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2010 AMC 8 Problems/Problem 16

Revision as of 03:00, 11 March 2012 by CYAX (talk | contribs) (Created page with "== Solution == Let the side length of the square be <math>s</math>, and let the radius of the circle be <math>r</math>. Thus we have <math>s^2=r^2\pi</math>. Dividing each side b...")
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Solution

Let the side length of the square be $s$, and let the radius of the circle be $r$. Thus we have $s^2=r^2\pi$. Dividing each side by $r^2$, we get $s^2/r^2=\pi$. Since $(s/r)^2=s^2/r^2$, we have $s/r=\sqrt{\pi}\Rightarrow \boxed{B}$

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