2011 AMC 10A Problems/Problem 18: Difference between revisions

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	~~== Solution ==~~		#redirect [[2011 AMC 12A Problems/Problem 11]]

	~~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 compute the shades area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{2 \ \mathbf{(C)}}</math>.