Difference between revisions of "1970 AMC 12 Problems/Problem 2"
(New page: A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is) |
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| + | == Problem == | ||
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A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is | A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is | ||
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| + | <math>\mathrm{(A)}\ \frac{2}{\pi}\qquad \mathrm{(B)}\ \frac{\pi}{\sqrt{2}}\qquad \mathrm{(C)}\ \frac{4}{5}\qquad \mathrm{(D)}\ \frac{5}{6}\qquad \mathrm{(E)}\ \frac{7}{8}</math> | ||
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| + | == Solution == | ||
| + | Let the side length of the square be s, and let the radius of the circle be r. | ||
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| + | <math>4s=2\pi r</math>, <math>2s=\pi r</math>, <math>\frac{2}{\pi}=\frac{r}{s}</math> | ||
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| + | We are looking for <math>\frac{\pi r^2}{s^2}=\frac{2^2\pi}{\pi^2}=\frac{4}{\pi} \Rightarrow \mathrm{(F)}</math>. | ||
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| + | ==See also== | ||
Latest revision as of 14:49, 9 January 2008
Problem
A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is
Solution
Let the side length of the square be s, and let the radius of the circle be r.
, 
, 
We are looking for 
.
