10.357. In trapezoid $ABCD$, the lengths of the bases $AD=24$ cm and $BC=8$ cm, and the diagonals $AC=13$ cm, $BD=5\sqrt{17}$ cm are known. Calculate the area of the trapezoid.
## Solution.

Draw $B H \perp A D$ and $C F \perp A D$ (Fig. 10.137). Let $A H=x$; then $A F=8+x, D H=24-x$. Considering that $B H=C F$, in $\triangle A F C$ and $\triangle B H D$ we have $A C^{2}-A F^{2}=B D^{2}-D H^{2} . \Rightarrow 64 x=256, x=4$ cm.

Then $C F=\sqrt{A C^{2}-A F^{2}}=\sqrt{13^{2}-12^{2}}=5 \text{ cm}$.

Thus, $S=\frac{1}{2}(24+8) \cdot 5=80 \text{ cm}^{2}$.

Answer: $\quad 80 \text{ cm}^{2}$.