What is the area of the minor segment?

Consider the following for the next two (02) items that follow:<br><br>A chord of length l of a circle makes an angle 90^\circ at the centre of the circle.

  1. A. \frac{l^{2}}{2}(\pi-\frac{1}{2})
  2. B. \frac{l^{2}}{4}(\pi-\frac{1}{2})
  3. C. \frac{l^{2}}{4}(\frac{\pi}{2}-1)
  4. D. \frac{l^{2}}{2}(\frac{\pi}{2}-\frac{1}{2})

Correct Answer: C. \frac{l^{2}}{4}(\frac{\pi}{2}-1)

Explanation

Let radius be r. In the right-angled triangle at the centre, r^2 + r^2 = l^2 \implies r^2 = \frac{l^2}{2}. Area of the minor segment = Area of sector - Area of triangle = \frac{90^\circ}{360^\circ}\pi r^2 - \frac{1}{2}r^2. Substituting r^2: \frac{\pi}{4}(\frac{l^2}{2}) - \frac{1}{2}(\frac{l^2}{2}) = \frac{l^2}{4}(\frac{\pi}{2}-1).

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