8,9

In trapezoid $ABCD$, side $AD$ is the larger base. It is known that $AD=CD=4 \frac{2}{3}$, $\angle BAD=90^{\circ}$, and $\angle BCD=150^{\circ}$. On base $AD$, triangle $AED$ is constructed, and points $B$ and $E$ lie on the same side of line $AD$ with $AE=DE$. The height of this triangle, drawn from vertex $E$, is $1 \frac{2}{5}$. Find the area of the common part of trapezoid $ABCD$ and triangle $AED$.
Prove that $\angle A D E>\angle A D C$ and apply the Law of Sines.

## Solution

Notice that

$$
\angle A D C=180^{\circ}-\angle B C D=30^{\circ}
$$

Let $P$ be the midpoint of $A D$. Since point $E$ lies on the perpendicular bisector of segment $A D$, then

$$
\operatorname{tg} \alpha=\operatorname{tg} \angle A D E=\frac{E P}{P D}=\frac{3}{5}>\frac{1}{\sqrt{3}}=\operatorname{tg} 30^{\circ}=\operatorname{tg} \angle A D C
$$

Therefore, $\angle A D C<\angle A D E$ and ray $D C$ lies between the sides of angle $A D E$. Moreover,

$$
E P=\frac{7}{5}<\frac{7}{3}=A B
$$

Thus, segment $A E$ intersects side $C D$. Let $F$ be the point of intersection. Then the desired common part is triangle $A F D$, in which

$$
A D=\frac{14}{3}, \angle A D F=30^{\circ}, \angle D A F=\alpha=\operatorname{arctg} \frac{3}{5}
$$

We will find side $D F$ using the Law of Sines:

$$
\begin{gathered}
D F=\frac{A D \sin \alpha}{\sin \left(30^{\circ}+\alpha\right)}=\frac{14 \sin \alpha}{3\left(\sin \alpha \cos 30^{\circ}+\cos \alpha \sin 30^{\circ}\right)}= \\
=\frac{14}{3\left(\cos 30^{\circ}+\operatorname{ctg} \alpha \sin 30^{\circ}\right)}=\frac{28}{3 \sqrt{3}+5}
\end{gathered}
$$

Therefore,

$$
S_{\triangle \mathrm{AFD}}=\frac{1}{2} A D \cdot D F \sin 30^{\circ}=\frac{49(3 \sqrt{3}-5)}{3}
$$

## Answer

$\frac{49(3 \sqrt{3}-5)}{3}$