A sector of $ 56^{\circ} $ cut out from a circle contains area $ 4.4 \mathrm{~cm}^{2} $. Find the radius of the circle.

Given:

A sector of \( 56^{\circ} \) cut out from a circle contains area \( 4.4 \mathrm{~cm}^{2} \).

To do:

We have to find the radius of the circle.

Solution:

Area of the sector $= 4.4\ cm^2$

Angle at the centre $= 56^o$.
Let $r$ be the radius of the circle.

This implies,

$\pi r^{2} \times \frac{\theta}{360^{\circ}}=4.4$

$\Rightarrow \frac{22}{7} \times r^{2} \times \frac{56^{\circ}}{360^{\circ}}=4.4$

$\Rightarrow \frac{22}{7} \times \frac{7}{45} r^{2}=4.4$

$\Rightarrow \frac{22}{45} r^{2}=\frac{44}{10}$

$\Rightarrow r^{2}=\frac{44}{10} \times \frac{45}{22}$

$\Rightarrow r^{2}=9$

$\Rightarrow r^{2}=(3)^{2}$

$\Rightarrow r=3$

The radius of the circle is $3\ cm$.

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Updated on: 10-Oct-2022

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