Two identical rectangles overlap, as shown above. Each rectangle has

The four triangles are identical -- each shares a side of length 1 with one of the rectangles, and using parallel lines and the fact lines cross in an X, we can see the angles are the same in all four.

Since the four triangles are identical, their hypotenuses are the same length, and the blue shape in the middle is a rhombus (a parallelogram with four equal sides). Say each side is x. Then looking at any triangle, say the one at the bottom, its vertical side has length 2-x, since it's just the remaining part of the side of the rectangle of length 2. So we know its three sides are 1, 2-x and x, and using Pythagoras we have

1^2 + (2 - x)^2 = x^2
1 + 4 - 4x + x^2 = x^2
4x = 5
x = 5/4

so our rhombus has sides of length 5/4, and since the area of any parallelogram (rhombus or not) is base*height, and since the height of our rhombus is the short side of a rectangle, the area is (5/4)(1) = 5/4.

I suspect there's a much faster way to do the problem that I'm not seeing at a glance.