1994 AHSME Problems/Problem 17

Problem

An $8$ by $2\sqrt{2}$ rectangle has the same center as a circle of radius $2$. The area of the region common to both the rectangle and the circle is

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

Solution

[asy] import cse5; import olympiad; real s=2*sqrt(2); pair A=(0,0),B=(0,s),C=(8,s),D=(8,0),O=(4,sqrt(2)),X; D(A--B--C--D--cycle); D(CR(O,2)); pair[] P; P=IPs(CR(O,2),box(A,C)); for(int i=0; i<4; i=i+1) { D(O--P[i],black); } X=foot(O,B,C); D(O--X); D(rightanglemark(O,X,C)); D(O); D(MP("O",O,S));[/asy]

We draw the diagram above. Dropping an altitude from the center of the circle to the top side of the rectangle, yields a segment of length $\sqrt{2}$. We can find that the two triangles formed by two radii of circle $O$ and the segment of the rectangle are $45-45-90$ triangles. The area of the overlapping region is the area of the circle minus the area of the two parts of the circle that are outside of the rectangle. The area of the circle is $4\pi$.

To find the area of the exterior parts, we find the area of the triangle and subtract it from the area of the quarter arc. The area of the triangle is $2$ and the area of the quarter arc is $\pi$. So the area of the exterior parts combined is $2(\pi-2)=2\pi-4$. So the area of the overlapping region is \[4\pi-(2\pi-4)=\boxed{\textbf{(D) }2\pi+4.}\]

--Solution by TheMaskedMagician

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