Ex.12.2 Q6 Areas Related to Circles Solution - NCERT Maths Class 10

Go back to  'Ex.12.2'


A chord of a circle of radius \(15\, \rm{cm}\) subtends an angle of \(60^\circ\) at the Centre. Find the areas of the corresponding minor and major segments of the circle.

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

Text Solution

What is known?

A chord of a circle with radius \({(r)} = 15\, \rm{cm}\) subtends an angle \(\begin{align}(\theta)=60^{\circ}\end{align}\) at the centre.

What is unknown?

Area of minor and major segments of the circle.


In a circle with radius r and angle at the centre with degree measure \(\rm{\theta }\);

(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 area to be found out.

Here, \(\begin{align}\rm{Radius, }r =15\,cm\,\,\,\,\,\,\,\,\,\,{\rm{ \theta }} = {60^{\rm{\circ}}}\end{align}\)

Visually it's clear from the figure that;

AB is the chord subtends \(\begin{align}{60^{\rm{\circ}}}\end{align}\) angle at the center.

(i) Area of minor segment APB = Area of sector \(\begin{align}\rm{OAPB} - {\text{ Area of }}\Delta AOB\end{align}\)

(ii) Area of major segment \(\begin{align}AQB{\rm{ }} = \pi {r^2} - \text{Area of minor segment}{\text{ }}APB\end{align}\)

\(\begin{align}\therefore \angle {{OAB}}\,= \,\angle {{OBA}} \end{align}\) (Angles opposite equal \(= \rm{}x^\circ\) (say) sides in a triangle)

\(\begin{align}\angle {{AOB + }}\angle {{OAB + }}\angle {{OBA = 18}}{{{0}}^{{o}}}\end{align}\) (angle sum of triangle)

\[\begin{align}{{{60}^\circ } + x^\circ + x^\circ }&= {{{180}^\circ }}\\{2{{x^\circ}}} &= {{{180}^\circ } - {{60}^\circ }}\\x^\circ &= {\frac{{{{120}^\circ }}}{2}}\\x^\circ &= {{{60}^\circ }}\end{align}\]

\[\begin{align}&= \frac{\theta }{{{{360}^\circ }}} \times \pi {r^2}\\ &= \frac{{{{60}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2}\end{align}\]

\[\begin{align} &= {\frac{{{{60}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2} - \frac{{{r^2}\sqrt 3 }}{4}}\\& = {\frac{{\pi {r^2}}}{6} - \frac{{{r^2}\sqrt 3 }}{4}}\end{align}\]

\(\begin{align}= \pi {r^2} \end{align} \) - \(\text{Area of minor segment}\)


Here,\(\begin{align}\rm{Radius, }r = 15\,cm\,\,\,\,\,\,\,\,{\rm{\theta }} = 60^\circ\end{align}\)

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

\[\begin{align}& = {\frac{{{{60}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2}}\\&= { \frac{1}{6} \times 3.14 \times {{15}^2}}\\& = {117.75\,\rm{{c}}{{{m}}^2}}\end{align}\]

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

\(\begin{align}OA = OB = r\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{}}\left( {{\text{radii of the circle}}} \right)\end{align}\)

\(\begin{align} \angle OBA=\angle OAB  \end{align}\) ( Angles opposite equal sides in a triangle)

\[\begin{align} \angle {AOB+}\angle {OBA+}\angle {OAB}&=180^\circ \quad\left( \text{Angle sum of a triangle} \right) \\\angle {OBA+}\angle {OAB}&=18{{{0}}^{{o}}}{-60^\circ} \\ {2}\angle {OAB}&=120^{\circ} \\ \angle {OAB}&={{60}^{\circ }} \end{align}\]

\(\therefore \,\Delta {{OAB}}\) is an equilateral triangle (as all its angles are equal)

\(\text{Area of }\Delta AOB\)\[\begin{align}&= \frac{{\sqrt 3 }}{2}{({{side}})^2}\\& = \frac{{\sqrt 3 }}{2}{r^2}\\& = \frac{{\sqrt 3 }}{4}{(15)^2}\\& = \frac{{225\sqrt 3 }}{4}\\ &= 56.25\sqrt 3 \\ &= 56.25 \times 1.73\\& = 97.3125\,\rm{{ cm}^2}\end{align}\]

Area of minor segment \(APB =\) Area of sector \(OAPB\,-\,\)Area of \(\Delta AOB\)

\[\begin{align}&= 117.75 - 97.3125\\ &= 20.4375\,\rm{{c}}{{{m}}^2}\end{align}\]

Area of major segment \(AQB\,=\,\) Area of circle \(-\,\)Area of minor segment \(APB\)

\[\begin{align}&= \pi {(15)^2} - 20.4375\\&= 3.14 \times 225 - 20.4375\\&= 706.5 - 20.4375\\&= 686.0625\,\rm{{ cm}}^2 \end{align}\]



Frequently Asked Questions

What are Class 10 NCERT Exemplars?
While getting good scores in school tests is a desirable outcome, it is not a reliable indicator of how strong your child’s math foundation really is. Many students who score well in school exams in their earlier years, might struggle with math in higher grades because of a weak foundation. At Cuemath, we evaluate your child’s grasp of math fundamentals, and take corrective actions immediately. Also, your child may have limited exposure in their school, and in most cases, may not feel challenged to learn more. Cuemath's customised learning plan ensures your child is challenged with varied difficulty levels of questions at every stage.
What is the difference between CBSE and NCERT syllabus for Class 10?
How will Class 10 NCERT books help in exam preparation?
How will Class 10 NCERT books help you understand basic math concepts?
Which is the best video solution for the class 10 maths NCERT?