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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.

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Hint- Here, we will be using the formula for area of the sector when the radius of the circle and the angle made by the arc is given.
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$= \left( {\dfrac{1}{4}} \right) \times \left( {\dfrac{{22}}{7}} \right) \times {\left( {10} \right)^2} = 78.5{\text{ c}}{{\text{m}}^2}$

Note- There are two sectors made by the given chord i.e., major sector and minor sector. Here, major sector refers to the sector whose arc subtends major angle at the centre of the circle and minor sector refers to the sector whose arc subtends minor angle at the centre of the circle.

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