Two identical circles intersect so that their centers , and their point of which they intersect , form a square of 1cm. Find the area of the portion which is common to the two circles.

Consider the following expression:

let O and R the centers of the two circles and OR,AB the two diagonals of the square whose area is 1 sq. cm.
Then edges of square is OA=OB=BR=AR=1 cm.

Let intersection point of the two diagonals is P.
Then OP=PR=1/2 cm.

Now we know that 2*pi creates pi*r*r sq. unit area of a circle.
Then θ creates (θ*r*r)/2 sq unit area.
here θ is the angle of the square i.e. angle BOA=angle ARB=θ=pi/2. and edge of square r=1 unit.
So area of OARB=((pi/2)*1*1/2)=pi/4 sq. cm.
Then area created by chord BR or AR or OA or OB is= (Area of OARB - area of square)/2.
So all areas created by these four chords(BR,AR,OA and OB)is 4*((Area of OARB - area of square)/2).
so the area of the portion which is common to the two circles is

4*((Area of OARB - area of square)/2) + area of the square.
=4*((pi/4-1)/2) + 1
=pi/2 - 1