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A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor sector.

Last updated date: 21st Mar 2023
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Given, radius of the circle $r = 10$ cm
Angle subtended by the chord at the centre or the angle made by the minor arc $\theta = {90^0}$
Since, we know that area of any sector whose arc make an angle of $\theta$ at the centre$= \left( {\dfrac{\theta }{{{{360}^0}}}} \right) \times \pi {r^2}$
$\Rightarrow$Required area of the minor sector whose minor arc makes an angle of ${90^0}$ at the centre$= \left( {\dfrac{{{{90}^0}}}{{{{360}^0}}}} \right) \times \pi {\left( {10} \right)^2}$
$\Rightarrow$Required area of the minor sector whose minor arc makes an angle of ${90^0}$ at the centre$= \left( {\dfrac{1}{4}} \right) \times \left( {\dfrac{{22}}{7}} \right) \times {\left( {10} \right)^2} = 78.5{\text{ c}}{{\text{m}}^2}$