2010 AMC 8 Problems/Problem 23: Difference between revisions
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy> | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy> | ||
<b> | <b>Solution</b> | ||
According to the pythagorean theorem, The radius of the larger circle is: | According to the pythagorean theorem, The radius of the larger circle is: | ||
Revision as of 20:32, 3 November 2012
Semicircles 
and 
pass through the center 
. What is the ratio of the combined areas of the two semicircles to the area of circle ?
![[asy] import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/c/2/4/c249e85286e9954a4b55cd1290e9ae9dbd749f05.png)
Solution
According to the pythagorean theorem, The radius of the larger circle is:
Therefore the area of the larger circle is:
Using the coordinate plane given we find that the radius of the two semicircles to be 1. Therefore the area of the two semicircles is:
Finally the ratio of the combined areas of the two semicircles to the area of circle is 
.
