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

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== Solution ==
 
== Solution ==
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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 by <math>r^2</math>, we get <math>\dfrac{s^2}{r^2}=\pi</math>. Since <math>(\dfrac{s}{r})^2=\dfrac{s^2}{r^2}</math>, we have <math>\dfrac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}</math>
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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 by <math>r^2</math>, we get <math>\frac{s^2}{r^2}=\pi</math>. Since <math>(\frac{s}{r})^2=\frac{s^2}{r^2}</math>, we have <math>\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}</math>
  
 
==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}}
 
{{MAA Notice}}

Revision as of 16:53, 12 October 2017

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 $\frac{s^2}{r^2}=\pi$. Since $(\frac{s}{r})^2=\frac{s^2}{r^2}$, we have $\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}$

See Also

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

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