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

## Question

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.

**Reasoning:**

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)

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

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

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

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

**Steps: **

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 \(APB\)

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