Question

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 $\overline{BC}?$

$7$

$7.5$

$8$

$8.5$

$9$

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Written solution:

The semicircle on $\overline{AB}$ has area $8\pi$ , so its full circle would have area $16\pi$ . Its radius is therefore $4$ , and $AB=8$ .

A semicircle of radius $r$ has arc length $\pi r$ . Since the arc on $\overline{AC}$ has length $8.5\pi$ , its radius is $8.5$ , so $AC=17$ .

Using the Pythagorean theorem in right triangle $ABC$ , $BC=\sqrt{17^2-8^2}=\sqrt{225}=15.$ The radius of the semicircle on $\overline{BC}$ is $15/2=7.5$ .

Thus, B is the correct answer.

2013 AMC 8 Problem 23

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Problem 23 in Other Years