Question

A chord of a circle of radius 30 cm makes an angle of 60^\circ at the centre of the circle. Find the areas of the minor and major segments.

[ $Take\pi =3.14and\sqrt{3}\mathrm{1.732.}$ ]

Answer 1

In a given circle,

radius, r = 30 cm

And $\theta ={60}^o$

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

Area of sector OAPB $=\frac{\theta }{360}\times \pi r^2=\frac{60}{360}\times 3.14\times (30{)}^2=471c{m}^2$

Finding the area of $△$ AOB

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

We draw OM $\perp$ AB

$\angle$ OMB = $\angle$ A OMA = 90°

In $△$ OMA & $△$ OMB

$\angle$ OMA = $\angle$ OMB (Both 90°)

OA = 0B (Both radius)
OM = OM (Common)

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

$\Rightarrow \angle$ AOM = $\angle$ BOM (CPCT)

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

$\Rightarrow \angle$AOM =$\angle$BOM = $\frac{1}{2}\times {60}^o={30}^o$

Also, since $△$ OMB $\cong △$ OMA
BM = AM (CPCT)

BM = AM = $\frac{1}{2}$AB -----(1)

In right triangle $△$OMA,

$sinO=\frac{AM}{AO}sin30=\frac{AM}{30}\frac{1}{2}=\frac{AM}{30}AM=\frac{30}{2}=15$

$cosO=\frac{OM}{AO}cos30=\frac{OM}{30}\frac{\sqrt{3}}{2}=\frac{OM}{30}OM=\frac{30\sqrt{3}}{2}=15\sqrt{3}$

From (1),

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

AB = 2AM = 2 $\times$ 15 = 30 cm

Area of $△$AOB $=\frac{1}{2}\times base\times height=\frac{1}{2}\times AB\times OM=\frac{1}{2}\times 30\times 15\sqrt{3}=389.25c{m}^2$

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

= 471 - 389.25 = 81.75 $c{m}^2$

Area of minor segment = 81.75 $c{m}^2$

Area of major segment = Area of circle - Area of minor segment

= $\pi r^2-81.75=3.14\times {30}^2-81.75=2826-81.75=2744.25c{m}^2$