Question

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

Answer 1

Area of segment $APB$ = Area of $OAPB$ - Area of $△OAB$

Area of $OAPB=\frac{\theta }{360}.\pi r^2=\frac{120}{360}\times 3.14\times {\left(12\right)}^2=150.72{cm}^2$

$OM$ bisects the angle $AOB$

$\sin 60=\frac{AM}{MO}\Rightarrow \frac{\sqrt{3}}{2}=\frac{AM}{12}$

$AM=6\sqrt{3}$

$\cos 60=\frac{OM}{AO}$

$OM=6$

$AM=\frac{AB}{2}$

$AB=2\times 6\sqrt{3}$

$AB=12\sqrt{3}$

Area of $△AOB=\frac{1}{2}\times Base\times height$

$=\frac{1}{2}\times 12\sqrt{3}\times 6=62.28$

Area of segment $APB=$ Area of $OAPB$ - Area of $△OAB$

$=150.72-62.28$

$=88.42c{m}^2$