Difference between revisions of "2005 AMC 8 Problems/Problem 23"
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<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi </math> | <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi </math> | ||
| − | == | + | ==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> |
| + | |||
| + | ==Solution 2== | ||
| + | We can figure out the radius of the semicircle because the question states that the area of the semicircle is <math> 2\pi</math> and we can multiply it by 2 to get <math> 4\pi </math> which we can see it is 2 from the formula. Draw line segment OD such that it is the midsegment of triangle ABC, using the midsegment theorem we can see that line segment AC = <math>2*2=4</math>. Since triangle ABC is an isosceles right triangle we can calculate the area to be <math>\frac{4^2}{2}</math> = <math>\boxed{8}</math> | ||
| + | |||
| + | ==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== | ==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 09:14, 18 October 2024
Contents
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 ?
![[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(circle(o, 2)); clip(a--b--c--cycle); draw(a--b--c--cycle); dot(o); label("$C$", c, NW); label("$A$", a, NE); label("$B$", b, SW);[/asy]](http://latex.artofproblemsolving.com/8/a/7/8a7fba91b0cdfee41e69cbd89d6b1389d4169877.png)
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 
Solution 2
We can figure out the radius of the semicircle because the question states that the area of the semicircle is and we can multiply it by 2 to get 
which we can see it is 2 from the formula. Draw line segment OD such that it is the midsegment of triangle ABC, using the midsegment theorem we can see that line segment AC = 
. Since triangle ABC is an isosceles right triangle we can calculate the area to be 
=
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. 
