2007 AMC 10B Problems/Problem 13

Problem

Two circles of radius $2$ are centered at $(2,0)$ and at $(0,2).$ What is the area of the intersection of the interiors of the two circles?

$\textbf{(A) } \pi -2 \qquad\textbf{(B) } \frac{\pi}{2} \qquad\textbf{(C) } \frac{\pi \sqrt{3}}{3} \qquad\textbf{(D) } 2(\pi -2) \qquad\textbf{(E) } \pi$

Solution

You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is a quarter of the circle with radius $2,$ and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is \[\frac{1}{4} 4\pi - \frac{1}{2}(2)(2) = \pi - 2\] That means the area of the whole intersection is $\boxed{\mathrm{(D) \ } 2(\pi-2)}$

See Also

2007 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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