Before I proceed on with the solution, you need to know how to find the area of a sector (of a circle).
Area of a Sector of a circle:
Formula is Sector Area = (1/2) r2 q
where "r" is the radius of the circle and "q " is the angle, in radian, defining the wideness of the sector.
Knowing the area of a sector, which is the key element in the question, you may check the below diagram A for a simplified version of the question (with the flipped back region moved back to the original location).
In diagram A, you can see that the flipped back area ACB is part of the sector area ACBD.
Within this sector is also a triangle defined by ABD.
Therefore to find the area ACB, you have to simply subtract 2 areas.
Area of flipped back ACB = Sector area ACBD - Triangle area ABD
For area of triangle, you can click here to reveal the answer (was done previously in another post). In that post, you can deduce the angle at corner D to be q = 2 x (600).
The sector angle q is determined, in radian, as 2 x (pi)/3.
So, area of sector ACBD = (1/2) r2 q = (1/2)(2 x 2) (2 x pi/3) = 4 x (pi)/3 unit 2
Therefore area of flipped back ACB = [4 x (pi)/3] - sqrt(3) unit2 (Answer)
Nothing difficult. Just some knowledge of various areas of shape, and you can solve basically any odd-shaped area.
For those who have the answer before seeing this, you are great! :)