1. Let $A B$ and $C D$ be the bases of trapezoid $A B C D$, and $E$ the midpoint of leg $B C$. If $A E=10, D E=8$ and $\measuredangle A E D=30^{\circ}$, determine the area of this trapezoid.
1. Let $a$ and $b$ be the bases of the observed trapezoid, and $h$ its height. Then the heights from vertex $E$ in $\triangle A B E$ and $\triangle D C E$ are $\frac{h}{2}$, so we have:

$$
\begin{aligned}
\frac{a+b}{2} h & =P(A B C D) \\
& =P(\triangle A B E)+P(\triangle C D E)+P(\triangle A D E) \\
& =\frac{a h}{4}+\frac{b h}{4}+P(\triangle A D E)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-57.jpg?height=221&width=523&top_left_y=401&top_left_x=981)

Dr 2017 1B 1

from which it follows that $P(\triangle A D E)=\frac{a+b}{4} h=\frac{P(A B C D)}{2}$. Further, if $h_{1}$ is the height of $\triangle A D E$ from vertex $D$, then we have $h_{1}=\frac{D E}{2}=4$ (because $\left.\measuredangle A E D=30^{\circ}\right)$, and then $P(\triangle A D E)=\frac{A E \cdot h_{1}}{2}=\frac{10 \cdot 4}{2}=20$. Finally, $P(A B C D)=$ $2 P(\triangle A D E)=40$.