2005 AMC 8 Problems/Problem 23

Problem

Isosceles right triangle $ABC$ encloses a semicircle of area $2\pi$ . The circle has its center $O$ on hypotenuse $\overline{AB}$ and is tangent to sides $\overline{AC}$ and $\overline{BC}$ . What is the area of triangle $ABC$ ?

$[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(circle(o, 2)); clip(a--b--c--cycle); draw(a--b--c--cycle); dot(o); label("$C$", c, NW); label("$A$", a, NE); label("$B$", b, SW);[/asy]$

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi$

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 $4\cdot 4 = 16.$ Divide 16 by 2 to get the original shape and you get $\boxed{8}$

Solution 2

We can figure out the radius of the semicircle because the question states that the area of the semicircle is $2\pi$ and we can multiply it by 2 to get $4\pi$ 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 = 2*2=4. Since triangle ABC is an isosceles right triangle we can calculate the area to be $\frac{4^2}{2}$ = $\boxed{8}$

Video Solution

https://www.youtube.com/watch?v=cNbXCQXUc6E ~David

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=1116

~ pi_is_3.14

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.

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