In the right triangle $ABC$, $AC=12$, $BC=5$, and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?

$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}$
[asy] draw((0,0)--(12,0)--(12,5)--cycle); draw((0,0)--(12,0)--(12,-5)--cycle); draw(circle((8.665,0),3.3333)); label("$A$", (0,0), hhhhhhhhhhhhhhhhhhhhhhhh); label("$C$", (12,0), E); label("$B$", (12,5), NE); label("$B'$", (12,-5), NE); label("$12$", (7, 0), S); label("$5$", (12, 2.5), E); label("$5$", (12, -2.5), E);[/asy]
We can see that our circle is the incircle of $ABB'.$ We can use a formula for finding the radius of the incircle. The area of a triangle $= \text{Semiperimeter} \cdot \text{inradius}$ .  The area of $ABB'$ is $12\times5 = 60.$ We use the pythagorean triples to find out that AB is 13. The semiperimeter is $\dfrac{10+13+13}{2}=18.$ Simplifying $\dfrac{60}{18} = \dfrac{10}{3}.$ Our answer is therefore $\boxed{\textbf{(D)}\ \frac{10}{3}}.$
~CHECKMATE2021