Mensuration

Mensuration​ is the branch of mathematics which studies the measurement

of the geometric figures and their parameters like length, volume, shape,
surface area, lateral surface area, etc. Here, the concepts of mensuration
are explained and all the important mensuration formulas and properties of
different geometric shapes and figures are covered.

Mensuration Maths- Definition

A branch of mathematics which talks about the length, volume or area of

different geometric shapes is called ​Mensuration​. These shapes exist in 2
dimension or 3 dimensions. Let’s learn the difference between the two.
Difference Between 2D and 3D shapes

2D Shape 3D Shape

If a shape is surrounded by If a shape is surrounded by a

three or more straight lines in a no. of surfaces or planes then
plane, then it is a 2D shape. it is a 3D shape.

These shapes have no depth or These are also called as solid

height. shapes and unlike 2D they
have both height or depth.
 These shapes have only 2-D These are called Three
length and breadth. dimensional as they have
depth, breadth and length.

We can measure their area and We can measure their volume,

Perimeter. CSA, LSA or TSA.

Mensuration in Maths- Important Terminologies

Let’s learn a few more definitions related to this topic.

Terms Abbreviati Unit Definition

on

Area A M2 or The area is the surface which is

Cm2 covered by the closed shape.

Perimet P Cm The measure of the continuous

er or m line along the boundary of the
given figure is called a Perimeter.

Volume V Cm3 In a 3D shape, the space included

or m3 is called a Volume.
Curved CSA M2 or If there’s a curved surface, then
Surface cm2 the total area is called a Curved
Area Surface area. Example: Sphere or
Cylinder.

Lateral LSA M2 or The total area of all the lateral

Surface cm2 surfaces that surrounds the figure
area is called the Lateral Surface area.

Total TSA M2 or If there are many surfaces like in

Surface Cm2 3D figures, then the sum of the
Area area of all these surfaces in a
closed shape is called Total
Surface area.

Square – M2 or The area covered by a square of

Unit cm2 side one unit is called a Square
unit.

Cube – M3 or The volume occupied by a cube of

Unit cm3 one side one unit

Mensuration Formulas
Now let’s learn all the important mensuration formulas involving 2D and 3D
shapes. Using this mensuration formula list, it will be easy to solve the
mensuration problems. Students can also download the mensuration
formulas list PDF from the link given above. In general, the most common
formulas in mensuration involve surface area and volumes of 2D and 3D
figures.

Mensuration Formulas For 2D Shapes

Area
(Squar Perimete
Shape e units) r (units) Figure

Square a2 4a
Rectangle l×b 2 ( l + b)

Circle πr2 2πr

Scalene √[s(s−a a+b+c

Triangle )(s−b)(s
−c)],

Where,
s=
 (a+b+c)
/2

Isosceles ½×b× 2a + b
Triangle h

Equilateral (√3/4) × 3a
Triangle a2

Right ½×b× b+
Angle h hypotenu
Triangle se + h
Rhombus ½ × d1 4 × side
× d2

Parallelogr b×h 2(l+b)

am

Trapezium ½ a+b+c+d
h(a+b)

Mensuration Formulas for 3D Shapes

Total
Curved Surface
Volum Surface Area Area
e (CSA) or (TSA)
(Cubic Lateral (Square
Shape units) Surface Area units) Figure
(LSA)
 (Square
units)

Cube a3 – 6 a2

Cuboid l×w× – 2 (lb +bh

h +hl)

Sphere (4/3) π 4 π r2 4 π r2
r3
Hemisphe (⅔) π 2πr2 3πr2
re r3

Cylinder πr2h 2π r h 2πrh +

2πr2
 Cone (⅓) π πrl πr (r + l)
r2 h

Mensuration Problems
Question:​ ​Find the area and perimeter of a square whose side is 5
cm?
Solution:
Given:
Side = 5 cm
Area of a square = a2 square units
Substitute the value of “a” in the formula, we get
Area of a square = 52
A = 5 x 5 = 25
Therefore, the area of a square = 25 cm2
The perimeter of a square = 4a units
P = 4 x 5 =20
Therefore, the perimeter of a square = 20 cm.
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Frequently Asked Questions

What is mensuration in Maths?

In maths, mensuration is defined as the study of the measurement of

various 2D and 3D geometric shapes involving their surface areas,
volumes, etc.

What is the difference between mensuration and geometry?

Mensuration refers to the calculation of various parameters of shapes like

the perimeter, area, volume, etc. whereas, geometry deals with the study of
properties and relations of points and lines of various shapes.

What are 2D and 3D Mensuration?

2D mensuration deals with the calculation of various parameters like area

and perimeter of 2-dimensional shapes like square, rectangle, circle,
triangles, etc.

3D mensuration is concerned with the study and calculation of surface

area, lateral surface area, and volume of 3-dimensional figures like cube,
sphere, cuboid, cone, cylinder, etc.
Formulas for Mensuration

In this article, we have discussed formulas of important 2D figures which

are frequently asked. We have compiled important information which are
mostly asked in SSC Exams at one place.
Rectangle

Let d​1​ and d​2​ are diagonals of the given rectangle ABCD.
then, both diagonals are equal but not perpendicular to each other.

Area of rectangle = length x breadth and perimeter =

2(length+breadth)

Path outside the rectangle

Suppose there is a park having length​ l​ and breadth ​b.​ There is a road of
width ​x​ outside of it.

Then, Area of path = 2x (​l​ + b + 2x)

Path inside the rectangle
Suppose there is a park having length​ l​ and breadth ​b.​ There is a road of
width ​x​ inside of it.

Then, Area of path = 2x (l + b – 2x)

When there is a road along both the length and breadth of the park.

Then, Remaining area of Rectangle (shaded region) = (l–x) (b-x)

Area of the path = lx + bx – x​2

Circle: Given a circle of radius ‘r’

We recommend you learn this table as it will save your time in calculating
these all.

If radius is ‘r’, then perimeter = 2πr and Area = πr​2

 Radius Perimeter (2πr) Area (πr​2​)

7 44 154

14 88 616

21 132 1386

28 176 2464

35 220 3850

42 264 5544

Length of Rope

Let ‘d’ is the diameter of pulley and ‘r’ is the radius, then d = 2r. All pulleys
are similar.

Length of rope = 2d + 2pr

Length of rope = 3d + 2pr

Length of rope = 4d + 2pr
Note: Trick to remember these formulas: number of pulleys x diameter +
Perimeter of one pulley

Sector

In this circle, ‘​r’​ is the radius, θ is the angle made by the arc of length ‘​l​’

Length of arc

Area of sector

Area of sector when ‘l’ is given

Segment

Area of minor segment

Area of major segment

Area of shaded portion

Inradius and Circumradius of Square:

There is a square of side ‘a’; ‘r’ is the inradius and ‘R’ is the circumradius.
Triangle:

Let ABC is a triangle and M​1​, M​2​ and M​3 are

​ medians of the given triangle.

Then,

Inradius of triangle:

Given, ABC is a triangle and a, b and c are the sides of given triangle. Let
‘r’ is the inradius of triangle.
Circumradius of triangle:

Given, ABC is a triangle and a, b and c are the sides of given triangle. Let
‘R’ is the circumradius of triangle.

Right angle triangle

Given ‘a’ is the base, ‘b’ is the perpendicular and ‘c’ is the hypotenuse of
triangle ABC.

Equilateral triangle:
Where, h is the height of triangle,
Hence, we can say that height of equilateral triangle is equal to the sum of
side perpendicular of the triangle.

Isosceles triangle
Regular Polygon

Let, n = no. of sides of regular polygon and a = length of side of regular

polygon

Internal angle of regular polygon =

Sum of internal angle of regular polygon

Angle made by centre =

Area of Regular polygon

or

External angle of regular polygon

sum of all external angle = 360º
For Regular Hexagon

Circumradius R = a

Inradius
 Cyclic Quadrilateral

Parallelogram

Let a and b are the sides, h is the height and d​1​ and d​2​ are the diagonals of
parallelogram

then,
Area of parallelogram = (i) Base × height
(ii)

(iii)
Imp. Relation
Imp. Note: In rectangle, parallelogram, square and Rhombus diagonals
bisect other.
Rhombus

In Rhombus, diagonals are not equal to each other but they bisect each
other at 90 degree.
Area = Base × height = a x h

Or Area

Trapezium

Case 1: If AD = BC, then DM = CN

Quadrilateral

Mensuration Questions and Answers:

Section A

Q 1: Find the volume and surface area of a cuboid 16m long, 14 m

broad and 7 m high.

A) 868 cm B) 920 cm C) 868 cm2 D) 920 cm2

Q 2: Find the length of the longest pole that can be placed in a room
12 m long, 8m broad and 9 m high.

A) 16 m B) 17 m C) 18 m D) 19 m
Q 3: The volume of a wall, 5 times as high as it is broad and 8 times as
long as it is high, is 12.8 cu. meters. Find the breadth of the wall.

A) 0.04m B) 4m C) 400 cm D) 40 cm

Q4: Find the number of bricks, each measuring 24 cm×12 cm × 8 cm,

required to construct a wall 24 m long, 8m high and 60 cm thick if 10%
of the wall is filled with mortar?

A) 450 B) 4500 C) 45000 D) 450000

Q5: The area of the base of a rectangular tank is 6500 cm2 and the
volume of ​water​ contained in it is 2.6 cubic meters. The depth of water
in the tank is:

A) 3.5 m B) 4 m C) 5 m D) 6 m

Q6: Given that one cubic cm of marble weighs 25 gms, the weight of a
marble block 28 cm in width and 5 cm thick is 112 kg. The length of
the block is:

A) 26.5 cm B) 32 cm C) 36 cm D) 37.5 cm
Q7: Half cubic meter of the gold sheet is extended by hammering so
as to cover an area of one hectare. The thickness of the sheet is:

A) 0.0005 cm B) 0.005 cm C) 0.05 cm D) 0.5 cm

Q8: In a shower, 5 cm of rain falls. The volume of water that falls on

1.5 hectares of the ground is:

A) 75 cu m B) 750 cu. m C) 7500 cu. m D)

75000 cu. m

Find Your Answers

Q1: C), Q2: B), Q3: D), Q4: C), Q5: B), Q6: B), Q7: B), Q8: B)
Section B

Q1: Water flows into a tank 200 m × 150 m through a rectangular pipe
1.5 m × 1.25 m @ 20 kmph. In what time (in minutes) will the water rise
by 2 meters?

A) 234 minutes B) 1.2 hours C) 9 hours D) 96 minutes

Q2: The dimensions of an open box are 50 cm, 40 cm, and 23 cm. Its
thickness is 3 cm. If 1 cubic cm of metal used in the box weighs 0.5
gms, find the weight of the box.

A) 6. 08 kg B) 8.04 kg C) 8.06 kg D) 6.04 kg

Q3: A cube of edge 15 cm is immersed completely in a rectangular

vessel containing water. If the dimensions of the base of the vessel
are 20 cm × 15 cm, find the rise in water level. [RRB 2003]

A) 11 cm B) 11.11 cm C) 11. 22 cm D) 11.25 cm

Q4: A conical vessel, whose internal radius is 12 cm and height 50
cm, is full of some liquid. The contents of this vessel are emptied into
a cylindrical vessel with an internal radius of 10 cm. Find the height to
which the liquid rises in the cylindrical vessel.

A) 22 cm B) 23 cm C) 24 cm D) 25 cm

Q5: How many spherical bullets can be made out of a lead cylinder 28
cm high and with base radius 6 cm, each bullet being 1.5 cm in
diameter? [RRB 2003]

A) 1600 B) 1793 C) 1601 D) 1792

Q6: If the radius of a sphere is increased by 50%, find the increase

percent in volume and the increase percent in the surface area.

A) 95% B) 100 % C) 115% D) 125%

Q7: Two metallic right circular cones having their heights 4.1 cm and
4.3 cm and the radii of their bases 2.1 cm each, have been melted
together and recast into a sphere. Find the diameter of the sphere.

A) 4 cm B) 4.1 cm C) 4.2 cm D) 4.3 cm

Q8: A hemispherical bowl of internal radius 9 cm contains a liquid.
This liquid is to be filled into cylindrical shaped small bottles of
diameter 3 cm and height 4 cm. How many bottles will be needed to
empty the bowl?

A) 23 B) 34 C) 54 D) 46

Find Your Answers Here

Q1: D), Q2: B), Q3: D), Q4: C), Q5: D), Q6: D), Q7: C), Q8: C)
Section C

Directions:​ Each of the questions given below consists of a statement

and/or a question and two statements numbered I and II given below it.
You have to decide whether the data provided in the statements are
sufficient to answer the given question. Read both the statements and give
an answer:

(a) when the data in Statement I alone are sufficient to answer the
question, while the data in Statement II alone are not sufficient to answer
the question.

(b) if the data in Statement II alone are sufficient to answer the question,
while the data in Statement I alone are not sufficient to answer the
question.

(c) the data either in Statement I or in Statement II alone are sufficient to

answer the question.

(d) if the data even in both Statements I and II together are not efficient to
answer the question.

(e) the data in both Statements I and II together are necessary to give the
answer.
Q1: What is the weight of the iron beam?

I. The beam is 9 m long, 40 cm wide and 20 cm high.

II. Iron weighs 50 kg per cubic meter.

Q2: What is the volume of 32 meter high cylindrical tank? [Bank

PO 2003]

I. The area of its base is 154 sq. m.

II. The diameter of the base is 14 m.

Q3: What is the volume of a cube?​ ​ [Bank PO 2003]

I. The area of each face of the cube is 64 sq. m.

II. Length of one side of the cube is 8 metres.

Find Your Answers Here

Q1: (e), Q2: (c), Q3: (c)

Volumes And Areas
1. Square

Area = S²

Perimeter = 4s

s = length of the sides, d = length of diagonal.

2. Rectangle

Area = base x height = b x h

Perimeter = 2 (b + h)

3. Triangle

Area = ½ x base x height

Perimeter = x + y + z ( summation of three sides of a triangle)

4. Rhombus

Area = ½ x product of the diagonals between the sides * sine of the angle
between the sides

Perimeter = 4 x side (any side)

Diagonal = 2 x area / diagonal

5. Parallelogram

Area = product of any two sides x sine of the included angle

Perimeter = 2 (a + b) (a and b are the two adjacent sides)

6. Trapezium

Area = ½ × sum of the parallel sides × height.

Volume

The magnitude or capacity of a solid space like a cube, cylinder, etc is

known as the volume of that solid.

Different solids and their volumes:

1. Cube

In a cube, length = breadth = height = s

The surface area of a cube = 6s²

The volume of a cube = S³

Diagonal of a cube = √3 s

2. Cuboid

Total surface area of a cuboid: 2 (lb + bh + lh)

The volume of a cuboid: lbh

3. Cylinder

Curved surface area of a cylinder: 2 πrh( r = radius of the base, h = height)

Total surface area of a right circular cylinder = 2 πrh + 2 πr²

The volume of the right circular cylinder: πr²h

4. Cone

Curved surface area of a cone: pirl (l is the slant height)

Total surface area of a cone: πrl + πr²

The volume of a cone: 1/3 πr²h

5. Sphere

The surface area of a sphere: 4 πr²

The volume of a sphere: 4/3 πr³

A half sphere is known as a hemisphere.

6. Hemisphere

Curved surface area of a hemisphere: 3 πr2

Total surface area of a hemisphere: 3 πr2

The volume of a hemisphere: 2/3 πr3

Examples Based on Area

Q. What must be the side of an equilateral triangle so that so that its

area may be equal to the area of an isosceles triangle with the base
and equal sides as 12 m and 10 m respectively?

A. 9m B. 10m C.10.5m D.11.5m

Ans: ​This question can be easily solved if you know the formula for the
area of an isosceles triangle.

Area of an isosceles triangle = b/4(4a² – b²)^½

= 12/4(4 x 10^2 – 12^2)^1/2

= 3 x 16 = 48m^2.

Now as per the given problem area of equilateral triangle = 48m^2

Therefore √3/4(a2) = 48 => a = 10.5m.

Hence the required side of the equilateral triangle = 10.5m. Thus C is the
correct answer.
Q. The length of the floor if a rectangle hall is 10 m more than its
breadth. If 34 carpets of size 6 × 4m are required to cover the floor of
the hall, then find the length and breadth of the hall.

A. 24, 24m B. 24, 34m C. 22, 32m D. 34, 34m

Ans: Let breadth = b then length = b + 10m.

Floor area of the rectangle hall = A = length × breadth = b × (b + 10)

Also area of each carpet = 6 × 4m and 34 pieces are required to cover the
floor

Therefore area of hall = 34 × 6 × 4

b × (b + 10) = 34 × 24

Therefore b = 24 m and length = b + 10 = 34m. Thus B is the correct

answer.
Examples Based on Volume

Q. Find the volume and the surface area of the sphere of radius
6.3cm.

Volume of the sphere = 4/3πr³ = 4/3 x 22/7 x 6.3 x 6.3 x 6.3 = 1047.82 cm²

Total surface area of the sphere = 4 πr³ = 4 x 22/7 x 6.3 x 6.3 = 498.96 cm²

Practice Questions

Q. The cost of carpeting a room 15m long with a carpet 75cm wide at
30paise per meter is Rs. 36. Find the width of the room.

A.6m B. 8m C. 1om D. 9m

The correct answer is A.

Q. The area of the right-angled triangle is 30 cm² and the length of the
hypotenuse is 13 cm. Find the length of the shorter leg.

A. 3cm B. 4cm C. 5cm D. 7 cm

The correct answer is C.

Q. The volume of the sphere is 38808 cm³, find the curved surface
area of the sphere.

A. 5202 cm³ B. 5544 cm³ C. 5562 cm³ D. 5542 cm³

The correct answer is B.

Q. The radii of two cylinders are in the ratio of 2:3 and their heights
are in the ratio 5:3. Find the ratio of their curved surface area.

A. 10:9 B. 10:11 C. 9:10 D. 9:11

The correct answer is B.