XAT 2020 — QA & DI Question 13
In the figure given below, the circle has a chord AB of length 12 cm, which makes an angle of 60° 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):
Answer & solution
- A
137
- B
63
- C
35
69
- E
215
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In a circle a chord which subtends an angle of 60° at the center is equal to the radius of the circle.
∴ OA = OB = AB = 12
âOAB is an equilateral triangle
OP (height of an equilateral triangle) = √3/2 × 12 = 6√3
∴ PM = OM = OP = 12 - 6√3
Area of AQBMA = Area of Triangle ABQ – Area of segment AMBP
Area of AQBMA = Area of Triangle ABQ – (Area of minor arc AMBO - Area of âOAB)
Now, PQ = MQ + PM = 12 + (12 - 6√3) = 24 - 6√3
Area of triangle ABQ = 1/2 × 12 × (24 - 6√3) = 6(24 - 6√3) = 81.64
Also, Area of minor arc AMB – Area of OAB
= 60/360 × π × 144 - 1/2 × 12× 6√3 = 24π - 36√3 = 13.07
∴ Area of AQBMA = 68.57 ≈ 69.
Hence, option (d).