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RD Sharma Solutions for Class 7 Maths Chapter 21 Mensuration II (Perimeter And Area of Rectilinear Figures)
By Ankit Gupta
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Updated on 24 Apr 2025, 12:32 IST
Understanding shapes and their measurements is a big part of learning maths, and Chapter 21 of RD Sharma Class 7 Maths — Mensuration II: Perimeter and Area of Rectilinear Figures — does just that. This chapter helps students learn how to measure the boundaries and spaces of different shapes, especially those with straight sides, like rectangles, squares, and other polygons. These shapes are called rectilinear figures because their sides are made of straight lines.
The perimeter of a shape means the total length around it. For example, if you walk around a garden in the shape of a rectangle, the total distance you walk is the perimeter. On the other hand, the area is the amount of space a shape covers. If you are laying tiles on the floor of a rectangular room, the number of tiles you need depends on the area.
In this chapter, students will learn how to find the perimeter and area of basic rectilinear figures like squares, rectangles, and other combined figures. It also teaches how to break complex shapes into smaller parts to make calculations easier. RD Sharma has explained all these concepts using clear steps, examples, and simple formulas. This helps students understand not only how to solve the questions but also why the method works.
Do Check: RD Sharma Solutions for Class 7 Maths
The RD Sharma Solutions for Chapter 21 are designed to make learning smooth and stress-free. The solutions are explained in an easy-to-understand way so that students can follow each step and understand the logic behind it. These solutions also include various practice questions with detailed answers to help students gain confidence and score well in their exams.
By studying this chapter, students will not only improve their problem-solving skills but also learn how maths is used in daily life. For instance, knowing how to calculate area and perimeter can help in real-world situations like measuring land, buying carpets, or building fences.
In short, RD Sharma Solutions for Class 7 Chapter 21 give students a strong foundation in mensuration, with clear concepts, simple language, and helpful examples. It is a great resource for understanding how to work with rectilinear figures and for building confidence in maths.
Download RD Sharma Solutions for Class 7 Maths Chapter 21 Mensuration II PDF Here
RD Sharma Class 7 Chapter 21 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 21 PDF.
Access Answers to RD Sharma Solutions for Class 7 Maths Chapter 21 Mensuration II
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. Calculate the area of a rectangle in square meters:
(i) Length = 5.5 m, Breadth = 2.4 m
To find the area of a rectangle, we multiply the length by the breadth.
Area = 5.5 × 2.4 = 13.2 m²
(ii) Length = 180 cm, Breadth = 150 cm
First, convert both measurements into meters: 180 cm = 1.8 m and 150 cm = 1.5 m.
Area = 1.8 × 1.5 = 2.7 m²
Q2. Calculate the area of a square in square centimetres:
(i) Side = 2.6 cm
Area = side × side = 2.6 × 2.6 = 6.76 cm²
(ii) Side = 1.2 dm
Convert decimetres to centimetres: 1.2 dm = 12 cm
Area = 12 × 12 = 144 cm²
Q3. Find the area of a square in square meters where the side is 16.5 decameters (dam):
1 dam = 10 meters, so 16.5 dam = 165 m
Area = 165 × 165 = 27225 m²
Q4. Find the area of a rectangular field in acres:
(i) Length = 200 m, Breadth = 125 m
Area = 200 × 125 = 25000 m². Since 100 m² = 1 acre,
Area in acres = 25000 ÷ 100 = 250 acres
(ii) Length = 75 m 5 dm = 75.5 m, Breadth = 120 m
Area = 75.5 × 120 = 9060 m²
Area in acres = 9060 ÷ 100 = 90.6 acres
Q5. Find the area of a rectangular field in hectares:
(i) Length = 125 m, Breadth = 400 m
Area = 125 × 400 = 50000 m². Since 10000 m² = 1 hectare,
Area in hectares = 50000 ÷ 10000 = 5 hectares
(ii) Length = 75.5 m, Breadth = 120 m
Area = 75.5 × 120 = 9060 m²
Area in hectares = 9060 ÷ 10000 = 0.906 hectares
Q6. A door of size 3 m × 2 m is built in a wall of size 10 m × 10 m. Calculate the cost to paint the remaining wall if painting costs Rs. 2.50 per square metre:
Wall area = 10 × 10 = 100 m²
Door area = 3 × 2 = 6 m²
Paintable area = 100 – 6 = 94 m²
Total painting cost = 94 × 2.5 = Rs. 235
Q7. A wire is bent into a rectangle of 40 cm by 22 cm. If reshaped into a square, find the length of one side and which shape has a larger area:
Perimeter of rectangle = 2 × (40 + 22) = 124 cm
Perimeter of square = 124 cm → Side = 124 ÷ 4 = 31 cm
Area of rectangle = 40 × 22 = 880 cm²
Area of square = 31 × 31 = 961 cm²
The square encloses a larger area.
Q8. A window has 12 glass panes, each 25 cm by 16 cm. How much glass is needed in square meters?
Area of one pane = 25 × 16 = 400 cm² = 0.04 m²
Total area for 12 panes = 12 × 0.04 = 0.48 m²
Q9. A marble tile is 10 cm × 12 cm. How many tiles are needed to cover a 3 m × 4 m wall? Also, find the cost if one tile costs Rs. 2.
Wall area = 3 × 4 = 12 m²
Tile area = 10 × 12 = 120 cm² = 0.012 m²
Number of tiles = 12 ÷ 0.012 = 1000 tiles
Total cost = 1000 × 2 = Rs. 2000
Q10. A table top is 9 dm 5 cm long and 6 dm 5 cm wide. Find the cost to polish it at 20 paise per square centimetre:
Length = 9 × 10 + 5 = 95 cm, Breadth = 6 × 10 + 5 = 65 cm
Area = 95 × 65 = 6175 cm²
Cost per cm² = 20 paise = Rs. 0.20
Total polishing cost = 6175 × 0.20 = Rs. 1235
FAQs on RD Sharma Solutions for Class 7 Maths Chapter 21 Mensuration II (Perimeter And Area of Rectilinear Figures)
What topics are covered in Chapter 21 of RD Sharma Class 7 Maths?
Chapter 21 focuses on Mensuration II, specifically the perimeter and area of rectilinear figures such as rectangles, squares, and composite figures made up of these shapes.
What is the main purpose of using RD Sharma Solutions for this chapter?
RD Sharma Solutions help students understand each concept with step-by-step explanations, making it easier to solve problems on perimeter and area and build a strong foundation in mensuration.
Are formulas provided in the solutions for easy learning?
Yes, all key formulas for perimeter and area are included and explained in simple language. These are used in examples and exercises to help students remember and apply them easily.
How do these solutions help in exam preparation?
The solutions offer a variety of solved problems, including important questions likely to appear in exams. This gives students more practice and helps them gain confidence in solving mensuration problems quickly and correctly.
Is it necessary to understand the concepts before using the solutions?
While the solutions are easy to follow, it's always helpful for students to first go through the theory and explanations in the textbook. This ensures better understanding and helps make the solutions more meaningful.
Can these solutions help in solving composite figure problems?
Yes, RD Sharma Solutions explain how to break down complex or composite rectilinear shapes into simpler parts (like rectangles or squares) to calculate the perimeter and area easily.
Are these solutions useful for competitive exams like Olympiads?
Absolutely! The detailed solutions not only help in school exams but also sharpen problem-solving skills, which are essential for math competitions and Olympiads.