Using a vertical line to break each parallelogram into a square and a triangle, and another vertical line to break the pentagon into a rectangle and a triangle, lots of congruent shapes emerge, as shown below.

The areas of the purple squares are $\tfrac{1}{4}$, because the horizontal distance from the centre to the edge of the large square if $\tfrac{1}{2}$ and half of the side length is $\frac{1}{2}$, and $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$. The four triangles are all congruent and their areas have been labelled $A$.

The area of one of the trapezia is $\frac{1}{4}+A$, so $\frac{1}{4}+A=\frac{1}{3}$, so $A=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$

The yellow area is $1\times\left(x-\frac{1}{2}\right)=x-\frac{1}{2}$, so $4A=x-\frac{1}{2}$. $$\begin{align}&4A=\tfrac{4}{12}=\tfrac{1}{3}\\

\Rightarrow&\tfrac{1}{3}=x-\tfrac{1}{2}\\

\Rightarrow&x=\tfrac{1}{3}+\tfrac{1}{2}=\tfrac{5}{6}\end{align}$$

The area of the trapezium coloured yellow in the diagram below is $\frac{1}{2}\times(a+b)\times h=\frac{1}{2}\times\left(x+\frac{1}{2}\right)\times\frac{1}{2}=\frac{1}{4}x+\frac{1}{8}$.

So $\frac{1}{4}x+\frac{1}{8}=\frac{1}{3}$, so $\frac{1}{4}x=\frac{1}{3}-\frac{1}{8}$, so $x=\frac{4}{3}-\frac{1}{2}=\frac{5}{6}$.

You can find more short problems, arranged by curriculum topic, in our short problems collection.