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

Go back to  'Ex.12.2'

Question

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.

Reasoning:

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\(\Delta AOB\)

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

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

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)

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

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

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

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

Steps:

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\(\Delta \,{OAB}\, \)

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