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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+ | ==Solution 2== | ||
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+ | The phrasing of the problem makes it clear that the rule would apply to all rational radii. So let r be equal to 1. <math>\pi</math> times {1^2} is equal to <math>\pi</math> which is irrational. Therefore, the answer is <math>\boxed{b}.</math> | ||
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+ | ~YJC64002776 | ||
==See also== | ==See also== | ||
{{AHSME 50p box|year=1952|before=First Question|num-a=2}} | {{AHSME 50p box|year=1952|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:32, 19 January 2021
Contents
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
Solution 2
The phrasing of the problem makes it clear that the rule would apply to all rational radii. So let r be equal to 1. times {1^2} is equal to which is irrational. Therefore, the answer is
~YJC64002776
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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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.