Incircle area of a 16-16-24 isosceles triangle

An isosceles triangle has sides $16$, $16$ and $24$. The biggest circle that fits inside it has area $\dfrac{p\pi}{q}$ with $p/q$ in lowest terms. Compute $p+q$.

$151$. The height onto the base of $24$ is $\sqrt{16^2-12^2}=4\sqrt7$, so the area is $48\sqrt7$. Dividing by the semiperimeter $28$ gives inradius $r=\tfrac{12\sqrt7}{7}$, and $\pi r^2=\tfrac{144\pi}{7}$, so $p+q=151$.