If we draw a vertical line through the middle of the figure it should be clear that the diameter of the large circle is the sum of the diameters of the other two circles. It should also be clear that this vertical line will pass through the midpoint of BD. Since we are told ABCD has side length of 2, BD has length sqrt(8) and each half of BD has length sqrt(2).

Using the intersecting chords theorem, we know that the product of the two smaller diameters must be the same as the product of the two halves of BD, namely 2.

If we call the RADIUS of the smallest circle x and the radius of the other inner circle y, then we know that 2x*2y =2, and so xy=1/2.

The outer circle has radius z = x+y

To find the shaded region we simply need to find the area of the outer circle and subtract the areas of the other two circles.

Shaded region S = (pi)z^2 – (pi)x^2 – (pi)y^2.

S/pi = z^2 – x^2 – y^2

= (x+y)^2 – x^2 – y^2

= x^2 + 2xy + y^2 – x^2 – y^2

= 2xy

But we know xy=1/2, and so S/pi = 1.

Therefore the shaded region S = pi, and is not dependent upon the size of the circles.