Difference between revisions of "2010 AMC 8 Problems/Problem 23"
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==Solution== | ==Solution== | ||
By the Pythagorean Theorem, the radius of the larger circle turns out to be <math>\sqrt{1^2 + 1^2} = \sqrt{2}</math>. Therefore, the area of the larger circle is <math>(\sqrt{2})^2\pi = 2\pi </math>. Using the coordinate plane given, we find that the radius of each of the two semicircles to be 1. So, the area of the two semicircles is <math>1^2\pi=\pi</math>. Finally, the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\boxed{\textbf{(B)}\ \frac{1}{2}}</math>. | By the Pythagorean Theorem, the radius of the larger circle turns out to be <math>\sqrt{1^2 + 1^2} = \sqrt{2}</math>. Therefore, the area of the larger circle is <math>(\sqrt{2})^2\pi = 2\pi </math>. Using the coordinate plane given, we find that the radius of each of the two semicircles to be 1. So, the area of the two semicircles is <math>1^2\pi=\pi</math>. Finally, the ratio of the combined areas of the two semicircles to the area of circle <math>O</math> is <math>\boxed{\textbf{(B)}\ \frac{1}{2}}</math>. | ||
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==Video Solution by OmegaLearn== | ==Video Solution by OmegaLearn== |
Revision as of 22:47, 31 July 2023
Contents
Problem
Semicircles and pass through the center . What is the ratio of the combined areas of the two semicircles to the area of circle ?
Solution
By the Pythagorean Theorem, the radius of the larger circle turns out to be . Therefore, the area of the larger circle is . Using the coordinate plane given, we find that the radius of each of the two semicircles to be 1. So, the area of the two semicircles is . Finally, the ratio of the combined areas of the two semicircles to the area of circle is .
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=903
~ pi_is_3.14
=Video by MathTalks
Video Solution by WhyMath
~savannahsolver
See Also
2010 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.