[ $[$ Trigonometry 

In an isosceles trapezoid $A B C D$, the lateral side $A B$ and the smaller base $B C$ are equal to 2, and $B D$ is perpendicular to $A B$. Find the area of this trapezoid.

#
Prove that the angles at the larger base of the trapezoid are $60^{\circ}$.

## Solution

Let $\angle A D B=\alpha$. Then

$$
\angle C B D=\angle A D B=\alpha
$$

and since triangle $B C D$ is isosceles,

$$
\angle C D B=\angle C B D=\alpha
$$

Therefore, $\angle A D C=2 \alpha$.

Since

$\angle B A D=90^{\circ}-\angle A D B=90^{\circ}-\alpha$ and

$$
\angle B A D=\angle A D C
$$

then $90^{\circ}-\alpha=2 \alpha$. From this, we find that

$\alpha=30^{\circ}$ and $\angle B A D=60^{\circ}$.

Let $B K$ be the height of the trapezoid.

Then

$$
B K=A B \sin 60^{\circ}=\frac{2 \sqrt{3}}{2}=\sqrt{3}
$$

and since $A D=2 A B=4$, then

$S_{\mathrm{ABCD}}=\frac{1}{2}(A D+B C) B K=\frac{1}{2}(4+2)$

$$
\frac{\sqrt{3}}{2}=3 \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-01.jpg?height=331&width=686&top_left_y=2467&top_left_x=0)

## Answer

## Problem