Problem 24

Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$, so $OC = 17$. By symmetry, $O$ is the midpoint of $DA$, so $OD=OA=\frac{16}{2}= 8$. By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$), we have that $CD$ (or $AB$) is $\sqrt{17^2-8^2}=15$. Accordingly, the area of $ABCD$ is $16\cdot 15=\boxed{\textbf{(A) }240}$.