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?

$4 \sqrt{3}$

$\dfrac{120}{17}$

$10$

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

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

Answer: B

Concepts:triangle area Pythagorean Theorem

Difficulty rating: 1610

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.

The right half of the isosceles triangle has area $[\triangle BOC]=\dfrac12\cdot8\cdot15=60.$ The radius $OE$ is perpendicular to the tangent side $BC$ , so the same area is $\dfrac12\cdot OE\cdot17.$ Hence $OE=\dfrac{120}{17}.$

Thus, B is the correct answer.

2016 AMC 8 Problem 25

Answer: B

Difficulty rating: 1610

Video solution:

Written solution:

Problem 25 in Other Years