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

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

We see 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 he original shape and you get 8.

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AJHSME/AMC 8 Problems and Solutions

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2005 AMC 8 Problems/Problem 23

Contents

Problem

Solution

Easier and More Logical Solution

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