Class 6 Exam  >  Class 6 Notes  >  Maths Olympiad Class 6  >  Chapter Notes: Mensuration

Mensuration Chapter Notes | Maths Olympiad Class 6 PDF Download

Introduction

Plane Figures: The closed 2-D shapes are referred to as plane figures.
Closed figure: A figure with no open ends is a closed figure.

Mensuration Chapter Notes | Maths Olympiad Class 6

  • The boundary of the figure is denoted by the letter "C".
  • The area enclosed within this boundary constitutes the region of the figure.
  • Point D is situated within the area defined by the given figure.

Perimeter

  • The perimeter of a closed figure is the total distance covered when moving along its boundary.
  • It represents the length of the boundary of the figure.
  • When a figure consists of line segments only, its perimeter is determined by adding the lengths of all the sides of that figure.

Example: Find the Perimeter of the given figure.

Mensuration Chapter Notes | Maths Olympiad Class 6

Sol: Perimeter = Sum of all the sides
= (12 + 3 + 7 + 6 + 10 + 3 + 15 + 12) m
= 68 m

Perimeter of a Rectangle

A rectangle is a closed figure with two pairs of equal opposite sides.

Mensuration Chapter Notes | Maths Olympiad Class 6

Perimeter of a rectangle = Sum of all sides
                                                        = length + breadth + length + breadth

Thus, Perimeter of a rectangle = 2 × (length + breadth)

Example 1: The length and breadth of a rectangular swimming pool are 16 and 12
meters respectively .find the perimeter of the pool.
Sol: Perimeter of a rectangle = 2 × (length + breadth)
Perimeter of the pool = 2 × (16 + 12)

              = 2 × 28

              = 56 meters

Example 2: Find the cost of fencing a rectangular farm of length 24 meters and breadth 18 meters at 8/- per meter.
Sol: Perimeter of a rectangle = 2 × (length + breadth)
Perimeter of the farm = 2 × (24 + 18)

               = 2 × 42

               = 84 meter

Cost of fencing = 84 × 8

            = Rs. 672

Thus, the cost of fencing the farm is Rs. 672/-.

Regular Closed Figure

  • Regular closed figures, or regular polygons, have sides of equal length and angles of equal measure.
  • To find the perimeter of a regular polygon, multiply the number of sides by the length of one side.

Perimeter of Square

Square is a regular polygon with 4 equal sides.

Mensuration Chapter Notes | Maths Olympiad Class 6

Perimeter of square = side + side + side + side

                                              = 4 × length of a side

Example: Find the perimeter of a square having side length 25 cm.
Sol: Perimeter of a square = 4 × length of a side
Perimeter of square = 4 × 25

           = 100 cm

Perimeter of an Equilateral Triangle

An equilateral triangle is a regular polygon with three equal sides and angles.

Mensuration Chapter Notes | Maths Olympiad Class 6

Perimeter of an equilateral triangle = 3 × length of a side

Example: Find the perimeter of a triangle having each side length 13 cm.
Sol: Perimeter of an equilateral triangle = 3 × length of a side
Perimeter of triangle = 3 × 13

                                                = 39 cm

Perimeter of a Regular Pentagon

A regular pentagon is a polygon with 5 equal sides and angles.

Mensuration Chapter Notes | Maths Olympiad Class 6

Perimeter of a regular pentagon = 5 × length of one side

Example: Find the perimeter of a pentagon having side length 9 cm.
Sol: Perimeter of a regular pentagon = 5 × length of one side
Perimeter of a regular pentagon = 5 × 9

                           = 45 cm

Perimeter of a Regular Hexagon

A regular hexagon is a polygon with 6 equal sides and angles.

Mensuration Chapter Notes | Maths Olympiad Class 6

Perimeter of a regular hexagon = 6 × Length of one side

Example: Find the perimeter of a hexagon having side length 15cm.
Sol: Perimeter of a regular hexagon = 6 × Length of one side
Perimeter of a regular hexagon = 6 × 15

                         = 90 cm

Perimeter of a Regular Octagon

A regular octagon is a polygon with 8 equal sides and angles.

Mensuration Chapter Notes | Maths Olympiad Class 6

Perimeter of a regular octagon = 8 × length of one side

Example: Find the perimeter of an octagon having side length 7cm.
Sol: Perimeter of a regular octagon = 8 × length of one side
Perimeter of a regular octagon = 8 × 7

                        = 56 cm

Area

Area refers to the surface enclosed by a closed figure.

Mensuration Chapter Notes | Maths Olympiad Class 6

  • To find the area of an irregular shape, you can place it on a grid paper with 1 cm × 1 cm squares.
  • Then, count how many squares are covered by the shape.
  • Each square is counted as 1 square unit.

Example: Find the area of the given figure. (1 square = 1 m2).Mensuration Chapter Notes | Maths Olympiad Class 6Sol: The given figure is made up of line segments and is covered with some full squares and some half squares.

Full squares in figure = 32

Half squares in figure = 21

Area covered by full squares = 32 × 1 sq. unit = 32 sq. unit.

Area covered by half squares = 21 × (1/2) sq. unit. = 10.5 sq. unit.

Total area covered by figure = 32 + 10.5 = 42.5 sq. unit.

Area of a Rectangle

Mensuration Chapter Notes | Maths Olympiad Class 6

Area of a rectangle = (length × breadth)
Example: Find the area of a rectangle whose length and breadth are 20 cm and 12 cm respectively.
Sol: Length of the rectangle = 20 cm
Breadth of the rectangle = 12 cm
Area of the rectangle = length × breadth

                     = 20 cm × 12 cm

                      = 240 sq cm.

To find the length of a rectangle if breadth and area are given:Mensuration Chapter Notes | Maths Olympiad Class 6

Example: What will be the length of the rectangle if its breadth is 6 m and the area is 48sq.m?
Sol: Mensuration Chapter Notes | Maths Olympiad Class 6

Length = 48/6

                = 8 m

To find the breadth of the rectangle if length and area are given:Mensuration Chapter Notes | Maths Olympiad Class 6

Example: What will be the breadth of the rectangle if its length is 8 m and the area is 81 sq.m?
Sol:Mensuration Chapter Notes | Maths Olympiad Class 6

Breadth = 81/8

        = 9 m

Area of a Square

Area of a square is the region covered by the boundary of a square.

Mensuration Chapter Notes | Maths Olympiad Class 6

Area of a square = side × side.

Example: Calculate the area of a square of side 13 cm.
Sol: Area of a square = side × side
                                               = 13 × 13
                                               = 169 cm2.

Area of a Triangle

Area of triangle = (1/2) × base × height = (1/2) × b × h 

Mensuration Chapter Notes | Maths Olympiad Class 6

Example: Find the area of a triangle with a base of 10 units and height of 4 units.
Sol: The formula for the area of a triangle is:
A = (1/2) × base × height.
A = (1/2) × 10 × 4
A = 20 square units

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