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Mathematics Level IC/IIC Subject Tests
223
ARCO
■ SAT II Subject Tests
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47. The correct answer is (A). First use the value of the circumference to solve for the radius
Then use the radius to find the area
or
r is now the length of the altitude of the triangle, and 2r the length of the base:
Again, the area of the circle is just 
π.
Finally, a little common sense gets you this problem without any math. The triangle, which has an
area of 1, takes up slightly less than half of the circle:
So the correct answer choice must be something that’s a bit larger than 2. Only one of the answer
choices qualifies. (B) says, for example, that the triangle is less than 
1
6
of the circle. So (B), and with
it the other choices that are larger, have to be incorrect.

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Lesson 8
224
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ARCO
■ SAT II Subject Tests
48. The correct answer is (E). The point of intersection will have the coordinates that satisfy both of
the equations. So treat the equations as a system of simultaneous equations. Using the first equation,
solve for x:
Substitute this into the second equation:
And solve for y:
(This allows you to eliminate (A) and (C). And that would be important if you were about to run out
of time.) Now use this value to find x:
49. The correct answer is (A). This is a good exercise in organized problem solving. Look at the figure
and ask yourself what you already know. You know the radius of the circle. In addition, you know that
the perimeter of the shaded area consists of two arcs. There must be some way to use the information
about the radius to find the length of the arcs. Arcs can be measured in terms of length or in terms of
degrees. Is it possible to find the degree measures of those arcs? Yes:
Since the sides of the triangles are all radii, the triangles must be equilateral. This means that the
degree measure of each arc is 120. Since the circles have a radius of 1, they have circumferences of
2
π (1) = 2π. And since each arc is a third that long: 
2
3
π
. And since there are two such arcs, the
perimeter of the shaded area is 2 
× 
2
3
π
= 
4
3
π
.