RD Sharma Solutions for Class 7 Maths Chapter 20 Mensuration I (Area of Circle)

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.

Download RD Sharma Solutions for Class 7 Maths Chapter 20 Mensuration I PDF Here

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.

Access Answers to RD Sharma Solutions for Class 7 Maths 20 Mensuration I

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:

(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