Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 4: Line 4:
  
 
== 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>\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>
+
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>- Timmy Deng ~ North Carolina
  
 
==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 21:48, 5 February 2018

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 $\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}$- Timmy Deng ~ North Carolina

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_16&oldid=90301"