Task A-2.7. (10 points)

In quadrilateral $A B C D$, the following holds:

$$
\angle A B C=\angle A D C=90^{\circ}, \quad|A B|=|B C|, \quad|C D|+|D A|=m .
$$

Determine the area of quadrilateral $A B C D$ in terms of $m$.
## Solution.

Let's introduce the notations

$$
|A B|=|B C|=a, \quad|C D|=c, \quad|D A|=d
$$

Due to the conditions of the problem, we have $c+d=m$.

![](https://cdn.mathpix.com/cropped/2024_05_30_d6d90a155ebaa296cbbeg-13.jpg?height=708&width=746&top_left_y=825&top_left_x=655)

The area of the quadrilateral $A B C D$ is

$$
P=\frac{|A B| \cdot|B C|}{2}+\frac{|C D| \cdot|D A|}{2}=\frac{a^{2}+c d}{2}
$$

From the Pythagorean theorem, we get

$$
|A C|^{2}=2 a^{2}=c^{2}+d^{2}
$$

We need to express the area of the triangle in terms of $m=c+d$,

so we substitute $a^{2}=\frac{1}{2}\left(c^{2}+d^{2}\right)$ into formula $(*)$.

Now the desired area is

$$
\begin{aligned}
P & =\frac{a^{2}+c d}{2}=\frac{\frac{c^{2}+d^{2}}{2}+c d}{2} \\
& =\frac{c^{2}+d^{2}+2 c d}{4}=\frac{(c+d)^{2}}{4} \\
& =\frac{m^{2}}{4}
\end{aligned}
$$

(1 point)