Ex.12.2 Q7 Areas Related to Circles Solution - NCERT Maths Class 10

Go back to  'Ex.12.2'


A chord of a circle of radius \(12\, \rm{cm} \)subtends an angle of \(120^\circ \) at the centre. Find the area of the corresponding segment of the circle.

(Use \(\pi =3.14\) and \(\sqrt{3}=1.73\))

Text Solution

What is known?

A chord of a circle with radius \(({r}) = 12 \,\rm{cm}\) subtends an angle \((\theta) =120^\circ \) at the centre.

What is unknown?

Area of segment of the circle.


In a circle with radius r and angle at the center with degree measure \(\theta\);

(i) Area of the sector\(\begin{align} = \frac{{\rm{\theta }}}{{360}^{\rm{o}}} \times \pi {r^2}\end{align}\)

(ii) Area of the segment = Area of the sector - Area of the corresponding triangle

Draw a figure to visualize the problem

Here,\(\begin{align}{\rm{Radius,}}r=12\rm{cm}\,\,\,\,\,\,{\rm{\theta }} = {120^{\rm{\circ}}}\end{align}\)

Visually it's clear from that figure that;

AB is the chord subtends\({120^{\rm{\circ}}}\) angle at the centre.

To find area of the segment AYB, We have to find area of the sector OAYB and area of the\(\Delta AOB\)

(i) Area of sector OAYB \(\begin{align}= \frac{{\rm{\theta }}}{{{{360}^\circ }}}{\rm{ \times }}\pi {r^2}\end{align}\)

(ii) Area of  \(\begin{align}\Delta {\rm{AOB}} = \frac{1}{2} \times \rm{}base \times \rm{}height\end{align}\)

For finding area of\(\Delta {AOB,}\)draw \({OM}\, \bot\, {AB}\)then find \({AB}\) and height\({OM.}\)

\(\begin{align}\Delta {{OMA}}\,& = \,\Delta {{OMB = 9}}{{{0}}^ \circ }\end{align}\)

\(\therefore \;\Delta \,{{OAM}}\, \cong \Delta \, {OBM}\)(By RHS Congruency)

\(\Rightarrow \quad \angle {{AOM}}= \angle {{BOM}}\)\[\begin{align} &= \frac{1}{2}\angle {{AOB}}\\& = \frac{1}{2} \times {120^ \circ }\\ &= {60^ \circ }\end{align}\]

\(\begin{align}\frac{AM}{OA}=\sin {{60}^\circ } = \frac{\sqrt 3 }{2}\end{align}\) \[\begin{align}& ={ AM = \frac{{OA\sqrt 3 }}{2}}\\ &= \frac{{12\sqrt 3 }}{2}\\ &= 6\sqrt 3 \end{align}\]

\(\begin{align}{\frac{{OM}}{{OA}}} {= \cos {{60}^\circ } = \frac{1}{2}}\end{align}\) \[\begin{align}OM& = \frac{{12\,\rm{{cm}}}}{2}\\&= 6\,\rm{{cm}}\end{align}\]

\(\text{Area of}\Delta OAB\)\[\begin{align}&=\frac{{{1}}}{{{2}}}\,\times \,{{base}}\,\times\,{{height}}\\&=\,\frac{{{1}}}{{{2}}}\times\,{{AB}}\, \times \,{{OM}}\end{align}\]


Here,\(\begin{align}{\rm{Radius, }}r = 12\,cm\,\,\,\,\,\,{\rm{\theta }} = {120^{\rm{\circ}}}\end{align}\)

\(\text{Area of the sector}\)\[\begin{align}&= \frac{{{{120}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2}\\ &= \frac{1}{3} \times 3.14 \times {(12)^2}\\& = \frac{{452.16}}{6}\,\rm{cm^2}\\ &= 150.72\,\rm {c m^2}\end{align}\]

Draw a perpendicular \({OM}\) from \({O}\) to chord \({AB}\)

In \(\Delta \,{AOM} \)and \(\Delta \,{BOM}\) 

\({AO}={BO}\)(radii of circle)

\({OM} = {OM}\)(common)

\(\angle {OMA} = \angle {OMB = {{90}^\circ}}\)

\(\therefore \;\Delta {AOM} \cong \Delta {BOM}\)(By RHS Congruency)

\(\begin{align}\therefore \angle {AOM} &= \angle {BOM} \\&= \frac{1}{2} \angle {AOB} = 60^\circ \end{align}\)

In \(\Delta {AOM}\)

\[\begin{align}\frac{{{{AM}}}}{{{{OA}}}}\, &=\,{{sin}}\,{{6}}{{{0}}^{{o}}}\\{{AM}}\, &= \,{{OA}}\frac{{\sqrt {{3}} }}{{{2}}}\\&=\frac{{{{12}}\sqrt {{3}} }}{{{2}}}\\&= 6\sqrt {{3}}\,\rm{{cm}}\\{AM} &= {2AM}\\&= 12\sqrt {{3}} \,\rm{{cm}}\end{align}\]

Area of \(\Delta \,{OAB}\) 

\[\begin{align}&=\frac{1}{2}\times AB\times OM \\&=\frac{1}{2}\times 12\sqrt{3}\times 6 \\&=36\sqrt{3} \\ &=36\times 1.73 \\&=62.28\,\rm{c}{{{m}}^{2}} \\\end{align}\]

Area segment \({APB} =\)Area of sector \({OAPB}\)\(-\)Area of\(\Delta \,{OAB}\, \)

\[\begin{align}&=150.72-62.28\\&=88.44\; \rm{cm^2} \end{align}\]