Difference between revisions of "1952 AHSME Problems/Problem 1"
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Let the radius of the circle be the common fraction <math>\frac{a}{b}.</math> Then the area of the circle is <math>\pi \cdot \frac{a^2}{b^2}.</math> | Let the radius of the circle be the common fraction <math>\frac{a}{b}.</math> Then the area of the circle is <math>\pi \cdot \frac{a^2}{b^2}.</math> | ||
Because <math>\pi</math> is irrational and <math>\frac{a^2}{b^2}</math> is rational, their product must be irrational. The answer is <math>\boxed{B}.</math> | Because <math>\pi</math> is irrational and <math>\frac{a^2}{b^2}</math> is rational, their product must be irrational. The answer is <math>\boxed{B}.</math> | ||
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Revision as of 12:30, 5 July 2013
Problem
If the radius of a circle is a rational number, its area is given by a number which is:
Solution
Let the radius of the circle be the common fraction 
Then the area of the circle is 
Because 
is irrational and 
is rational, their product must be irrational. The answer is 
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
