Overlapping Semicircles

Some elaboration : We cut the shaded region in the square $BDFE$ through the diagonal $BF$ and we move each of these halves into the arcs $AF$ and $FC$ . So we have moved the shaded region to the entire arc $AC$ . Thus the area of the shaded region simply the area of a quarter of a circle with radius 2 cm minus area of right triangle $ABC$ , to get $\dfrac{\pi r^2}4 - \dfrac12 \cdot 2^2 = \pi - 2$ . Hence our answer is $\boxed2$ .

Area of Semi-circles ADBF and BECF

(pix1x1)/2 = pi/2, Total i.e (area of ADBF and BECF) = pi ..............(1)

Area of quarter with radius 2 cm

(pix2x2)/4 = pi .......................................... (2)

From equation 1 and equation 2 we can say that shaded area inside square of 1 cm and outside it should be equal

So let the area of shaded area under square of 1 cm be Z

Considering semi circle ADBF

Area of the square = 1 sq.cm

Area of quarter with radius 1 cm = (pix1x1)/4

(1-Z)/2 + pi/4 = pi/2 -Z

=> Z=pi/2 - 1

=> 2Z = pi - 2,

So A = 2