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NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles 2025-26

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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.

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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

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Current Syllabus Exercises of Class 10 Maths Chapter 11

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NCERT Solutions of Class 10 Maths Areas Related to Circles Exercise 11.1

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NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles 2025-26
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AREAS RELATED TO CIRCLES in One Shot (Complete Chapter) | CBSE 10 Math Chap 12 [Term 1 Exam] Vedantu
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AREAS RELATED TO CIRCLES L-1 (Perimeter and Area of a Circle) CBSE 10 Math Chap 12 [Term 1] Vedantu
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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:

the area of a sector of a circle with radius 6 cm


\[\frac{{132}}{7}c{m^2}\]

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:

the area of a quadrant of a circle whose circumference is 22 cm


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:

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

                  

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:


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


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:

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


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:

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


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:

A chord of a circle of radius 12 cm subtends an angle of 120° at the center. Find the area of the corresponding segment of the circle


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). \[\mathbf{Use}\text{ }\pi =3.14\]

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.png

         

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:

it is clear that the horse can graze a sector of


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

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


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 \[ = r\] \[ = \frac{{diameter}}{2}\]\[ = \frac{{35}}{2}mm\]

ii.The area of each sector of the brooch


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

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.


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

A car has two wipers which do not overlap. Each wiper has blade of length 25 cm sweeping through an angle of 115°


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:

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle


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. \[\mathbf{Use}\text{ }\sqrt{3}=1.7\text{ }\]

A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm.png


Ans:


The designs are segments of the circle


Given in the figure,

The designs are segments of the circle. 

Radius of circle is 28cm.

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



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.



Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

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FAQs on NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles 2025-26

1. What is a sector of a circle as per Class 10 Maths Chapter 11?

A sector is the region of a circle enclosed by two radii and their corresponding arc, much like a slice of pizza. Its area is calculated using the formula (θ/360) × πr², where θ is the angle between the two radii.

2. How do you find the area of a segment of a circle?

The area of a segment is found by subtracting the area of the corresponding triangle from the area of its sector. For a minor segment, the formula is (Area of Sector) – (Area of Triangle). This is a key concept in areas related to circles.

3. What is included in the NCERT Solutions for Class 10 Maths Chapter 11?

NCERT Solutions for this chapter provide clear, step-by-step answers for every exercise question. They cover all formulas for finding the area of sectors, segments, and combined plane figures, helping students solve problems accurately and efficiently for their board exams.

4. Where can I find a free PDF for Areas Related to Circles Class 10 NCERT Solutions?

You can download a Free PDF of the NCERT Solutions for this chapter from Vedantu’s website. This PDF contains detailed answers to all textbook questions, allowing you to study offline and revise the chapter anytime without an internet connection.

5. What is the formula for the length of an arc of a sector?

The length of an arc of a sector with an angle θ and radius r is given by the formula (θ/360) × 2πr. This formula calculates the length of the portion of the circle's circumference that forms the sector's curved boundary.

6. How can NCERT Solutions for Areas Related to Circles Class 10 help with exams?

These solutions explain the correct application of formulas for every problem in the NCERT textbook. Practising with them helps in understanding the logic, managing time better, and familiarising yourself with the types of questions asked in board exams.

7. How do you calculate the area of combinations of plane figures involving circles?

To find the area of a combined figure, first, break it down into basic shapes like sectors, segments, triangles, and squares. Then, calculate the area of each individual shape and add or subtract them as required by the figure's geometry.

8. What is the difference between a major sector and a minor sector?

A minor sector is the region corresponding to the smaller arc, where the central angle is less than 180°. A major sector is the region corresponding to the larger arc, where the central angle is greater than 180°.

9. Are the Class 10 Maths Chapter 11 question answers aligned with the latest syllabus?

Yes, the NCERT Solutions provided by Vedantu for Class 10 Maths Chapter 11 are fully updated and aligned with the latest CBSE syllabus. They cover all topics and exercises present in the current rationalised NCERT textbook.

10. What are the main formulas to learn for Areas Related to Circles questions and answers?

The three key formulas are: Area of a sector = (θ/360)×πr², Length of an arc = (θ/360)×2πr, and Area of a segment = Area of the corresponding sector – Area of the corresponding triangle. Mastering these is crucial.