Question

A chord of a circle of radius $12cm$ subtends an angle of ${120}^{\circ }$ at the centre. Find the area of the corresponding segment of the circle.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$)

Answer 1

In the mentioned figure,
$O$ is the centre of circle,
$AB$ is a chord
$AYB$ is the minor arc,
$OA=OB=$ Radius $=12$ cm
Arc $AYB$ subtends an angle ${120}^o$ at $O.$
Area of Sector $AOB=\frac{120}{360}\times \pi r^2$

$=\frac{1}{3}\times 3.14\times 12\times 12c{m}^2$

$=150.72c{m}^2$
By trigonometry,

In $\Delta AOC$
$AC=AO\sin {60}^o=12\times \frac{\sqrt{3}}{2}cm=10.38cm$

So, $AB=2AC$$=2\times 10.38cm=20.76cm$

And, $OC=AO\cos {60}^o=12\times \frac{1}{2}cm=6cm$
$\therefore$ Area of $△AOB=\frac{1}{2}\times AB\times OC=\frac{1}{2}\times 20.76\times 6c{m}^2=62.28c{m}^2$

Area of the segment (Area of Shaded region) $=$ Area of sector $AOB-$ Area of $△AOB$

$=(150.72-62.28)c{m}^2$

$=88.44c{m}^2$