Difference between revisions of "2010 AMC 8 Problems/Problem 16"
(Created page with "== 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 b...") |
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| + | == 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? | ||
== 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> | 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> | ||
Revision as of 11:33, 11 March 2012
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?
Solution
Let the side length of the square be 
, and let the radius of the circle be 
. Thus we have 
. Dividing each side by 
, we get 
. Since 
, we have 
