NCERT Solutions Class 10 Maths Chapter 11 Areas Related to Circles

Welcome to our comprehensive guide on NCERT Solutions for Class 10 Maths, specifically focusing on Chapter 11: Areas Related to Circles. In this chapter, we embark on an exciting journey through the fascinating world of circles, exploring their properties and the various formulas that govern their area and circumference. 

Understanding these concepts is crucial not only for mastering your Class 10 curriculum but also for laying a strong foundation for advanced mathematics in the future.

Our NCERT Solutions provide step-by-step explanations that simplify complex problems, making it easier for you to tackle exercises with confidence. These NCERT Solutions for Class 10 Maths help students understand the types of questions that will be asked in the CBSE Class 10 Maths board exams. 

1.0Download NCERT Solutions Class 10 Chapter 11 Areas Related to Circles : Free PDF

In chapter 11 of class 10 Maths, students will get to know the parameters of circles, arcs, polygons, and segments and then they will be able to solve different problems combining all these concepts. Below, we have mentioned the link to free access, as per your comfort, to the NCERT questions, along with solutions for chapter 11 class 10 Maths “Areas Related to Circle.” 

2.0Importance of NCERT Solutions Class 10 Chapter 11 Areas Related To Circles

Practising NCERT Solutions of Chapter 11 Areas Related to Circles is important from an exam point of view and for understanding their applications in real-life geometry situations. This chapter covers concepts, which are crucial and considered advanced mathematical topics. What else the reasons that it seems important to keep practising with NCERT solutions are as follows:

• This helps students grasp fundamental ideas like the area of a circle, sector, and segment, which are crucial for solving geometry problems.
• Offers a variety of exercises that build a systematic approach to tackling circle-related questions.
• NCERT solutions provide step-by-step guidance, which is necessary for scoring well.
• Regular practice strengthens the mathematical base required for advanced geometry and related subjects in higher education.

3.0Exercise-wise NCERT Solutions Chapter 11 Areas Related to Circles

Applying the NCERT Solutions for Class 10 Maths - Chapter 11 is necessary because these help you understand the concepts and apply them practically. This chapter is an important lesson for various industries, especially architecture, to build strong structures. So what is the exercise given in this chapter’s NCERT Solutions is given below:

4.0Important Topics of Class 10 Math Chapter 11 Areas Related To Circles Solutions

Chapter 11 Areas Related to Circles in Class 10 Math focuses on calculating the area and perimeter of circles, sectors, and segments. You will learn how to find the area of a circle, sector, and segment using formulas, as well as how to compute the perimeter of a sector. The chapter also covers problems involving combinations of circular figures with other shapes, providing practical applications of these concepts in real-life situations.

1. Area of a Circle
2. Circumference of a Circle
3. Length of arc of a Sector 
4. Area of a Sector of a Circle
5. Area of a Segment of a Circle

5.0General Outline for Class 10 Maths Chapter 11 Areas Related to Circles

Understanding Key Concepts:

• Circle: A circle is a two-dimensional shape consisting of all points that are at a fixed distance (radius) from a central point.
• Radius (r): The distance from the center of the circle to any point on the circle.
• Diameter (d): The longest distance across the circle, passing through the center. d = 2r.
• Circumference: The perimeter or boundary length of the circle. It is calculated as: $\text{Circumference}=2\pi r$
• Sector: A region in a circle bounded by two radii and the arc between them.
• Segment: A region in a circle bounded by a chord and the arc subtended by it.

Key Formulas

• Area of a Circle: The area enclosed by a circle is given by: $\text{Area of Circle}=\pi r^2$
• Circumference of a Circle: The distance around the circle: $\text{Circumference}=2\pi r$
• Area of a Sector: The area of a sector with central angle $\theta$ (in degrees) is given by: $\text{Area of Sector}=\frac{\theta }{360}\times \pi r^2$
• Area of a Segment: The area of a segment is the area of the sector minus the area of the triangular portion: $\text{Area of Segment}=\text{Area of Sector}-\text{Area of Triangle}$
  For a segment with a central angle θ\theta, the area of the triangle can be calculated using: $\text{Area of Triangle}=\frac{1}{2}{r}^2\sin \theta$
• Length of an Arc: The length of an arc of a sector with central angle $\theta$ (in degrees) is: $\text{Length of Arc}=\frac{\theta }{360}\times 2\pi r$

6.0Sample NCERT Solutions Class 10 Maths Chapter 11 Areas Related To Circles

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Solution:

The circumference of the circle having radius 19 cm = 2π × 19 cm = 38π cm

The circumference of the circle having radius 9 cm = 2π × 9 cm = 18π cm

Sum of the circumferences of the two circles = (38π + 18π) cm = 56π cm

Therefore, if r cm be the radius of the circle which has circumference equal to the sum of the circumference of the two given circles, then

2πr = 56π => r = 28

Hence, the required radius is 28 cm.

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

Solutions:

Radius, r = 6 cm; sector angle, θ = 60 degrees.

Area of the sector = (θ/360) × πr²

= (60/360) × (22/7) × (6)² cm²

= (1/6) × (22/7) × 36 cm²

= (132/7) cm²

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

Solutions:

Let the radius of the circle = r.

Therefore, 2πr = 22.

=> 2 × (22/7) × r = 22.

=> r = 22 × (7/22) × (1/2) = 7/2 cm.

Here, θ = 90°.

Therefore, Area of the (1/4)th part of the circle or area of quadrant of the circle = (θ/360) × πr²

= (90/360) × (22/7) × (7/2)² cm²

= (1/4) × (22/7) × (7/2) × (7/2) cm²

= (77/8) cm²

4. Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:

The area of the square ABCD = (14)² cm² = 196 cm² (since the side of the square is 14 cm)

The sum of the areas of the semicircles APD and BPC 

= 2 × {area of semicircle APD} 

(since the areas of the two semicircles are equal)

= 2 × (1/2 πr²) = π × (AD/2)² = π × (14/2)² 

(since AD is the diameter of the semicircle APD)

= (22/7) × 49 cm² = 154 cm²

The area of the shaded region 

= The area of the square ABCD - The sum of the areas of the semicircles APD and BPC.

= 196 cm² - 154 cm² = 42 cm²

5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in figure. Find the area of the remaining portion of the square.

Solution:

Side of the square = 4 cm

Therefore, Area of the square ABCD = 4 × 4 cm² = 16 cm²

Each corner has a quadrant of a circle of radius 1 cm.

Therefore, Area of all the 4 quadrants = 4 × (1/4)πr² = πr²

= (22/7) × 1 × 1 cm² = 22/7 cm²

Diameter of the middle circle = 2 cm

Therefore, Radius of the middle circle = 1 cm

Therefore, Area of the middle circle = πr² = (22/7) × 1 × 1 cm² = 22/7 cm²

Now, area of the shaded region = [Area of the square ABCD] - [(Area of the 4 quadrants of a circle) + (Area of the middle circle)]

= [16 cm²] - [(22/7 + 22/7) cm²]

= 16 cm² - 2 × (22/7) cm²

= 16 cm² - 44/7 cm²

= (112 - 44)/7 cm²

= 68/7 cm²

= 9.714 cm²

7.0Benefits of Introduction to Areas Related to Circles

The Areas Related to Circles chapter in Class 10 Maths offers several key benefits:

1. Real-World Applications: It helps in calculating areas of circular objects, which is useful in fields like architecture, engineering, and design.
2. Conceptual Understanding: Students gain a deeper understanding of the area of a circle, sectors, and segments, which is foundational for further geometry topics.
3. Problem-Solving Skills: The chapter strengthens the ability to solve practical problems involving areas of circles and related figures.
4. Foundation for Advanced Topics: It prepares students for more complex concepts in calculus and other higher-level mathematics.