A trapezoid has height 12 and diagonals of length 13 and 15 . What is its area?
Place trapezoid $A B C D$ in the coordinate plane so that $A=(a, 12)$, $B=(b, 12)$. If $A C=13$, by the Pythagorean theorem we then have $C=(a+5,0)$, $D=(b-9,0)$. Thus, the average length of the bases is $\frac{(b-a)+((a+5)-(b-9))}{2}=7$, so the area of $A B C D$ is $7 \cdot 12=84$.