝ ⎠
AE 8
1 3 36. = ⇒ AE = 3.2
21. V = Bh = (226.0 )(34.0 )(14.0) =6190 cm 4 10

1
b 2 + ( 2a 2 ) + π ( 2a ) = b +
2
22. V = π r 2 h = π ( 36.0 ) 2 ( 2.40 ) = 9770 in.3 37. P = b + b 2 + 4a2 + π a
2

1 1
23. V = Bh = (3850 )(125 ) =160,000 ft3
1
3 3 38. p = (2π s) + 4s = π s + 4s
3
2

4 3 4 ⎛ 2.21 ⎞ 1 1 1
24. V = π r = π = 5.65 mm3 2 2

b( 2a ) + ⋅ π ( a ) = ab + π a
3 ⎜ 2 ⎟
39. A =
3
2 2 2

⎝ ⎠
25. A = 6e = 6 (0.520 ) =1.62 m2
2
1
40. A =
2
(π s2 ) + s2

⎛ ⎛ 12.02 ⎛12.0 ⎞
26. A = 2πr 2 + 2πrh = 2π + (58.0) 41. A square is a rectangle with four equal sides and a

⎜ ⎜⎝2 ⎜ 2
⎝ ⎟
⎝ ⎠ rectangle is a parallelogram with perpendicular
⎟ ⎟⎠
A = 2410 ft2 ⎠ intersecting sides so a square is a parallelogram.
A rhombus is a parallelogram with four equal sides
27. s2 = r2 + h2 = 1.822 +11.52 ⇒ s = 1.822 +11.52 and since a square is a parallelogram, a square is

S = π rs = π (1.82) 1.822 +11.52 a rhombus.

2
S = 66.6 in.
42. If two angles are equal then so is the third and
2 the triangles are similar.
⎛ 12, 760 ⎞
28. A = 4πr 2 = 4π = 5.115×108 km2
⎜ ⎟
⎝ 2 ⎠
43. A = π r 2 , r ⇒ nr ⇒ A = π ( nr ) = n 2 ( π r 2 )
2

50
29. ∠ BTA = = 25
°
The area is multiplied by n2 .
2

3
° ° ° ° 44. V = e 3; e ⇒ ne ⇒ V = ( ne ) = n3 e3
30. ∠ TBA = 90 , ∠ BTA = 25 ⇒ ∠ TAB = 90 − 25
3

= 65° The volume is multiplied by n .

 ° ° ° °
31. ∠ BTC = 90 − ∠ BTA = 90 − 25 = 65

°
32. ∠ ABT = 90
°
(any angle inscribed in a semi-circle is 90 )
° ° °
33. ∠ ABE = 90 − 37 = 53
 ⎛ 18.0 ⎞ 2 ⎛ 18.0 ⎞ 2
45. 52. A= ⎜ ⎟ + 2π ⎜ ⎟ = 52.1 cm
2

4
⎝ ⎠ ⎝8 ⎠
B
C
a
b 53. AB = 14
E 13 18
c
d AB = 10 m
D
54.

(BEC = (AED, vertical ( ' s. h

B

120 ft
(BCA = (ADB, both are inscribed in p
AB A
p
(CBE = (CAD, both are inscribed in CD
140 ft 84 ft
a b
which shows ΔAED ∼ ΔBEC ⇒ =
d c h 120
= ⇒ h = 192
140 + 84 140
46. (B + 2 ( 90° ) + 36° = 180 ° 1

°
area of A = (140)(120) = 8400 ft2
(B = 144 2
1 2

C area of B = (120 +192 )( 84 ) = 13,000 ft

2
FB 1.60
55. = ⇒ FB = 6.0 m
A B
4.5 1.20

DE 33
56. = ⇒ DE = 22 in.
16 24

57. The longest distance in inches between points on

47. 2(base angle) + 38 = 180
° ° the photograph is,
°
base angle = 71 8.002 + 10.02 = 12.8 in. from which
x 18,450
=
48. The two volumes are equal. 12.8 1

3 2 ⎛ 1 ft ⎞⎛ mi ⎞
x = 12.8 18, 450 in.
4 ⎛ 1.50 ⎞ ⎛ 14.0 ⎞ ( )( )
π =π ⋅ t ⎜ ⎟⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎝ 12 in. ⎠⎝ 5280 ft ⎠
3 2 2
⎝ ⎠ ⎝ ⎠
t = 0.0115 in. x = 3.73 mi
2
49. L = 2
0.48 + 7.8 2 = 7.8 m π ⎛ 3.10 ⎞
⎜ ⎟
58. MA = ⎠2 = 1.90
⎝
2

π⎛
2.25 ⎞
50. c = 2100 2 +95002 =9700 ft
⎜ 2 ⎟

51.
 ⎝
⎛ 2.4
p =6 = 10 cm ⎠
⎞
⎜
2
⎟ 59. c = π D = π ( 7920 + 2( 210 ) ) = 26,200 mi
⎝ ⎠
 651
60. c = 2π r = 651⇒ r = 67. ( 500.10 )
2
− 500 2 = 10 ft
2π

2
⎛ 651 68.
A = π r2 = π ⎜ =33,700 m2
⎝ 2π x + 4.0 ft
⎞
⎟ x
2

1.0
61. A = ( 4.0 )(8.0 ) − 2π ⋅ = 30 ft 2
4 15.6 ft

62.
( x + 4.0 ) 2 = x 2 + 15.6 2
6
1.38 ×10 x = 28.4
2 1.27 ×10 4 x + 4 = 32.4 ft, length of guy wire
2
1.50 ×108
69.
d 1500 m

600 m
8
d = d +1.50 ×10
1.27×104 1.38×106

2 2

d = 1.39 ×106
1700 m
250
63. A = 3 ⎡ 220 + 4(530) + 2(480)

⎡ ⎣ +4 ( 320 + 190 + 260 ) + 2( 510 ) + 4 ( 350 ) 2 2 2 2

d = 1700 −1500 + 600

⎡ ⎣ +2 ( 730 ) + 4( 560 ) + 240 d = 1000 m
A = 1, 000, 000 m2

250 70. w + 44 ft
64. V = ⎡ 560+2 (1780 ) +2 (4650 ) +2 (6730 )
2 ⎣ w
+ 2( 5600 ) + 2( 6280 ) + 2( 2260 ) + 230]

V = 6,920,000 ft3
p = 2l + 2w

⎛ 4.3 ( )
65. V = π r 2 h = π (13) =190 m3
⎞

2
288 = 2 w + 44 + 2w
⎜ 2
w = 50 ft
⎟ l = w + 44 = 94 ft
⎝
⎠
66. Area of cross section = area of six equilateral
triangles with sides of 2.50 each triangle has
1 4 3
71. V = π r h + ⋅ π r
2
2.50 (3)
23
 3
⎤
semi-perimeter = = 3.75 ⎡ ⎛ 2 2.50 ⎞
⋅ ⋅ π
⎞ ⎛ 1
2 = ⎢ π 2.50 4.75 2.50 ⎞ ⎥
+ 4 ⎛
−
 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
V = area of cross section × 6.75 2 2 2 3 2
⎝ ⎠ ⎝ ⎠ ⎝

3
= 6 3.75 ( 3.75 − 2.50 ) ×6.75 ⎛ 7.48 gal ⎞
⎠
⎜ ⎟
3 ⎝ ft3
=110 m
=159 gal
72. 75.
rr r
h
3.25 - 2.50
s r
1 4
Vcyl = π r 2h =V = ⋅ πr3
hemisphere
23
3h
r=
2
76.

tent surface area 1620 1590

= surface area of pyramid + surface area of cube
1 r
= ps + 4e2
2
2

1 2 ⎛ 2.50 ⎞
= (4)(2.50) ( 3.25 − 2.50 ) + + 4( 2.50 )
2

⎜
2 ⎝2
= 32.3 m2 ⎛ ?
2⎞
A = πr
2
2
⎜162−0 1590 ⎟ = 303, 000 km
⎝

w 16 16h
73. = ⇒w =

h 9 9
⎛16h
2
77. Label the vertices of the pentagon ABCDE. The
2
152 = w + h = 2 2
+ h 2 ⇒ h = 74.5 cm

⎜ ⎟ area is the sum of the areas of three triangles, one

⎝ 9
16h with sides 921, 1490, and 1490 and two with sides
w= =132 cm
9 921, 921, and 1490. The semi-perimeters are given
by
74. 921+ 921 + 1490
s1 = = 1666 and
2
h
5 +2 921+ 1490 + 1490

4 k+3 s2 = = 1950.5.
2

A = 2 1666 (1666−921)( 1666 −921)( 1666−1490)

3k -1 + 1950.5(1950.5 −1490)(1950.5 −1490)

+ (1950.5 − 921)

2 2 2
( 5k + 2 ) = ( 4k + 3) + ( 3k − 1)
k=3
4k + 3 = 15, 3k −1 = 8
1 2
A = (8)(15) = 60 ft
2
1
Note: 1) 3k −1 > 0 ⇒ k >
 3
1
2) There is a solution for < k < 1.
3
3) For k = 1 the triangle solution is an
isosceles, but not right triangle.
= 1, 460, 000 ft2