In the figure given below, the circle has a chord AB of length 12 cm, which makes an angle of 60 degree at the center of the circle, O. ABCD, as shown in the diagram, is a rectangle. OQ is the perpendicular bisector of AB, intersecting the chord AB at P, the arc AB at M and CD at Q. OM = MQ. The area of the region enclosed by the line segments AQ and QB,and the arc BMA is closest to(in cm square) –

A. 215

B. 137

C. 35

D. 69

D. 69

In triangle OAP, since AP=6, OA will be 12

Area of AQBMA=Area of Triangle ABQ- (Area of minor arc AMB-Area of OAB)

Length of PQ=MQ+PM = 12+(12- 6$\sqrt{3}$​)=24- 6$\sqrt{3}$

Area of triangle ABQ= $\frac{1}{2}$*12*24- 6$\sqrt{3}$= 81.64

Area of minor arc AMB-Area of OAB=$\frac{60}{360}$*$\Pi$*144 – $\frac{1}{2}$*12*6$\sqrt{3}$ =13.07

Area of AQBMA= 68.57 =69(approx.)