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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 $\sqrt3= 1.73$).
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 $88.44\ cm^{2}$.