The picture shows that the distance OO1 is 6+6=12 and the radius of each circle = OA = O1A = 12. So the triangle OO1A is equilateral and the angle O1OA is 60 degrees. By symmetry so is the angle O1OB and so AOB is 120 degrees.

Now let's consider the area of the intersection on the right of the line AB.

(You'll want twice this in the end.) This is a sector of a circle with a triangle removed: specifically, the sector of the circle centre O between the radii OA and OB, minus the triangle ABO.

The area of a sector of a circle of radius 12 and angle 120 degrees is one-third (120/360) of the area of the whole circle which is pi.12^2: so pi.12^2/3 = 48.pi

The area of a triangle with angle 120 and sides 12, 12 enclosing that angle is computed as follows. It is isosceles and has the other angles 30,30. So the height (from O down to AB) is 12 sin 30 = 12. 1/2 = 6. The base is twice 12 cos 30 = 2.12.sqrt(3)/2 = 12 sqrt(3). The area is half base times height = (1/2).6.12.sqrt(3) = 36 sqrt(3).

Putting these together gives you the area you already mentioned.