A chord of a circle of the radius 12 cm subtends an angle of $120^o$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt3= 1.73$).

AcademicMathematicsNCERTClass 10

Given:

A chord of a circle of the radius 12 cm subtends an angle of $120^o$ at the centre.

To do: 

We have to find the area of the corresponding segment of the circle.

Solution:

Let us say there is a circle with centre O, and AB is the given chord as shown in the figure.


Here given radius of the given circle, $r=12\ cm$

Angle subtended by the chord AB, $\angle AOB=120^{o}$.

Area of the sector $AOB=\frac{\theta }{360^{o}} \pi r^{2}$

Here $\theta=120^{o}$ and$\ r=12\ cm$

$=\frac{120^{o}}{360^{o}} \times 3.14 \times ( 12)^{2}$

$=\frac{1}{3} \times 3.14 \times 12\times 12$

$=150.72 \ cm^{2}$

Area of the sector $AOB=150.72\ cm^{2}$

Area of the $\vartriangle ABC=\frac{1}{2} \times OA\times OB\times sin120^{o}$

Here $OA=OB=$radius of the given circle $=12\ cm$

And we know $sin120^{o} =\frac{\sqrt{3}}{2}$, On submitting these values in the formula,

Area of $\vartriangle ABC=\frac{1}{2} \times 12\times 12\times \frac{\sqrt{3}}{2}$

$=36\sqrt{3}$

$=36\times 1.73$ 

$=62.28\ cm^{2}$

Area of the corresponding segment of the circle $=$ Area of the sector AOB$-$Area of $\vartriangle AOB$

$\ =150.72-62.28$

$=88.44\ cm^{2}$

Therefore, the area of the corresponding segment is $88.44\ cm^{2}$. 

raja
Updated on 10-Oct-2022 13:23:55

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