Download the latest CBSE Class 6 Mathematics Mensuration Chapter Notes in PDF format. These Class 6 Mathematics revision notes are carefully designed by expert teachers to align with the 2025-26 syllabus. These notes are great daily learning and last minute exam preparation and they simplify complex topics and highlight important definitions for Class 6 students.
Chapter-wise Revision Notes for Class 6 Mathematics Chapter 10 Mensuration
To secure a higher rank, students should use these Class 6 Mathematics Chapter 10 Mensuration notes for quick learning of important concepts. These exam-oriented summaries focus on difficult topics and high-weightage sections helpful in school tests and final examinations.
Chapter 10 Mensuration Revision Notes for Class 6 Mathematics
CBSE Class 6 Mensuration Chapter Concepts. Learning the important concepts is very important for every student to get better marks in examinations. The concepts should be clear which will help in faster learning. The attached concepts made as per NCERT and CBSE pattern will help the student to understand the chapter and score better marks in the examinations.
Mensuration
1. More Problems Related To Square And Rectangle
Formulae
a) Perimeter of a square = 4 × side
b) Perimeter of a rectangle = 2 × (length + breadth)
c) Area of a square = side × side
d) Area of a rectangle = length × breadth
Problems
Example 1: A door-frame of dimensions 3 m × 2 m is fixed on the wall of dimension 10 m × 10 m. Find the total labour charges for painting the wall if the labour charges for painting 1 m2 of the wall is Rs 2.50.
Solution: Painting of the wall has to be done excluding the area of the door.
Area of the door = l × b = 3 × 2 m2 = 6 m2
Area of wall including door = side × side = 10 m × 10 m = 100 m2
Area of wall excluding door = (100 − 6) m2 = 94 m2
Total labour charges for painting the wall = Rs 2.50 × 94 = Rs 235
Example 2: The area of a rectangular sheet is 500 cm2. If the length of the sheet is 25 cm, what is its width? Also find the perimeter of the rectangular sheet.
Solution: Area of the rectangular sheet = 500 cm2
Length (l) = 25 cm
Area of the rectangle = l × b (where b = width of the sheet)
Therefore, width b =1/A r e a -5 0 0/25 = 40 cm
Perimeter of sheet = 2 × (l + b) = 2 × (25 + 20) cm = 90 cm
So, the width of the rectangular sheet is 20 cm and its perimeter is 90 cm.
Example 3: Anu wants to fence the garden in front of her house (Fig 11.5), on three sides with lengths 20 m, 12 m and 12 m. Find the cost of fencing at the rate of Rs 150 per metre.
Solution: The length of the fence required is the perimeter of the garden (excluding one side) which is equal to 20 m + 12 m + 12 m, i.e., 44 m.
Cost of fencing = Rs 150 × 44 = Rs 6,600.
Example 4: A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 12 cm, find its breadth. Which encloses more area, the square or the rectangle?
Solution: Side of the square = 10 cm
Length of the wire = Perimeter of the square = 4 × side = 4 × 10 cm = 40 cm
Length of the rectangle, l = 12 cm. Let b be the breadth of the rectangle.
Perimeter of rectangle = Length of wire = 40 cm
Perimeter of the rectangle = 2 (l + b)
Thus, 40 = 2 (12 + b)
or
4 0/ 2 = 12 + b
Therefore, b = 20 − 12 = 8 cm
The breadth of the rectangle is 8 cm.
Area of the square = (side)2
= 10 cm × 10 cm = 100 cm2
Area of the rectangle = l × b
= 12 cm × 8 cm = 96 cm2
So, the square encloses more area even though its perimeter is the same as that of
the rectangle.
Example 5: The area of a square and a rectangle are equal. If the side of the square is 40 cm and the breadth of the rectangle is 25 cm, find the length of the rectangle. Also, find the perimeter of the rectangle.
Solution Area of square = (side) 2 = 40 cm × 40 cm = 1600 cm2 It is given that, The area of
the rectangle = The area of the square
Area of the rectangle = 1600 cm2, breadth of the rectangle = 25 cm
Area of the rectangle = l × b
or 1600 = l × 25
or 1 6 0 0/ 2 5 = l or l = 64 cm
So, the length of rectangle is 64 cm.
Perimeter of the rectangle = 2 (l + b) = 2 (64 + 25) cm
= 2 × 89 cm = 178 cm
So, the perimeter of the rectangle is 178 cm even though its area is the same as
that of the square
2. Circle
A circle is a shape with all points the same distance from the center. It is named by the center. The circle to the left is called circle A since the center is at point A. If you measure the distance around a circle and divide it by the distance across the circle through the center, you will always come close to a particular value, depending upon the accuracy of our measurement. This value is approximately 3.14159265358979323846... We use the Greek letter ∏ (pronounced Pi) to represent this value. The number ∏ goes on forever. However, using computers, has been calculated to over 1 trillion digits past the decimal point.
Circumference: The distance around a circle is called the circumference. The distance across a circle through the center is called the diameter. ∏ is the ratio of the circumference of a circle to the diameter. Thus, for any circle, if you divide the circumference by the diameter, you get a value close to ∏ . This relationship is expressed in the following formula:
c/d- ∏
where C is circumference and d is diameter. You can test this formula at home with a round dinner plate. If you measure the circumference and the diameter of the plate and then divide C by d, your quotient should come close to π. Another way to write this formula is: C=π·d where means multiply. This second formula is commonly used in problems where the diameter is given and the circumference is not known
Radius: The radius of a circle is the distance from the center of a circle to any point on the circle. If you place two radii end-to-end in a circle, you would have the same length as one diameter. Thus, the diameter of a circle is twice as long as the radius. This relationship is expressed in the following formula: d = 2·r , where d is the diameter and is the radius.
Note: Circumference, diameter and radii are measured in linear units, such as inches and centimeters. A circle has many different radii and many different diameters, each passing through the center. A real-life example of a radius is the spoke of a bicycle wheel. A 9-inch pizza is an example of a diameter: when one makes the first cut to slice a round pizza pie in half, this cut is the diameter of the pizza. So a 9-inch pizza has a 9-inch diameter. Let's look at some examples of finding the circumference of a circle. In these examples, we will use = 3.14 to simplify our calculations.
Example 1: The radius of a circle is 2 inches. What is the diameter?
Solution: d = 2 × 4
d = 2 × (2 in)
d = 4 in
Example 2: The diameter of a circle is 3 centimeters. What is the circumference?
Solution: C = π × d
C = 3.14 × (3 cm) = 9.42 cm
Example 3: The radius of a circle is 2 inches. What is the circumference?
Solution: d = 2 × r
d = 2 × (2 in)
d = 4 in
C = π × d
C = 3.14 × (4 in)
C = 12.56 in
Example 4: The circumference of a circle is 15.7 centimeters. What is the diameter?
Solution: C = π × d
15.7 cm = 3.14 × d
15.7 cm ÷ 3.14 = d
d = 15.7 cm ÷ 3.14
d = 5 cm
Area of a Circle: The area of a circle is the number of square units inside that circle. If each square in the circle to the left has an area of 1 cm2, you could count the total number of squares to get the area of this circle. Thus, if there were a total of 28.26 squares, the area of this circle would be 28.26 cm2 However, it is easier to use one of the following formulas: A = π× r2 or A = π × r × r where A is the area, and r is the radius. Let's look at some examples involving the area of a circle. In each of the three examples below, we will use π = 3.14 in our calculations
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Important Practice Resources for Class 6 Mathematics
CBSE Class 6 Mathematics Chapter 10 Mensuration Notes
Students can use these Revision Notes for Chapter 10 Mensuration to quickly understand all the main concepts. This study material has been prepared as per the latest CBSE syllabus for Class 6. Our teachers always suggest that Class 6 students read these notes regularly as they are focused on the most important topics that usually appear in school tests and final exams.
NCERT Based Chapter 10 Mensuration Summary
Our expert team has used the official NCERT book for Class 6 Mathematics to design these notes. These are the notes that definitely you for your current academic year. After reading the chapter summary, you should also refer to our NCERT solutions for Class 6. Always compare your understanding with our teacher prepared answers as they will help you build a very strong base in Mathematics.
Chapter 10 Mensuration Complete Revision and Practice
To prepare very well for y our exams, students should also solve the MCQ questions and practice worksheets provided on this page. These extra solved questions will help you to check if you have understood all the concepts of Chapter 10 Mensuration. All study material on studiestoday.com is free and updated according to the latest Mathematics exam patterns. Using these revision notes daily will help you feel more confident and get better marks in your exams.
You can download the teacher prepared revision notes for CBSE Class 6 Mathematics Mensuration Chapter Notes from StudiesToday.com. These notes are designed as per 2025-26 academic session to help Class 6 students get the best study material for Mathematics.
Yes, our CBSE Class 6 Mathematics Mensuration Chapter Notes include 50% competency-based questions with focus on core logic, keyword definitions, and the practical application of Mathematics principles which is important for getting more marks in 2026 CBSE exams.
Yes, our CBSE Class 6 Mathematics Mensuration Chapter Notes provide a detailed, topic wise breakdown of the chapter. Fundamental definitions, complex numerical formulas and all topics of CBSE syllabus in Class 6 is covered.
These notes for Mathematics are organized into bullet points and easy-to-read charts. By using CBSE Class 6 Mathematics Mensuration Chapter Notes, Class 6 students fast revise formulas, key definitions before the exams.
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