10.134. In an isosceles trapezoid, one base is equal to $40 \mathrm{~cm}$, and the other is 24 cm. The diagonals of this trapezoid are perpendicular to each other. Find its area.
## Solution.

The area of an isosceles trapezoid, whose diagonals are perpendicular to each other, is equal to the square of its height: $S=h^{2}$. On the other hand, $S=\frac{a+b}{2} h$, from which $h=\frac{a+b}{2}=\frac{40+24}{2}=32$ (cm). And the area

$S=32^{2}=1024\left(\mathrm{~cm}^{2}\right)$.

Answer: $1024 \mathrm{~cm}^{2}$.