NCERT Solutions for Class 10 Mathematics Chapter 11
NCERT solutions for Class 10 Mathematics Chapter 11 Areas related to circles – complete answers & explanations
This chapter in Class 10 Mathematics focuses on areas related to circles and helps students understand how to calculate the area of sectors, segments, and different parts of a circle using formulas. It is an important chapter because it builds strong concepts required for geometry and also appears frequently in exams. Students learn how to apply formulas in real-life situations such as finding areas of fields, paths, and designs. This blog provides clear and reliable NCERT solutions that help students follow the correct approach to answering questions. Download the worksheet and practice alongside solutions for better clarity. Book a free trial now to get expert guidance and strengthen your understanding of concepts step by step.
What this NCERT chapter covers?
1. Understanding the concept of circles
and their different parts such as radius, diameter, sector, and segment
2. Learning formulas to calculate the area of a sector and segment of a circle
3. Applying the concept of angles in calculating different parts of a circle
4. Finding the length of arcs and their relation to the circumference
5. Solving problems involving real-life applications of circular regions
6. Understanding the difference between minor and major segments
7. Using π (pi) values like 22/7 and 3.14 in calculations
8. Learning how to calculate areas of shaded regions
9. Combining concepts of triangles and circles in problem-solving
10. Practicing calculations involving multiple steps and formulas
11. Developing accuracy in numerical problem solving
12. Strengthening exam preparation through application-based questions
How to use these NCERT solutions?
1. First, carefully read each question
from the worksheet and try solving it on your own
2. Use the solutions only after attempting the question to check your understanding
3. Compare your steps with the provided answers to identify mistakes
4. Focus on the formulas used in each step and understand their application
5. Practice similar questions to improve speed and accuracy
6. Parents and teachers can guide students by discussing mistakes and clarifying doubts
7. Follow the solutions in the exact order to match the worksheet flow
8. Revise important formulas regularly for better retention
Important tips & tricks for students
1. Always remember the formulas for area of sector and segment
2. Be careful while converting angles into fractions of 360°
3. Use the correct value of π as given in the question
4. Avoid calculation errors by solving step by step
5. Practice diagram-based questions carefully
6. Understand the difference between minor and major segments clearly
7. Do not skip steps in calculations during exams
8. Double-check units like cm², m², or mm²
9. Read questions properly before applying formulas
10. Practice regularly to gain confidence and accuracy
NCERT solutions – complete answer key
Exercise No. 11.1
Area of sector = (θ/360) × πr²
= (60/360) × (22/7) × 6 × 6
= (1/6) × (22/7) × 36
= (22 × 6) / 7
= 132/7
= 18.86 cm²
Circumference = 22 cm
2πr = 22
r = 22 / (2 × 22/7) = 3.5 cm
Area of quadrant = (1/4) × πr²
= (1/4) × (22/7) × (3.5)²
= (1/4) × (22/7) × 12.25
= 9.625 cm²
Time = 5 minutes → angle = (5/60) × 360 = 30°
Area = (30/360) × (22/7) × 14 × 14
= (1/12) × (22/7) × 196
= (22 × 28) / 12
= 616/12
= 51.33 cm²
r = 10 cm, θ = 90°
Area of sector = (90/360) × 3.14 × 100 = 78.5 cm²
Area of triangle = (1/2) × 10 × 10 = 50 cm²
(i) Minor segment = 78.5 – 50 = 28.5 cm²
(ii) Major sector = 3.14 × 100 – 78.5 = 235.5 cm²
(i) Area grazed = (1/4) × 3.14 × 5 × 5 = 19.63 m²
(ii) New area = (1/4) × 3.14 × 10 × 10 = 78.5 m²
Increase = 78.5 – 19.63 = 58.88 m²
r = 21 cm, θ = 60°
(i) Length of arc = (60/360) × 2 × (22/7) × 21 = 22 cm
(ii) Area of sector = (60/360) × (22/7) × 21 × 21 = 231 cm²
(iii) Area of triangle = (√3/4) × (21)² = (441√3)/4
Area of segment = 231 – (441√3 / 4) cm²
r = 15 cm, θ = 60°
Area of sector = (60/360) × 3.14 × 225 = 117.75 cm²
Area of triangle = (√3/4) × 225 = (1.73/4) × 225 = 97.31 cm²
Minor segment = 117.75 – 97.31 = 20.44 cm²
Area of circle = 3.14 × 225 = 706.5 cm²
Major segment = 706.5 – 20.44 = 686.06 cm²
r = 12 cm, θ = 120°
Area of sector = (120/360) × 3.14 × 144 = 150.72 cm²
Area of triangle = (√3/4) × (12)² = (1.73/4) × 144 = 62.28 cm²
Segment = 150.72 – 62.28 = 88.44 cm²
Diameter = 35 mm → r = 17.5 mm
(i) Circumference = 2 × (22/7) × 17.5 = 110 mm
5 diameters = 5 × 35 = 175 mm
Total wire = 110 + 175 = 285 mm
Area between ribs = (360/8 = 45°)
Area = (45/360) × (22/7) × 45 × 45
= (1/8) × (22/7) × 2025
= 794.81 cm²
(ii) Area of each sector = πr² / 10
= (22/7) × 17.5 × 17.5 / 10
= 96.25 mm²
Area of one wiper = (115/360) × 3.14 × 25 × 25 = 627.9 cm²
Total = 2 × 627.9 = 1255.8 cm²
Area = (80/360) × 3.14 × (16.5)²
= (2/9) × 3.14 × 272.25
= 190 km²
Total area of circle = (22/7) × 28 × 28 = 2464 cm²
Area of 6 designs = 2464 × (1/6) = 410.67 cm²
Cost = 410.67 × 0.35 = 143.73
Area of a sector of angle p (in degrees) of a circle with
radius R is
The standard formula is shown below:
Area of sector=(p/360)πR2
Correct option is(C)
Why NCERT solutions help students?
NCERT solutions help students prepare effectively for exams by providing clear steps and correct approaches to solving questions. They improve understanding of concepts, build confidence, and ensure students practice answers that are aligned with NCERT expectations.
Build strong Mathematics fundamentals with structured practice and expert learning support.