NCERT Solutions For Class 10 Maths Chapter 11 Areas Related To Circles

Class 10 Maths Chapter 11 Areas Related To Circles

NCERT Solutions for Class 10 Maths Chapter 11 – Areas Related to Circles serve as essential study resources for Class 10 students. The solutions enable students to see the various question formats that emerge from this chapter during board examinations. The explanation system enables students to solve problems which involve finding areas of circles and sectors and segments through step-by-step guidance.

The students need to answer all questions from the chapter to enhance their conceptual comprehension which will help them prepare for their upcoming tests. The solutions provide clear answers together with structured methods which enable users to review material while their numerical problem-solving accuracy improves.

Chapter 11 – Areas Related to Circles Solutions with Step-by-Step Explanations

The NCERT Solutions for Class 10 Maths provided by ToppersSky help students prepare confidently for their board exams well in advance. The structured solutions enable students to resolve all their conceptual uncertainties while they study the complete subject matter. The step-by-step explanations together with simple solutions enable students to enhance their basic knowledge and develop their ability to solve problems.

 

The solutions help Class 10 students achieve improved academic performance because they follow the most recent NCERT syllabus standards. ToppersSky App provides all necessary material for students to study essential concepts and practice different question types which will help them succeed in their exams.

The perimeter and area measurement of circles is the main subject of Chapter 11 in NCERT Class 10 Maths. The chapter teaches students to calculate the area of two specific circular parts which are called sector and segment. The concepts hold fundamental importance because they enable students to solve actual measurement challenges through the application of mathematical formulas and logical reasoning skills.

The board examination requires students to study Areas Related to Circles because it serves as an essential component of the Mensuration unit. The chapter tests students ability to understand three elements of the subject which include formulas and diagrams and calculations. Students who practice correctly and understand core concepts will perform well in this unit.

ToppersSky offers a complete set of NCERT Solutions for Class 10 Maths Chapter 11 which includes detailed explanations of each solution from start to finish. The solutions have been developed according to the current NCERT syllabus requirements to assist students in their exam preparation.

Topics Covered in Chapter 11 – Areas Related to Circles

• Introduction: Students review fundamental circle equations to learn their practical application in mathematical computations.
• Perimeter and Area of a Circle: The section provides details about circumference and area calculation through established mathematical formulas. Students learn how to apply formulas correctly in numerical problems.
• Areas of Sector and Segment of a Circle: Students understand how to calculate the area of a sector (a portion of a circle formed by two radii) and a segment (a region between a chord and an arc). Diagram understanding is very important in this part.
• Areas of Combinations of Plane Figures: This section includes problems where different shapes are combined together. Students learn how to break complex figures into simpler shapes and calculate total area.
• Summary: The summary provides important formulas and key concepts for quick revision before exams.

Key Features of ToppersSky NCERT Solutions for Chapter 11

• Strengthens understanding of circle-related formulas and concepts
• Step-by-step solutions for all textbook questions
• Clear diagram explanations for better visualization
• Simple and easy-to-understand language
• Helps students solve complex mensuration problems confidently
• Structured learning for effective revision

ToppersSky also provides Animation learning videos, chapter notes, mind maps, practice sets, and exam-writing guidance. These resources help students understand the logic behind formulas and improve accuracy in calculations.

By using ToppersSky study materials, students can build strong fundamentals in Areas Related to Circles and prepare confidently for their Class 10 exams.

Access Question & Answers of Maths NCERT Class 10 Chapter 11 – Areas Related to Circles

Q.1 The problem requires finding the radius of a third circle which has a circumference that matches the total of two given circles whose radii measure 19 cm and 9 cm.

Solution:

The first circle has a radius of 19 cm and its circumference measures 38π cm which is calculated by 2π times 19.

The second circle has a radius of 9 cm and its circumference measures 18π cm which is calculated by 2π times 9.

The total length of the two circles’ circumferences amounts to 56π cm since 38π and 18π combine to form that total.

Let the radius of the third circle be R.

The circumference of the third circle can be expressed as 2πR.

According to the question:

2πR = 56π

R = 56π / 2π
R = 28 cm

Therefore, the third circle has a radius measuring 28 centimeters.

Q.2 The problem requires finding the radius of a third circle which has an area that matches the total area of two given circles whose radii measure 8 cm and 6 cm.

Solution:

The first circle has a radius of 8 cm and its area calculates to 64π cm² because of the formula π times 8².

The second circle has a radius of 6 cm and its area calculates to 36π cm² because of the formula π times 6².

The total area between the two circles amounts to 100π cm² because 64π and 36π combine to form that total.

Let the radius of the third circle be R.

The area of the third circle can be calculated using the formula πR².

According to the question:

πR² = 100π

R² = 100

R = 10 cm

The third circle has a radius of 10 centimeters.

Q.3 The diameter of each wheel of a car is 80 cm. Find how many complete revolutions each wheel makes in 10 minutes if the car is moving at a speed of 66 km per hour.

Solution:

Diameter of the wheel = 80 cm

Radius = 80 ÷ 2 = 40 cm

Circumference of the wheel = 2πr
= 2π × 40
= 80π cm

This means in one revolution, the wheel covers a distance of 80π cm.

Now, speed of the car = 66 km per hour

Convert km to cm:
1 km = 100000 cm

So, distance covered in 1 hour = 66 × 100000
= 66,00,000 cm

Distance covered in 10 minutes:

10 minutes = 10/60 hour = 1/6 hour

Distance covered in 10 minutes = 66,00,000 ÷ 6
= 11,00,000 cm

Now, number of revolutions = Distance covered ÷ Circumference

= 11,00,000 ÷ (80π)

Taking π = 22/7,

Number of revolutions = 4375

Therefore, each wheel makes 4375 complete revolutions.

Q.4 Choose the correct option and justify your answer. If the perimeter and the area of a circle are numerically equal, then what is the radius of the circle?

(A) 2 units
(B) π units
(C) 4 units
(D) 7 units

Solution:

Perimeter (circumference) of circle = 2πr
Area of circle = πr²

According to the question:

2πr = πr²

Divide both sides by π:

2r = r²

r² − 2r = 0

r(r − 2) = 0

So, r = 2

Therefore, the correct answer is option (A), and the radius of the circle is 2 units.

Q.5 Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.

Solution:

Given:
Radius (r) = 6 cm
Angle (θ) = 60°

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

Area = (θ / 360°) × πr²

Substituting the values:

Area = (60 / 360) × π × 6²

= (1 / 6) × π × 36

= 6π cm²

Taking π = 22/7,

Area = 6 × 22/7

= 132/7 cm²

Therefore, the area of the sector is 132/7 cm².

Q.6 Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:

Given:
Circumference (C) = 22 cm

We know:

C = 2πr

So,

2πr = 22

r = 22 / (2π)

Taking π = 22/7,

r = 22 / (2 × 22/7)
= 7/2 cm

A quadrant is a sector with angle 90°.

Area of quadrant = (90 / 360) × πr²

(1 / 4) × π × (7/2)²

(1 / 4) × π × 49/4

49π / 16

Taking π = 22/7,

Area = 77/8 cm²

≈ 9.6 cm²

Therefore, the area of the quadrant is 77/8 cm².

Q.7 The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:

Length of minute hand = Radius (r) = 14 cm

In 60 minutes, the minute hand sweeps 360°.

So, in 5 minutes, the angle swept is:

(360 × 5) / 60
= 30°

Now, using the formula for area of a sector:

Area = (θ / 360°) × πr²

(30 / 360) × π × 14²

(1 / 12) × π × 196

49π / 3

Taking π = 22/7,

Area = 154/3 cm²

Therefore, the area swept by the minute hand in 5 minutes is 154/3 cm².

Q.8 A car has two wipers that do not overlap. Each wiper has a blade of length 25 cm and sweeps through an angle of 115°. Find the total area cleaned in one sweep.

Solution:

Given:
Radius (r) = 25 cm
Angle (θ) = 115°

The area cleaned by one wiper is the area of a sector.

Area of sector = (θ / 360°) × πr²

For one wiper:

Area = (115 / 360) × π × 25²

= (115 / 360) × π × 625

Since there are two wipers and they do not overlap, total area cleaned is:

Total area = 2 × (115 / 360) × π × 625

Taking π = 22/7,

Total area = 2 × (115 / 360) × (22 / 7) × 625

After simplifying,

Total area = 158125 / 126

≈ 1254.96 cm²

Therefore, the total area cleaned by the two wipers in one sweep is approximately 1254.96 cm².

Q.9 Tick the correct option: The area of a sector of angle p (in degrees) of a circle with radius R is:

(A) (p / 180) × 2πR
(B) (p / 180) × πR²
(C) (p / 360) × 2πR
(D) (p / 720) × 2πR²

Solution:

We know that the formula for the area of a sector is:

Area = (θ / 360°) × πr²

Here, θ = p and r = R

So, the area becomes:

Area = (p / 360) × πR²

Now, multiply and divide by 2:

(p / 360) × (2 / 2) × πR²

(2p / 720) × πR²

(p / 720) × 2πR²

This matches option (D).

Therefore, the correct answer is option (D).

Q.10 On a square handkerchief, nine circular designs are made. Each circular design has a radius of 7 cm. Find the area of the remaining portion of the handkerchief.

Solution:

Number of circular designs = 9
Radius of each circle = 7 cm

Since there are 3 circles along each side of the square,

Diameter of one circle = 2 × 7 = 14 cm

Side of the square = 3 × 14
= 42 cm

Area of the square = 42 × 42
= 1764 cm²

Area of one circle = πr²
= (22/7) × 7 × 7
= 154 cm²

Total area of 9 circles = 9 × 154
= 1386 cm²

Area of the remaining portion = Area of square − Total area of circles

= 1764 − 1386
= 378 cm²

Therefore, the area of the remaining portion of the handkerchief is 378 cm².

FAQs

1. Why is it important to study NCERT Solutions for Class 10 Maths Chapter 11?

Students learn about board exam question patterns through section 12 of NCERT Solutions. The chapter presents three types of evaluation formats which include short answer questions and long answer questions and application-based problems. The practice activities assist students in developing their skills because they help students gain self-confidence while their accuracy improves. Your confidence in taking exams increases as you solve more questions.

2. Which concepts are most important from an exam point of view in Chapter 11?

The key concepts from an exam perspective include: Introduction to area of a circle Perimeter circumference and area of a circle Areas of sector and segment of a circle Areas of combinations of plane figures Summary and important formulas These topics are frequently tested in exams and require strong formula understanding.

3. Why are diagrams important in this chapter?

Diagrams play an important role in solving problems related to sectors, segments, and combined figures. Understanding the figure correctly helps in applying formulas accurately and avoiding calculation mistakes.

4. How can students score better marks in Chapter 11?

Students will achieve better marks by continuously studying essential formulas and practicing numerical problems and developing their understanding of step-by-step solution procedures. The two factors which help students improve their exam scores include their ability to manage time during tests and their precision in performing calculations.