jmc

Since $△AOB$ is isosceles with $AO=OB$ and $OP$ is perpendicular to $AB$, point $P$ 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 $24$, its area is $$\frac{1}{2}\times \text{mbox}Height\times \text{mbox}Sumofparallelsides=\frac{1}{2}(8)(12+24)=\boxed{144}.$$

Since $XY$ cuts $AD$ and $BC$ each in half, then it also cuts the height $PO$ in half. Thus, each of the two smaller trapezoids has height 4. Next, we find the length of $XY$. The sum of the areas of trapezoids $ABYX$ and $XYCD$ must equal that of trapezoid $ABCD$. Therefore, $$\begin{array}{rl}\frac{1}{2}(4)(AB+XY)+\frac{1}{2}(4)(XY+DC)& =144\\ 2(12+XY)+2(XY+24)& =144\\ 4(XY)& =72\\ XY& =18\end{array}$$ Thus, the area of trapezoid $ABYX$ is $\frac{1}{2}(4)(12+18)=60$ and the area of trapezoid $XYCD$ is $\frac{1}{2}(4)(18+24)=84$. Thus, the ratio of their areas is $60:84=5:7$. Our answer is then $5+7=\boxed{12}$.