jmc

Label points $O,A,B,C,D,E$ as follows.



The shaded area is the area of $△CDE$. To find this area, we examine pairs of similar triangles to find desired side lengths.

First, we have $△EOA\sim △COB$, so we have $$\frac{EA}{CB}=\frac{OA}{OB}=\frac{3}{3+9}=\frac{1}{4},$$ and since we know $CB=9$, we can find that $EA=9/4$. This means that $DE=9-9/4=27/4$.

Since we know $DE$ and $DC$, we can now find the area of triangle $CDE$. The desired area is $\frac{27/4\cdot 9}{2}=\frac{243}{8}=30.375$. This value, rounded to the nearest integer as requested, is $\boxed{30}$.