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UPSC CDS 2 2022 Math Paper

Question Numbers: 74-75

Consider the following for the next two (02) items that follow:

A chord of length l of a circle makes an angle 90° at the centre of the circle.

Question : 74 of 100

Marks: +1, -0

What is the area of the minor segment ?

Solution:

Concept:

The area of a minor segment is found using the formula

A = \frac{1}{2}r^2(\theta - \sin\theta)

, where

\theta

is the central angle in radians.

Explanation:

Given that the chord length

l

subtends a right angle at the centre.

\triangle OAB

OA = OB = r

and

AB = l

By Pythagoras:

l^2 = r^2 + r^2 = 2r^2

Thus

r = \frac{l}{\sqrt{2}}

The central angle

\theta = 90^{\circ} = \frac{\pi}{2}

radians.

Area of segment =

\frac{1}{2}r^2\left(\frac{\pi}{2} - \sin\frac{\pi}{2}\right)

Since

\sin\frac{\pi}{2} = 1

, we get:

A = \frac{1}{2}\left(\frac{l}{\sqrt{2}}\right)^2\left(\frac{\pi}{2} - 1\right) = \frac{1}{2} \cdot \frac{l^2}{2} \left(\frac{\pi}{2} - 1\right)

Simplifying:

A = \frac{l^2}{4}\left(\frac{\pi}{2} - 1\right)

Answer:

\frac{l^{2}}{4}\left(\frac{\pi}{2} - 1\right)

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