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Last updated date: 01st Dec 2023
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MVSAT Dec 2023

Find the area of the minor segment of a circle of radius 14cm, when its central angle is ${60^0}$.Also find the area of the corresponding major segment. $\left[ {use{\text{ }}\pi {\text{ = }}\dfrac{{22}}{7}} \right]$

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Hint-Simply use formulae of area of segment. This is the case of a circular segment which is cut off from the rest of the circle.
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Now it has been given that the radius of the circle is 14cm and the central angle of the minor segment is ${60^0}$.
Now using the formulae that area of segment = $\pi {r^2}\left( {\dfrac{c}{{360}}} \right)$ where c is the central angle of segment
Area of minor segment = $\dfrac{{22}}{7} \times {(14)^2} \times \dfrac{{60}}{{360}}$
$ \Rightarrow \dfrac{{22 \times 2 \times 14}}{6} = 102.67c{m^2}$
Area of minor segment + area of major segment =Total area of the circle…………………….. (1)
Total area of circle $ = \pi {r^2}$
$ \Rightarrow \dfrac{{22}}{7} \times {\left( {14} \right)^2} = 22 \times 2 \times 14 = 616c{m^2}$
Using equation (1) area of major segment is
Total area of circle – area of minor segment
$ \Rightarrow 616 - 102.67 = 513.33c{m^2}$

Note – Whenever we face such problems the key concept is having the basic understanding of the formula for the area of segment when the central angle corresponding to a segment is given.