Solution. Since the diagonals of the trapezoid are perpendicular, we get that $\text{A S B=90}$, 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,

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

!

Therefore,

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

## VIII Section