# Difference between revisions of "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 $\sqrt{2}$. Since the hypotenuse is the radius of the circle, it has length 2 and by Pythagoras the other leg of the triangle also has length $\sqrt{2}$. We deduce that the two triangles formed by two radii of circle $O$ and the segment of the rectangle are 45-45-90 triangles.

The overlapping area consists of two sectors with central angle 90 and four 45-45-90 triangles with base $\sqrt{2}$ and height $\sqrt{2}$. The area of the circle is $4\pi$ so the area of the sectors is $2 \cdot 4\pi/4 = 2\pi$. The area of the triangles is $4\cdot (\sqrt{2})^2 /2 = 4$. The combined area is $$\boxed{\textbf{(D) }2\pi+4.}$$