Notes
lengths in a crossed trapezium solution

Lengths in a Crossed Trapezium

Solution by Similar Triangles and Pythagoras' Theorem

In the above diagram, let $a$ and $b$ be the lengths of the parallel sides in the trapezium, so that $G E$ has length $a$ and $A D$ has length $b$. Triangles $G H E$ and $A H B$ are similar, so $F H$ and $H C$ are in the ratio $a : b$. Since $F C$ has length $3$, this means that $F H$ has length $\frac{3 a}{a + b}$ and $H C$ has length $\frac{3 b}{a + b}$.

The area of triangle $G H C$ is $\frac{3 a^2}{2(a + b)}$ and of $A H D$ is $\frac{3 b^2}{2(a + b)}$ so:

Therefore $b - a = 4$.

Applying Pythagoras' theorem to triangle $G B A$ then shows that

and so $x = 5$.