The chord AB of a circle with centre at O is 2\sqrt{3} times the height of the minor segment. If P is the area of the sector OAB and Q is the area of the minor segment of the circle, then what is the approximate value of \frac{P}{Q}? (Take \sqrt{3}=1.7 and \pi=3.14)

  1. A. 1.4
  2. B. 1.7
  3. C. 2.2
  4. D. 2.6

Correct Answer: B. 1.7

Explanation

Let central angle be 2\theta. Chord = 2R \sin \theta. Height of segment = R - R\cos \theta. Given 2R \sin \theta = 2\sqrt{3}R(1 - \cos \theta), which simplifies to \tan(\theta/2) = 1/\sqrt{3} \implies \theta=60^{\circ}. So central angle is 120^{\circ}. P = \frac{1}{3}\pi R^2, Q = R^2(\frac{\pi}{3} - \frac{\sqrt{3}}{4}). \frac{P}{Q} = \frac{4\pi}{4\pi - 3\sqrt{3}} = \frac{12.56}{12.56 - 5.1} \approx 1.68 \approx 1.7.

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