Question

A chord of a circle of radius $15$ cm subtends an angle of ${60}^o$ at the centre. Find the area of the corresponding minor and major segments of the circle.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$)

Answer 1

In a given circle, Radius$r=15$cm

And $\theta ={60}^{\circ }$

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

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

$\frac{60}{360}\times 3.14\times 15\times 15=117.75{cm}^2$

Area of $△AOB=\frac{1}{2}\times b\times h$

We draw $OM$ perpendicular to $AB$

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

In $△OMA$ and $△OMB$

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

$OA=OB$(both radius)

$OM=OM$(common)

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

$\Rightarrow \angle AOM=\angle BOM$ by C.P.C.T

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

$\Rightarrow \angle AOM=\angle BOM=\frac{1}{2}\times {60}^{\circ }$

Also, since $△OMB\cong △OMA$

$\therefore BM=AM$ by CPCT

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

In rightangled triangle $OMA$

$\sin {30}^{\circ }=\frac{AM}{AO}=\frac{AM}{15}\Rightarrow AM=\frac{15}{2}$

In $△OMA$

$\cos {30}^{\circ }=\frac{OM}{AO}\Rightarrow OM=\frac{\sqrt{3}}{2}\times 15$

From $\left(1\right)$

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

$\Rightarrow 2AM=AB$

Put the value of $AM$

$\Rightarrow AB=2\times \frac{15}{2}=15$cm

Now, Area of $△AOB=\frac{1}{2}bh=\frac{1}{2}\times AB\times OM$

$=\frac{1}{2}\times 15\times \frac{\sqrt{3}}{2}\times 15=97.3125{cm}^2$

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

$=117.75-97.3125=20.4375{cm}^2$

Thus, area of minor segment$=20.4375{cm}^2$

Now, Area of major segment$=$Area of circle$-$Area of minor segment

$=\pi r^2-20.4375=3.24\times 15\times 15-20.4375=706.5-10.4375=686.0625{cm}^2$