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

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

Edited By Komal Miglani | Updated on Jun 28, 2025 09:10 AM IST | #CBSE Class 10th
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CBSE Class 10th  Exam Date : 10 Jul' 2025 - 15 Jul' 2025

From the wheels we ride to the plates we use, areas of circles are everywhere, and math shows us how to measure them. Have you ever wondered how one determines the amount of material required for a round tablecloth, or how engineers calculate the area of a circular stadium? Areas related to circles are used to solve problems in real life involving circles. As per the latest NCERT syllabus, this chapter contains the basic concepts of calculating the circumference, area, and components of a circle, such as sectors and segments. They find extensive usage in fields such as architecture, sports, and engineering, and therefore are extremely practical to use in everyday life. These NCERT solutions for class 10 prove that even endless curves are measurable with a finite formula.

This Story also Contains
  1. NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles PDF Free Download
  2. NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles (Exercise)
  3. Areas Related to Circles Class 10 Chapter 5: Topics
  4. Class 10 Maths Chapter 11 NCERT Solutions: Important Formulae
  5. NCERT Solutions for Class 10 Maths Chapter Wise
NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles
NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles

Defining the area of a circle means capturing the space within its boundary, turning curves into clear measures. The NCERT solutions of class 10 Maths Chapter 11 Areas related to circles, are designed by our experienced subject experts to offer a systematic, structured approach and step-by-step solutions to these important concepts and help students to prepare well for exams and to gain knowledge about all the natural processes happening around them by a series of solved questions given in the NCERT textbook exercise. It covers questions from all the topics and will help you improve your speed and accuracy. Get syllabus details, key notes, and downloadable PDFs directly from the following link: NCERT

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles PDF Free Download

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NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles (Exercise)

Class 10 Maths chapter 11 solutions Exercise: 11.1
Page number: 158-159
Total questions: 14
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Q1: Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.

Answer:

We know that the area of a sector having radius r and angle Θ is given by:-

Area=Θ360×πr2

Thus, the area of the given sector is:-

= 60360×π×62

= 1327cm2

Q2: Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer:

We are given the circumference of the circle.

Thus,

2πr = 22

r=11πcm

Also, we know that the area of a sector is given by :

Area=Θ360×πr2

It is given that we need to find the area of a quadrant, thus Θ = 90∘

Hence, the area becomes:-

= 90360×π(11π)2

= 778cm2

Q3: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer:

The minute hand rotates 360° in one hour.

We need to find a rotation in 5 minutes. :-

= 36060×5=30

The area of the sector is given by :

Area=Θ360×πr2

Area=30360×π142

Area=1543cm2

Hence, the area swept by the minute hand in 5 minutes is 1543cm2.

Q4: A chord of a circle of radius 10 cm subtends a right angle at the center.
Find the area of the corresponding :
(i) minor segment

Answer:

The angle in the minor sector is 90o.

Thus, the area of the sector is given by:-

Area=Θ360×πr2

Area=90360×π102

Area=110014cm2 = 78.5 cm2

Now the area of a triangle is:-

Area=12bh=12×10×10 = 50 cm2

Thus, the area of the minor segment
= Area of the sector - Area of a triangle
= 78.5 − 50 = 28.5 cm2

Q4: A chord of a circle of radius 10 cm subtends a right angle at the center.
Find the area of the corresponding :
(ii) major sector. (Use π = 3.14)

Answer:

The area of the major sector can be found directly by using the formula :

Area=Θ360×πr2

In this case, the angle is 360o - 90o = 270°.

Thus, the area is: -

= 270360×π×10×10

= 330014 = 235.7 cm2

Q5: In a circle of radius 21 cm, an arc subtends an angle of 60° at the center.
Find:
(i) the length of the arc

Answer:

The length of the arc is given by:-

Length of arc = Θ360×2πr

= 60360×2×π×21

=22cm

Hence, the length of the arc is 22 cm.

Q5: In a circle of radius 21 cm, an arc subtends an angle of 60° at the center.
Find:
(ii) the area of the sector formed by the arc

Answer:

We know that the area of the sector is given by:-

Area=Θ360×πr2

= 60360×π×21=231cm

Thus, the area of the sector is 231 cm2.

Q5: In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.
Find:
(iii) area of the segment formed by the corresponding chord

Answer:

For the area of the segment, we need to subtract the area of the triangle attached from the area of the arc.

Thus, consider the triangle:-

It is given that the angle of arc is 60o, or we can say that all angles are 60o (since two sides are equal). Hence, it is an equilateral triangle.

Area of the triangle is:-

= 34×a2

= 34×212

= 44134

Hence, the area of the segment is:-

= (23144134)cm2

Q6: A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use π=3.14 and 3=1.73 )

Answer:

The area of the sector is :

Area=Θ360×πr2

Area=60360×π152

= 117.85 cm2

Now consider the triangle; the angle of the sector is 600.

This implies it is an equilateral triangle. (As two sides are equal, so will have the same angle. This is possible only when all angles are equal, i.e., 60o .)

Thus, the area of the triangle is:-

= 34×152

=56.253=97.31cm2

Hence area of the minor segment: = 117.8597.31=20.53cm2

And the area of the major segment is :

=πr220.53

=π×15220.53

=707.1420.53

=686.6cm2

Q7: 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.
(Use π=3.14 and $\sqrt 3=1.73$ )

Answer:

For the area of the segment, we need the area of the sector and the area of the associated triangle.

So, the area of the sector is :

= 120360×π×122

= 150.72cm2

Now, consider the triangle:-

Draw a perpendicular from the centre of the circle to the base of the triangle (let it be h).

Using geometry, we can write,

hr=cos60

or h = 6 cm

Similarly, b2r=sin60

or b=123 cm

Thus, the area of the triangle is :

= 12×123×6

= 62.28 cm2

Hence, the area of segment is: = 150.72 − 62.28 = 88.44 cm2.

Q8: 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 Fig.). Find (i) the area of that part of the field in which the horse can graze.

1636091304687

Answer:

The part grazed by the horse is given by = Area of the sector

Area=Θ360×πr2

= 90360×π×52

= 19.62 m2

Q8: 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 Fig.). Find (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π=3.14 )

1636091352998

Answer:

When the length of the rope is 10 m, the area grazed will be:-

Area=Θ360×πr2

90360×π×102

= 25π m2

Hence, the change in the grazing area is given by :

= 25π25π4=58.85m2

Q9: A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. Find: (i) the total length of the silver wire required.

1636091383511

Answer:

The total wire required will be for 5 diameters and the circumference of the brooch.

The circumference of the brooch:-

= 2πr=2×π×352

= 110mm

Hence, the total wire required will be = 110mm+5×35mm=285mm.

Q9: A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find : (ii) the area of each sector of the brooch.

1636091389686

Answer:

The total number of lines present in the brooch is 10 (lines starting from the centre).

Thus, the angle of each sector is 36o.

The area of the sector is given by:

Area=Θ360×πr2

Area=36360×π×(352)2

Area=3854mm2

Q10: An umbrella has 8 ribs that are equally spaced (see Fig. ). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

1636091407785

Answer:

It is given that the umbrella has 8 ribs, so the angle of each sector is 45o.

Thus, the area of the sector is given by:-

Area=Θ360×πr2

Area=45360×π×452

Area=2227528

Hence, the area between two consecutive ribs is 2227528cm2

Q11: A car has two wipers that do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Answer:

The area cleaned by one wiper is:-

Area=Θ360×πr2

Area=115360×π×252

Area=158125252cm2

Hence, the required area (area cleaned by both blades) is given by:-

= 2×158125252cm2

= 158125126cm2

Q12: To warn ships of underwater rocks, a lighthouse spreads a red-colored light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Answer:

The area of the sector is given by:-

Area=Θ360×πr2

In this case, the angle is 80o.

Thus, the area is:-

Area=80360×π×16.52

=189.97Km2

Q13: A round table cover has six equal designs as shown in Fig. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm 2. (Use 3=1.7 )

Answer:

The angle of each of the six sectors is 60° at the centre. (∵3606=60)

Area of the sector is given by:-

Area=Θ360×πr2

Area=60360×π×782

Area=410.66cm2

And the area of the equilateral triangle associated with the segment:-

= 34×a2

=34×282=333.2cm2

Hence, the area of segment is: = 410.66333.2=77.46cm2

Thus, the total area of design is: = 6×77.46=464.76cm2

So, the total cost for the design is: = 0.35×464.76=Rs.162.66

Q14: Tick the correct answer in the following :

Area of a sector of angle p (in degrees) of a circle with radius R is

(A) p180×2πR

(B) p180×πR2

(C) p360×2πR

(D) p720×2πR2

Answer:

We know that the area of the sector is given by:-

Area=Θ360×πr2

= p360×πr2

Hence, option (d) is correct.

Areas Related to Circles Class 10 Chapter 5: Topics

The topics discussed in the NCERT Solutions for class 10, chapter 11, Areas Related to Circles are:

  • Areas of Sector and Segment of a Circle
  • Summary

Class 10 Maths Chapter 11 NCERT Solutions: Important Formulae

  • Circumference of a Circle: 2πr, where r represents the radius of the circle.
  • Area of a Circle: πr2, where r represents the radius of the circle.
  • Length of an Arc of a Sector:
    To find the length of an arc within a sector of a circle, given the radius 'r' and the angle in degrees 'θ', the formula is:
    Arc Length = θ360°×2πr
  • Area of a Sector of a Circle:
    For calculating the area of a sector within a circle with radius 'r' and angle in degrees 'θ', the formula is:
    Sector Area = θ360°×πr2
  • Area of a Segment of a Circle: The area of a segment of a circle can be found by subtracting the area of the corresponding triangle from the area of the corresponding sector.
    Segment Area = Sector Area – Triangle Area
    ⇒ Segment Area = θ360°×πr2(12×r2sinθ)
  • When considering the distance (d) moved by a wheel in a single revolution, it is equal to the circumference of the wheel.
    Number of Revolutions = Distance Travelled circumference = d2πr

NCERT Solutions for Class 10 Maths Chapter Wise

The links below allow students to access all the Maths solutions from the NCERT book.

Also, read,

NCERT Exemplar Solutions: Subject Wise

After completing the NCERT textbooks, students should practice exemplar exercises for a better understanding of the chapters and clarity. The following links will help students find exemplar exercises.

NCERT Solutions of Class 10: Subject Wise

Students can check the following links for more in-depth learning.

NCERT Books and NCERT Syllabus

Before planning a study schedule, always analyse the latest syllabus. Here are the links to the latest NCERT syllabus and some of the important books that will help students in this cause.

Frequently Asked Questions (FAQs)

1. What is the formula for the length of an arc in Class 10 Maths?

The length of an arc in a circle with radius r and central angle θ (in degrees) is:
Arc Length = $\frac{θ}{360°}×2πrIftheangleθisinradians,theformulais:ArcLength=rθ$

2. How to calculate the area of a segment in a circle?

The area of a segment is found by subtracting the area of the triangle from the sector's area:

Segment Area = Sector Area − Triangle Area
Using the sector formula:
Sector Area = $\frac{θ}{360°}×πr^2$
The triangle's area can be determined using trigonometry or Heron's formula.

3. How to find the area of a sector in a circle?

A sector is a portion of a circle (like a pizza slice). Its area is given by:

Area of sector = $\frac{θ}{360°}×πr^2$ 

where θ is the central angle in degrees.

4. What is the difference between a sector and a segment in a circle?
SectorSegment

A sector includes a central angle and two radii.

A segment includes a chord and its corresponding arc.

A sector looks like a pizza slice or wedge.

It is crescent-shaped.

Centre of a circle is included in a sector.

Centre of a circle is not necessarily included in a segment.

$\frac{θ}{360°}×πr^2$, where θ= central angle and r= radius

Segment Area
= Sector Area - The Area of the corresponding triangle
= $\frac{θ}{360°}×πr^2-(\frac12×r^2\sin⁡θ)$ 

5. What type of questions are asked from Chapter 11 in board exams?

 Some of the common types of questions asked in this chapter are:

  • Finding the area of a sector
  • Finding the area of a segment
  • Finding the length of an arc
  • Finding the number of revolutions of a wheel or distance traveled
  • Using the area and circumference of a circle to find the cost or quantity of material
  • Finding the radius of a circle given that the area or arc length

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

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Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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