Class 8

Chapter-11 (MENSURATION)

Mensuration
It deals with the measurement of area, perimeter and volume
of the plane and solid figures.

Area
• The surface covered by the border line of the figure is the
area of the plain shape.
• Unit of the area is square if the length unit.

Perimeter
• The perimeter is the length of the boundary of the plane
shape.
• The unit of the perimeter is same as the length unit.
Eg.

The green part is the area of the square and

the distance all the way around the outside
is the perimeter.
Area and Perimeter of Some 2D Shapes

Shape Image Area Perimeter

Square (Side)2 4 × Side

Rectangle Length × Breadth 2(Length +

Breadth)
Triangle (1/2)×Base × Height a+b+c
(where, a,
b and c
are the
three
sides of
the
triangle)
Parallelogram Base × Height 2(sum of
adjacent
sides)
Circle πr2 2πr
Where, r =
radius of
the circle
Q. A square and a rectangular field with measurements as given
in the figure have the same perimeter.

Which field has a larger area?

Answer

Given: The side of a square = 60 m and the length of rectangular

field = 80 m
According to question,
Perimeter of rectangular file = Perimeter of square field

⇒ 2(l+b) = 4 × Side

⇒ 2(80 + b) = 4 × 60

⇒ (80 + b) = 240/2

⇒ (80 + b) = 120
⇒ b = 120 - 80

⇒ b = 40 m
Hence, the breadth of the rectangular field is 40 m.
Now, Area of Square field= (Side)2
= (60)2 sq.m = 3600 sq.m
Area of Rectangular field = (length × breadth)
= 80 × 40 sq. m = 3200 sq. m
Hence, area of square field is larger.
Q. Mrs. Kaushik has a square plot with the measurement as
shown in the figure. She wants to construct a house in the
middle of the plot. A garden is developed around the house.
Find the total cost of developing a garden around the house at
the rate of Rs. 55 per m2.
Answer

Side of a square plot = 25 m

∴ Area of square plot = (Side)2 = (25)2 = 625 m2

Length and Breadth of the house is 20 m and 15 m respectively

∴ Area of the house = (length x breadth )

= 20 × 15 = 300 m2
Area of garden = Area of square plot – Area of house
= (625 – 300) = 325 m2

∵ Cost of developing the garden around the house is Rs.55

∴ Total Cost of developing the garden of area 325 sq. m =

Rs.(55 × 325)
= Rs.17,875

Q. A flooring tile has the shape of a parallelogram whose base is

24 cm and the corresponding height is 10 cm. How many such
tiles are required to cover a floor of area 1080 m2 ? [If required
you can split the tiles in whatever way you want to fill up the
corners]
Answer

Base of flooring tile = 24 cm

⇒ 0.24 m
height of a flooring tile = 10 cm
⇒0.10 m [1cm = 1/100 m]
Now, Area of flooring tile= Base × Altitude
= 0.24 × 0.10 sq. m
= 0.024 m2

∴ Number of tiles required to cover the floor = Area of floor/area

of one tile
= 1080/0.024
= 45000 tiles
Hence 45000 tiles are required to cover the floor.

Q. An ant is moving around a few food pieces of different shapes

scattered on the floor. For which food-piece would the ant have
to take a longer round? Remember, circumference of a circle can
be obtained by using the expression c = 2πr, where r is the radius
of the circle.
(a)

(b)

(c)

Answer

(a) Radius = Diameter/2 = 2.8/2

= 1.4 cm
Circumference of semi circle = πr
= 22/7 × 1.4 = 4.4 cm
Total distance covered by the ant= (Circumference of semi circle +
Diameter)
=( 4.4 + 2.8 )cm
= 7.2 cm

(b) Diameter of semi circle = 2.8 cm

Radius = Diameter/2 = 2.8/2 = 1.4cm

Circumference of semi circle = πr
= 22/7 × 1.4 = 4.4 cm
Total distance covered by the ant= (1.5 + 2.8 + 1.5 + 4.4) 10.2
cm

(c) Diameter of semi circle = 2.8 cm

Radius = Diameter/2 = 2.8/2
= 1.4 cm
Circumference of semi circle = πr
= 22/7 × 1.4 = 4.4 cm
Total distance covered by the ant= (2 + 2 + 4.4) = 8.4 cm
Hence for figure (b) food piece, the ant would take a longer
round.
Area of Trapezium
• A trapezium is a quadrilateral whose two sides are
parallel.
• Note: if its non-parallel sides are equal then it is said to be
an isosceles trapezium.

Area of Trapezium formula

Where , a and b = parallel sides

h= perpendicular distance between two parallel sides.

Example
Find the area of the trapezium whose parallel sides are 6 cm
and 16 cm, with a height of 5 cm. Calculate the area.
Solution:

Alternative method: Splitting the trapezium we get –

Area of the trapezium = Area of rectangle + Area of a triangle

= (6 x 5) + (1/2) x 5 x 10
= 30 + 25
= 55 cm 2
Remark: We should use the formula most of the time if
possible as it is the quick and easy method.
Area of a General Quadrilateral

To find the area of any quadrilateral we can divide it into two

triangles and then the area can be easily calculated by
calculating the area of both the triangles separately.
Area of ABCD = Area of ∆ABC + Area of ∆ACD
= (1/2) × AC × h 1 + (1/2) × AC× h 2

The formula for the Area of a General Quadrilateral

Where
• h1 and h2 are the height of both the triangles and
• d is the length of common diagonal i.e.AC.
Example
Find the area of quadrilateral ABCD.

Solution:
In the quadrilateral ABCD,
BD is the common diagonal so d = 5 cm.
Height of the two triangles are h 1 = 2 cm and h 2 = 1 cm.

Area of Special Quadrilaterals (Rhombus)

A rhombus is a quadrilateral with all the sides are equal
Formula of Area of Rhombus

Area of rhombus is half of the product of its two diagonals.

Q. The shape of the top surface of a table is a trapezium. Find its

area if its parallel sides are 1 m and 1.2 m and perpendicular
distance between them is 0.8 m.
Answer

Parallel side of the trapezium AB =1m , CD = 1.2 m and

height (h) of the trapezium (AM) = 0.8 m

Area of top surface of the table

= 1/2 × (sum of parallel sides) × Height
= 1/2 × (AB + CD) × AM
= 1/2 × ( 1 + 1.2) × 0.8
= 1/2 × 2.2 × 0.8
= 0.88 m2
Thus surface area of the table is 0.88 m2

Q. The floor of a building consists of 3000 tiles which are

rhombus shaped and each of its diagonals are 45 cm and 30 cm
in length. Find the total cost of polishing the floor, if the cost
per m2 is ₹4.

Answer

Here, d1 = 45 cm and d2 = 30 cm

∵ Area of one tile = ( ½) x (d1 x d2 )

= ½ x (45 x 30 )
= ½ (1350)
= 675 cm2
So, the area of one tile is 675 cm2
Area of 3000 tiles = 675 × 3000 cm2
= 2025000 cm2

= >2025000/10000× 1m2 [ since 1 cm2 = 1/10000 m2 ]

= 202.50 m2

∵ Cost of polishing the floor per sq. meter = Rs. 4

∴ Cost of polishing the floor per 202.50 sq. meter =Rs. 4 × 202.50
= Rs. 810
Hence the total cost of polishing the floor is Rs. 810.

Area of a Polygon
There is no particular formula for the area of the polygon so
we need to divide it in a possible number of figures like a
triangle, rectangle, trapezium and so on. By adding the area of
all the split figures we will get the area of the required
polygon.
Example
Find the area of the given octagon.

Solution:
We can divide the given octagon into three parts.

Two trapezium A and B and one rectangle shown by part B.

Two trapezium A and B and one rectangle shown by part B.
Area of A = Area of B = (1/2) × (a + b) × h
= (1/2) x (10 + 3) × 2
= 13 cm 2.
Area of B = Length x Breadth
= 10 x 3
= 30 cm 2.
So, the area of Octagon = 2A + B
= 2 × 13 + 30
= 56 cm 2.

Q. Mohan wants to buy a trapezium shaped field. Its side along

the river is parallel to and twice the side along the road. If the
area of this field is 10500 m2 and the perpendicular distance
between the two parallel sides is 100 m, find the length of the
side along the river.
Solution

Given: Perpendicular distance (h) AM = 100 m

Area of the trapezium shaped field = 10500 m2
Let side along the road AB= x m
side along the river CD = 2x m

∴Area of the trapezium field = 12× (AB + CD) x AM

10500 = 1/2 ( x + 2x) × 100
10500 = 3x × 50
3x = 10500/50
x = 10500/(50 × 3)
x = 70 m
Hence the side along the river = 2m = (2 ×70) = 140 m.

Q. Diagram of the adjacent picture frame has outer dimensions

= 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the
area of each section of theframe, if the width of each section is
same.
Solution

Here two of given figures (I) and (II) are similar in dimensions. And
also figures (III) and (IV) are similar in dimensions.
Area of figure (I) = Area of trapezium
= 1/2 (a+b) × h = 1/2 (28 + 20) × 4
= 1/2 ×48 × 4 = 96 cm2
Also Area of figure (II) = 96 cm2
Now Area of figure (III)
Area of trapezium = 1/2 (a + b) × h
= 1/2 (24 + 16) × 4
= 1/2 × 40 × 4
= 80 cm2
Also Area of figure (IV) = 80 cm2
Solid Shapes
The 3-dimensional shapes which occupy some space are called
solid shapes. Example- Cube, Cylinder, Sphere etc.

Surface Area

Volume
Space occupied by any solid shape is the capacity or volume of
that figure. The unit of volume is a cubic unit.
Surface Area
The area of all the faces of the solid shape is the total surface area
of that figure. The unit of surface area is a square unit.
Lateral or Curved Surface Area
The surface area of the solid shape after leaving the top and
bottom face of the figure is called the lateral surface of the shape.
The unit of lateral surface area is square unit.
Surface Area of Cube, Cuboid and Cylinder

Name Figure Lateral Total Surface Nomenclature

or Area
Curved
Surface
Area
Cube 4l2 6l2 l = Edge of the
cube

Cuboid 2h(l + b) 2(lb + bh + lh) l = Length,

b = Breadth,
h = Height
Cylinder 2πrh 2πr2+ 2πrh r = Radius,
= 2πr(r + h) h = Height

Volume of Cube, Cuboid and Cylinder

Name Volume Nomenclature

Cube l3 l = Edge of the cube
Cuboid lbh l = Length, b = Breadth, h = Height
Cylinder πr2h r = Radius, h = Height

Example 1
There is a shoe box whose length, breadth and height is 9 cm,
3 cm and 4 cm respectively. Find the surface area and volume
of the shoe box.

Solution:
Given,
length = 9 cm
Breadth = 3 cm
Height = 4 cm
Area of cuboid = 2(lb + bh + lh)
= 2(9 × 3 + 3×4 + 9 × 4)
= 2(27 + 12 + 36)
= 2(75)
= 150 cm 2
Volume of cuboid = lbh
=9×3×4
= 108 cm 3

Example 2
If there is a cold drink can whose height is 7 cm and the
radius of its round top is 3 cm then what will be the lateral
surface area and volume of that cylinder? (π = 3.14)

Solution:
Given,
radius = 3 cm
Height = 7 cm
Lateral surface area of cylinder = 2πrh
= 2 × 3.14 × 3 × 7
= 131.88 cm 2
Volume of cylinder = πr 2h
= 3.14 × 3 × 3 × 7
= 197.82 cm 3
Example 3
If there is a box of cube shape with the length of 4 cm then
what will be the capacity of this box. Also, find the surface
area of the box if it is open from the top.

Solution:
Given, side = 4 cm
Capacity or volume of the box = s 3
= 4 3 = 64 cm 3
The total surface area of the box = 6s 2
But, if the box is open from the top then the surface area will
be total surface area minus the area of one face of the cube.
Surface Area = Total Surface Area - Area of one face
= 6s2 – s2
= 5s2 = 5 × 4 2
= 80 cm 2
Volume and Capacity
• Volume and capacity are one and the same thing.
• Volume is the amount of space occupied by a shape.

• Capacity is the quantity that a container can hold.

• Capacity can be measured in form of litres.

We can see the relation between litre and cm 3 as,

➢ 1 L = 1000 mL
➢ 1 mL = 1 cm 3,
➢ 1 L = 1000 cm 3.
➢ Thus, 1 m 3 = 1000000 cm 3 = 1000 L

Q. There are two cuboidal boxes as shown in the adjoining

figure. Which box requires the lesser amount of material to
make?
Answer

(a) Length of cuboidal box (l)= 60 cm

Breadth of cuboidal box (b) = 40 cm
Height of cuboidal box (h) = 50 cm
∴ Total surface area of cuboidal box = 2(lb+bh+hl)

= 2 (60 × 40 + 40 × 50 + 50 × 60) cm2

= 2 (2400 + 2000 + 3000) cm2
= 2 × 7400 cm2
= 14800 cm2

(b) Length of the cube is 50 cm

∴ Total surface area of cuboidal box =6(side)2

=6 ( 50)2 cm2
= 6 (2500) cm2
= 15000 cm2
Thus, the cuboidal box (a) requires the lesser amount of materal.

Q. Rukshar painted the outside of the cabinet of measure 1

m × 2 m × 1.5 m. How much surface area did she cover if she
painted all except the bottom of the cabinet?
Answer

Length of cabinet (l) = 2 m

Breadth of cabinet (b) = 1 m
Height of cabinet (h) = 1.5 m
Surface area of cabinet = (Area of Base of cabinet (Cuboid) +
Area of four walls)
=lb+2(l+b)h
={2 × 1 + 2 (1 + 2) 1.5 } m2
= 2 + 2 (3) 1.5 m2
=2+6 (1.5) m2
= (2 + 9.0) m2
= 11 m2
Hence required surface area of cabinet is 11 m2
Q. A road roller takes 750 complete revolutions to move once
over to level a road. Find the area of the road if the diameter of
a road roller is 84 cm and length 1 m.

Answer

Diameter of road roller = 84 cm

∴ Radius of road roller (r) = d/2 = 84/2
= 42 cm
Length of road roller (h) = 1 m = 100 cm
Curved surface area of road roller = 2πrh
= 2 × 22/7 × 42 × 100
= 26400 cm2
∴ Area covered by road roller in 750 revolutions = 26400 × 750
cm2
= 1,98,00,000 cm2
= 1980 m2 [∵ 1 m2= 10,000 cm2]
Thus, the area of the road is 1980 m2.
Q. A company packages its milk powder in cylindrical container
whose base has a diameter of 14 cm and height 20 cm. Company
places a label around the surface of the container (as shown in
figure). If the label is placed 2 cm from top and bottom, what is
the area of the label?

Answer

Diameter of cylindrical container = 14 cm

∴ Radius of cylindrical container (r) = d/2 = 14/2 = 7 cm

Height of cylindrical container = 20 cm
Height of the label (h) = (20 – 2 – 2)
= 16 cm
Curved surface area of label = 2πrh
= 2×22/7×7×16
= 704 cm2
Hence the area of the label of 704 cm2.
Q. A milk tank is in the form of cylinder whose radius is 1.5 m
and length is 7 m. Find the quantity of milk in liters that can be
stored in the tank.

Answer
Given: Radius of cylindrical tank (r) = 1.5 m
Height of cylindrical tank (h) = 7 m

Volume of cylindrical tank = πr2h

= 22/7 × 1.5 × 1.5 × 7
= 49.5 m3
= 49.5 × 1000 liters [∵ 1 m3 = 1000 liters]
= 49500 liters
Hence required quantity of milk is 49500 liters that can be stored
in the tank.