# 1952 AHSME Problems/Problem 37

## Problem

Two equal parallel chords are drawn $8$ inches apart in a circle of radius $8$ inches. The area of that part of the circle that lies between the chords is: $\textbf{(A)}\ 21\frac{1}{3}\pi-32\sqrt{3}\qquad \textbf{(B)}\ 32\sqrt{3}+21\frac{1}{3}\pi\qquad \textbf{(C)}\ 32\sqrt{3}+42\frac{2}{3}\pi \qquad\\ \textbf{(D)}\ 16\sqrt {3} + 42\frac {2}{3}\pi \qquad \textbf{(E)}\ 42\frac {2}{3}\pi$

## Solution $[asy] pair A,B,C,D,E,F,G; A=(0,0); B=(-5,4sqrt(3)+2); C=(5,4sqrt(3)+2); D=(-5,-4sqrt(3)-0.5); E=(5,-4sqrt(3)-0.5); F=(-4,0); G=(4,-sqrt(2)/2); label("A",(0,-0.5),S); label("B",B,SE); label("C",C,SW); label("D",D,SE); label("E",E,SW); label("F",F,W); label("G",G,NE); draw(circle(A,8)); draw((-4,-4sqrt(3))--(-4,4sqrt(3))); draw((4,-4sqrt(3))--(4,4sqrt(3))); draw((-4,-4sqrt(3))--(4,4sqrt(3))); draw((4,-4sqrt(3))--(-4,4sqrt(3))); draw((4,0)--(-4,0)); label("4",(-2.5,0.5)); label("4",(2.5,0.5)); label("8",(-2.5,3)); label("8",(2.5,3)); label("8",(-2.5,-3)); label("8",(2.5,-3)); [/asy]$ Draw in the diameter through A perpendicular to chords BD and CE. Label the intersection points of the diameter with the chords F and G. Then, AF=4, AG=4, and AFB ,AFD, AEG. and AGC are right triangles. We're going to find the area inside sectors BAD and CAE but outside triangles BAD and CAE.

By pythagorean theorem, BF= $4\sqrt{3}$, as are DF, EG, and GC.

It then follows that the area of triangles BAD and CAE are $16\sqrt{3}$.

Since AFB and AFD are 30-60-90 triangles, the area of sector BAD is $\frac{64\pi}{3}$, as is sector CAE.

Thus, the area outside of the two chords is $\frac{128\pi}{3}-32\sqrt{3}$. Since we want the area inside the two chords, you can subtract the outside from the whole circle, which is $64\pi-(\frac{128\pi}{3}-32\sqrt{3})$ $\Rightarrow \frac{64\pi}{3}+32\sqrt{3}$, or $\fbox{B}$.

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