RD Sharma Solutions Class 6 Maths Chapter 20 Mensuration

RD Sharma Solutions Class 6 Maths Chapter 20: Mensuration is an important chapter in the Class 6 Maths syllabus, and RD Sharma Solutions for Class 6 Chapter 20 make it easier for students to understand and master this topic. This chapter introduces students to the basics of measuring the area and perimeter of different shapes such as squares, rectangles, and triangles. It forms the foundation for advanced topics in higher classes, so having a clear understanding is essential.

The RD Sharma Solutions Class 6 Maths Chapter 20 are carefully prepared to match the latest CBSE syllabus. These step-by-step solutions help students solve textbook problems with confidence and improve their problem-solving skills. Whether you're revising for an exam or practicing daily, these solutions make learning more structured and less stressful.

In this chapter, students learn how to calculate:

• Perimeter of closed figures
• Area of square and rectangle
• Basic units of measurement

These concepts are not just useful for exams, but also play a role in real-life situations like finding the area of a garden or the boundary of a playground. The solutions come with clear explanations and examples that help in better understanding and application.

To make your preparation even better, you can use NCERT Solutions Class 6 Maths along with Mensuration Worksheet Class 6 Maths practice. These resources together will strengthen your base in Mensuration and improve your accuracy and speed in solving questions.

By using RD Sharma Solutions and regular practice through worksheets and NCERT exercises, students can gain a strong grip on the topic and be well-prepared for upcoming exams.

Download RD Sharma Class 6 Chapter 20 PDF with Solutions

The RD Sharma Class 6 Chapter 20 PDF is a valuable resource designed to help students understand and practice the topic of Mensuration with ease. This PDF includes detailed solutions, step-by-step examples, and a wide range of extra questions that cover every important concept from the chapter. With a clear and structured format, the PDF helps students revise efficiently and build a strong foundation in measuring area and perimeter of various shapes like rectangles, squares, and triangles.

By using the RD Sharma Class 6 Maths Chapter 20 PDF, students can learn at their own pace, clarify doubts quickly, and improve their problem-solving skills.

RD Sharma Solutions Class 6 Maths Chapter 20 - Advanced-Level Mensuration Questions 

Q1. Find the perimeter of a rectangle whose length is 12.5 cm and breadth is 8 cm.
Answer:
Perimeter = 2 × (length + breadth)
= 2 × (12.5 + 8) = 2 × 20.5 = 41 cm

Q2. A square park has a side of 25 m. Find the cost of fencing it at ₹15 per metre.
Answer:
Perimeter of square = 4 × side = 4 × 25 = 100 m
Cost = 100 × 15 = ₹1500

Q3. The area of a rectangle is 96 cm². If its length is 12 cm, find its breadth.
Answer:
Area = length × breadth
96 = 12 × breadth ⇒ breadth = 96 ÷ 12 = 8 cm

Q4. A triangle has base = 10 cm and height = 6 cm. Find its area.
Answer:
Area = ½ × base × height = ½ × 10 × 6 = 30 cm²

Q5. A rectangular sheet is 16 cm long and 10 cm wide. Find its area and perimeter.
Answer:
Area = 16 × 10 = 160 cm²
Perimeter = 2 × (16 + 10) = 2 × 26 = 52 cm

Q6. A farmer wants to fence a square field of area 2500 m². What is the total cost if fencing costs ₹20 per metre?
Answer:
Side = √2500 = 50 m
Perimeter = 4 × 50 = 200 m
Cost = 200 × 20 = ₹4000

Q7. Convert 3.5 m² to cm².
Answer:
1 m² = 10,000 cm²
3.5 m² = 3.5 × 10,000 = 35,000 cm²

Also Check: NCERT Solutions Class 6 Maths Mensuration

Q8. A rectangular plot is 80 m long and 60 m wide. Find how many square tiles of side 4 m are needed to cover the plot.
Answer:
Area of plot = 80 × 60 = 4800 m²
Area of 1 tile = 4 × 4 = 16 m²
Number of tiles = 4800 ÷ 16 = 300 tiles

Q9. A square and a rectangle have the same perimeter. The square's side is 20 cm. The rectangle's length is 25 cm. Find its breadth.
Answer:
Perimeter of square = 4 × 20 = 80 cm
Rectangle perimeter = 2 × (25 + b) = 80
⇒ 25 + b = 40 ⇒ b = 15 cm
Breadth = 15 cm

Q10. Find the perimeter of an equilateral triangle with side 7.5 cm.
Answer:
Perimeter = 3 × side = 3 × 7.5 = 22.5 cm

Q11. A rectangular garden is 90 m long and 60 m wide. Find the area not covered if a 2 m wide path runs along its boundary.
Answer:
Outer dimensions = 90 × 60 = 5400 m²
Inner rectangle = (90−4) × (60−4) = 86 × 56 = 4816 m²
Area of path = 5400 − 4816 = 584 m²

Q12. A square has area 324 cm². Find its perimeter.
Answer:
Side = √324 = 18 cm
Perimeter = 4 × 18 = 72 cm

Q13. Find the length of a rectangle whose area is 200 cm² and breadth is 10 cm.
Answer:
Area = length × breadth
200 = length × 10 ⇒ length = 20 cm
Length = 20 cm

Q14. The perimeter of a triangle is 36 cm. Two of its sides are 12 cm and 14 cm. Find the third side.
Answer:
Third side = 36 − (12 + 14) = 10 cm

Q15. A square and rectangle have the same area. Square's side is 14 cm. Rectangle's breadth is 7 cm. Find its length.
Answer:
Area = 14 × 14 = 196 cm²
Length = 196 ÷ 7 = 28 cm

Q16. A floor is 10 m long and 8 m wide. How many square tiles of side 50 cm are needed to cover it?
Answer:
Area = 10 × 8 = 80 m² = 80,000 cm²
Tile area = 50 × 50 = 2500 cm²
No. of tiles = 80000 ÷ 2500 = 32 tiles

Q17. A path of 1.5 m width is made inside a rectangular park of 20 m × 15 m. Find the area of the path.
Answer:
Inner rectangle = (20−3) × (15−3) = 17 × 12 = 204 m²
Outer area = 20 × 15 = 300 m²
Area of path = 300 − 204 = 96 m²

Q18. A triangular garden has sides 40 m, 32 m, and 24 m. Find its perimeter.
Answer:
Perimeter = 40 + 32 + 24 = 96 m

Q19. A wire is bent to form a square of side 22 cm. Find the length of the wire.
Answer:
Length = perimeter = 4 × 22 = 88 cm

Q20. A room is 6 m long and 4 m wide. What is the cost of carpeting it at ₹100 per m²?
Answer:
Area = 6 × 4 = 24 m²
Cost = 24 × 100 = ₹2400

Q21. The area of a rectangle is 120 cm². If its length is double its breadth, find both dimensions.
Answer:
Let breadth = x, length = 2x
Area = x × 2x = 2x² = 120 ⇒ x² = 60 ⇒ x = √60 ≈ 7.75 cm
Length ≈ 15.5 cm, Breadth ≈ 7.75 cm

Q22. A square garden is surrounded by a 1 m wide path. If the area of the whole field is 121 m², find the side of the garden.
Answer:
Total area = 121 m² ⇒ side = √121 = 11 m
Inner garden = 11 − 2 = 9 m
Area = 9 × 9 = 81 m²
Area of path = 121 − 81 = 40 m²

Q23. If the perimeter of a square is 60 cm, find its area.
Answer:
Side = 60 ÷ 4 = 15 cm
Area = 15 × 15 = 225 cm²

Q24. A rectangular playground has an area of 960 m². If its length is 40 m, find its width and perimeter.
Answer:
Width = 960 ÷ 40 = 24 m
Perimeter = 2 × (40 + 24) = 2 × 64 = 128 m

Q25. A field is 100 m long and 50 m wide. A path 2 m wide is built around the inside edge. Find the area of the path.
Answer:
Inner dimensions = (100−4) × (50−4) = 96 × 46 = 4416 m²
Outer area = 100 × 50 = 5000 m²
Area of path = 5000 − 4416 = 584 m²

1. Builds Strong Basics in Mensuration
   RD Sharma Solutions Class 6 Maths Chapter 20 helps students clearly understand the basics of area and perimeter. This foundation is key for scoring well in future classes.
2. Aligned with NCERT Curriculum
   These solutions follow the CBSE pattern and work well with NCERT Solutions Class 6 Maths, ensuring you are always studying the right content.
3. Improves Problem-Solving Speed
   Regular practice with these solutions and Worksheet Class 6 Maths can make students faster and more accurate in solving Mensuration problems.
4. Easy Step-by-Step Explanations
   RD Sharma provides clear steps, which helps students learn the logic behind every formula and method, making revision easier before exams.
5. Boosts Exam Confidence
   Solving a wide range of questions from RD Sharma Class 6 Chapter 20 gives students the confidence to attempt any type of Mensuration question in exams.
6. Supports Homework and Self-Study
   These solutions are great for completing homework independently and can be used for daily self-study along with NCERT exercises and class worksheets.