Geometry

First, let's note that PXQ and RXS are similar triangles. 

The scale factor of the two is $\sqrt{8/4} = \sqrt{2}$

This means that the height of RXS is $\frac{h\sqrt{2}}{ 1 + \sqrt{2}}$ where h is the height of the trapezoid. 

We have $\frac{ h}{1 + \sqrt{2} }$ for the height of PXQ

The base of PXQ is $\text{base of RXS}/\sqrt{2}$

Now, using this information from PXQ, we can write

$\text{Area} = \frac{(1/2) (\text{base of RXS})}{\sqrt{2} * h ( 1 + \sqrt{2})} \\ 4 = (1/2) (\text{base RXS}) * h / ( 2 +\sqrt{ 2}) \\ 8(2 + \sqrt{2}) / \text{base RXS} = h$

We know the area of the trapezoid is $(1/2) h (\text{sum of bases})$

We have 

$4 [ 2 +\sqrt {2}] [ 2 + \sqrt {2} ] / / 2 \\ 2 [ 2 + \sqrt {2}] ^2 \\ 2 [ 4 + 4\sqrt {2}+ 2] = \\ 2 [ 6 + 4\sqrt {2}] = \\ 12 + 8\sqrt {2}$

Thanks! :)