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UPSC CDS 2 2021 Math Paper

Directions (Q.Nos. 49 and 50) Consider the following for next two question that follow.

The chord of a circle of radius

2.1 \text{cm}

is inclined the angle of

120^{\circ}

on the centre of circle.

(Take

\pi=\frac{22}{7}

and

\sqrt{3}=1.732

)

Question : 49 of 100

Marks: +1, -0

What is the approximate area of minor segment of the circle ?

Solution: 👈: Video Solution

Given,

Radius

(r)

of the circle

=2.1\,\text{cm};\angle AOB=\theta=120^{\circ}

Area of minor segment of circle

=

Area of minor sector - Area of triangle formed

=\frac{\theta}{360^{\circ}}\pi r^{2}-\frac{1}{2}\times\left(2r\sin\left(\frac{\theta}{2}\right)\right)

\times r\cos\left(\frac{\theta}{2}\right)

=\frac{\theta}{360^{\circ}}\pi r^{2}-r^{2}\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)

=\frac{120^{\circ}}{360^{\circ}}\times\frac{22}{7}\times(2.1)^{2}-(2.1)^{2}

=\frac{1}{3}\times\frac{22}{7}\times\frac{21}{10}\times\frac{21}{10}

-\frac{21}{10}\times\frac{21}{10}\times\sin60^{\circ}\cos60^{\circ}

=\frac{462}{100}-\frac{441}{100}\times\frac{\sqrt{3}}{2}\times\frac{1}{2}

=\frac{1}{100}(462-110.25\times1.732)

=\frac{1}{100}(462-190.953)

=

Area of minor sector

-

Area of triangle formed

=\frac{271.047}{100}\approx2.71\,\text{cm}^{2}

Hence, approximate area of minor segment of the circle is

2.71\,\text{cm}^{2}

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