Question

A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

$4 \sqrt{3}$

$\dfrac{120}{17}$

$10$

$\dfrac{17\sqrt{2}}{2}$

$\dfrac{17\sqrt{3}}{2}$

Video solution:

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Written solution:

Let $O$ be the center of the circle, which is the midpoint of $\overline{AB}.$

We then get that $BC = 17$ via the Pythagorean theorem.

In addition, we can also calculate the area of $\triangle BOC$ as: $\begin{align*}\triangle BOC &= \dfrac{1}{2} \cdot 8 \cdot 15 \\&= 60.\end{align*}$ As the area of $\triangle BOC=60=\dfrac{1}{2} OE \cdot CB$ we can see that $OE = \dfrac{120}{17}.$

Thus, B is the correct answer.

2016 AMC 8 Problem 25

Video solution:

Written solution:

Problem 25 in Other Years