Question

A chord of a circle of radius 12 cm subtends an angle of${120}^0$ at the centre. Find the area of the corresponding segment of the circle.
$[take\pi =3.14and\sqrt{3}=1.73]$

Answer 1

We have,

In a circle,

Radius

$r=12cm.$

$\theta ={120}^o$

Area of segment $APB=$ Area of sector $OAPB$ -Area of $\Delta OAB$

Area of sector $OAPB$

$=\frac{\theta }{{360}^o}\times \pi r^2$

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

$=\frac{1}{3}\times 3.14\times 144$

$=3.14\times 4\times 12$

$=3.14\times 48$

$=150.72c{m}^2$

Finding area of triangle $\Delta AOB$

$Areaof\Delta AOB=\frac{1}{2}\times Base\times Height$

We draw

$\begin{align}

$OM\perp AB$

$\therefore \angle OMB=\angle OMA={90}^o$

In

$\Delta OMAand\Delta OMB$

$\angle OMA=\angle OMB\left(Both{90}^o\right)$

$OM=OM\left(Bothradius\right)$

$OA=OB\left(Commonline\right)$

$\therefore \Delta OMA\cong \Delta OMB(S.A.S.Congurency)$

$\Rightarrow \angle AOM=\angle BOM$

Then,

$BM=AM=\frac{1}{2}AB.......\left(1\right)$

Now,

$\angle AOM=\angle BOM=\frac{1}{2}\angle BOA$

$=\frac{1}{2}\times {120}^0$

$={60}^o$

In right angle triangle $OMA$

$\sin O=\frac{AM}{AO}$

$\sin {60}^o=\frac{AM}{12}$

$\frac{\sqrt{3}}{2}=\frac{AM}{12}$

$AM=6\sqrt{3}$

Again, In right angle triangle $OMA$

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

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

$\frac{1}{2}=\frac{OM}{12}$

$OM=6cm.$

Now, from equation (1) and we get,

$AM=\frac{1}{2}AB$

$2AM=AB$

$AB=2AM$

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

$AB=12\sqrt{3}cm.$

Now,

$Areaof\Delta AOB=\frac{1}{2}\times Base\times Height$

$=\frac{1}{2}\times AB\times OM$

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

$=36\sqrt{3}$

$=36\times 1.73$

$62.28c{m}^2$

Hence, the Area of segment of

$APB=\text{Area}\text{of}\text{sector}OAPB-Areaof\Delta OAB$

$=150.72-62.28$

$=88.44c{m}^2$

Hence, this is the answer.