Observe that $AD+BC=AF+FB=20$, and that $ABCD$ is a trapezoid with height $BC=14$. Hence the answer is $\frac{1}{2}(AD+BC)(14)=140$.
Let $\mathcal{P}$ be a parabola with focus $F$ and directrix $\ell$. A line through $F$ intersects $\mathcal{P}$ at two points $A$ and $B$. Let $D$ and $C$ be the feet of the altitudes from $A$ and $B$ onto $\ell$, respectively. Given that $AB=20$ and $CD=14$, compute the area of $ABCD$.