Difference between revisions of "2010 AMC 8 Problems/Problem 23"
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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> | ||
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+ | ==Solution== | ||
According to the pythagorean theorem, The radius of the larger circle is: | According to the pythagorean theorem, The radius of the larger circle is: | ||
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Finally the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\frac{1}{2}</math>. | Finally the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\frac{1}{2}</math>. | ||
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+ | ==See Also== | ||
+ | {{AMC8 box|year=2011|num-b=22|num-a=24}} |
Revision as of 18:54, 4 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
?
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
.
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |