2012 AMC 10A Problems/Problem 18

Revision as of 21:22, 10 February 2012 by Fibonacci97 (talk | contribs) (Solution)

Problem 18

The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?

$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}$

Solution

Draw the hexagon between the centers of the circles, and compute its area (6*.5*2*sqrt3 = 6sqrt3). Then add the areas of the three sectors outside the hexagon (2pi) and subtract the areas of the three sectors inside the hexagon (pi) to get the area enclosed in the curved figure (6sqrt3 + pi), which is E.

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