Problem:
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?
Answer Choices:
A. 43​
B. 17120​
C. 10
D. 2172​​
E. 2173​​
Solution:
Let O be the midpoint of base AB of △ABC and the center of the semicircle. Triangle △OBC is a right triangle with OB=8 and OC=15, and so, by the Pythagorean Theorem, BC=17. Let E be the point where the semicircle intersects BC, so radius OE is perpendicular to BC. Then △OEB and △COB are similar, and therefore, OE:CO=OB:CB. Hence, 15OE​=178​ and so OE=17120​.
OR
Let O be the center of the semicircle, which is also the midpoint of base AB. Since OB=8 and OC=15, then by the Pythagorean Theorem BC=17. Let E be the point where the semicircle intersects BC, so radius OE is perpendicular to BC. Since the area of △OBC is 21​(BC)(OE)=21​(OB)(OC), then 21​(17)(OE)=21​(8)(15) and so OE=120/17.
