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NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles

Edited By Irshad Anwar | Updated on Mar 15, 2025 09:01 PM IST | #CBSE Class 10th
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Chapter 11 maths class 10 Areas related to circles are a key concept in mathematics that play an important role in understanding various geometric properties. It involves calculating the area of circles and solving related problems using formulas for circumference, radius, and diameter. This topic of class10 math chapter 11 also extends to applications like sectors, segments, and the area of a circle within different types of planes. Whether you're a beginner or looking to strengthen your understanding, this article will break down complex topics into simpler parts and help you solve a variety of questions related to class 10 chapter 11 area related to circle.

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 - Important Formulae
  3. NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles (Questions and Exercise)
  4. NCERT Solutions of Class 10 - Subject Wise
  5. Key features of Areas Related To Circles Class 10 NCERT solutions
  6. NCERT Exemplar Solutions - Subject Wise
  7. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles

Additionally, we will address some common questions students often face in class 10th math chapter 11, such as:

  • What is the formula for the area of a circle?
  • How do we find the area of a sector or segment of a circle?
  • Can we apply the properties of a circle to solve real-life problems?

NCERT Solutions or Class 10 Maths Chapter 11, area related to Circles, created by experts at Careers360, provides detailed and step-by-step solutions that make learning easier for students preparing for the CBSE Class 10 board exams. The solutions cover all exercises in the NCERT solutions for Class 10 Maths and offer accessible, downloadable study material for a thorough understanding of the topic of class 10th maths chapter 11. By following the provided solutions, students can master the concepts related to circles and solve any problems of class 10 chapter 11 maths with confidence.

Also Read,

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 - Important Formulae

Circumference of a Circle: Circumference = 2πr

Where r represents the radius of the circle.

Area of a Circle: Area = πr2

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×πr212×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

Free download NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles PDF for CBSE Exam.


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

NCERT Solutions for Maths Chapter 11 Area Related To Circle Class 10 Exercise: 11.1

Q1. Find the area of a sector of a circle with a radius of 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= 1327 cm2.

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 = 22r = 11π cm

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= 778 cm2.

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 degrees in one hour.

We need to find rotation in 5 min. :-

= 36060×5 = 30

The area of the sector is given by :

Area = Θ360×πr2
Area= 30360×π×142
Area= 1543 cm2

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

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

(ii) major sector. (Use π = 3.14)

Answer :

(i)The angle in the minor sector is 90 o

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

Area = Θ360×πr2
Area= 90360×π×102
Area= 110014 cm2 = 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 - The area of a triangle

= 78.5  50 = 28.5 cm2.

(ii)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 = 270o.

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

(ii) area of the sector formed by the arc

(iii) the area of the segment formed by the corresponding chord.

Answer:

(i)The length of the arc is given by:-

Length of arc = Θ360×2πr

= 60360×2×π×21= 22 cm

Hence, the length of the arc is 22 cm.

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

Area = Θ360 ×πr2

= 60360 ×π×212= 231 cm2

Thus, the area of the sector is 231 cm2.

(iii) For the area of the segment, we need to subtract the area of the triangle attached to the area of the arc.

Thus, consider the triangle:-

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

The area of the triangle is:-

= 34×a2= 34×212= 44134 cm2

Hence, the area of the segment is:-

= (231  44134) cm2.

Q6. 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.
(Use π=3.14 and 3=1.73 )

Answer:

The area of the sector is :

= Θ360×πr2= 60360×π×152= 117.85 cm2

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

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

Thus, the area of the triangle is:-

= 34×152

= 56.253 = 97.31 cm2

Hence area of the minor segment : =117.85  97.31 = 20.53 cm2

The area of the major segment is :

= πr2  20.53

= π×152  20.53

= 707.14  20.53

= 686.6 cm2.

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 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

= 150.72 cm2

Now, consider the triangle:-

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

Using geometry, we can write,

hr = cos60

h = 6 cm

Similarly, b2r = sin60

b = 123 cm

Thus, the area of the triangle is :

= 12×123×6

= 62.28 cm2

Hence, the area of the 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 using a 5 m long rope (see Fig). 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. (Use π=3.14 )

16360913046871742052266598

Answer:

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

Area = Θ360 ×πr2

= 90360 ×π×52

= 19.62 m2

(ii)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 :

= 25π  25π4 = 58.85 m2.

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.

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

16360913835111742052266632

Answer:

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

The circumference of the brooch:-

= 2πr= 2×π×352= 110 mm

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

(ii)

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

Thus, the angle of each sector is 36°.

The area of the sector is given by:-

Area = Θ360 ×πr2
Area= 36360 ×π×(352)2
Area= 3854 mm2.

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.

16360914077851742052266655

Answer:

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

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

Area = Θ360 ×πr2
Area= 45360 ×π×452
Area= 2227528 cm2

Hence, the area between two consecutive ribs is 2227528 cm2.

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

= 115360 ×π×252

= 158125252 cm2

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

= 2×158125252 = 158125126 cm2.

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 80 o .

Thus, the area is:-

= 80360 ×π×16.52

= 189.97 Km2

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 cm2. (Use 3=1.7 )

17420522680981742052266684

Answer:

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

The area of the sector is given by:-

Area = Θ360 ×πr2

= 60360 ×π×782

= 410.66 cm2

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

= 34×a2

= 34×282 = 333.2 cm2

Hence the area of segment is : = 410.66  333.2 = 77.46 cm2

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

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

Q14. Tick the correct answer in the following :

The 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.


NCERT Solutions of Class 10 - Subject Wise

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Key features of Areas Related To Circles Class 10 NCERT solutions

Enhanced Conceptual Understanding: These ch 11 class 10 maths solutions serve as a valuable tool for students seeking to grasp circle-related concepts thoroughly.

Visual Clarity: The class 10 maths ch 11 solutions are thoughtfully supplemented with diagrams, facilitating a more interactive and comprehensive learning journey.

Accessible Language: The language employed in these chapter 11 maths class 10 solutions is intentionally straightforward, ensuring that students can easily grasp the content.

Step-by-Step Guidance: The chapter 11 maths class 10 solutions follow a methodical, step-by-step approach, aiding students in building a strong foundation and understanding the fundamentals with ease.

Individualized Learning: Students have the flexibility to tackle complex problems at their own pace, which fosters self-directed learning and builds problem-solving skills.

Resource Diversity: These ch 11 class 10 maths solutions open doors for students to explore NCERT solutions across various classes and subjects.

Expertly Crafted: These ch 11 class 10 maths solutions are meticulously prepared by experienced educators at Careers360, who prioritize providing clarity on key concepts and nurturing students' problem-solving abilities.

Also, students can get solutions for extracurricular-

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NCERT Exemplar Solutions - Subject Wise

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Students can also fiareasrea related to circle class 10 questions with solutions. ions

NCERT Books and NCERT Syllabus

NCERT Class 10 Maths Solutions - Chapter Wise

Chapter No.

Chapter Name

Chapter 1

Real Numbers

Chapter 2

Polynomials

Chapter 3

Pair of Linear Equations in Two Variables

Chapter 4

Quadratic Equations

Chapter 5

Arithmetic Progressions

Chapter 6

Triangles

Chapter 7

Coordinate Geometry

Chapter 8

Introduction to Trigonometry

Chapter 9

Some Applications of Trigonometry

Chapter 10

Circles

Chapter 11

Constructions

Chapter 12

Areas Related to Circles

Chapter 13

Surface Areas and Volumes

Chapter 14

Statistics

Chapter 15

Probability


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 =θ360×2πr
If the angle θ is in radians, the formula is:
Arc Length =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 =θ360×πr2
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 =θ360×πr2

where θ is the central angle in degrees.

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

A segment is the part of a circle enclosed by a chord and its corresponding arc, whereas a sector includes the central angle and two radii.

The area of a segment is found by subtracting the area of the triangle formed by the chord from the area of the sector:

 Area of segment =θ360×πr212r2sinθ

5. What are real-life applications of areas related to circles?

- Gardens \& parks - Calculating circular flower beds and paths.
- Clocks \& dials - Dividing time into equal parts.
- Pizza \& cakes - Cutting equal slices.
- Sports fields - Designing semicircular goal areas in stadiums.
- Wheels \& tires - Measuring rolling distance and coverage area.

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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

https://medicine.careers360.com/neet-college-predictor

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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