2013 AMC 8 Problems/Problem 23

Problem

Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$ , and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$ . What is the radius of the semicircle on $\overline{BC}$ ? $[asy] import graph; draw((0,8)..(-4,4)..(0,0)--(0,8)); draw((0,0)..(7.5,-7.5)..(15,0)--(0,0)); real theta = aTan(8/15); draw(arc((15/2,4),17/2,-theta,180-theta)); draw((0,8)--(15,0)); label("$A$", (0,8), NW); label("$B$", (0,0), SW); label("$C$", (15,0), SE); [/asy]$

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9$

Solution 1

If the semicircle on AB were a full circle, the area would be 16pi. Therefore the diameter of the first circle is 8. The arc of the largest semicircle would normally have a complete diameter of 17. The Pythagorean theorem says that the other side has length 15, so the radius is $\boxed{\textbf{(B)}\ 7.5}$ .

Solution 2

We go as in A, finding the diameter of the circle on AC and AB. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is $\frac{289\pi}{8}$ , and the middle one is $\frac{289\pi}{8}-\frac{64\pi}{8}=\frac{225\pi}{8}$ , so the radius is $\frac{15}{2}=\boxed{\textbf{(B)}\ 7.5}$ .

2013 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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2013 AMC 8 Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

See Also