In the figure given below, the circle has a chord AB of length 12 cm, which makes an angle of $$60^\circ$$ 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$$^2$$):

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=$$\frac{1}{2}\cdot12.\left(24-6\sqrt{\ 3}\right)=6\left(24-6\sqrt{\ 3}\right)$$.=81.64

Area of minor arc AMB-Area of OAB=$$\frac{60}{360}\cdot\pi\ \cdot144-\frac{1}{2}\cdot12\cdot6\sqrt{\ 3}$$ = $$24\pi\ -36\sqrt{\ 3}$$.=13.07

.'.Area of AQBMA=68.57$$\approx\ 69$$