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?
[asy]draw((0,0)--(8,15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,180));[/asy]
$\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}$
First, we drop a perpendicular, shown above, to the base of the triangle, cutting the triangle into two congruent right triangles. This triangle is isosceles, which means perpendiculars are medians and vice versa. The base of the resulting right triangle is $8$ for both sides, and the height is $15,$ as given. Using the Pythagorean theorem, we can find the length of the hypotenuse, or $17.$ Using the two legs of the right triangle, we find the area of the right triangle, $60$. $\frac{60}{17}$ times $2$ results in the radius, which is the height of the right triangle when using the hypotenuse as the base. Hence, the answer is $\boxed{\textbf{(B) }\frac{120}{17}}$.

Note — There are several other following solutions below that use similar methods of finding the area two different ways shown in this solution. Try to eliminate any words (such as 'can' or 'would be') that make your solution less concise. Try to write 'sure of yourself.'