2005 AMC 8 Problems/Problem 23


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]

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


The semi circle has an area of $\pi r^2 /2 = 2\pi$ and a radius of $2$.

Because this is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments. They also create a square in the top left corner. From this, we can conclude the legs of the triangle are twice the length of the radii, $4$. The area of the triangle is $(4)(4)/2 = \boxed{\textbf{(B)}\ 8}$.

Easier and More Logical 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 $\boxed{8}$

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS