Random Math Question

Find the shaded area.
Let the midpoint of line AD be E. (Centre of semi-circle). Let the arc of AC and AD cut at point F.
AE = EF = 3.5 (radius of circle)
BA = BF = 7 (radius of circle)
EB is common. Then,
Triangle ABE is congruent to Triangle FBE. Hence,
angle ABE = angle FBE = tan-1(AE/AB) = tan-1(1/2) (Note: tan-1 means tangent inverse)
angle ABF = angle ABE + angle FBE = 2 tan-1(1/2) (get this angle in degree)
Area of sector ABF = (2 tan-1 (1/2))/360 x 7 x 7 x pi = (49 pi tan-1(1/2))/180

angle AEB = angle FEB = tan-1(AB/AE) = tan-1(2)
angle AEF = angle AEB + angle FEB = 2 tan-1(2) (get this angle in degree too)
Area of sector AEF = (2 tan-1(2))/360 x 3.5 x 3.5 x pi = (49 pi tan-1(2))/720

Area of quadrilateral ABFE = 3.5x7 = 24.5

Area of shaded region = Area of sector ABF + Area of sector AEF - Area of quadrilateral ABFE
= (49 pi tan-1(1/2))/180 + (49 pi tan-1(2))/720 - 24.5
= 11.8cm2 (round off to 1 decimal point)
(exact pi is used, not 22/7)