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 $\sqrt{3}=1.732$).

Answer 1

Area of segment $\overline{AB}=$ Area of sector $AOB$ $-$ area of ${\Delta }^{\prime \prime }AOB$

Area of sector $AOB=\frac{\theta }{{360}^0}\times \pi \times {\left(12\right)}^2$

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

$=48\times 3.14=150.72$

$OD=12\cos {60}^0$, $AD=12\sin {60}^0\Rightarrow AB=2\times 12\sin {60}^0$

area of ${\Delta }^{\prime \prime }AOB=\frac{1}{2}\times 12\sin {60}^0\times 2\times \cos {60}^0\times 12$

$=6\times 12\times \sin {120}^0$

$=72\times \frac{\sqrt{3}}{2}=36\times \sqrt{3}=62.35$

area of segment $=150.72-62.35=88.4$ ${cm}^2$