Difference between revisions of "2013 AMC 8 Problems/Problem 20"
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A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. The area is <math>\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}</math>. | A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. The area is <math>\frac{2\pi}{2}=\boxed{\textbf{(C)}\ \pi}</math>. | ||
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==Video Solution== | ==Video Solution== | ||
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https://youtu.be/tdh0u9_xjN0 ~savannahsolver | https://youtu.be/tdh0u9_xjN0 ~savannahsolver | ||
Revision as of 09:51, 16 July 2024
Problem
A rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?
Solution
A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, . The area is
.
Video Solution
https://youtu.be/tdh0u9_xjN0 ~savannahsolver
Video Solution 2
https://youtu.be/0g14IJJ2Z-8 Soo, DRMS, NM
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.
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