Question

Find the area of the segment $AYB$ show in the figure, if radius of the circle is $21cm$ and $\angle AOB={120}^{\circ }$(Use $n=\frac{22}{7})$

Answer 1

In a given circle, Radius$\left(r\right)=21$cm and $\theta ={120}^{\circ }$

Area of segment $AYB=$Area of sector $OAYB-$Area of $△OAB$

Area of sector $OAYB=\frac{\theta }{360}\times \pi r^2$

$=\frac{120}{360}\times \frac{22}{7}\times 21\times 21$

$=462$sq.cm

Area of $△OAB=\frac{1}{2}\times base\times height$

$OM\perp AB$

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

In $△OMA$ and $△OMB$

$\angle OMA=\angle OMB$ both ${90}^{\circ }$

$OA=OB$(both radius)

$OM=OM$(common)

$\therefore △OMA\cong △OMB$ by R.H.S congruency

$\Rightarrow \angle AOM=\angle BOM$ by CPCT

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

Also, since $△OMB\cong △OMA$

$\therefore BM=AM$ by CPCT

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

In rightangled triangle $OMA$

$\sin O=\sin {60}^{\circ }=\frac{AM}{AO}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{AM}{21}$

$\therefore AM=\frac{21\sqrt{3}}{2}$

In rightangled triangle $OMA$

$\cos O=\cos {60}^{\circ }=\frac{AM}{AO}$

$\Rightarrow \frac{1}{2}=\frac{OM}{21}$

$\therefore OM=\frac{21}{2}$

From $\left(1\right)$

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

$\Rightarrow 2AM=AB$

Putting the value of $AM$

$AB=2\times \frac{\sqrt{3}}{2}\times 21=\sqrt{3}\times 21=21\sqrt{3}$

Now, Area of $△AOB=\frac{1}{2}\times AB\times OM=\frac{1}{2}\times 21\sqrt{3}\times \frac{21}{2}=\frac{441\sqrt{3}}{4}$

Area of the segment $AYB=$Area of sector$-$Area of $△AOB$

$=(462-\frac{441\sqrt{3}}{4})$sq.cm