Solution. Since the diagonals of the trapezoid are perpendicular, we get that $\Varangle A S B=90^{\circ}$, so $\triangle A B S$ is an isosceles right triangle and $\triangle A M S$ is an isosceles right triangle. Therefore, $h_{1}=\frac{a}{2}=6 \mathrm{~cm}$.

Similarly, $h_{2}=\frac{b}{2}=4 \mathrm{~cm}$. Thus,

$$
h=h_{1}+h_{2}=10 \mathrm{~cm}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_4fa80d9916883368a7a6g-5.jpg?height=322&width=390&top_left_y=1810&top_left_x=1074)

Therefore,

$$
P=\frac{a+b}{2} \cdot h=\frac{12+8}{2} \cdot 10=100 \mathrm{~cm}^{2}
$$

## VIII Section
Problem 4. The diagonals of an isosceles trapezoid with perpendicular diagonals measure 12 cm and $8 \mathrm{~cm}$. Calculate the area of the trapezoid.