jmc

We first connect point $B$ with point $D$. We can see that $△BCD$ is a 3-4-5 right triangle. We can also see that $△BDA$ is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of $△BDA$ is $\frac{5\cdot 12}{2}$, and the area of $△BCD$ is $\frac{3\cdot 4}{2}$. Thus, the area of quadrilateral $ABCD$ is $30-6=\boxed{24}.$