Mensuration

Mensuration is the branch of mathematics that deals with the measurement of length, area or volume of various geometric shapes.

Perimeter

If we go around the figure along its boundary to form a closed figure then the distance covered is the perimeter of that figure. Hence the Perimeter refers to the length of the boundary of a closed figure.

If a figure is made up of line segments only then we can find its perimeter by adding the length of all the sides of the given figure.

Example

Find the Perimeter of the given figure.

Solution

Perimeter = Sum of all the sides

= (12 + 3 + 7 + 6 + 10 + 3 + 15 + 12) m

= 68 m

• Perimeter of a circle is also called as the circumference of the circle.

Perimeter of a Triangle

 Perimeter of triangle = Sum of lengths of all sides = a + b + c.

• If the given triangle is equilateral that is if all the sides are equal (a = b = c),
  then its perimeter is equal to 3 × length of one side of the triangle.

Perimeter of a Rectangle

 Perimeter of the rectangle = length ( l ) + length ( l ) + width (w) + width (w)

= 2 × [length ( l ) + width (w)]

Example: 1

The length and breadth of a rectangular swimming pool are 16 and 12 meters respectively .find the perimeter of the pool.

Solution: 

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.

Solution:

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

Figures with equal length of sides and an equal measure of angles are known as Regular Closed Figures or Regular Polygon.

Perimeter of Regular Polygon = Number of sides × Length of one side

Perimeter of a Square

 Perimeter of square = 4 × length of a side = 4a

Example

Find the perimeter of a square having side length 25 cm.

Solution

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.

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

Example

Find the perimeter of a triangle having each side length 13 cm.

Solution

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.

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

Example

Find the perimeter of a pentagon having side length 9 cm.

Solution

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.

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

Example

Find the perimeter of a hexagon having side length 15cm.

Solution

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.

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

Example

Find the perimeter of an octagon having side length 7cm.

Solution

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

Perimeter of a regular octagon = 8 × 7

= 56 cm

Perimeter of a ‘n’ sided polygon

• A polygon is a closed shape made up of line segments.
• Perimeter of n sided polygon = n × length of one side.
• Example: Length of each side of a decagon is a cm, then:
  Perimeter of the decagon = 10a cm

Perimeter of irregular shapes

• Irregular shapes are the shapes which do not have all sides and angles equal.
• The perimeter of irregular shapes is equal to total length covered by the shape.
  In the figure given below, perimeter is the sum of all sides.

Area

• Area is the total amount of surface enclosed by a closed figure.
• To find the area of any irregular closed figure, we can put them on a graph paper with the square of 1 cm × 1 cm .then estimate the area of that figure by counting the area of the squares covered by the figure.
• Here one square is taken as 1 sq.unit.

Area of Square

• Area of a square = Side × Side = Side² = a², where a is the length of each side.

Example

Calculate the area of a square of side 13 cm.

Solution

Area of a square = side × side

= 13 × 13

= 169 cm².

Area of Rectangle

•  Area = length ( l ) × breadth ( b)

Example

Find the area of a rectangle whose length and breadth are 20 cm and 12 cm respectively.

Solution

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.

Length of a rectangle if breadth and area are given:

Example

What will be the length of the rectangle if its breadth is 6 m and the area is 48sq.m?

Solution

Length = 48/6

 = 8 m

Breadth of the rectangle if length and area are given:

Example

What will be the breadth of the rectangle if its length is 8 m and the area is 81 sq.m?

Solution

Breadth = 81/8

= 9 m

Area of a triangle

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

Areas of different types of triangles

• Consider an acute and an obtuse triangle.
  Area of each triangle = (1/2) × base × height = (1/2) × b × h

Visualisation of Area

• In the given graph, if the area of each small square is 1 cm², then
  Area of rectangle = l × b = 5 × 2 = 10 cm²
  Area of square = a × a = 2 × 2 = 4 cm²

Area of irregular shapes

• Area of an irregular figure can be calculated :
  Step 1: Divide the irregular shape into regular shapes that you can recognize (eg. triangles, rectangles, circles and squares)
  Step 2: Find the area of these individual shapes and add them. Sum will be the area of the irregular figure.

Example: Area of the given figure

         = Area of MNCB + Area of AMGH + Area of EFND

= [ 5 × 9 + 4 × 2 + 3 × 3 ] cm²
= [45 + 8 + 9 ] cm²
= 62 cm²

Example

Find the area of the given figure. (1 square = 1 m²)

Solution

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.