8. In $\triangle A B C$, $A B=16, B C=5 \sqrt{5}, C A=9$. Outside $\triangle A B C$, the area covered by the set of points that are less than 6 units away from points $B, C$ is


$$
\begin{array}{l}
\cos \angle B A C=\frac{16^{2}+9^{2}-(5 \sqrt{5})^{2}}{2 \cdot 16 \cdot 9}=\frac{53}{72}, \\
\cos \angle B D C=\frac{6^{2}+6^{2}-(5 \sqrt{5})^{2}}{2 \cdot 6 \cdot 6}=-\frac{53}{72},
\end{array}
$$

Thus, points $A, B, C, D$ are concyclic $\Rightarrow \alpha+\beta=\pi$.
$$
S_{\triangle B D C}=\frac{1}{2} \cdot 5 \sqrt{5} \cdot \frac{\sqrt{19}}{2}=\frac{5 \sqrt{95}}{4} \text {, }
$$

Therefore, the area of the shaded region is $54 \pi+\frac{5 \sqrt{95}}{4}$.