Class Notes

Mensuration

Grade 08
 Topics to be Covered

1. Mensuration of
2D Shapes
• 1.1. Perimeter and Area
of Basic Shapes
• 1.2. Area of a Trapezium
• 1.3. Area of a General
Quadrilateral
• 1.4. Area of a Rhombus
• 1.5. Area of a Polygon

2. Mensuration of
3D Shapes
• 2.1. Surface Area
• 2.2. Volume
 Mind Map

Area of a general
quadrilateral

Area of a
Area of a rhombus
trapezium

Perimeter and area

Area of a polygon
of basic shapes

Mensuration of
2D shapes

Mensuration

Mensuration of
3D shapes

Surface Area Volume

Total surface Area

Lateral surface Area

 1. Mensuration of 2D Shapes

1.1. Perimeter and Area of Basic Shapes

Triangle

• Perimeter: Sum of all

ℎ the three sides
1
• Area: 2 𝑏ℎ

Parallelogram

• Perimeter: Twice the sum

ℎ of adjacent sides
• Area: 𝑏ℎ
𝑏

Rectangle

• Perimeter: 2(𝑎 + 𝑏)
𝑎 • Area: 𝑎𝑏

𝑏
 1. Mensuration of 2D Shapes

1.1. Perimeter and Area of Basic Shapes

Square

• Perimeter: 4𝑎
𝑎 • Area: 𝑎 2

Circle

𝑟 • Perimeter: 2𝜋𝑟
• Area: 𝜋𝑟 2
 1. Mensuration of 2D Shapes

1.2. Area of a Trapezium

𝑎
D C

A E F B
𝑐 𝑑
𝑏
Consider a trapezium ABCD

Area of trapezium ABCD

= Area of △AED + Area of DEFC + Area of △CFB

1 1
= 2 𝑐ℎ + 𝑎ℎ + 2
𝑑ℎ

1
= 2 ℎ 𝑐 + 2𝑎 + 𝑑

1
= 2ℎ 𝑐 + 𝑎 + 𝑑 + 𝑎

1
= 2ℎ 𝑎 + 𝑏

Area of a trapezium
1
= 2 × height × sum of two parallel sides
 1. Mensuration of 2D Shapes

1.3. Area of a General Quadrilateral

• A general quadrilateral can be split into two

triangles by drawing one of its diagonals.
• This triangulation helps us to find the formula for
area any general quadrilateral.

B
𝑑
ℎ1
ℎ2

A
D

Consider a quadrilateral ABCD.

Here, 𝑑 denotes the length of the diagonal AC. ℎ1
and ℎ2 are the length of the perpendiculars from
point B and point D on AC, respectively.

Area of quadrilateral ABCD

= Area of △ABC + Area of △ADC

1 1
= 2 ℎ1 𝑑 + 2 ℎ2 𝑑

1
= 2 𝑑 ℎ1 + ℎ2
 1. Mensuration of 2D Shapes

1.4. Area of a Rhombus

D C

A B

Consider a rhombus ABCD.

Here, 𝑑1 denotes the length of the longer diagonal
AC and 𝑑2 is the length of the shorter diagonal BD.

Area of rhombus ABCD

= Area of △ABC + Area of △ADC

1 𝑑2 1 𝑑2
= 2 𝑑1 + 2 𝑑1
2 2

1 𝑑2
= 2 × 2 𝑑1 2

1
= 2 𝑑1 𝑑2

1
Area of a rhombus = × product of its diagonals
2
 1. Mensuration of 2D Shapes

1.5. Area of a Polygon

• While finding the area of a polygon, the polygon

needs to be divided into triangles and
quadrilaterals and their individual areas should be
added.

E A

D F B
G

Consider a polygon ABCDE.

Here, DB is a diagonal. Line segments EF and AG
are the perpendiculars from points E and A on DB,
respectively.

Area of polygon ABCDE

= Area of △DFE + Area of △AGB + Area of △DCB
+ Area of EFGA
 2. Mensuration of 3D Shapes

2.1. Surface Area

• The total surface area of a 3D shape is the sum of

the area of all the faces.

• The lateral surface area of a 3D shape is the sum

of the area of all the faces excluding the top and
the bottom face.

Surface area of a Cuboid

𝑏

𝑙
𝑙
ℎ
ℎ ℎ
𝑙 𝑏 𝑏 𝑙
𝑙

𝑏
• Total surface area = 2(𝑙𝑏 + 𝑏ℎ + 𝑙ℎ)

𝑙 𝑙
ℎ ℎ ℎ
𝑙 𝑏 𝑏 𝑏

• Lateral surface area = 2(𝑙ℎ + 𝑏ℎ)

 2. Mensuration of 3D Shapes

2.1. Surface Area

Surface area of a Cube

𝑎2

𝑎 𝑎2 𝑎2 𝑎2 𝑎2

𝑎
𝑎 𝑎2

• Total surface area = 6𝑎2

𝑎 𝑎2 𝑎2 𝑎2 𝑎2

𝑎
𝑎
• Lateral surface area = 4𝑎2

Surface area of a Cylinder

𝑟
𝑟 2𝜋𝑟

ℎ ℎ
𝑟

• Total surface area = 𝜋𝑟 2 + 2𝜋𝑟ℎ + 𝜋𝑟 2 = 2𝜋𝑟(𝑟 + ℎ)

• Lateral (Curved) surface area = 2𝜋𝑟ℎ
 2. Mensuration of 3D Shapes

2.2. Volume

The amount of space occupied by a 3D object is

called its volume. It is calculated as the area of the
base of a 3D object multiplied by its height.

There is not much difference between volume and

capacity.
• Volume refers to the amount of space occupied by
an object.
• Capacity refers to the quantity that a container
holds. It is also measured in litres.
• 1 litre = 1000 cm3 and 1 m3 = 1000 litres

Volume of a Cuboid

𝑏
𝑙

Volume = Area of the base × Height

= 𝑙 × 𝑏 ×ℎ
=𝑙 × 𝑏 × ℎ
 2. Mensuration of 3D Shapes

2.2. Volume

Volume of a Cube

𝑙
𝑙
Volume = Area of the base × Height
= 𝑙 ×𝑙 ×𝑙
= 𝑙3

Volume of a Cylinder

Volume = Area of the base × Height

= 𝜋𝑟 2 × ℎ
= 𝜋𝑟 2 ℎ
 Important Questions

Question 1 2 marks

If the area of a parallelogram is 60 cm2 and

one of its altitude is 5 cm, what is the length of
its corresponding side?

Solution

Let the corresponding side be 𝑎 cm.

Let the altitude be ℎ cm. 0.5 mark

We know that,
Area of parallelogram = Side × Altitude
0.5 mark
⇒ 60 = 𝑎 × ℎ
⇒ 𝑎 × ℎ = 60
⇒ 𝑎 × 5 = 60
60
⇒ 𝑎= = 12
5
Hence, 𝑎 = 12 cm
1 mark
 Important Questions

Question 2 2 marks

If the area of a quadrilateral ABCD is 20 cm2 and

perpendiculars on BD from opposite vertices are
1 cm and 1.5 cm, find the length of BD.

Solution

Area of the given quadrilateral

1
= 2 × sum of the altitudes × corresponding diagonal
0.5 mark
1
⇒ 20 = 2 × (1 + 1.5) × BD

1
⇒ × 2.5 × BD = 20
2

⇒ 2.5 × BD = 40

⇒ BD = 16 cm
1.5 marks
 Important Questions

Question 3 2 marks

Find the area of a rhombus whose side is 6 cm

and whose altitude is 4 cm. If one of the diagonals
is 8 cm long, find the length of the other diagonal.

Solution

Area of the rhombus whose one side and altitude is

given = side × length of the altitude
= 6 cm × 4 cm
= 24 cm2
1 mark

Now, let the length of the other diagonal be 𝑥.

It is known that area of the rhombus is half of the
product of the diagonals.

1
⇒ × 8 × 𝑥 = 24
2

⇒ 4𝑥 = 24

⇒𝑥=6

Hence, the length of the other diagonal is 6 cm.

1 mark
 Important Questions

Question 4 5 marks

The length of a fence of a trapezium-shaped field

ABCD is 120 m. If BC = 48 m, CD = 17 m, and AD
= 40 m, find the area of this field. The side AB is
perpendicular to the parallel sides AD and BC.

Solution

Perimeter of the trapezium-shaped field

= AB + BC + CD + DA 0.5 mark

Now according to the problem:

⇒ 120 = AB + 48 + 17 + 40
⇒ 120 = AB + 105
⇒ AB = 15 cm 1.5 marks
1
Area of the trapezium = 2 × ℎ × (𝑎 + 𝑏)
Here, ℎ = 15 cm.
Let 𝑎 be 40 cm and 𝑏 be 48 cm.
1 mark
1
Area of the trapezium = × 15 × (40 + 48)
2
1
= × 15 × 88
2
= 15 × 44
= 660 cm2
2 marks
 Important Questions

Question 5 3 marks

A cuboid is of dimensions 60 cm × 54 cm × 30 cm.

How many small cubes with side 6 cm can be
placed in the given cuboid?

Solution

Given:
Length of the cuboid (𝑙) = 60 cm
Breadth of the cuboid (𝑏) = 54 cm
Height of the cuboid (ℎ) = 30 cm

Volume of the cuboid = 𝑙 × 𝑏 × ℎ

= (60 × 54 × 30) cm3
1 mark

Volume of the cube = 𝑙 3

= (6 × 6 × 6) cm3
1 mark

Number of small cubes that can be placed

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑢𝑏𝑜𝑖𝑑 60 × 54 × 30
= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑢𝑏𝑒 = = 450
6×6×6
1 mark
 Important Questions

Question 6 2 marks

A closed cylindrical tank of radius 7 m and height

3 m is made from a sheet of metal. How much
22
sheet of metal is required? Take 𝜋 = 7 .

Solution

Given:
Radius of the cylindrical tank (𝑟) = 7 m
Height of the cylindrical tank (ℎ) = 3 m
Total surface area of the cylindrical tank
= 2𝜋𝑟(𝑟 + ℎ) 1 mark

22
= 2 × 7 × 7(7+3)
= 44 × 10
= 440 m2
1 mark
 Important Questions

Question 7 2 marks

Water is pouring into a cuboidal reservoir at the

rate of 60 litres per minute. If the volume of
reservoir is 108 m3, find the number of hours it
will take to fill the reservoir.

Solution

Given:
Volume of cuboidal reservoir = 108 m3
Rate of pouring water into cuboidal reservoir
= 60 𝑙/min

60 1
Now, 60 𝑙/min = m3/min [ 1𝑙= m3]
∴
1000 1000

60 × 60 3
= m /hour
1000
1 mark

Time taken to fill 1 m3 of water

1000
= hours
60 × 60

Time taken to fill 108 m3 of water

108 × 1000
= hours
60 × 60
1 mark
= 30 hours
 Important Questions

Question 8 3 marks

Find the height of the cylinder whose volume is

1.54 m3 and diameter of the base is 140 cm.
22
Take 𝜋 = 7 .

Solution

Given:
Volume of cylinder = 1.54 m3
Diameter of cylinder = 140 cm
𝑑 140
Radius (𝑟) = = = 70 cm = 0.7 m
2 2
0.5 mark
Volume of cylinder = 𝜋𝑟 2 ℎ
0.5 mark
22
⇒ 1.54 = × (0.7)2 × ℎ
7

1.54 × 7
⇒ℎ=
22 × 0.7 × 0.7

⇒ℎ=1m
2 marks