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?

Answer 1

7 1/17

Explanation:

A figure can be helpful.

The inscribed semicircle has its center at the midpoint of th base. It is tangent to the side of the isosceles triangle, so a radius makes a 90° angle there.

The long side of the isosceles triangle can be found from the Pythagorean theorem to be ...

BC² = BD² +CD²

BC² = 8² +15² = 289

BC = √289 = 17

The radius mentioned (DE) creates right triangles that are similar to ∆BCD. In particular, we have ...

(long side)/(hypotenuse) = DE/BD = CD/BC

DE = BD·CD/BC = 8·15/17

DE = 7 1/17 ≈ 7.059