# 1969 AHSME Problems/Problem 6

## Problem

The area of the ring between two concentric circles is $12\tfrac{1}{2}\pi$ square inches. The length of a chord of the larger circle tangent to the smaller circle, in inches, is: $\text{(A) } \frac{5}{\sqrt{2}}\quad \text{(B) } 5\quad \text{(C) } 5\sqrt{2}\quad \text{(D) } 10\quad \text{(E) } 10\sqrt{2}$

## Solution $[asy] draw(circle((0,0),50)); draw(circle((0,0),40)); draw((-30,40)--(30,40),dotted); draw((-30,40)--(0,0)--(0,40),dotted); draw((-5,40)--(-5,35)--(0,35)); dot((-30,40)); dot((0,40)); dot((30,40)); dot((0,0)); [/asy]$ Let $a$ be radius of larger circle, and $b$ be radius of smaller circle. The area of the ring can be determined by subtracting the area of the smaller circle from the area of the larger circle, so $$\pi a^2 - \pi b^2 = \frac{25 \pi}{2}$$ $$a^2 - b^2 = \frac{25}{2}$$ From the diagram, by using the Pythagorean Theorem, half of the chord is $\frac{5}{\sqrt{2}} = \frac{5 \sqrt{2}}{2}$ units long, so the the length of the entire chord is $\boxed{\textbf{(C) } 5 \sqrt{2}}$.

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