Difference between revisions of "1969 AHSME Problems/Problem 6"
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Rockmanex3 (talk | contribs) (Solution to Problem 6) |
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== Solution == | == Solution == | ||
| − | <math>\ | + | |
| + | <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 <math>a</math> be radius of larger circle, and <math>b</math> 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 | ||
| + | <cmath>\pi a^2 - \pi b^2 = \frac{25 \pi}{2}</cmath> | ||
| + | <cmath>a^2 - b^2 = \frac{25}{2}</cmath> | ||
| + | From the diagram, by using the [[Pythagorean Theorem]], half of the chord is <math>\frac{5}{\sqrt{2}} = \frac{5 \sqrt{2}}{2}</math> units long, so the the length of the entire chord is <math>\boxed{\textbf{(C) } 5 \sqrt{2}}</math>. | ||
| + | |||
== See also == | == See also == | ||
Latest revision as of 02:29, 7 June 2018
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
