Difference between revisions of "2010 AMC 8 Problems/Problem 16"

(Solution)
Line 8: Line 8:
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=15|num-a=17}}
 
{{AMC8 box|year=2010|num-b=15|num-a=17}}
 +
{{MAA Notice}}

Revision as of 00:51, 5 July 2013

Problem

A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle? $\textbf{(A)}\ \frac{\sqrt{\pi}}{2}\qquad\textbf{(B)}\ \sqrt{\pi}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ 2\pi\qquad\textbf{(E)}\ \pi^{2}$

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 $s/r=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}$

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png