Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$, and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$. What is the radius of the semicircle on $\overline{BC}$?

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9$
If the semicircle on $\overline{AB}$ were a full circle, the area would be $16\pi$. 
$\pi r^2=16 \pi \Rightarrow r^2=16 \Rightarrow r=+4$, therefore the diameter of the first circle is $8$. 
The arc of the largest semicircle is $8.5 \pi$, so if it were a full circle, the circumference would be $17 \pi$. So the $\text{diameter}=17$.
By the Pythagorean theorem, the other side has length $15$, so the radius is $\boxed{\textbf{(B)}\ 7.5}$