Question

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

Answer 1

Given that:-

Radius of circle $\left(r\right)=12cm$

$\theta =120°$

To find:-

Area of segment $APB=?$

Solution:-

Area of sector $OAPB\left(A_1\right)=\frac{\theta }{360°}\times \pi r^2$

$\Rightarrow A_1=\frac{120}{360}\times 3.14\times {12}^2=150.72{cm}^2$

Let M be the point on $AB$ such that $AB\perp OM$

$\therefore \angle OMA=\angle OMB=90°$

Now in $△OMA$ and $△OMB$,

$\angle OMA=\angle OMB\left[\text{each}90°\right]$

$OA=OB[\because OA=OB=r]$

$OM=OM\left[\text{common}\right]$

By R.H.S. congruency,

$△OMA\cong △OMB$

Now by CPCT,

$\angle AOM=\angle BOM$

$AM=BM$

Therefore,

$\angle AOM=\angle BOM=\frac{1}{2}\angle AOB=60°$

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

Now in right angled $△AOM$

$\sin 60=\frac{AM}{OA}[\because \sin \theta =\frac{\text{Perpndicular}}{\text{hypotenuse}}]$

$\Rightarrow \sin 60=\frac{AM}{12}$

$\Rightarrow AM=6\sqrt{3}cm$

$\cos 60=\frac{OM}{OA}[\because \cos \theta =\frac{\text{base}}{\text{hypotenuse}}]$

$\Rightarrow \cos 60=\frac{OM}{12}$

$\Rightarrow OM=6cm$

Now, from ${eq}^n\left(1\right)$, we have

$AB=2AM=12\sqrt{3}cm$

Now, area of $△AOB\left(A_2\right)$ will be-

$A_2=\frac{1}{2}\times AB\times OM=\frac{1}{2}\times 12\sqrt{3}\times 6=36\sqrt{3}{cm}^2=62.28cm$

$\therefore$ Area of segment $APB=A_1-A_2=150.72-62.28=88.44{cm}^2$

Hence the aea of the corresponding segment of the circle is $88.44{cm}^2$.