[Projections about [

Find the area of an isosceles trapezoid if its diagonal is equal to $l$, and the angle between this diagonal and the larger base is $\alpha$.

#
The projection of the diagonal of an isosceles trapezoid onto the base is equal to the midline of the trapezoid.

## Solution

Let $K$ be the projection of vertex $C$ of the smaller base $BC$ of trapezoid $ABCD$ onto the larger base $AD$, diagonal $AC = l$, and $\angle CAD = \alpha$. Then,

$CK = AC \sin \angle CAD = l \sin \alpha, AK = l \cos \alpha$

Since segment $AK$ is equal to the midline of the trapezoid, we have

$$
S_{\mathrm{ABCD}} = CK \cdot AK = l^2 \sin \alpha \cos \alpha = \frac{1}{2} l^2 \sin 2 \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-02.jpg?height=306&width=504&top_left_y=1631&top_left_x=91)

## Answer

$l^2 \sin \alpha \cos \alpha = \frac{1}{2} l^2 \sin 2 \alpha$.

## Problem