Examples - Chapter 11 Areas Related To Circles class 10 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 11 - Areas Related To Circles - Ncert Solutions class 10 - Maths

Example 1 :

Find the area of the sector of a circle with radius $4 \mathrm{~cm}$ and of angle $30^{\circ}$. Also, find the area of the corresponding major sector (Use $\pi=3.14$ ).
Solution :

Given sector is OAPB (see Fig. 11.5).

$
\begin{aligned}
\text { Area of the sector } & =\frac{\theta}{360} \times \pi r^2 \\
& =\frac{30}{360} \times 3.14 \times 4 \times 4 \mathrm{~cm}^2
\end{aligned}
$

$
=\frac{12.56}{3} \mathrm{~cm}^2=4.19 \mathrm{~cm}^2 \text { (approx.) }
$

Area of the corresponding major sector
$
\begin{aligned}
& =\pi r^2-\text { area of sector OAPB } \\
& =(3.14 \times 16-4.19) \mathrm{cm}^2 \\
& =46.05 \mathrm{~cm}^2=46.1 \mathrm{~cm}^2 \text { (approx.) } \\
& \text { Alternatively, area of the major sector }=\frac{(360-\theta)}{360} \times \pi r^2 \\
& =\left(\frac{360-30}{360}\right) \times 3.14 \times 16 \mathrm{~cm}^2 \\
& =\frac{330}{360} \times 3.14 \times 16 \mathrm{~cm}^2=46.05 \mathrm{~cm}^2 \\
& =46.1 \mathrm{~cm}^2 \text { (approx.) } \\
&
\end{aligned}
$

Example 2 :

Find the area of the segment AYB shown in Fig. 11.6, if radius of the circle is $21 \mathrm{~cm}$ and $\angle \mathrm{AOB}=120^{\circ}$. (Use $\pi=\frac{22}{7}$ )

Solution :

Area of the segment AYB
$
=\text { Area of sector OAYB }- \text { Area of } \triangle \mathrm{OAB}
$

Now, area of the sector OAYB $=\frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^2=462 \mathrm{~cm}^2$
For finding the area of $\triangle \mathrm{OAB}$, draw $\mathrm{OM} \perp \mathrm{AB}$ as shown in Fig. 11.7.
Note that $\mathrm{OA}=\mathrm{OB}$. Therefore, by $\mathrm{RHS}$ congruence, $\Delta \mathrm{AMO} \cong \Delta \mathrm{BMO}$.
So, $\mathrm{M}$ is the mid-point of $\mathrm{AB}$ and $\angle \mathrm{AOM}=\angle \mathrm{BOM}=\frac{1}{2} \times 120^{\circ}=60^{\circ}$.

Let
$
\mathrm{OM}=x \mathrm{~cm}
$

So, from $\triangle O M A$,
$
\frac{\mathrm{OM}}{\mathrm{OA}}=\cos 60^{\circ}
$
or,
$
\frac{x}{21}=\frac{1}{2} \quad\left(\cos 60^{\circ}=\frac{1}{2}\right)
$
or,
$
x=\frac{21}{2}
$

So, $\quad \mathrm{OM}=\frac{21}{2} \mathrm{~cm}$
Also,
$
\frac{\mathrm{AM}}{\mathrm{OA}}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}
$

So,
$
\mathrm{AM}=\frac{21 \sqrt{3}}{2} \mathrm{~cm}
$

Therefore,
$
\mathrm{AB}=2 \mathrm{AM}=\frac{2 \times 21 \sqrt{3}}{2} \mathrm{~cm}=21 \sqrt{3} \mathrm{~cm}
$

So,
$
\text { area of } \begin{aligned}
\triangle \mathrm{OAB} & =\frac{1}{2} \mathrm{AB} \times \mathrm{OM}=\frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \mathrm{~cm}^2 \\
& =\frac{441}{4} \sqrt{3} \mathrm{~cm}^2
\end{aligned}
$

Therefore, area of the segment AYB $=\left(462-\frac{441}{4} \sqrt{3}\right) \mathrm{cm}^2$ [From (1), (2) and (3)]
$
=\frac{21}{4}(88-21 \sqrt{3}) \mathrm{cm}^2
$