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

Latest revision as of 14:16, 11 October 2024

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

Solution 2

Let the area of both the circle and square be $\pi$ . We know that the radius of a circle with area $\pi$ is 1 and the side length of the square is $\sqrt{\pi}$ . Now we can calculate the ratio $\frac{\sqrt{\pi}}{1} \Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}$

Video by MathTalks

https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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

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