Question

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 $	ext{cm}^2$):

Accepted Answer

E. Since $	riangle OAB$ is equilateral with $AB = 12$, radius $R = 12$. $OP = R \cos 30^\circ = 12 	imes \frac{\sqrt{3}}{2} = 6\sqrt{3}$. $PM = R - OP = 12 - 6\sqrt{3}$. $OM = 12$. Since $MQ = OM = 12$, $OQ = 24$. Height of $	riangle ABQ$ with base $AB = 12$ is $PQ = OQ - OP = 24 - 6\sqrt{3}$. Area of region = Area(	riangle ABQ) - Area(segment BMA). Area(	riangle ABQ) = $\frac{1}{2} 	imes 12 	imes (24 - 6\sqrt{3}) = 6(24 - 6\sqrt{3}) = 144 - 36\sqrt{3} \approx 144 - 62.35 = 81.65$. Area(segment BMA) = Area(sector OAB) - Area(	riangle OAB) $= \frac{60^\circ}{360^\circ} 	imes \pi 	imes 12^2 - \frac{\sqrt{3}}{4} 	imes 12^2 = 24\pi - 36\sqrt{3} \approx 75.40 - 62.35 = 13.05$. Area of region = $81.65 - 13.05 = 68.6 \approx 69	ext{ cm}^2$.

Area of Enclosed Region in Circle

Explanation

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