jmc

Since $△AOB$ is isosceles with $AO=OB$ and $OP$ is perpendicular to $AB$, then $P$ is the midpoint of $AB$, so $AP=PB=\frac{1}{2}AB=\frac{1}{2}(12)=6$. By the Pythagorean Theorem, $OP=\sqrt{A{O}^2-A{P}^2}=\sqrt{10^2-6^2}=\sqrt{64}=8$.

Since $ABCD$ is a trapezoid with height of length 8 ($OP$ is the height of $ABCD$) and parallel sides ($AB$ and $DC$) of length $12$ and $24$, then its area is $$\frac{1}{2}\times \text{mbox}Height\times \text{mbox}Sumofparallelsides=\frac{1}{2}(8)(12+24)=\boxed{144}.$$