No matter the proportions of the rectangle, it can be stretched by some factor in the horizontal and by the reciprocal of that same factor in the vertical, preserving the are of each of the regions. That being the case we can assume the rectangle is a square.

The upper right region will have sides of sqrt(2). The other two area 1 regions will have sides of ‘s’ and ‘s’-sqrt(2). We can therefore show that the side length ‘s’ must be equal to (sqrt(2)+sqrt(10))/2, or sqrt(2) times the golden ration.

The overall area of the square will then be sqrt(5)+3, and so the central area is sqrt(5).