CBSE Class 8 Maths Chapter 11: Mensuration NCERT Solutions

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NCERT Solutions PDF Class 8 PDF

This chapter focuses on Mensuration, covering the calculation of area and perimeter for various geometric shapes. Students will learn to apply formulas for squares, rectangles, and composite figures that combine these basic shapes. The NCERT Solutions provide step-by-step guidance for solving problems related to finding areas and perimeters, including practical applications like calculating the cost of developing a garden or comparing areas of different fields. These solutions are designed to help students understand the concepts thoroughly and build confidence in tackling exam questions related to mensuration. By working through these problems, students will develop a strong foundation in geometric measurement, which is crucial for further mathematical studies. The detailed explanations and clear methods presented in the solutions aid in effective revision, ensuring students can recall and apply the learned formulas and techniques during their examinations. This resource aims to make learning mensuration accessible and straightforward for all Class 8 students.

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Quick info

BoardCBSE
ClassClass 8
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 11

Chapter summary

Chapter 11, Mensuration, for Class 8 Maths NCERT Solutions, deals with the concepts of area and perimeter. It includes problems involving squares, rectangles, and composite shapes formed by combining these. Students will practice calculating the area and perimeter of such figures and apply these concepts to real-world scenarios, such as finding the cost of developing a garden or comparing the areas of fields with the same perimeter. The exercises focus on reinforcing the understanding of geometric formulas and their practical applications.

Learning outcomes

  • Understand the concepts of perimeter and area for squares and rectangles.
  • Calculate the perimeter and area of square and rectangular fields.
  • Determine the area of composite shapes by subtracting areas.
  • Apply mensuration formulas to solve practical problems involving cost calculation.
  • Compare the areas of different shapes with the same perimeter.

Topics covered

Paper topics

  • Perimeter of a square
  • Area of a square
  • Perimeter of a rectangle
  • Area of a rectangle
  • Composite shapes
  • Area of a garden
  • Cost calculation based on area
  • Comparison of areas
  • Mensuration

Important topics

  • Area and Perimeter of Squares and Rectangles
  • Calculating Area of Composite Shapes
  • Practical Applications of Mensuration (Cost Calculation)
  • Comparing Areas of Different Fields

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Questions and Solutions

Question 1

A square and a rectangular field have the same perimeter. If the side of the square field is 60 m and the length of the rectangular field is 80 m, which field has a larger area?
Solution:

We are given the side of the square field and the length of the rectangular field, and that their perimeters are equal.

Step 1: Calculate the perimeter of the square field.

The formula for the perimeter of a square is 4 \times \text{side}.

Perimeter of square = 4 \times 60 \text{ m} = 240 \text{ m}.

Step 2: Use the perimeter to find the breadth of the rectangular field.

We know the perimeter of the rectangular field is equal to the perimeter of the square field, so it is also 240 m.

The formula for the perimeter of a rectangle is 2(l+b), where l is the length and b is the breadth.

Given length l = 80 \text{ m}.

So, 2(80 + b) = 240 \text{ m}.

Divide both sides by 2: 80 + b = 120 \text{ m}.

Subtract 80 from both sides to find b: b = 120 - 80 = 40 \text{ m}.

The breadth of the rectangular field is 40 m.

Step 3: Calculate the area of the square field.

The formula for the area of a square is \text{side}^2.

Area of square = (60 \text{ m})^2 = 3600 \text{ m}^2.

Step 4: Calculate the area of the rectangular field.

The formula for the area of a rectangle is l \times b.

Area of rectangle = 80 \text{ m} \times 40 \text{ m} = 3200 \text{ m}^2.

Step 5: Compare the areas.

Area of square field = 3600 \text{ m}^2.

Area of rectangular field = 3200 \text{ m}^2.

Since 3600 > 3200, the square field has a larger area.

Answer: The square field has a larger area.

Question 2

Mrs. Kaushik has a square plot of land with a side measuring 25 m. She wants to construct a house in the middle of the plot, leaving a garden area around the house. The house has dimensions of 20 m in length and 15 m in breadth. Calculate the total cost of developing the garden around the house at a rate of ₹ 55 per square meter.
Solution:

This problem involves finding the area of the garden, which is the space between the square plot and the rectangular house, and then calculating the cost based on this area.

Step 1: Calculate the area of the square plot.

The side of the square plot is given as 25 m.

Area of square plot = \text{side}^2 = (25 \text{ m})^2 = 625 \text{ m}^2.

Step 2: Calculate the area of the house.

The house is rectangular with a length of 20 m and a breadth of 15 m.

Area of house = \text{length} \times \text{breadth} = 20 \text{ m} \times 15 \text{ m} = 300 \text{ m}^2.

Step 3: Calculate the area of the garden.

The garden is the area of the plot that is not occupied by the house. Therefore, the area of the garden is the difference between the area of the square plot and the area of the house.

Area of garden = Area of square plot – Area of house

Area of garden = 625 \text{ m}^2 - 300 \text{ m}^2 = 325 \text{ m}^2.

Step 4: Calculate the total cost of developing the garden.

The cost of developing the garden is given as ₹ 55 per square meter.

Total cost = Area of garden × Cost per square meter

Total cost = 325 \text{ m}^2 \times ₹ 55/\text{m}^2 = ₹ 17,875.

Answer: The total cost of developing the garden around the house is ₹ 17,875.

Question 3

The shape of a garden is a rectangle in the middle with semi-circular ends. The length of the rectangular part is 20 m, and the radius of each semi-circular end is 7 m. Find the total area and the perimeter of this garden.
Solution:

This garden is a composite shape made up of a rectangle and two semi-circles. We need to calculate the area and perimeter of this combined shape.

Part 1: Calculate the Area of the Garden

The garden consists of a rectangular section and two semi-circular sections at both ends. The two semi-circular ends together form a full circle.

Step 1.1: Find the area of the rectangular part.

The length of the rectangle is given as 20 m.

The radius of the semi-circular ends is 7 m. This radius is also the breadth of the rectangular part.

Area of rectangle = \text{length} \times \text{breadth} = 20 \text{ m} \times 7 \text{ m} = 140 \text{ m}^2.

Step 1.2: Find the area of the two semi-circular ends (which form a full circle).

The radius of each semi-circle is r = 7 \text{ m}.

Area of a circle = \pi r^2.

Area of the two semi-circles = Area of one full circle = \pi (7 \text{ m})^2 = \frac{22}{7} \times 49 \text{ m}^2 = 22 \times 7 \text{ m}^2 = 154 \text{ m}^2.

Step 1.3: Calculate the total area of the garden.

Total Area = Area of rectangle + Area of two semi-circles

Total Area = 140 \text{ m}^2 + 154 \text{ m}^2 = 294 \text{ m}^2.

Part 2: Calculate the Perimeter of the Garden

The perimeter of the garden consists of the two straight sides of the rectangle and the curved boundaries of the two semi-circles.

Step 2.1: Find the length of the two straight sides of the rectangle.

The length of the rectangle is 20 m. The two straight sides contribute 20 \text{ m} + 20 \text{ m} = 40 \text{ m} to the perimeter.

Step 2.2: Find the perimeter of the two semi-circular ends (which form the circumference of a full circle).

The circumference of a circle is 2\pi r.

Perimeter of the two semi-circles = Circumference of one full circle = 2 \times \frac{22}{7} \times 7 \text{ m} = 2 \times 22 \text{ m} = 44 \text{ m}.

Step 2.3: Calculate the total perimeter of the garden.

Total Perimeter = Length of two straight sides of rectangle + Perimeter of two semi-circles

Total Perimeter = 40 \text{ m} + 44 \text{ m} = 84 \text{ m}.

Answer: The total area of the garden is 294 \text{ m}^2 and the total perimeter is 84 \text{ m}.

Common mistakes

  • Confusing perimeter and area formulas.
  • Errors in calculating the area of composite shapes (e.g., subtraction errors).
  • Incorrectly identifying the dimensions of shapes within composite figures.
  • Calculation mistakes in multiplication or subtraction.

Revision tips

  • Memorize the formulas for the area and perimeter of squares and rectangles.
  • Practice drawing diagrams for composite shapes to visualize the problem.
  • Work through each step of the provided solutions to understand the logic.
  • Attempt to solve the problems independently before checking the solutions.

Practice MCQs

Q1. A square field has a side of 60 m. What is its area?

Q2. If a rectangular field has a length of 80 m and a breadth of 40 m, what is its perimeter?

Q3. Mrs. Kaushik's house has dimensions 20 m by 15 m. What is the area of the house?

Q4. A garden is developed around a house within a square plot. The area of the garden is the difference between the plot's area and the house's area. This is an example of finding the area of a:

Frequently asked questions

What is the main topic of CBSE Class 8 Maths Chapter 11?

Chapter 11 of CBSE Class 8 Maths focuses on Mensuration, which involves calculating the area and perimeter of various geometric shapes like squares, rectangles, and composite figures.

How do these NCERT Solutions help students?

These solutions provide clear, step-by-step explanations for each problem in Chapter 11, helping students understand the concepts of area and perimeter and how to apply them to solve exercises and real-world problems.

What kind of problems are covered in Chapter 11?

The chapter covers problems related to finding the area and perimeter of squares and rectangles, calculating the area of composite shapes (like a garden around a house), and determining costs based on area.

Are the formulas for area and perimeter included in the solutions?

Yes, the solutions demonstrate the application of standard formulas for the area and perimeter of squares and rectangles, and how to derive or use them for composite shapes.

Important links

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