$ABCD$ is perpendicular trapezoid. $|AB| = 5, |DC| = 1, A(EFCD) = S_1 \space cm^2, A(ABFE) = S_2 \space cm^2$ and $S_1 = S_2$

Question

$ABCD$ is perpendicular trapezoid. $|AB| = 5, |DC| = 1, A(EFCD) = S_1 \space cm^2, A(ABFE) = S_2 \space cm^2$ and $S_1 = S_2$ What is the $|EF| = x?$

This question seems too confused.

Answer 1

For the sake of simplicity, we can say $S_1 = S_2 = 12S$. Then if we unite the extension of $AD$ and $BC$ at point $G$, we have $$\frac{A(GDC)}{A(GAB)} = \bigg(\frac{|DC|}{|AB|}\bigg)^2 = \frac{1}{25}$$ So we can say $A(GDC) = S$ (That's why in the beginning, I defined $S_1 = S_2 = 12S$). Now by a similar argument, since $\Delta GDC \sim \Delta GFE$, we have $$\bigg(\frac{|DC|}{|EF|}\bigg)^2 = \frac{A(GDC)}{A(GFE)} \implies \frac{1}{x^2} = \frac{1}{13} \implies x = \sqrt{13}$$

Answer 2

This is not an answer but a comment showing an aspect that could look a little paradoxical.

This will be understood by looking at the two displayed figures.

We see that in fact, triangle $ABG$ can be arbitrarily "high", with the same base $5$ and a same segment length $EF=\sqrt{13}$ as established in the nice proof by @ArsenBerk. More precisely, $EF$ is (apart fixed lengths $AB=5$ and $DC=1$) the unique length in the figure which can be given a value.