
Solution: Produce the legs of the trapezoid until they intersect at point $E$. The triangles $A B H$ and $E C D$ are congruent (ASA). The area of $A B C D$ is equal to area of triangle $E A H$ of hypotenuse

$$
A E=A D+D E=A D+A H=8
$$

Let $M$ be the midpoint of $A E$. Then

$$
M E=M A=M H=4
$$

and $\angle A M H=30^{\circ}$. Now, the altitude from $H$ to AM equals one half of $M H$, namely 2. Finally, the area is 8 .

![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-13.jpg?height=802&width=300&top_left_y=1689&top_left_x=1272)


G2 In a right trapezoid $A B C D(A B \| C D)$ the angle at vertex $B$ measures $75^{\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $B C$. If $B H=D C$ and $A D+A H=8$, find the area of $A B C D$.
