Exercise 11.1 (Revised) - Chapter 11 - Areas Related To Circles - Ncert Solutions class 10 - Maths

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NCERT Class 10 Maths Solutions: Chapter 11 - Areas Related To Circles

Unless stated otherwise, take $\pi=\frac{22}{7}$.
Ex 11.1 Question 1.

Find the area of a sector of a circle with radius $6 \mathbf{~ c m}$, if angle of the sector is $60^{\circ}$.

Answer.

Here, $r=6 \mathrm{~cm}$ and $\theta=60^{\circ}$
Area of sector $=\frac{\theta}{360^{\circ}} \times \pi r^2$
$=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6 \times 6=\frac{132}{7} \mathrm{~cm}^2$
Ex 11.1 Question 2.

Find the area of a quadrant of a circle whose circumference is $22 \mathrm{~cm}$.

Answer 

Given, $2 \pi r=22 \mathrm{~cm}$
$\Rightarrow 2 \times \frac{22}{7} \times r=22$
$\Rightarrow r=\frac{7}{2} \mathrm{~cm}$
We know that for quadrant of circle, $\theta=90^{\circ}$
$\therefore$ Area of quadrant $=\frac{\theta}{360^{\circ}} \times \pi r^2$
$=\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}=\frac{77}{8} \mathrm{~cm}^2$

Ex 11.1 Question 3.

The length of the minute hand of a clock is $14 \mathrm{~cm}$. Find the area swept by the minute hand in 5 minutes.

Answer.

Here, $r=14 \mathrm{~cm}$ and since the minute hand rotates through $\frac{360^{\circ}}{60^{\circ}}=6^{\circ}$ in one minute, therefore, angle swept by minute hand in 5 minutes $=\theta=6^{\circ} \times 5=30^{\circ}$.
$
\begin{aligned}
& \therefore \text { Area swept }=\frac{\theta}{360^{\circ}} \times \pi r^2 \\
& =\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14=\frac{154}{3} \mathrm{~cm}^2
\end{aligned}
$
Ex 11.1 Question 4.

A chord of a circle of radius $10 \mathrm{~cm}$ subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment, (ii) major segment. (Use $\pi=3.14$ )

Answer.

(i)Here, $r=10 \mathrm{~cm}$ and $\theta=90^{\circ}$

$\begin{aligned}
& \therefore \text { Area of minor sector }=\frac{\theta}{360^{\circ}} \times \pi r^2 \\
& =\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times 10 \times 10=78.5 \mathrm{~cm}^2 \\
& \text { Area of } \Delta \mathrm{OAB}=\frac{1}{2} \times \text { Base } \times \text { Height } \\
& =\frac{1}{2} \times 10 \times 10=50 \mathrm{~cm}^2
\end{aligned}$

$\therefore$ Area of minor segment $=$ Area of minor sector - Area of $\triangle \mathrm{OAB}$
$
=78.5-50=28.5 \mathrm{~cm}^2
$
(ii) For major sector, radius $=10 \mathrm{~cm}$ and $\theta=360^{\circ}-90^{\circ}=270^{\circ}$
$\therefore$ Area of major sector $=\frac{\theta}{360^{\circ}} \times \pi r^2$
$
=\frac{270^{\circ}}{360^{\circ}} \times 3.14 \times 10 \times 10=235.5 \mathrm{~cm}^2
$
Ex 11.1 Question 5.

In a circle of radius $21 \mathrm{~cm}$, an arc subtends an angle of $60^{\circ}$ at the centre. Find:
(i) the length of the arc.
(ii) area of the sector formed by the arc.
(iii) area of the segment formed by the corresponding chord.

Answer.

Given, $r=21 \mathrm{~cm}$ and $\theta=60^{\circ}$

$\begin{aligned}
& \text { (i) Length of arc }=\frac{\theta}{360^{\circ}} \times 2 \pi r \\
& =\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21=22 \mathrm{~cm}
\end{aligned}$

(ii) Area of the sector $=\frac{\theta}{360^{\circ}} \times \pi r^2$
$
=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 21 \times 21=231 \mathrm{~cm}^2
$
(iii) Area of segment formed by corresponding chord
$
\begin{aligned}
& =\frac{\theta}{360^{\circ}} \times \pi r^2-\text { Area of } \Delta \mathrm{OAB} \\
& \Rightarrow \text { Area of segment }=231-\text { Area of } \triangle \mathrm{OAB}
\end{aligned}
$

In right angled triangle OMA and OMB,
$\mathrm{OM}=\mathrm{OB}$ [Radii of the same circle]
$\mathrm{OM}=\mathrm{OM}[$ Common $]$
$\therefore \triangle \mathrm{OMA} \cong \triangle \mathrm{OMB}[$ RHS congruency]
$\therefore \mathrm{AM}=\mathrm{BM}[$ By C.P.C.T.]
$\therefore \mathrm{M}$ is the mid-point of $\mathrm{AB}$ and $\angle \mathrm{AOM}=\angle \mathrm{BOM}$
$
\begin{aligned}
& \Rightarrow \angle \mathrm{AOM}=\angle \mathrm{BOM} \\
& =\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2} \times 60^{\circ}=30^{\circ}
\end{aligned}
$

Therefore, in right angled triangle OMA,
$
\cos 30^{\circ}=\frac{\mathrm{OM}}{\mathrm{OA}} \Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{OM}}{21}
$

$\begin{aligned}
& \Rightarrow \mathrm{OM}=\frac{21 \sqrt{3}}{2} \mathrm{~cm} \\
& \text { Also, } \sin 30^{\circ}=\frac{\mathrm{AM}}{\mathrm{OA}} \Rightarrow \frac{1}{2}=\frac{\mathrm{AM}}{21}
\end{aligned}$

$
\begin{aligned}
& \Rightarrow \mathrm{AM}=\frac{21}{2} \mathrm{~cm} \\
& \therefore \mathrm{AB}=2 \mathrm{AM}=2 \times \frac{21}{2}=21 \mathrm{~cm} \\
& \therefore \text { Area of } \triangle \mathrm{OAB}=\frac{1}{2} \times A B \times O M \\
& =\frac{1}{2} \times 21 \times \frac{21 \sqrt{3}}{2}=\frac{441 \sqrt{3}}{4} \mathrm{~cm}^2
\end{aligned}
$

Using eq. (i),
Area of segment formed by corresponding chord $=\left(231-\frac{441 \sqrt{3}}{4}\right) \mathrm{cm}^2$
Ex 11.1 Question 6.

A chord of a circle of radius $15 \mathrm{~cm}$ subtends an angle of $60^{\circ}$ at the centre. Find the area of the corresponding minor and major segment of the circle.
$
\text { (Use } \pi=3.14 \text { and } \sqrt{3}=1.73 \text { ) }
$

Answer.

Here, $r=15 \mathrm{~cm}$ and $\theta=60^{\circ}$

$\begin{aligned}
& \text { Area of minor sector }=\frac{\theta}{360^{\circ}} \times \pi r^2 \\
& =\frac{60^{\circ}}{360^{\circ}} \times 3.14 \times 15 \times 15
\end{aligned}$

$
=117.75 \mathrm{~cm}^2
$

For, Area of $\triangle A O B$,
Draw $\mathrm{OM} \perp \mathrm{AB}$.
In right triangles OMA and OMB,
$\mathrm{OA}=\mathrm{OB}$ [Radii of same circle]
$\mathrm{OM}=\mathrm{OM}[$ Common $]$
$\therefore \triangle \mathrm{OMA} \cong \triangle \mathrm{OMB}[\mathrm{RHS}$ congruency]
$\therefore \mathrm{AM}=\mathrm{BM}[$ By C.P.C.T. $]$
$\Rightarrow \mathrm{AM}=\mathrm{BM}=\frac{1}{2} \mathrm{AB}$ and
$\angle \mathrm{AOM}=\angle \mathrm{BOM}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2} \times 60^{\circ}=30^{\circ}$
In right angled triangle $\mathrm{OMA}, \cos 30^{\circ}=\frac{\mathrm{OM}}{\mathrm{OA}}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{OM}}{15}$
$\Rightarrow \mathrm{OM}=\frac{15 \sqrt{3}}{2} \mathrm{~cm}$
Also, $\sin 30^{\circ}=\frac{\mathrm{AM}}{\mathrm{OA}}$
$\Rightarrow \frac{1}{2}=\frac{\mathrm{AM}}{15} \Rightarrow \mathrm{AM}=\frac{15}{2} \mathrm{~cm}$

$\Rightarrow 2 \mathrm{AM}=2 \times \frac{15}{2}=15 \mathrm{~cm}$

$
\begin{aligned}
& \Rightarrow \mathrm{AB}=15 \mathrm{~cm} \\
& \therefore \text { Area of } \Delta \mathrm{AOB}=\frac{1}{2} \times A B \times O M \\
& =\frac{1}{2} \times 15 \times \frac{15 \sqrt{3}}{2}=\frac{225 \sqrt{3}}{3} \\
& =\frac{225 \times 1.73}{4}=97.3125 \mathrm{~cm}^2 \\
& \therefore \text { Area of minor segment }=\text { Area of minor sector }- \text { Area of } \Delta \mathrm{AOB} \\
& =117.75-97.3125=20.4375 \mathrm{~cm}^2 \\
& \text { And, Area of major segment }=\pi \mathrm{r}^2-\text { Area of minor segment } \\
& =3.14 \times 15 \times 15-20.4375=706.5-20.4375=686.0625 \mathrm{~cm}^2
\end{aligned}
$
Ex 11.1 Question 7.

A chord of a circle of radius $12 \mathrm{~cm}$ subtends an angle of $120^{\circ}$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi=3.14$ and $\sqrt{3}=1.73$ )

Answer.

Here, $r=12 \mathrm{~cm}$ and $\theta=120^{\circ}$
$
\text { Area of corresponding sector }=\frac{\theta}{360^{\circ}} \times \pi r^2
$

$=\frac{120^{\circ}}{360^{\circ}} \times 3.14 \times 12 \times 12$

$=150.72 \mathrm{~cm}^2$
For, Area of $\triangle \mathrm{AOB}$,
Draw $\mathrm{OM} \perp \mathrm{AB}$.
In right triangles OMA and OMB,
$\mathrm{OA}=\mathrm{OB}[$ [Radii of same circle]
$\mathrm{OM}=\mathrm{OM}$ [Common]
$\therefore \triangle \mathrm{OMA} \cong \triangle \mathrm{OMB}$ [RHS congruency]
$\therefore \mathrm{AM}=$ BM [By C.P.C.T.]
$\Rightarrow \mathrm{AM}=\mathrm{BM}=\frac{1}{2} \mathrm{AB}$ and
$\angle \mathrm{AOM}=\angle \mathrm{BOM}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2} \times 120^{\circ}=60^{\circ}$
In right angled triangle $\mathrm{OMA}, \cos 60^{\circ}=\frac{\mathrm{OM}}{\mathrm{OA}}$
$
\begin{aligned}
& \Rightarrow \frac{1}{2}=\frac{O M}{12} \\
& \Rightarrow O M=6 \mathrm{~cm}
\end{aligned}
$

Also, $\sin 60^{\circ}=\frac{\mathrm{AM}}{\mathrm{OA}}$
$
\begin{aligned}
& \Rightarrow \frac{\sqrt{3}}{2}=\frac{A M}{12} \\
& \Rightarrow A M=6 \sqrt{3} \mathrm{~cm}
\end{aligned}
$

$\Rightarrow 2 \mathrm{AM}=2 \times 6 \sqrt{3}=12 \sqrt{3} \mathrm{~cm}$

$
\begin{aligned}
& \Rightarrow \mathrm{AB}=12 \sqrt{3} \mathrm{~cm} \\
& \therefore \text { Area of } \triangle \mathrm{AOB}=\frac{1}{2} \times A B \times O M \\
& =\frac{1}{2} \times 12 \sqrt{3} \times 6=36 \sqrt{3} \\
& =36 \times 1.73=62.28 \mathrm{~cm}^2 \\
& \therefore \text { Area of corresponding segment = Area of corresponding sector }- \text { Area of } \triangle \mathrm{AOB} \\
& =150.72-62.28=88.44 \mathrm{~cm}^2
\end{aligned}
$
Ex 11.1 Question 8.

A horse is tied to a peg at one corner of a square shaped grass field of side $15 \mathrm{~m}$ by means of a $5 \mathrm{~m}$ long rope (see figure). Find:

(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were $10 \mathrm{~m}$ long instead of $5 \mathrm{~cm}$.
(Use $\pi=3.14$ )
Answer.

(i) Area of quadrant with $5 \mathrm{~m}$ rope
$
=\frac{\theta}{360^{\circ}} \times \pi r^2
$

$
=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times 5 \times 5=19.625 \mathrm{~m}^2
$
(ii) Area of quadrant with $10 \mathrm{~m}$ rope
$
\begin{aligned}
& =\frac{\theta}{360^{\circ}} \times \pi r^2 \\
& =\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times 10 \times 10=78.5 \mathrm{~m}^2
\end{aligned}
$
$\therefore$ The increase in grazing area
$
\begin{aligned}
& =78.5-19.625 \\
& =58.875 \mathrm{~m}^2
\end{aligned}
$
Ex 11.1 Question 9.

A brooch is made with silver wire in the form of a circle with diameter $35 \mathrm{~mm}$. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find:

(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Answer.

(i) Diameter of wire $=35 \mathrm{~mm}$
$
\Rightarrow \text { Radius }=\frac{35}{2} \mathrm{~mm}
$

$
\begin{aligned}
& \text { Circumference }=2 \pi r=2 \times \frac{22}{7} \times \frac{35}{2} \\
& =110 \mathrm{~mm} . \ldots . . . . \text { (i) }
\end{aligned}
$

Length of 5 diameters $=35 \times 5=175 \mathrm{~mm}$
$\therefore$ Total length of the silver wire required
$
=110+175=285 \mathrm{~mm}
$
(ii) $r=\frac{35}{2} \mathrm{~mm}$ and $\theta=\frac{360^{\circ}}{10}=36^{\circ}$
$\therefore$ The area of each sector of the brooch $=\frac{\theta}{360^{\circ}} \times \pi r^2$
$
=\frac{36^{\circ}}{360^{\circ}} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}=\frac{385}{4} \mathrm{~mm}^2
$
Ex 11.1 Question10.

An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius $45 \mathrm{~cm}$, find the area between the two consecutive ribs of the umbrella.

Answer.

Here, $r=45 \mathrm{~cm}$ and
$
\theta=\frac{360^{\circ}}{8}=45^{\circ}
$

Area between two consecutive ribs of the umbrella

$
\begin{aligned}
& =\frac{\theta}{360^{\circ}} \times \pi r^2 \\
& =\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 45 \times 45 \\
& =\frac{22275}{28} \mathrm{~cm}^2
\end{aligned}
$
Ex 11.1 Question 11.

A car has two wipers which do not overlap. Each wiper has a blade of length $25 \mathrm{~cm}$ sweeping through an angle of $115^{\circ}$. Find the total area cleaned at each sweep of the blades.

Answer.

Here, $r=25 \mathrm{~cm}$ and $\theta=115^{\circ}$
The total area cleaned at each sweep of the blades
$
\begin{aligned}
& =2 \times\left(\frac{\theta}{360^{\circ}} \times \pi r^2\right) \\
& =2 \times\left(\frac{115^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 25 \times 25\right) \\
& =\frac{158125}{126} \mathrm{~cm}^2
\end{aligned}
$
Ex 11.1 Question 12.

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^{\circ}$ to a distance of $16.5 \mathbf{~ k m}$. Find the area of the sea over which the ships are warned. (Use $\pi=3.14$ )

Answer.

Here, $r=16.5 \mathrm{~km}$ and $\theta=80^{\circ}$

$\begin{aligned}
& \text { The area of sea over which the ships are warned }=\frac{\theta}{360^{\circ}} \times \pi r^2 \\
& =\frac{80^{\circ}}{360^{\circ}} \times 3.14 \times 16.5 \times 16.5=189.97 \mathrm{~km}^2
\end{aligned}$

Ex 11.1 Question 13.

A round table cover has six equal designs as shown in figure. If the radius of the cover is $28 \mathrm{~cm}$, find the cost of making the designs at the rate of Rs. 0.35 per $\mathrm{cm}^2$. $\text { (Use } \sqrt{3}=1.7 \text { ) }$

$\text { Ans. } r=28 \mathrm{~cm} \text { and } \theta=\frac{360^{\circ}}{6}=60^{\circ}$

$
\begin{aligned}
& \text { Area of minor sector }=\frac{\theta}{360^{\circ}} \times \pi r^2 \\
& =\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 28 \times 28 \\
& =\frac{1232}{3}=410.67 \mathrm{~cm}^2
\end{aligned}
$

For, Area of $\triangle \mathrm{AOB}$, Draw $\mathrm{OM} \perp \mathrm{AB}$.

In right triangles OMA and OMB,
$\mathrm{OA}=\mathrm{OB}$ [Radii of same circle]
$\mathrm{OM}=\mathrm{OM}[$ Common $]$
$\therefore \triangle \mathrm{OMA} \cong \triangle \mathrm{OMB}$ [RHS congruency]
$\therefore \mathrm{AM}=\mathrm{BM}[$ By C.P.C.T. $]$
$\Rightarrow \mathrm{AM}=\mathrm{BM}=\frac{1}{2} \mathrm{AB}$ and $\angle \mathrm{AOM}=\angle \mathrm{BOM}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2} \times 60^{\circ}=30^{\circ}$
In right angled triangle OMA,
$
\begin{aligned}
& \cos 30^{\circ}=\frac{\mathrm{OM}}{\mathrm{OA}} \\
& \Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{OM}}{28} \\
& \Rightarrow \mathrm{OM}=14 \sqrt{3} \mathrm{~cm}
\end{aligned}
$

Also, $\sin 30^{\circ}=\frac{\mathrm{AM}}{\mathrm{OA}}$
$
\begin{aligned}
& \Rightarrow \frac{1}{2}=\frac{\mathrm{AM}}{28} \\
& \Rightarrow \mathrm{AM}=14 \mathrm{~cm} \\
& \Rightarrow 2 \mathrm{AM}=2 \times 14=28 \mathrm{~cm} \\
& \Rightarrow \mathrm{AB}=28 \mathrm{~cm}
\end{aligned}
$
$\therefore$ Area of $\triangle \mathrm{AOB}=\frac{1}{2} \times A B \times O M$

$=\frac{1}{2} \times 28 \times 14 \sqrt{3}=196 \sqrt{3}$

$
=196 \times 1.7=333.2 \mathrm{~cm}^2
$
$\therefore$ Area of minor segment $=$ Area of minor sector - Area of $\triangle A O B$
$
=410.67-333.2=77.47 \mathrm{~cm}^2
$
$\therefore$ Area of one design $=77.47 \mathrm{~cm}^2$
$\therefore$ Area of six designs $=77.47 \times 6=464.82 \mathrm{~cm}^2$
Cost of making designs $=464.82 \times 0.35=$ Rs. 162.68
Ex 11.1 Question 14.

Tick the correct answer in the following:

Area of a sector of angle $p$ (in degrees) of a circle with radius $\mathbf{R}$ is:
(A) $\frac{p}{180^{\circ}} \times 2 \pi \mathrm{R}$
(B) $\frac{p}{180^{\circ}} \times \pi \mathrm{R}^2$
(C) $\frac{p}{360^{\circ}} \times 2 \pi \mathrm{R}$
(D) $\frac{p}{720^{\circ}} \times 2 \pi \mathrm{R}^2$

Answer.

(D) Given, $r=\mathrm{R}$ and $\theta=p$
$
\begin{aligned}
& \text { Area of sector }=\frac{\theta}{360^{\circ}} \times \pi r^2 \\
& =\frac{p}{360^{\circ}} \times \pi \mathrm{R}^2 \\
& =\frac{p}{2 \times 360^{\circ}} \times 2 \pi \mathrm{R}^2
\end{aligned}
$

$=\frac{p}{720^{\circ}} \times 2 \pi \mathrm{R}^2$