Two identical circles of area 36π overlap as shown above. If the dista

ibb.co/iyL3zz (copy-paste this to see the picture I drew for you, it won't let me attach it for some reason...)

Basically, AB = 6 which happens to be = r (you can reverse engineer "r" looking at the area that is 36pi. Area of a circle = pi*r^2, therefore 36*pi means that r = 6).

You can now build an equilateral triangle with side = 6, and find its area since it's a 30-60-90 triangle.

You can also compute the "sector's area" using the formula in the image I posted (it's just a proportional fraction of the whole circle's area).

Now you can find the difference between the two, which will give you the area of half the shaded region. Multiply that by 2 and you get the result.

For only one circle

Area = 36\(\pi\)
So Radius = 6
Distance between A to B = 6

So Origin,A & B form a equilateral Triangle
Area of \(\triangle\) = 9\(\sqrt{3}\)

\(\angle\) AOB= 60\(^{o}\) , Since Equilateral Triangle [ O= Origin]

Area of Circular segment OAB =(36\(\pi\) / \(360^{\circ}\)) x \(60^{\circ}\) = 6\(\pi\)

Shaded Portion by only one Circle = 6\(\pi\)-9\(\sqrt{3}\)

Shaded Portion by two circle = 12\(\pi\)-18\(\sqrt{3}\)