1.69. In trapezoid $ABCD$, the lengths of the bases $AD=24 \, \text{cm}$, $BC=8 \, \text{cm}$, and the diagonals $AC=13 \, \text{cm}$, $BD=5 \sqrt{17} \, \text{cm}$ are known. Calculate the area of the trapezoid.
1.69. Draw $B H \perp A D$ and $C F \perp A D$ (Fig. 1.55). 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

$$
A C^{2}-A F^{2}=B D^{2}-D H^{2}
$$

Or

$13^{2}-(8+x)^{2}=(5 \sqrt{17})^{2}-(24-x)^{2}$,

from which $64 x=256$, i.e., $x=4$ (cm).

Then $C F=\sqrt{A C^{2}-A F^{2}}=\sqrt{13^{2}-12^{2}}=5$ (cm). Therefore,

$$
\begin{aligned}
S= & \frac{1}{2}(24+8) \cdot 5=80\left(\mathrm{~cm}^{2}\right) \\
& \text { Answer: } 80 \mathrm{~cm}^{2}
\end{aligned}
$$