7. The diagonals of a trapezoid are perpendicular to each other, and one of them is 17. Find the area of the trapezoid if its height is 15.

Answer. $\frac{4335}{16}$.
Let $ABCD$ be the given trapezoid, $AB \| CD, BD=17$, its height $BH$ is 15. Mark a point $M$ on the line $CD$ such that $BM \| AC$. Then $ABMC$ is a parallelogram and $CM=AB$. Triangles $ABD$ and $BCM$ are equal in area (since they have equal heights and bases). Therefore, the area of trapezoid $ABCD$ is equal to the area of triangle $BDM$. Since $AC \perp BD$ and $AC \| BM$, then $BM \perp BD$. Thus, triangle $BDM$ is a right triangle. Then $DH=\sqrt{BD^2-BH^2}=8$, $MH=\frac{BH^2}{DH}=\frac{225}{8}, DM=DH+HM=\frac{289}{8}, S_{\triangle BDM}=\frac{1}{2} \cdot 15 \cdot \frac{289}{8}=\frac{4335}{16}$.