## Task A-1.2.

In an isosceles trapezoid, the midline is of length $l$, and the diagonals are mutually perpendicular. Determine the area of the trapezoid.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_a9c12f6adaba3761a080g-02.jpg?height=446&width=788&top_left_y=591&top_left_x=631)

Let: $|A B|=a, |C D|=c, |B D|=d$, and the height of the trapezoid be $v$.

Notice that in an isosceles trapezoid, the diagonals are congruent, so $|A C|=|B D|=d$. According to the problem, $l=\frac{a+c}{2}$.

Extend the line segment $\overline{A B}$ to point $E$ such that $|B E|=|D C|$.

Then $B E C D$ is a parallelogram, and $|C E|=|D B|$, and $\overline{C E} \| \overline{D B}$, so $|C E|=d$.

Triangle $A E C$ is an isosceles right triangle with legs of length $d$, because $|A C|=|C E|=d$ and $A C \perp C E (C E \| B D, B D \perp A C)$.

Therefore, the height $v$ is equal to half the length of the hypotenuse of the right triangle $A C E$, i.e., $v=\frac{1}{2}|A E|=\frac{a+c}{2}=l$.

(5 points)

Finally,

$$
P(A B C D)=\frac{a+c}{2} \cdot v=l \cdot v=l^{2}
$$