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2011 AMC 10A Problems/Problem 18

Revision as of 00:39, 16 February 2011 by Flyingpenguin (talk | contribs) (Created page with '== Solution == Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can c…')
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Solution

Draw a rectangle with vertices at the centers of $A$ and $B$ and the intersection of $A, C$ and $B, C$. Then, we can compute the shades area as the area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$. This is $\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{2 \ \mathbf{(C)}}$.

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