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

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== Problem ==
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== Problem 16 ==
 
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?
 
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?
 
<math> \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} </math>
 
<math> \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} </math>

Revision as of 21:04, 22 October 2018

Problem 16

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 $\frac{s^2}{r^2}=\pi$. Since $\left(\frac{s}{r}\right)^2=\frac{s^2}{r^2}$, we have $\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}$- Timmy Deng ~ North Carolina

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

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