Download 1999 BC EXAM SOLUTIONS 1. If the length of the rectangle is x

1999 BC EXAM SOLUTIONS
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1. If the length of the rectangle is x meters, then the width of the rectangle is
meters.
x
The perimeter P in meters is
32
16
= 2x +
.
P = 2l + 2w = 2x + 2
x
x
2. The length of rope required is equal to the circumference of the Earth. This is about
2π(3960) = 7920π miles.
3. The sum of the interior angles of a pentagon is 540◦ . Thus
6x + 4x + (5x + 5) + (6x − 20) + (7x − 5) = 540
28x − 20 = 540
28x = 560
x = 20◦ .
4. Let x be the length and y be the width of the rectangle. The area of the rectangle is
xy . The length of the rectangle increases to 1.2x and the width decreases to 0.8y .
The rectangle now has area (1.2x)(0.8y) = 0.96xy . The percentage change in the
area of the rectangle is
−0.04xy
0.96xy − xy
× 100 =
× 100 = −4% .
xy
xy
5. Let r be the radius of the circle. The area of the circle is A = πr 2 and the circumference of the circle is C = 2πr . Then
A = 4C
πr 2 = 4(2πr)
πr 2 = 8πr
r 2 − 8r = 0
r(r − 8) = 0
r=8.
6. Let x be the length (in inches) of an edge. Each of the six faces of the cube has
surface area x2 . The surface area of the cube is 6x2 = 24 , which has the positive
solution x = 2 . The volume of the cube is x3 = 23 = 8 in3 .
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7. If√x is the radius
√ (in inches) of the shaded circle, then the larger square has length
2 2x . Then (2 2x)2 = 8x2 = 108 and consequently, x2 = 27/2 . The area of the
27π 2
in .
shaded circle is πx2 =
2
2x
x
2 2 x
8. The length of the elastic band is equal to the sum of the perimeter of a square of
length 12 inches and the perimeter of a circle of radius 6 inches. The length of the
elastic band is (4 · 12) + (2π6) = 48 + 12π inches.
1 √ √
( 2)( 2) = 1 .
2
9. The area of the shaded triangle is
(0,3)
2
(1,2)
2
2
1
1
10. Let x and y be the two numbers. Now
x2 + 2xy + y 2 = (x + y)2 = 64
x2 + y 2 = 34
Subtracting the second equation from the first equation gives 2xy = 30 , so xy = 15 .
11. Let 1 unit be the length of a square as shown. The shaded area is the same as the
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total area of 8 such squares. The triangle shown has area (16)(16) = 128 units2 .
2
8
1
8
=
=
.
The ratio of the shaded area to the unshaded area is
128 − 8
120
15
2
1
1
16
√
√
12. Triangle OPQ has area (1/2)(2)( 3) = 3 in2 . The area of the unshaded region
contained in this triangle is the same as the area of a semicircle of radius 1 inch.
√
π
Consequently, the area of the shaded region is 3 − in2 .
2
O
1
P
1
o
o
60
60
1
1
1 60o
1
Q
13. A slice through the vertex of the cone that is perpendicular to its base results in the
following cross-section.
3
r
5
10
Using similar triangles,
3
r
= ,
10
8
so r = 15/4 inches.
14. Let x be the length (in feet) of the square base. The suface area of the bottom of
the box is x2 . A side panel of the box has suface area (x)(2x) = 2x2 . The cost
of the box in cents is [(15)(x2 )] + [(10)(4)(2x2)] = 95x2 . The volume of the box is
(x)(x)(2x) = 2x3 = 54 in3 , so x = 3 in. The cost of the box is 95x2 = 95(32 ) = $8.55 .
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15. The incoming beam of light has slope −1 , the first reflected beam of light has slope
1 and the second reflected beam of light has slope −1 . The figure below shows that
H = 3 ft.
(3,0)
y= -x + 3
45 o
(2,1)
y= x - 1
45
45 o
(0,0)
45
o
o
(1,0)
(2,0)
16. If (0, y) is equidistant from (5, −5) and (1, 1) , then
p
(5 − 0)2 + (−5 − y)2 =
p
(1 − 0)2 + (1 − y)2
25 + (5 + y)2 = 1 + (y − 1)2
25 + y 2 + 10y + 25 = 1 + y 2 − 2y + 1
y 2 + 10y + 50 = y 2 − 2y + 2
12y = −48
y = −4 .
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