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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?
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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>
 
<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>
  
 
== 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>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{\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>\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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==Solution 2==
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Let the area of both the circle and square be <math>\pi</math>. We know that the radius of a circle with area <math>\pi</math> is 1 and the side length of the square is <math>\sqrt{\pi}</math>. Now we can calculate the ratio <math>\frac{\sqrt{\pi}}{1} \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==
 
==See Also==
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{{AMC8 box|year=2010|num-b=15|num-a=17}}
 
{{AMC8 box|year=2010|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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


See Also

2010 AMC 8 (Problems • Answer Key • Resources)
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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