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

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== Problem 16 ==
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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?
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<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>
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== Solution ==
 
== 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 by <math>r^2</math>, we get <math>s^2/r^2=\pi</math>. Since <math>(s/r)^2=s^2/r^2</math>, we have <math>s/r=\sqrt{\pi}\Rightarrow \boxed{B}</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>\left(\frac{s}{r}\right)^2=\frac{s^2}{r^2}</math>, we have <math>\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}</math>
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==Video by MathTalks==
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https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
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==See Also==
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{{AMC8 box|year=2010|num-b=15|num-a=17}}
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{{MAA Notice}}

Revision as of 19:55, 15 January 2023

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

Video by MathTalks

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


See Also

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

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