Question

In the right triangle $ABC,$ $AC=12,$ $BC=5,$ and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?

$\dfrac{7}{6}$

$\dfrac{13}{5}$

$\dfrac{59}{18}$

$\dfrac{10}{3}$

$\dfrac{60}{13}$

Video solution:

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Written solution:

Let $O$ be the center of the inscribed semicircle and $D$ be the tangent point of the semicircle on $\overline{AB}.$ Then $BD = 5$ since $\overline{BD}$ and $\overline{BC}$ are tangents to the semicircle. Then $AD = 8$ and $OD = r.$ $\overline{OD}$ is perpendicular to $\overline{AB}$ so $\triangle ADO \sim \triangle BCA,$ so $\dfrac{r}{8} = \dfrac{5}{12}.$ Solving this, we get $r = \dfrac{10}{3}.$

Thus, D is the correct answer.

2017 AMC 8 Problem 22

Video solution:

Written solution:

Problem 22 in Other Years