# CBSE Class 6 Mathematics Mensuration Chapter Notes

Download CBSE Class 6 Mathematics Mensuration Chapter Notes in PDF format. All Revision notes for Class 6 Mathematics have been designed as per the latest syllabus and updated chapters given in your textbook for Mathematics in Class 6. Our teachers have designed these concept notes for the benefit of Class 6 students. You should use these chapter wise notes for revision on daily basis. These study notes can also be used for learning each chapter and its important and difficult topics or revision just before your exams to help you get better scores in upcoming examinations, You can also use Printable notes for Class 6 Mathematics for faster revision of difficult topics and get higher rank. After reading these notes also refer to MCQ questions for Class 6 Mathematics given on studiestoday

## Revision Notes for Class 6 Mathematics Chapter 10 Mensuration

Class 6 Mathematics students should refer to the following concepts and notes for Chapter 10 Mensuration in Class 6. These exam notes for Class 6 Mathematics will be very useful for upcoming class tests and examinations and help you to score good marks

### Chapter 10 Mensuration Notes 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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### CBSE Class 6 Mathematics Chapter 10 Mensuration Notes

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### Notes for Mathematics CBSE Class 6 Chapter 10 Mensuration

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#### Chapter 10 Mensuration Notes for Mathematics CBSE Class 6

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#### Chapter 10 Mensuration CBSE Class 6 Mathematics Notes

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#### Notes for CBSE Mathematics Class 6 Chapter 10 Mensuration

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