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

## Video Solution

https://youtu.be/PvNpudgB8LI Soo, DRMS, NM

## Video Solution by OmegaLearn

~ pi_is_3.14

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