1969 AHSME Problems/Problem 6
Problem
The area of the ring between two concentric circles is 
square inches. The length of a chord of the larger circle tangent to the smaller circle, in inches, is:
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]](http://latex.artofproblemsolving.com/e/2/3/e23e2a47e2965fb4b5fdb0d2f14e891ed8af6ecc.png)
Let 
be radius of larger circle, and 
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
![\[\pi a^2 - \pi b^2 = \frac{25 \pi}{2}\]](http://latex.artofproblemsolving.com/9/b/c/9bc588251a784316614c2296856ac75668d4c2c8.png)
![\[a^2 - b^2 = \frac{25}{2}\]](http://latex.artofproblemsolving.com/5/f/7/5f702c6a6e58c9e80d6af739fce628e219915d0e.png)
From the diagram, by using the Pythagorean Theorem, half of the chord is 
units long, so the the length of the entire chord is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
