1999 BC EXAM SOLUTIONS
1. If the length of the rectangle is x meters, then the width of the rectangle is 16
x meters.
The perimeter P in meters is
P = 2l+ 2w= 2x+ 2
16
x
= 2x+ 32
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
0.96xy−xy
xy ×100 =
−0.04xy
xy ×100 =−4% .
5. Let r be the radius of the circle. The area of the circle is A =πr2
and the circum-ference of the circle is C = 2πr. Then
A= 4C πr2
= 4(2πr)
πr2
= 8πr r2
−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 .
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
shaded circle is πx2
= 27π 2 in
2 .
x
2 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.
9. The area of the shaded triangle is 1 2(
√
2)(√2) = 1 .
1
1 2
2
2 (1,2) (0,3)
10. Let x and y be the two numbers. Now
x2
+ 2xy+y2
= (x+y)2 = 64
x2 +y2
= 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 total area of 8 such squares. The triangle shown has area 1
2(16)(16) = 128 units 2
.
The ratio of the shaded area to the unshaded area is 8 128−8 =
8 120 =
1 15.
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. Consequently, the area of the shaded region is √3− π
2 in 2
.
1 1
1
1 1
1
O P
Q
60 60
60
o o
o
13. A slice through the vertex of the cone that is perpendicular to its base results in the following cross-section.
3
5 r
10 Using similar triangles,
r
10 = 3 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 .
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.
y= -x + 3
y= x - 1
(3,0)
(0,0) (1,0) (2,0) (2,1)
45 o
o o
45o
45 45
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 +y2
+ 10y+ 25 = 1 +y2
−2y+ 1
y2
+ 10y+ 50 =y2
−2y+ 2
12y=−48
y=−4 .
