Compute the area of a parallelogram defined by a particular construction I got stuck with this mathematical task. Can someone help me how to solve this problem?
I need to find the F(area) value. It is kind of a thinking task

Context
The problem is extracted from a German book "Paul Eigenmann, Geometrische Wiederholungs- und Denkaufgaben. Ernst-Klett-Verlag, 1964."
I do not know where to start exactly. Can I assume that the base side is split on a half which gives us two sides with 25cm each?
There are no heights given. It is only given that what you see in the picture, the length of two parallel base lines $50$ and $57$.


If you don't want to rely on any geometric knowledge, or at least almost on none, then you can attack such a problem using coordinates.
$$
A=\begin{pmatrix}0\\0\end{pmatrix}\qquad
B=\begin{pmatrix}50\\0\end{pmatrix}\qquad
C=\begin{pmatrix}57\\h\end{pmatrix}\qquad
D=\begin{pmatrix}7\\h\end{pmatrix}
$$
The circular arcs in the figure indicate $AD=AP=CQ=CB=r$ for some length $r$ which is the radius of these arcs. So you have coordinates
$$
P=\begin{pmatrix}r\\0\end{pmatrix}\qquad
Q=\begin{pmatrix}57-r\\h\end{pmatrix}
$$
and the condition
$$AD^2 = 7^2+h^2 = r^2\tag1$$
You also have the marked right angles. The slope of $DP$ is
$$\frac{7-r}{h}$$
while the slope of $AQ$ is
$$\frac{57-r}{h}$$
If they are to be orthogonal, the slopes must multiply to $-1$:
\begin{align*}
\frac{7-r}{h} \cdot \frac{57-r}{h} &= -1 \\
(7-r)(57-r) &= -h^2 \\
399 - 64r + r^2 + h^2 &= 0 \tag2
\end{align*}
Now subtract $(1)$ from $(2)$ and you obtain
\begin{align*}
399 - 64r + r^2 - 7^2 &= -r^2 \\
2r^2 - 64r + 350 &= 0 \\
r^2 - 32r + 175 &= 0 \\
r &= 16\pm 9 \\
r_1 &= 7 \quad r_2 = 25
\end{align*}
If you plug $r_1$ into $(1)$ you get $h=0$ which is not the solution you want. So take $r_2=25$ instead. This tells you that indeed the points $P$ and $Q$ are midpoints of their respective sides. Plugging them into $(1)$ you get
$$h = \sqrt{r_2^2-7^2} = \sqrt{25^2-7^2} = 24$$
I'll leave computing the area from this to you as an excercise.
