Splitting the square into rectangles and triangles
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}
Using the formula for the area of a trapezium
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}.