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 one-fourth of the area 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 $\boxed{\mathrm{(D) \ } 2(\pi-2)}$

2007 AMC 10B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2007 AMC 10B Problems/Problem 13

Problem

Solution

See Also