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Area of Enclosed Region in Circle

XAT 2020 · Quantitative Ability & Data Interpretation · Hard · Quantitative Aptitude

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In the figure given below, the circle has a chord AB of length 12 cm, which makes an angle of 60∘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 cm2\text{cm}^2):

  1. A.

    215

  2. B.

    137

  3. C.

    35

  4. D.

    63

  5. E.

    69

Answer

E

Explanation

Since △OAB\triangle OAB is equilateral with AB=12AB = 12, radius R=12R = 12. OP=Rcos⁡30∘=12×32=63OP = R \cos 30^\circ = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}. PM=R−OP=12−63PM = R - OP = 12 - 6\sqrt{3}. OM=12OM = 12. Since MQ=OM=12MQ = OM = 12, OQ=24OQ = 24. Height of △ABQ\triangle ABQ with base AB=12AB = 12 is PQ=OQ−OP=24−63PQ = OQ - OP = 24 - 6\sqrt{3}.

Area of region = Area(\triangle ABQ) - Area(segment BMA). Area(\triangle ABQ) = 12×12×(24−63)=6(24−63)=144−363≈144−62.35=81.65\frac{1}{2} \times 12 \times (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(\triangle OAB) =60∘360∘×π×122−34×122=24π−363≈75.40−62.35=13.05= \frac{60^\circ}{360^\circ} \times \pi \times 12^2 - \frac{\sqrt{3}}{4} \times 12^2 = 24\pi - 36\sqrt{3} \approx 75.40 - 62.35 = 13.05.

Area of region = 81.65−13.05=68.6≈69 cm281.65 - 13.05 = 68.6 \approx 69\text{ cm}^2.

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