2016 AMC 8 Problem 25

Attempt Problem 25 of the 2016 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

A semicircle is inscribed in an isosceles triangle with base 1616 and height 1515 so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?
43 4 \sqrt{3}
12017 \dfrac{120}{17}
10 10
1722 \dfrac{17\sqrt{2}}{2}
1732 \dfrac{17\sqrt{3}}{2}
Answer: B
Concepts:triangle areaPythagorean Theorem
Small hint:
Draw the radius to the point where the semicircle touches a side of the triangle.
Big hint:
Compare the area of the right half of the triangle using two different bases.
Video solution:
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Written solution:
Let OO be the center of the circle, which is the midpoint of AB‾.\overline{AB}.
We then get that BC=17BC = 17 via the Pythagorean theorem.
The right half of the isosceles triangle has area [△BOC]=12⋅8⋅15=60.[\triangle BOC]=\dfrac12\cdot8\cdot15=60. The radius OEOE is perpendicular to the tangent side BCBC, so the same area is 12⋅OE⋅17.\dfrac12\cdot OE\cdot17. Hence OE=12017.OE=\dfrac{120}{17}.
Thus, B is the correct answer.
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