1998 AHSME Problems/Problem 16
Problem
The figure shown is the union of a circle and two semicircles of diameters and , all of whose centers are collinear. The ratio of the area, of the shaded region to that of the unshaded region is
Solution
To simplify calculations, double the radius of the large circle from to . Each region is similar to the old region, so this should not change the ratio of any areas.
In other words, relabel and to and .
The area of the whole circle is
The area of the white area is about , which is the bottom half of the circle. However, you need to subtract the little shaded semicircle on the bottom, and add the area of the big unshaded semicircle on top. Thus, it is actually
Factoring gives
Simplfying the inside gives
With similar calculations, or noting the symmetry of the situation,
The desired ratio is thus , which is option .
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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