NCERT solutions for Class 10 Mathematics Chapter Areas Related to Circles – complete answers & explanations

This blog provides complete NCERT solutions for Class 10 Mathematics Chapter Areas Related to Circles. This chapter helps students understand how to calculate areas of sectors, segments, and other parts of a circle using different formulas and concepts. It is an important chapter for building strong geometry skills and applying mathematics to real-life situations such as designs, measurements, and practical calculations. Students learn how to use formulas correctly, solve problems step by step, and improve accuracy in calculations. In this blog, you will find 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. You can also book a free trial now to get expert guidance and strengthen your understanding of mathematical concepts. The answers are presented in a simple and structured way to help students learn easily and confidently.

What this NCERT chapter covers?

1. The chapter explains different parts of a circle such as radius, diameter, arc, sector, and segment.
2. It teaches how to calculate the area of a sector using the angle and radius.
3. Students learn to find the area of segments by subtracting triangle areas from sector areas.
4. It covers the difference between minor and major segments.
5. The chapter includes problems involving arc length and sector area.
6. Students learn to apply trigonometric values in triangle-based calculations.
7. It explains how to handle questions involving combinations of figures.
8. The chapter develops problem-solving skills using step-by-step calculations.
9. It includes real-life applications such as wire length, designs, and wiper movement.
10. Students learn to use values of π correctly in different situations.
11. It improves accuracy and speed in solving geometry-based problems.
12. Overall, it strengthens understanding for exams and practical applications.

How to use these NCERT solutions?

1. Start by attempting each question on your own before checking the answers.
2. Carefully read and understand what each question is asking.
3. Try solving the problems step by step using formulas.
4. After solving, compare your answers with the given solutions.
5. Identify mistakes and understand where you went wrong.
6. Focus on how formulas are applied in each step.
7. Practice similar questions to build confidence.
8. Parents and teachers can guide students in difficult steps.
9. Revise formulas regularly to improve speed and accuracy.
10. Follow the same order of questions as given for better understanding.

Important tips & tricks for students

1. Always write the correct formula before starting the solution.
2. Use the correct value of π (22/7 or 3.14) as required.
3. Be careful with calculations to avoid mistakes.
4. Understand the difference between sector and segment clearly.
5. Practice trigonometric values used in triangle calculations.
6. Always include units in your final answer.
7. Draw diagrams wherever necessary for better clarity.
8. Double-check your answers before finalising them.
9. Manage time properly during exams.
10. Practice regularly to improve confidence and accuracy.

NCERT solutions – complete answer key

Exercise No. 11.1

1. 6π = 6 × 22/7 = 132/7 = 18.86 cm² (approx.)

2. Radius = 22 / (2π) = 22 / (2 × 22/7) = 3.5 cm  
  Area of quadrant = (1/4) × πr² = (1/4) × (22/7) × (3.5)² = 9.625 cm²

3. Angle in 5 minutes = (5/60) × 360 = 30°  
  Area = (30/360) × π × 14² = (1/12) × (22/7) × 196 = 154/3 = 51.33 cm²

4. (i) Area of sector = (90/360) × 3.14 × 100 = 78.5 cm²  
  Area of triangle = (1/2) × 10 × 10 = 50 cm²  
  Minor segment = 78.5 – 50 = 28.5 cm²  

  (ii) Major sector = 3.14 × 100 – 78.5 = 235.5 cm²

5. (i) Length of arc = (60/360) × 2π × 21 = 22 cm  
  (ii) Area of sector = (60/360) × π × 21² = 231 cm²  
  (iii) Area of triangle = (1/2) × 21 × 21 × sin60°  
  = (1/2) × 21 × 21 × (√3/2) = 441√3/4  
  Segment = 231 – (441√3/4) cm²  

6. Area of sector = (60/360) × 3.14 × 225 = 117.75 cm²  
  Area of triangle = (1/2) × 15 × 15 × (√3/2)  
  = (1/2) × 225 × (1.73/2) = 97.31 cm²  
  Minor segment = 117.75 – 97.31 = 20.44 cm²  
  Major segment = 3.14 × 225 – 20.44 = 706.5 – 20.44 = 686.06 cm²  

7. Area of sector = (120/360) × 3.14 × 144 = 150.72 cm²  
  Area of triangle = (1/2) × 12 × 12 × sin120°  
  = (1/2) × 144 × (√3/2)  
  = 72 × (1.73/2) = 62.28 cm²  
  Segment = 150.72 – 62.28 = 88.44 cm²  

8. (i) Area = (1/4) × π × 5² = (1/4) × 3.14 × 25 = 19.625 m²  
  (ii) New area = (1/4) × π × 10² = 78.5 m²  
  Increase = 78.5 – 19.625 = 58.875 m²  

9. (i) Total wire = circumference + 5 diameters  
  = π × 35 + 5 × 35  
  = 110 + 175 = 285 mm  

  (ii) Area of each sector = (1/10) × π × (17.5)²  
  = (1/10) × (22/7) × 306.25 = 96.25 mm²  

10. Area between ribs = (1/8) × π × 45²  
   = (1/8) × 3.14 × 2025 = 794.81 cm²  

11. Area cleaned by one wiper = (115/360) × π × 25²  
   = (115/360) × 3.14 × 625 = 627.43 cm²  
   Total area = 2 × 627.43 = 1254.86 cm²  

12. Area = (80/360) × 3.14 × (16.5)²  
   = (2/9) × 3.14 × 272.25 = 189.97 km²  

13. Area of circle = π × 28² = (22/7) × 784 = 2464 cm²  
   Area of one design = (1/6) × 2464 = 410.67 cm²  
   Cost = 6 × 410.67 × 0.35 = ₹ 862.41  

14. Correct Answer: C

Why NCERT solutions help students?

NCERT solutions help students prepare effectively for exams by improving concept clarity and showing the correct way to solve problems. They build confidence, reduce mistakes, and help students understand how to write accurate answers in exams.

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