jmc

Let $E$ and $F$ be the intersections of lines $AB$ and $BC$ with the circle. One can prove that $BCDE$ is a rectangle, so $BE=CD$. In order for the area of trapezoid $ABCD$ to be an integer, the expression $\frac{(AB+CD)BC}{2}=(AB+CD)BF$ must be an integer, so $BF$ must be rational. By Power of a Point, $AB\cdot BE=B{F}^2⟹AB\cdot CD=BF$, so $AB\cdot CD$ must be a perfect square. Among the choices, the only one where $AB\cdot CD$ is a perfect square is $\boxed{AB=9,CD=4}$.