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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| + | ==See also== | ||
| + | {{AHSME 50p box|year=1952|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:25, 1 January 2014
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 
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
