The area of the pentagon ABCDE, given that $$AB = BC = CD = AE = 2, DE = 2\sqrt{2}$$ and $$\angle BAE = \angle BCD = 90^\circ$$ is 

Draw a rough figure, write length of sides alongside and draw diagonals BE and BD. 
You can notice that BE is hypotenuse of triangle BAE and hence, BE = 2*(root 2)
Similarly, BD is also the hypotenuse of BCD and BD = 2*(root 2) 

Next, we can see that BDE becomes an equilateral triangle of side 2*(root 2)
Area of the entire pentagon = Area of the 2 right angled triangles BAE and BCD + Area of equilateral triangle BED
= 4 + 2*(root 3) sq units