ABCD is a square with AB = 1 cm. Arcs CD and BC are semicircles.

This seems trickier  than it really is

Look  at the following figure  :

Note  that   since the sides of the  square = 1....then DB = sqrt 2 = AC

E is the  center of the  square so  BE  = CE  =  sqrt (2)/2

And angle CEB = 90°

And the circle has a radius of 1/2

So....if we take  the area of 1/2 of this  circle  and  subtract  the  area of triangle CEB we will have the same area  as  the shaded region

Area of half-circle  with radius of 1/2  =  pi (1/2)^2  / 2 =  pi/8

Area of triangle CEB  =  area of an isosceles triangle  with equal sides of sqrt (2)/2  and an included angle of 90° =

(1/2) [ sqrt (2) / 2 ] ^2 *sin 90° =  (1/2) ( 2/4) * 1  =  2/8  = 1/4

So....the shaded area =   pi/8  - 1/4  =    [ pi - 2 ] / 8