In an isosceles trapezoid, the midline is equal to $m$, and the diagonals are perpendicular to each other. Find the area of this trapezoid.
The projection of the diagonal of an isosceles trapezoid onto the larger base is equal to the midline of the trapezoid.

## Solution

Through vertex $C$ of the smaller base $BC$ of the given isosceles trapezoid $ABCD$, draw a line parallel to the diagonal $BD$ until it intersects the extension of the base $AD$ at point $E$. Then, $ACE$ is an isosceles right triangle. Its area is equal to the area of the given trapezoid, the base $AE$ is the sum of the bases of the trapezoid, i.e., twice the midline, and since $\angle CAD = 45^{\circ}$, the height $CH$ is equal to the segment $AH$, which, by the property of an isosceles trapezoid, is equal to its midline. Therefore,

$$
S_{\mathrm{ABCD}} = S_{\triangle \mathrm{ACE}} = \frac{1}{2} AE \cdot CH = \frac{1}{2} \cdot 2m \cdot m = m^2
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-11.jpg?height=460&width=834&top_left_y=1411&top_left_x=612)

## Answer

$m^2$