Answers:
1. B) 21
2. A) 10
3. D) 11:4
4. B) 360 degrees
5. B) 52.5 square inches
6. Figure D
7. C) 10 cm
8. A) 157 ft
9. B) 156 square meters
10. C) 4.2 cubic yards
11. A) 16 cubic meters
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Work Shown:
Problem 1)
ABCD ~ EFGH
AB/EF = BC/FG
12/20 = BC/35
0.6 = BC/35
0.6*35 = BC
BC = 21
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Problem 2)
The triangles are similar
AC/MN = AB/MB = BC/NB
AB/MB = BC/NB
30/18 = (x+15)/15
(30/18)*15 = x+15
25 = x+15
25-15 = x+15-15
x = 10
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Problem 3)
QR:YZ
55:20
55/5:20/5 <--- divide both parts by 5
11:4
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Problem 4)
For ANY convex polygon, the sum of the exterior angles is 360 degrees. It doesn't matter how many sides there are.
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Problem 5)
A = 0.5*b*h
A = 0.5*12.2*8.6
A = 52.46
A = 52.5
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Problem 6)
Figure D has both line symmetry and rotational symmetry
We can reflect figure D over a vertical line through the center to have it line up with itself
We can rotate figure D 120 degrees to have it line up with itself
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Problem 7)
a^2 + b^2 = c^2
8^2 + 6^2 = r^2
64 + 36 = r^2
100 = r^2
r^2 = 100
sqrt(r^2) = sqrt(100)
r = 10
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Problem 8)
C = 2*pi*r
C = 2*pi*6
C = 12pi
C = 37.6991118430776
The wheel travels roughly 37.6991118430776 inches in one revolution
1 rev : 37.6991118430776 inches
50*1 rev : 50*37.6991118430776 inches
50 rev : 1,884.95559215389 inches
Divide by 12 to convert to ft
1,884.95559215389/12 = 157.07963267949
That rounds to 157
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Problem 9)
P = perimeter of triangle
LSA = lateral surface area
LSA = P*H
LSA = (3+4+5)*12
LSA = 12*12
LSA = 144 square meters
base area = 0.5*b*h
base area = 0.5*3*4
base area = 6
SA = 2*(base area)+LSA
SA = 2*(6)+144
SA = 156 square meters
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Problem 10)
V = (4/3)*pi*r^3
V = (4/3)*pi*1^3
V = (4/3)*pi
V = 4.1887902047864
V = 4.2
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Problem 11)
(height of solid A):(height of solid B)
6:15
6/3:15/3
2:5
Cube both parts of the ratio
2^3:5^3
8:125
The ratio 8:125 can be used to figure out the volume of solid A
(volume of solid A)/(volume of solid B) = 8/125
(volume of solid A)/250 = 8/125
volume of solid A = 250*(8/125)
volume of solid A = 16