Difference between revisions of "2005 AMC 8 Problems/Problem 23"
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==Solution== | ==Solution== | ||
− | First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be 4 | + | First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be <math>4\cdot 4 = 16.</math> Divide 16 by 2 to get the original shape and you get <math>\boxed{8}</math> |
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+ | ==Video Solution== | ||
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+ | https://www.youtube.com/watch?v=cNbXCQXUc6E ~David | ||
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+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/j3QSD5eDpzU?t=1116 | ||
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+ | ~ pi_is_3.14 | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=22|num-a=24}} | {{AMC8 box|year=2005|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:52, 17 June 2024
Problem
Isosceles right triangle encloses a semicircle of area
. The circle has its center
on hypotenuse
and is tangent to sides
and
. What is the area of triangle
?
Solution
First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be Divide 16 by 2 to get the original shape and you get
Video Solution
https://www.youtube.com/watch?v=cNbXCQXUc6E ~David
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=1116
~ pi_is_3.14
See Also
2005 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.