10.184. Calculate the area of a trapezoid, the parallel sides of which are 16 and 44 cm, and the non-parallel sides are 17 and 25 cm.
Solution.

Let $B K$ and $C L$ be the perpendiculars dropped onto $A D$ (Fig. 10.152). $B C=16$ (cm), $A D=44$ (cm), $A B=17$ (cm), $C D=25$ (cm). Denote $A K=x, L D=y$. Then $A B^{2}-A K^{2}=B K^{2}, C L^{2}=C D^{2}-L D^{2}$. This means that $17^{2}-x^{2}=B K^{2}, C L^{2}=25^{2}-y^{2}$. We have the system $\left\{\begin{array}{l}17^{2}-x^{2}=25^{2}-y^{2}, \\ x+y+16=44,\end{array}\right.$ from which $x=8, y=20$. We get $C L=\sqrt{625-400}=15$, $S_{A B C D}=\frac{B C+A D}{2} \cdot C L=\frac{16+44}{2} \cdot 15=450\left(\mathrm{~cm}^{2}\right)$.

Answer: $450 \mathrm{~cm}^{2}$.