Difference between revisions of "2013 AMC 8 Problems/Problem 23"
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==Solution 2== | ==Solution 2== | ||
We go as in Solution 1, finding the diameter of the circle on AC and AB. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is <math>\frac{289\pi}{8}</math>, and the middle one is <math>\frac{289\pi}{8}-\frac{64\pi}{8}=\frac{225\pi}{8}</math>, so the radius is <math>\frac{15}{2}=\boxed{\textbf{(B)}\ 7.5}</math>. | We go as in Solution 1, finding the diameter of the circle on AC and AB. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is <math>\frac{289\pi}{8}</math>, and the middle one is <math>\frac{289\pi}{8}-\frac{64\pi}{8}=\frac{225\pi}{8}</math>, so the radius is <math>\frac{15}{2}=\boxed{\textbf{(B)}\ 7.5}</math>. | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/abSgjn4Qs34?t=2584 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=22|num-a=24}} | {{AMC8 box|year=2013|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:52, 2 January 2023
Contents
Problem
Angle of is a right angle. The sides of are the diameters of semicircles as shown. The area of the semicircle on equals , and the arc of the semicircle on has length . What is the radius of the semicircle on ?
Video Solution for Problems 21-25
Video Solution
https://youtu.be/crR3uNwKjk0 ~savannahsolver
Solution 1
If the semicircle on AB were a full circle, the area would be .
, therefore the diameter of the first circle is .
The arc of the largest semicircle is , so if it were a full circle, the circumference would be . So the .
By the Pythagorean theorem, the other side has length , so the radius is
Solution 2
We go as in Solution 1, finding the diameter of the circle on AC and AB. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is , and the middle one is , so the radius is .
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=2584
~ pi_is_3.14
See Also
2013 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.