7. In trapezoid $A B C D, A D$ is parallel to $B C$. If $A D=52, B C=65, A B=20$, and $C D=11$, find the area of the trapezoid.
Answer: 594
Solution: Extend $A B$ and $C D$ to intersect at $E$. Then $\sqrt{\frac{[E A D]}{[E B C]}}=\frac{A D}{B C}=\frac{4}{5}=\frac{E A}{E B}=\frac{E D}{E C}$. This tells us that $E B=5 A B=100$, and $E C=5 C D=55$. Triangle $E B C$ has semiperimeter 110 , and so by Heron's formula, the area of triangle $E B C$ is given by $\sqrt{110(10)(55)(45)}=1650$. Since $\frac{[A E D]}{[B E D]}=\frac{16}{25}$, the area of $A B C D$ is exactly $\frac{9}{25}$ of that, or 594 .