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By Ankit Gupta
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Updated on 11 Jun 2025, 18:18 IST
Mathematics is a subject that helps us understand the world around us in a logical and practical way. One important branch of mathematics is Mensuration, which deals with the measurement of different shapes and figures. In Class 7, Chapter 20 of the RD Sharma Maths textbook focuses on Mensuration I – Area of Circle, a topic that introduces students to the concept of measuring the space inside a circular shape.
A circle is a very common shape we see around us every day – in plates, wheels, clocks, and coins. But have you ever thought about how much space is inside a circle? This space is called the area of the circle, and Chapter 20 helps students learn how to calculate it using simple formulas and steps. The chapter starts by explaining the important parts of a circle, such as the radius, diameter, and circumference, and then moves on to how we can use the value of π (pi) to find the area.
Understanding how to find the area of a circle is not only useful in school but also in real life. Whether it’s designing a round table, laying circular carpets, or even in sports like cricket where the pitch is circular – this concept plays an important role. RD Sharma’s Chapter 20 makes this topic easy to learn by offering a variety of examples, practice problems, and clear explanations.
Do Check: RD Sharma Solutions for Class 7 Maths
The RD Sharma Solutions for Class 7 Chapter 20 are designed to help students grasp these concepts better. These solutions provide step-by-step answers to all the textbook questions, making it easier for students to understand the logic and method behind each problem. Whether you are trying to solve a direct question or a tricky word problem, these solutions guide you in a clear and simple way.
By studying this chapter with the help of RD Sharma Solutions, students can improve their problem-solving skills, gain confidence in handling geometry-based questions, and build a strong foundation in Mensuration. The solutions are also helpful during exam preparation, as they allow students to revise important concepts quickly and effectively.
In conclusion, Chapter 20 of RD Sharma Class 7 Maths is a fun and important topic that teaches students how to measure circular areas in a practical and interesting way. The RD Sharma Solutions make learning this topic easy and enjoyable.
RD Sharma Class 7 Chapter 20 PDF includes detailed solutions, examples, and extra questions to help you master real numbers and other topics. Click here to download the RD Sharma Class 7 Chapter 20 PDF.
In this chapter, students will learn about decimals and how to perform basic operations with them. The solutions provided here are detailed and easy to follow, helping students understand each concept thoroughly.
Q1. Find the area of a circle if its radius is:
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(i) 7 cm
Area = π × r² = (22/7) × 7 × 7 = 154 cm²
(ii) 2.1 m
Area = π × r² = (22/7) × (2.1 × 2.1) = (22/7) × 4.41 = 13.86 m²
(iii) 7 km
Area = π × r² = (22/7) × 7 × 7 = 154 km²
Q2. Find the area of a circle if its diameter is:
(i) 8.4 cm
Radius = 8.4 ÷ 2 = 4.2 cm
Area = π × r² = (22/7) × (4.2 × 4.2) = 55.44 cm²
(ii) 5.6 m
Radius = 5.6 ÷ 2 = 2.8 m
Area = π × r² = (22/7) × (2.8 × 2.8) = 24.64 m²
(iii) 7 km
Radius = 7 ÷ 2 = 3.5 km
Area = π × r² = (22/7) × (3.5 × 3.5) = 38.5 km²
Q3. The area of a circle is 154 cm². What is the radius?
Area = π × r² = 154
r² = (154 × 7) ÷ 22 = 49
Radius = √49 = 7 cm
Q4. Find the radius of a circle if its area is:
(i) 4π cm²
π × r² = 4π → r² = 4 → r = 2 cm
(ii) 55.44 m²
(22/7) × r² = 55.44 → r² = (55.44 × 7) ÷ 22 = 17.64 → r = √17.64 = 4.2 m
(iii) 1.54 km²
(22/7) × r² = 1.54 → r² = (1.54 × 7) ÷ 22 = 0.49 → r = √0.49 = 0.7 km or 700 m
Q5. The circumference of a circle is 3.14 m. Find its area.
Circumference = 2 × π × r → 3.14 = 2 × 3.14 × r → r = 0.5 m
Area = π × r² = (22/7) × 0.5 × 0.5 = 0.785 m²
Q6. If the area of a circle is 50.24 m², what is its circumference?
Area = (22/7) × r² = 50.24 → r² = (50.24 × 7) ÷ 22 = 15.985 → r = √15.985 ≈ 3.998 m
Circumference = 2 × π × r = 2 × (22/7) × 3.998 ≈ 25.12 m
Q7. A horse is tied with a rope 28 m long. What is the area it can graze?
Radius = 28 m
Area = π × r² = (22/7) × 28 × 28 = 2464 m²
Q8. A steel wire forms a square with area 121 cm². If reshaped into a circle, what will be the area?
Square area = 121 → side = √121 = 11 cm → perimeter = 4 × 11 = 44 cm
Circumference of circle = 44 → 2 × (22/7) × r = 44 → r = 7 cm
Area = π × r² = (22/7) × 7 × 7 = 154 cm²
Q9. A circular park has a circumference of 352 m. A road 7 m wide runs around it. What is the area of the road?
Circumference = 2 × (22/7) × r = 352 → r = (352 × 7) ÷ 44 = 56 m
Outer radius = 56 + 7 = 63 m
Area of road = π × (63² – 56²) = (22/7) × (3969 – 3136) = 2618 m²
Q10. Show that the area of a circular ring with width h around a circle of radius r is πh(2r + h)
Outer radius = r + h
Area of ring = π × (r + h)² – π × r² = π(r² + 2rh + h² – r²) = πh(2r + h)
Q11. If the perimeter of a circle is 4πr, what is its area?
Perimeter = 4πr = 2π × (2r) → radius = 2r
Area = π × (2r)² = 4πr²
Q12. A 5024 m long wire is first shaped into a square and then into a circle. What is the ratio of the square’s area to the circle’s area?
Square perimeter = 5024 → side = 5024 ÷ 4 = 1256 m → area = 1256 × 1256
Same length used to form circle → circumference = 5024
2πr = 5024 → r = 2512 ÷ π
Area of circle = π × r² = π × (2512/π)²
Ratio = (1256²) : (π × (2512² / π²)) = 11:14
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Chapter 20 focuses on Mensuration I – Area of Circle. It teaches students how to calculate the area and perimeter (circumference) of a circle using formulas involving the radius, diameter, and the constant π (pi).
RD Sharma Solutions provide step-by-step answers to all questions in the textbook. They help students understand the correct methods, improve problem-solving skills, and prepare well for exams.
You should remember these key formulas:
Area of Circle = π × r²
Circumference of Circle = 2πr
Where r is the radius and π (pi) is taken as 22/7 or 3.14.
Yes, these solutions are perfect for quick revision as they provide concise explanations and help you recall important concepts and formulas easily.
Absolutely! RD Sharma Solutions include detailed solutions for word problems so that students can learn how to apply the formulas in real-life situations, such as calculating the area of circular fields, plates, or rings.
Yes, the solutions are written in simple and easy-to-understand language, making them ideal for students who want to study independently and clear their doubts without external help.
Definitely! By regularly practicing with RD Sharma Solutions, you can master the topic, avoid common mistakes, and perform better in your school exams and class tests.