CBSE Class 5 Mathematics NCERT Solutions: Chapter 11 Area and Its Boundary

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NCERT Solutions PDF Class 5 PDF

CBSE Class 5 Mathematics chapter 11, Area and Its Boundary, helps students understand the concepts of perimeter and area. The NCERT Solutions offer clear explanations and step-by-step guidance for problems involving irregular shapes, rectangles, and squares. Students will learn practical ways to compare sizes, calculate how much space a shape covers, and find the length of its boundary. This includes figuring out which shape is bigger, how to tile surfaces, and the connection between a shape's size and its dimensions. These solutions aim to build strong spatial reasoning and measurement skills, showing how these ideas apply in real life and helping students prepare for their exams.

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BoardCBSE
ClassClass 5
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
Chapter11 Area and Its Boundary

Chapter summary

Chapter 11, "Area and Its Boundary," focuses on introducing Class 5 students to the concepts of area and perimeter. The NCERT Solutions cover methods for comparing the sizes of different shapes, calculating the area of rectangles by counting squares or using multiplication, and understanding the perimeter. It includes practical examples like tiling a floor and covering a rectangle with stamps, reinforcing the practical application of these geometric concepts.

Learning outcomes

  • Understand the concept of area as the space occupied by a 2D shape.
  • Learn to compare the areas of different shapes.
  • Calculate the area of rectangles using the formula length × width.
  • Understand the concept of perimeter as the boundary length of a shape.
  • Calculate the perimeter of rectangles and squares.
  • Apply area and perimeter concepts to solve real-world problems involving tiles and fencing.

Topics covered

Paper topics

  • Comparing Areas
  • Measuring Area with Unit Squares
  • Area of Rectangles
  • Area of Irregular Shapes
  • Perimeter of Shapes
  • Calculating Area and Perimeter
  • Tiling a Floor
  • Covering with Stamps
  • Wire Formed into Shapes

Important topics

  • Understanding and comparing areas
  • Calculating area of rectangles
  • Calculating perimeter of squares and rectangles
  • Applying area and perimeter to practical problems

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Questions and Solutions

Whose Slice is Bigger?

Parth and Gini bought aam paapad (dried mango slice) from a shop. Their pieces looked like these. Both could not make out whose piece was bigger. Piece A is 11 cm by 6 cm. Piece B is 5 cm by 3 cm.

1. Suggest some ways to find out whose piece is bigger. Discuss.

Solution:

To determine which piece of aam paapad is bigger, we can use a couple of methods:

  1. Method 1: Cutting and Comparing We can try to cut Piece B into smaller parts and see if they can cover Piece A. For example, if Piece B (5 cm x 3 cm) is cut into three parts: two parts of 5 cm x 1 cm and one part of 5 cm x 1 cm. If we place these parts over Piece A (11 cm x 6 cm), we can visually compare. However, this method can be imprecise.
  2. Method 2: Using Unit Squares A more accurate method is to divide both pieces into small squares of equal size, for instance, 1 cm by 1 cm squares. We can then count the number of squares in each piece. The piece with more squares has a larger area.

Let's apply Method 2 to find the exact areas.

Area Calculation for Pieces

1. Altogether how many squares can be arranged on it?

2. So the area of piece A =-----Square cm

3. In the same way find the area of piece B.

4. Who had the bigger piece? How much bigger?

Solution:

Let's calculate the area of Piece A and Piece B using 1 cm squares.

For Piece A (11 cm x 6 cm):

We can arrange 11 squares along the length and 6 squares along the width. The total number of 1 cm squares that can be arranged on Piece A is the product of its length and width.

Number of squares = Length × Width = 11 cm × 6 cm = 66 squares.

Therefore, the area of piece A is 66 square cm.

For Piece B (5 cm x 3 cm):

Similarly, we can arrange 5 squares along the length and 3 squares along the width.

Number of squares = Length × Width = 5 cm × 3 cm = 15 squares.

Therefore, the area of piece B is 15 square cm.

Comparison:

Comparing the areas, Piece A has an area of 66 square cm and Piece B has an area of 15 square cm.

Piece A is bigger than Piece B.

The difference in area is 66 sq cm - 15 sq cm = 51 sq cm.

Answer Summary:

1. 66 squares can be arranged on Piece A.

2. The area of piece A = 66 square cm.

3. The area of piece B = 15 square cm.

4. Piece A had the bigger piece. It was bigger by 51 square cm.

Cover With Stamps

The stamp has an area of 4 square cm. Guess, how many such stamps will cover this big rectangle.

Check Your Guess

a) Measure the yellow rectangle. It is ...... cm long.

b) How many stamps can be placed along its length?.....

c) How wide is the rectangle?..... cm

d) How many stamps can be placed along its width?.....

e) How many stamps are needed to cover the rectangle? .....

f) How close was your earlier guess? Discuss

g) What is the area of the rectangle?..... square cm.

h) What is the perimeter of the rectangle?..... cm

Solution:

Let's analyze the problem step-by-step.

Guess: My guess is 20 stamps.

Checking the Guess:

a) Measuring the yellow rectangle, its length is 12 cm.

b) Since each stamp is 4 sq cm, and we assume it's a square stamp with side length 2 cm (as 2 cm x 2 cm = 4 sq cm), we can place 12 cm / 2 cm = 6 stamps along its length.

c) The width of the rectangle is 8 cm.

d) Along the width, we can place 8 cm / 2 cm = 4 stamps.

e) To cover the entire rectangle, we multiply the number of stamps along the length by the number of stamps along the width: 6 stamps × 4 stamps = 24 stamps.

f) My earlier guess was 20 stamps, and the actual number needed is 24 stamps. So, my guess was quite close.

g) The area of the rectangle can be calculated in two ways:

- Using the dimensions: Area = Length × Width = 12 cm × 8 cm = 96 square cm.

- Using the stamps: Area = Number of stamps × Area of one stamp = 24 stamps × 4 sq cm/stamp = 96 square cm.

So, the area of the rectangle is 96 square cm.

h) The perimeter of the rectangle is the total length of its boundary.

Perimeter = 2 × (Length + Width) = 2 × (12 cm + 8 cm) = 2 × (20 cm) = 40 cm.

Practice Time: Tiling and Fencing

a) Arbaz plans to tile his kitchen floor with green square tiles. Each side of the tile is 10 cm. His kitchen is 220 cm in length and 180 cm wide. How many tiles will he need?

b) The fencing of a square garden is 20 m in length. How long is one side of the garden?

c) A thin wire 20 cm long is formed into a rectangle. If the width of this rectangle is 4 cm, what is its length?

Solution:

Let's solve each problem step-by-step:

a) Tiling the kitchen floor:

First, calculate the area of the kitchen floor:

Area of floor = Length × Width = 220 cm × 180 cm = 39600 square cm.

Next, calculate the area of one tile:

Area of a tile = Side × Side = 10 cm × 10 cm = 100 square cm.

To find the number of tiles needed, divide the total area of the floor by the area of one tile:

Number of tiles = Area of floor / Area of a tile = 39600 sq cm / 100 sq cm = 396 tiles.

Arbaz will need 396 tiles.

b) Fencing a square garden:

The length of the fencing represents the perimeter of the square garden.

Perimeter of the square garden = 20 m.

The formula for the perimeter of a square is 4 × side length.

So, 4 × side length = 20 m.

To find the length of one side, divide the perimeter by 4:

Side length = 20 m / 4 = 5 m.

One side of the garden is 5 m long.

c) Forming a rectangle with a wire:

The length of the wire represents the perimeter of the rectangle.

Perimeter of the rectangle = 20 cm.

The width of the rectangle is given as 4 cm.

The formula for the perimeter of a rectangle is 2 × (Length + Width).

So, 2 × (Length + 4 cm) = 20 cm.

Divide both sides by 2:

Length + 4 cm = 20 cm / 2 = 10 cm.

Now, subtract the width from this sum to find the length:

Length = 10 cm - 4 cm = 6 cm.

The length of the rectangle is 6 cm.

Common mistakes

  • Confusing area with perimeter.
  • Incorrectly counting squares when calculating area for irregular shapes.
  • Errors in applying the perimeter formula for rectangles.
  • Calculation mistakes when dealing with larger numbers in area and perimeter problems.

Revision tips

  • Visualize area by imagining filling shapes with unit squares.
  • Practice calculating perimeter by tracing the boundary of objects.
  • Use the provided examples to understand the difference between area and perimeter.
  • Work through the practice problems to reinforce calculation skills.

Practice MCQs

Q1. What is the area of a rectangle with length 6 cm and width 5 cm?

Q2. If a square garden has a perimeter of 20 m, what is the length of one side?

Q3. Which concept measures the length of the boundary of a shape?

Q4. How many 1 cm squares are needed to cover a rectangle of 11 cm by 3 cm?

Q5. A rectangle is 12 cm long and 8 cm wide. How many stamps of 4 sq cm area would be needed to cover it?

Frequently asked questions

What is the main concept covered in CBSE Class 5 Maths Chapter 11?

Chapter 11, 'Area and Its Boundary,' introduces students to the concepts of area (the space inside a shape) and perimeter (the length of the boundary of a shape) and how to calculate them for simple shapes.

How do the NCERT Solutions for this chapter help students?

These solutions provide step-by-step explanations for all exercises, helping students understand how to compare areas, calculate area and perimeter of rectangles and squares, and solve related practical problems.

What is the difference between area and perimeter?

Area measures the space covered by a flat shape, usually in square units (like square cm). Perimeter measures the total length of the outline or boundary of a shape, usually in linear units (like cm).

Are there practical examples in this chapter?

Yes, the chapter includes practical examples such as comparing pieces of 'aam paapad', covering a rectangle with stamps, and calculating tiles needed for a kitchen floor, making the concepts relatable.

How can I use these solutions for exam preparation?

You can use these solutions to review the concepts, understand the methods for solving problems, and practice the calculations. Ensure you can solve each problem independently after reviewing the solution.

Important links

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