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# NCERT Solutions for Class 10 Maths Chapter 11 - Areas Related to Circles

Last updated date: 13th Sep 2024
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NCERT Chapter 11 Area Related to Circles Class 10 Solutions by Vedantu offers a comprehensive guide to understanding and solving problems related to the areas of circles and their parts. This chapter delves into important concepts like finding the area of a circle, the area of sectors and segments, and applying these concepts in solving real-life problems. By breaking down complex formulas into simple steps, Vedantu's Class 10 Maths NCERT Solutions help students grasp the techniques required to calculate the areas of different circular shapes, ensuring a solid foundation for their exams.

Table of Content
1. NCERT Class 10 Maths Chapter 11: Complete Resource for Areas Related to Circles
2. Glance on Maths Chapter 11 Class 10 - Area Related to Circles
3. Access Exercise Wise NCERT Solutions for Chapter 11 Maths Class 10
4. Exercises Under NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles
5. Access NCERT Solutions for Class 10 Maths Chapter 11 – Areas Related to Circles
6. NCERT Solutions for Class10 Maths Chapter 11 PDF
7. Overview of Deleted Syllabus for CBSE Class 10 Maths Areas Related to Circles
8. Class 10 Maths Chapter 11: Exercises Breakdown
9. Other Study Material for CBSE Class 10 Maths Chapter 11
10. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance on Maths Chapter 11 Class 10 - Area Related to Circles

• Basic Definitions of understanding terms like radius, diameter, circumference, and area of a circle

• Determining the Area of a Circle, Perimeter (Circumference) of a circle by using the formulas:

• The area of a circle is given by A = $\pi r^2$

• The circumference of a circle is given by 𝐶=2𝜋𝑟

•  A sector is a 'slice' of a circle bounded by two radii and the corresponding arc. Finding area of a sector with formula $\text{Area} = \dfrac{\theta}{360^\circ} \times \pi r^2$

• A segment is a region bounded by a chord and the corresponding arc. A Segment is calculated by finding the area of the sector and subtracting the area of the triangle formed by the two radii and the chord.

• This article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 11 - Areas Related to Circles, which you can download as PDFs.

• Class 10 Chapter 11 has only one exercise (14 fully solved questions) in class 10th Maths Chapter 11 Areas Related to Circles.

## Access Exercise Wise NCERT Solutions for Chapter 11 Maths Class 10

 S.No. Current Syllabus Exercises of Class 10 Maths Chapter 11 1 NCERT Solutions of Class 10 Maths Areas Related to Circles Exercise 11.1
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Exercise 11.1: This exercise involves questions based on finding the area of shaded regions, which consist of circles, triangles, and rectangles. The exercise contains a total of five questions that test the student's understanding of the concepts related to the area of circles and their related figures. The solutions to each question provide step-by-step instructions and diagrams to help students understand the concept better.

## Access NCERT Solutions for Class 10 Maths Chapter 11 – Areas Related to Circles

Exercise (11.1)

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is ${60^ \circ }$.$pi =\dfrac{22}{7}$

Ans:

$\frac{{132}}{7}c{m^2}$

Given that,

Radius of the circle = $r = 6cm$

Angle made by the sector with the center, $\theta = {60^ \circ }$

Let OACB be a sector of the circle making ${60^ \circ }$ angle at center O of the circle.

We know that area of sector of angle, $= \frac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$

Thus, Area of sector OACB $= \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \frac{{22}}{7} \times {(6)^2}$

$= \frac{1}{6} \times \frac{{22}}{7} \times 6 \times 6$

$= \frac{{132}}{7}c{m^2}$

Therefore, the area of the sector of the circle making 60° at the center of the circle is $\frac{{132}}{7}c{m^2}$.

2. Find the area of a quadrant of a circle whose circumference is 22 cm. $pi =\dfrac{22}{7}$

Ans:

Given that,

Circumference = 22 cm

Let the radius of the circle be $r$.

According to the given condition,

$2\pi r = 22$

$\Rightarrow r = \frac{{22}}{{2\pi }}$

$= \frac{{11}}{\pi }$

We know that, quadrant of circle subtends ${90^ \circ }$ angle at the center of the circle.

Area of such quadrant of the circle $= \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {r^2}$

$= \frac{1}{4} \times \pi \times {\left( {\frac{{11}}{\pi }} \right)^2}$

$= \frac{{121}}{{4\pi }}$

$= \frac{{77}}{8}c{m^2}$

Hence, the area of a quadrant of a circle whose circumference is 22 cm is $= \frac{{77}}{8}c{m^2}$.

3.The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

$pi =\dfrac{22}{7}$

Ans:

Given that,

Radius of clock or circle = $r$ = 14 cm.

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates ${360^ \circ }$.

Thus, in 5 minutes, minute hand will rotate $= \frac{{{{360}^ \circ }}}{{{{60}^ \circ }}} \times 5$

$= {30^ \circ }$

Now,

the area swept by the minute hand in 5 minutes = the area of a sector of ${30^ \circ }$ in a                                      circle of 14 cm radius.

Area of sector of angle $\theta = \frac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$

Thus, Area of sector of ${30^ \circ } = \frac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \frac{{22}}{7} \times 14 \times 14$

$= \frac{{11 \times 14}}{3}$

$= \frac{{154}}{3}c{m^2}$

Therefore, the area swept by the minute hand in 5 minutes is $\frac{{154}}{3}c{m^2}$.

4. A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding:

$[\text{Use }\pi =3.14]$

Ans:

Given that,

Radius of the circle $= r = 10cm$

Angle subtended by the cord = angle for minor sector$= {90^ \circ }$

Angle for minor sector $= {360^ \circ } - {90^ \circ } = {270^ \circ }$

i) Minor segment

Ans: It is evident from the figure that,

Area of minor segment ACBA = Area of minor sector OACB − Area of ΔOAB

Thus,

Area of minor sector OACB $= \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$ $= \frac{1}{4} \times 3.14 \times {(10)^2}$ $= 78.5c{m^2}$

Area of ΔOAB $= \frac{1}{2} \times OA \times OB = \frac{1}{2} \times {(10)^2} = 50c{m^2}$

Area of minor segment ACBA $= 78.5 - 50 = 28.5c{m^2}$

Hence, area of minor segment is $28.5c{m^2}$

ii) Major sector

Ans: It is evident from the figure that,

Area of major sector OADB $= \frac{{{{270}^ \circ }}}{{{{360}^ \circ }}} = \frac{3}{4} \times 3.14 \times {(10)^2} = 235.5c{m^2}$.

Hence, area of major sector is $235.5c{m^2}$.

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the center.

$pi =\dfrac{22}{7}$

Find:

Ans:

Given that,

Radius of circle = $r =$ 21 cm

Angle subtended by the given arc = $\theta = {60^ \circ }$

i) The length of the arc

Ans: We know that, Length of an arc of a sector of angle $\theta = \frac{\theta }{{{{360}^ \circ }}} \times 2\pi r$

Thus, Length of arc ACB $= \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times 2 \times \frac{{22}}{7} \times 21$

$= 22cm$

Hence, length of the arc of given circle is $22cm$.

ii) Area of the sector formed by the arc

Ans: We know that, Area of sector OACB $= \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= 231c{m^2}$

Hence, area of the sector formed by the arc of the given circle is $231c{m^2}$.

iii) Area of the segment formed by the corresponding chord

Ans: In $OAB$,

As radius $OA = OB$

$\Rightarrow \angle OAB = \angle OBA$

$\angle OAB + \angle AOB + \angle OBA = {180^ \circ }$

$2\angle OAB + {60^ \circ } = {180^ \circ }$

$\angle OAB = {60^ \circ }$

Therefore, $OAB$ is an equilateral triangle.

Now, area of $OAB$ $= \frac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}$

$= \frac{{\sqrt 3 }}{4} \times {\left( r \right)^2}$

$= \frac{{\sqrt 3 }}{4} \times {\left( {21} \right)^2}$

$= \frac{{441\sqrt 3 }}{4}c{m^2}$

We know that, Area of segment ACB = Area of sector OACB − Area of

$= \left( {231 - \frac{{441\sqrt 3 }}{4}} \right)c{m^2}$.

Hence, Area of the segment formed by the corresponding chord in circle is

$\left( {231 - \frac{{441\sqrt 3 }}{4}} \right)c{m^2}$.

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the center. Find the areas of the corresponding minor and major segments of the circle.

[\text{Use}\text{ }\pi =\dfrac{22}{7}\text{ }\mathbf{and}\text{ }\sqrt{3}=1.73\text{  }]

Ans:

Given that,

Radius of circle = $r =$ 15 cm

Angle subtended by chord $= \theta = {60^ \circ }$

Area of circle $= \pi {r^2}$ $= 3.14{\left( {15} \right)^2}$

$= 706.5c{m^2}$

Area of sector OPRQ $= \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \dfrac{1}{6} \times 3.14{(15)^2} = 117.75c{m^2}$

Now, for the area of major and minor segments,

In $OPQ$,

Since, $OP = OQ$

$\Rightarrow \angle OPQ = \angle OQP$

$\angle OPQ = {60^ \circ }$

Thus, $OPQ$ is an equilateral triangle.

Area of $OPQ$ $= \dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}$

$= \dfrac{{\sqrt 3 }}{4} \times {(r)^2}$

$= \dfrac{{225\sqrt 3 }}{4}$ $= 97.3125c{m^2}$.

Now,

Area of minor segment PRQP = Area of sector OPRQ − Area of $OPQ$

$= 117.75 - 97.3125$

$= 20.4375c{m^2}$

Area of major segment PSQP = Area of circle − Area of minor segment PRQP

$= 706.5 - 20.4375$

$= 686.0625c{m^2}$

Therefore, the areas of the corresponding minor and major segments of the circle are $20.4375c{m^2}$ and $686.0625c{m^2}$ respectively.

7. A chord of a circle of radius 12 cm subtend an angle of 120° at the center. Find the area of the corresponding segment of the circle. [\text{  }\mathbf{Use}\text{ }\pi =\frac{22}{7}\text{ }\mathbf{and}\text{ }\sqrt{3}=1.73\text{  }]

Ans:

Draw a perpendicular OV on chord ST bisecting the chord ST such that SV = VT

Now, values of OV and ST are to be found.

Therefore,

In $OVS$,

$\cos {60^ \circ } = \frac{{OV}}{{OS}}$

$\Rightarrow \frac{{OV}}{{12}} = \frac{1}{2}$

$\Rightarrow OV = 6cm$

Also, $\frac{{SV}}{{SO}} = \sin {60^ \circ }$

$\Rightarrow \frac{{SV}}{{12}} = \frac{{\sqrt 3 }}{2}$

$\Rightarrow SV = 6\sqrt 3$

Now, $ST = 2SV$

$= 2 \times 6\sqrt 3 = 12\sqrt 3 cm$

Area of $OST = \dfrac{1}{2} \times ST \times OV$

$= \dfrac{1}{2} \times 12\sqrt 3 \times 6$

$= 62.28c{m^2}$

Area of sector OSUT $= \dfrac{{{{120}^ \circ }}}{{{{360}^ \circ }}} \times \pi {(12)^2}$

$= 150.42c{m^2}$

Area of segment SUTS = Area of sector OSUT − Area of $OVS$

$= 150.72 - 62.28$

$= 88.44c{m^2}$

Hence, the area of the corresponding segment of the circle is $88.44c{m^2}$.

8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see the given figure). [\mathbf{Use}\text{ }\pi =3.14]

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 m long instead of 5 m.

Ans:

From the above figure, it is clear that the horse can graze a sector of ${90^ \circ }$ in a circle of 5 m radius.

Hence,

$\theta \text{ }\!\!~\!\!\text{ }={{90}^{{}^\circ }}$

$r=5m$

(i) The area of that part of the field in which the horse can graze.

Ans: It is evident from the figure,

Area that can be grazed by horse = Area of sector OACB

$= \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}}\pi {r^2}$

$= \dfrac{1}{4} \times 3.14 \times {(5)^2}$ $= 19.625{m^2}$

(ii) The increase in the grazing area if the rope were 10 m long instead of 5 m.

Ans: It is evident from the figure,

Area that can be grazed by the horse when length of rope is 10 m long

$= \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {(10)^2}$

$= 78.5{m^2}$

Therefore, the increase in grazing area for horse $= (78.5 - 19.625){m^2} = 58.875{m^2}$.

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. $pi =\dfrac{22}{7}$

Find:

1. The total length of the silver wire required.

2. The area of each sector of the brooch.

Ans:

Given that,

Radius of the circle $= r$ $= \frac{{diameter}}{2}$$= \frac{{35}}{2}mm$

It can be observed from the figure that each of 10 sectors of the circle is subtending 36° (i.e., 360°/10=36°) at the center of the circle.

(i) The total length of the silver wire required.

Ans:

Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Circumference of brooch $= 2\pi r$

$= 2 \times \frac{{22}}{7} \times \left( {\frac{{35}}{2}} \right)$

$= 110mm$

Length of wire required $= 110 + \left( {5 \times 35} \right)$

$= 285mm$

Therefore, The total length of the silver wire required is $285mm$.

(ii) The area of each sector of the brooch.

Ans:

Area of each sector $= \frac{{{{36}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \frac{1}{{10}} \times \frac{{22}}{7} \times {\left( {\frac{{35}}{2}} \right)^2}$

$= \frac{{385}}{4}m{m^2}$

Hence, The area of each sector of the brooch is $\frac{{385}}{4}m{m^2}$.

10. An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. $pi =\dfrac{22}{7}$

Ans:

Given that,

Radius of the umbrella $= r$$= 45cm$

There are 8 ribs in an umbrella.

The angle between two consecutive ribs is subtending $\frac{{{{360}^ \circ }}}{8} = {45^ \circ }$ at the center of the assumed flat circle.

Area between two consecutive ribs of the assumed circle $= \frac{{{{45}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \frac{1}{8} \times \frac{{22}}{7} \times {\left( {45} \right)^2}$

$= \frac{{22275}}{{28}}c{m^2}$

Hence, the area between the two consecutive ribs of the umbrella is $\frac{{22275}}{{28}}c{m^2}$.

11. A car has two wipers which do not overlap. Each wiper has blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at${115^ \circ }$ each sweep of the blades. $pi =\dfrac{22}{7}$

Ans:

Given that,

Each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.

Area of sector $= \frac{{{{115}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {\left( {25} \right)^2}$

$= \frac{{158125}}{{252}}c{m^2}$

Area swept by 2 blades $= 2 \times \frac{{158125}}{{252}}$

$= \frac{{158125}}{{126}}c{m^2}$.

Therefore, the total area cleaned at each sweep of the blades is $\frac{{158125}}{{126}}c{m^2}$.

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 km. Find the area of the sea over which the ships warned. [\mathbf{Use}\text{ }\pi =3.14]

Ans:

Given that,

The lighthouse spreads light across a sector (represented by shaded part in the figure) of ${80^ \circ }$ in a circle of 16.5 km radius.

Area of sector OACB $= \frac{{{{80}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \frac{2}{9} \times 3.14 \times {(16.5)^2}$

$= 189.97k{m^2}$

Hence, the area of the sea over which the ships are warned is $189.97k{m^2}$.

13. A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs.0.35 per cm2. [\mathbf{Use}\text{ }\sqrt{3}=1.7\text{ }]

Ans:

Given in the figure,

The designs are segments of the circle.

Consider segment APB and chord AB is a side of the hexagon.

Each chord will substitute at $\frac{{{{360}^ \circ }}}{6} = {60^ \circ }$ at the center of the circle.

In $OAB$,

Since, $OA = OB$

$\Rightarrow \angle OAB + \angle OBA + \angle AOB = {180^ \circ }$

$2\angle OAB = {180^ \circ } - {60^ \circ } = {120^ \circ }$

$\angle OAB = {60^ \circ }$

Therefore, $OAB$ is an equilateral triangle.

Area of $OAB$ $= \frac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}$

$= \frac{{\sqrt 3 }}{4} \times {\left( {28} \right)^2}$

$= 333.2c{m^2}$

Area of sector OAPB $= \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}$

$= \frac{1}{6} \times \frac{{22}}{7} \times {(28)^2}$

$= \frac{{1232}}{3}c{m^2}$

Area of segment APBA = Area of sector OAPB − Area of ∆OAB

$= \left( {\frac{{1232}}{3} - 333.2} \right)c{m^2}$

Therefore, area of designs $= 6 \times \left( {\frac{{1232}}{3} - 333.2} \right)c{m^2}$

$= 464.8c{m^2}$

Now, given that Cost of making 1 $c{m^2}$ designs = Rs 0.35

Cost of making 464.76 $c{m^2}$ designs $= 464.8 \times 0.35 =$162.68

Therefore, the cost of making such designs is Rs 162.68.

14. Tick the correct answer in the following: Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is

$(A)\frac{P}{{180}} \times 2\pi R$ $(B)\frac{P}{{180}} \times 2\pi {R^2}$ $(C)\frac{P}{{180}} \times \pi R$ $(D)\frac{P}{{720}} \times 2\pi {R^2}$

Ans:

We know that area of sector of angle $\theta = \frac{\theta }{{{{360}^ \circ }}} \times \pi {R^2}$

So, Area of sector of angle $P = \frac{P}{{{{360}^ \circ }}}\left( {\pi {R^2}} \right)$

$= \left( {\frac{P}{{{{720}^ \circ }}}} \right)\left( {2\pi {R^2}} \right)$

Hence, (D) is the correct answer.

## NCERT Solutions for Class10 Maths Chapter 11 PDF

In Class 10 Maths, the chapter "Areas Related to Circles" helps students understand how to calculate the area and perimeter (circumference) of circles and their parts. This includes learning to find the area of a whole circle, as well as parts of circles like sectors and segments. Understanding these concepts is useful not only for solving math problems but also for real-life applications, such as determining the size of circular objects like wheels, gardens, or pizzas. The chapter breaks down complex shapes into simpler parts to make calculations easier and more intuitive.

## Area and Perimeter of Circle

NCERT Class 10 Maths Chapter 11 highlights various applicational problems of the concept ‘Area and Perimeter of Circle’. The usage of the formula of area and perimeter of the circle and the same analytical skill is vital for this part of the chapter. The types of problems covered in Chapter 11 Maths Class 10 are finding radii of a circle that covers the area or perimeter of the sum area or perimeter covered by two other circles, finding the area of rings, and also the calculation of distance covered given the number of revolutions.

## Area of Sector and Segment of a Circle

The formula for finding the area of sector and segment has been used in Class 10 Areas Related to Circle in the form of application-based sums.

The part of the circular region enclosed by the two radii and the corresponding arc is called the sector of the circle. The part of the circular region enclosed between a chord and the corresponding arc is called the segment of the circle. The segment is classified as a minor segment and a major segment. The major segment covers a larger area corresponding to the minor segment. Similarly, the sector is classified into minor and major where the major sector covers a larger area.

The area and perimeter significantly depend on the angle subtended by the arc at the centre of the circle. When applicational problems are to be solved, the ability to pull out the angle subtended is crucial. For example, the angle subtended by a minute and hour hand at a certain time in the wall clock is repetitive. Also, a lot of questions are based on how the circle is divided equally, thus the technique to find the angle of each equally divided arc is to be known by the student for earning better marks. These are asked in the forms of segments by umbrella or sectors when a regular polygon is inscribed in a circle.

## Overview of Deleted Syllabus for CBSE Class 10 Maths Areas Related to Circles

 Chapter Dropped Topics Areas Related to Circles 11.1 Introduction 11.2 Perimeter and area of a circle - A Review 11.4 Areas of combinations of Plane figures

## Class 10 Maths Chapter 11: Exercises Breakdown

 Exercise Number of Questions Exercise 11.1 14 Questions and Solutions

## Conclusion

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles by Vedantu provides students with clear and detailed explanations to tackle problems involving the areas of circles, sectors, and segments. Understanding these concepts is crucial as they form the basis for various practical applications and higher-level geometry. Key points to focus on include mastering the formulas for the area of a circle, the area of a sector, and the area of a segment.

Additionally, practising the problems regularly will help solidify these concepts. In previous years' exams, around 3–4 questions have been asked from this chapter, often focusing on real-life applications of calculating areas. Ensuring a strong grasp of these topics will help students perform well in their exams.

## Other Study Material for CBSE Class 10 Maths Chapter 11

 S.No. Important Links for Chapter 11 Areas Related to Circles 1 Class 10 Areas Related to Circles Important Questions 2 Class 10 Areas Related to Circles Revision Notes 3 Class 10 Areas Related to Circles NCERT Exemplar Solution 4 Class 10 Areas Related to Circles RD Sharma Solutions

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 11 - Areas Related to Circles

1. What are the topics included in the CBSE prescribed NCERT book of Class 10 Maths Chapter 11?

The topics included in the CBSE prescribed NCERT book of Class 10 Maths Chapter 11:

• Introduction

• Perimeter and Area of a Circle

• Sector of a Circle

• Segment of a Circle

• Combinations of Circles and Their Parts

2. What is the advantage of using Vedantu NCERT Solutions for Class 10 Maths Areas Related to Circles?

Class 10 Board exams are a crucial part of your future, the preparation would be next to perfection with guidance from subject experts. Vedantu's reference materials satisfy all the needs of a student who is determined to excel in their exams. The solutions provided in the notes are accurate and involve various techniques that help students.

3. What are the important formulae used in this chapter?

The important formulae used in the chapter are:

• Circumference of a circle = 2 π r.

• Area of a circle = π r

• Length of an arc of a sector of a circle with radius r and angle with degree measure θ is= theta/360 *2*π *r

• Area of a sector of a circle with radius r and angle with degree measure θ is theta/360 *π*r2

• Area of a segment of a circle = Area of the corresponding sector – Area of the corresponding triangle.

4. How many exercises are there in Chapter 11 of Class 10 Maths?

There is only one exercise in Chapter 11 Areas Related to Circles of Class 10 Maths. Exercise 11.1 has fourteen questions. You can get answers to the textbook questions on Vedantu. On this website, you will find the NCERT Solutions of Chapter 11, that is, Areas Related To Circles written in simple language. You can also access the study material on the Vedantu app. All the resources are available free of cost.

5. How can I prepare Chapter 11 of Class 10 Maths?

To get well-versed with Chapter 11 - Areas Related to Circles, you are required to attend the lectures related to this chapter at your school. While understanding the concepts of the chapter or solving questions, if you face any doubts then ask for help from your subject teacher. Additionally, you should practice this chapter at your home. Making a separate notebook for important formulas will help you to revise the entire chapter within a short period of time.

6. What is the area of a circle according to Chapter 11 of Class 10 Maths?

The area of a circle is defined as the space occupied by a circle in a one-dimensional plane. The other definition of the area of a circle is the space occupied by a circle within its circumference or boundary. It is represented with the formula A= πr2 where A is the area, and r is the radius of the circle.

The area of a circle can also be written as A= πd2/4. Here, d is the diameter of the circle.

It is the mathematical constant having a value of 3.14 or 22/7.

7. What is the perimeter of the circle according to Chapter 11 of Class 10 Maths?

The perimeter of the circle is its circumference or boundary. It is the complete arc that makes the length or the periphery of the circle. Hence, the perimeter of the circle is represented by the formula of the circumference of the circle. The formula for the perimeter of the circle has one variable that is the radius of the circle and two constants. The formula is represented as C = 2π r orC= π d. Here, d is the diameter of the circle, r is the radius of the circle, and pi is the mathematical constant having a value of 22/7 or 3.14.

8. What do you understand by the segment of a circle according to Chapter 11 of Class 10 Maths?

The segment of a circle is the region or area surrounded by a chord or an arc of the circle. Arc is the small portion for the circumference of the circle and chord is a line that connects two points on the circumference of the circle. There are two types of segments. One is the major segment and the second is the minor segment. The major segment is created by a major arc of a circle whereas the minor segment is formed by the minor arc of the circle.

9. What are common real-life applications of areas related to circles class 10?

The concepts in Ch 11 Maths Class 10 are used in real life to calculate areas and perimeters of circular objects like gardens, wheels, clocks, and pizzas.

10. What are the key formulas to remember in area related to circle Class 10?

The key formulas to remember are:

The area of a circle is given by A = $\pi r^2$

The circumference of a circle is given by 𝐶=2𝜋𝑟

The  area of a sector $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$