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Difference between revisions of "1970 AMC 12 Problems/Problem 2"

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== Solution ==
 
== Solution ==
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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

$\mathrm{(A)}\ \frac{2}{\pi}\qquad \mathrm{(B)}\ \frac{\pi}{\sqrt{2}}\qquad \mathrm{(C)}\ \frac{4}{5}\qquad \mathrm{(D)}\ \frac{5}{6}\qquad \mathrm{(E)}\ \frac{7}{8}$

Solution

Let the side length of the square be s, and let the radius of the circle be r.

$4s=2\pi r$, $2s=\pi r$, $\frac{2}{\pi}=\frac{r}{s}$

We are looking for $\frac{\pi r^2}{s^2}=\frac{2^2\pi}{\pi^2}=\frac{4}{\pi} \Rightarrow \mathrm{(F)}$.

See also

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