# Ex.12.2 Q4 Areas Related to Circles Solution - NCERT Maths Class 10

## Question

A chord of a circle of radius \(10\, \rm{cm}\) subtends a right angle at the centre. Find the area of the corresponding

(i) minor segment (ii) major sector. (Use \(\pi = 3.14\))

## Text Solution

**What is known?**

Radius of the circle and angle subtended by the chord at the center.

**What is unknown?**

(i) Area of minor segment

(ii) Area of major segment

**Reasoning:**

In a circle with radius r and angle at the centre 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.

Area of the triangle\(\begin{align}= \frac{1}{2} \times \rm base \times height\end{align}\)

Draw a figure to visualize the area to be found out.

Here,\(\begin{align}{\rm{Radius, }}r = 10\rm \,cm{\rm{\theta }} = {90^{\rm{o}}}\end{align}\)

Visually it's clear from the figure that;

Let \({AB}\) is the chord subtends a right angle at the centre.

(i) Area of minor segment \({APB} =\) \(\text{Area of sector}\) \({OAPB}\, -\, \)\(\text{Area of right}\)\(\Delta {AOB}\)

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

Area of the right triangle\(\begin{align}\Delta AOB = \frac{1}{2} \times OA \times OB\end{align}\)

\[\begin{align}&= {\frac{\theta }{{{{360}^\circ }}} \times \pi {r^2}}\\&= {\frac{{{{90}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2}}\\&= {\frac{{\pi {r^2}}}{4}}\end{align}\]

\[\begin{align} &= \frac{1}{2} \times {\text{ base }} \times \,\text{height}\\ &= \frac{1}{2} \times {OA} \times {OB}\,\,\,\,\,\,\,\,\,\,\left( {\because {{OA}} \bot {{OB}}} \right)\\& = \frac{1}{2} \times {\text{radius}} \times \rm{radius}\end{align}\]

**Steps:**

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

\[\begin{align} &= {\frac{{{{90}^\circ }}}{{{{360}^\circ }}} \times \pi {{r^2}} - \frac{1}{2} \times OA \times OB\;(\because OA \bot OB)} \\ &= { \frac{{\pi \rm{r^2}}}{4} - \frac{1}{2} \times {r} \times {r}} \\&= \rm{r^2}\left( {\frac{\pi }{4} - \frac{1}{2}} \right) \hfill \\ &= {r^2}\left( {\frac{{3.14 - 2}}{4}} \right) \\ &= {\left( {\frac{{{r^2} \times 1.14}}{4}} \right)} \\ & = {\frac{{10 \times 10 \times 1.14}}{4}{\text{ cm}^2}} \\ &= {28.5\,\,{\text{cm}^2}} \end{align}\]

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

\[\begin{align} &= \pi \times 10 \times 10 - 28.5\\&= 3.14 \times 100 - 28.5\\&= 314 - 28.5\\&= 285.5{\text{ cm}^2}\end{align} \]