Introduction to Mensuration | Formulas for 2D and 3D Shapes, Examples

Last Updated : 23 Jul, 2025

Mensuration is the branch of mathematics that deals with the measurement of various geometric figures and shapes. This includes calculating areas, volumes, and perimeters of two-dimensional shapes like squares, rectangles, circles, and triangles, as well as three-dimensional figures like cubes, cylinders, spheres, and cones.

These shapes can exist in 2 ways:

  • Two-Dimensional Shapes - circle, triangle, square, etc.
  • Three-Dimensional Shapes - cube, cuboid, cone, etc.

Difference Between 2D and 3D Shapes

2-Dimensional vs 3-Dimensional Shapes

2D Shape3D Shape
Any shape is 2D if it is bound by three or more straight lines in a plane.A shape is a three-dimensional shape if there are several surfaces or planes around it.
There is no height or depth in these shapes.In contrast to 2D forms, these are sometimes known as solid shapes and have height or depth.
These shapes just have length and width as their dimensions.Since they have depth (or height), breadth, and length, they are referred to as three-dimensional objects.
We can calculate their perimeter and area.Their volume, curved surface area, lateral surface area, or total surface area can all be calculated.

Mensuration Terminologies

Here is the list of terms you will come across in mensuration class. We have provided the term, it's abbreviation, unit and definition for easy understanding.

TermsAbbreviationUnitDefinition
AreaAm2 or cm2The surface that the closed form covers is known as the area.
PerimeterPcm or mA perimeter is the length of the continuous line that encircles the specified figure.
VolumeVcm3 or m3A 3D shape's space is referred to as its volume.
Curved Surface AreaCSAm2 or cm2The overall area is known as a Curved surface area if there is a curved surface. Example: Sphere.
Lateral Surface areaLSAm2 or cm2The term "Lateral Surface area" refers to the combined area of all lateral surfaces that encircle the provided figure.
Total Surface AreaTSAm2 or cm2The total surface area is the total of all the curved and lateral surface areas.
Square Unit-m2 or cm2A square unit is the area that a square of side one unit covers.
Cube Unitm3 or cm3The space taken up by a cube with a single side.

Mensuration Formula For 2D Shapes

The following table provides a list of all mensuration formulas for 2D shapes:

ShapeArea (Square units)Perimeter (units)Figure
Squarea24a
Shapes-1
Square dimensions
Rectanglel × b2(l + b)
Shapes-2
Rectangle dimensions
Circleπr22πr
Shapes-4
Circle with radius
Scalene Triangle√[s(s-a)(s-b)(s-c)],
Where,
s = (a+b+c)/2
a + b + c
Shapes-6
Scalene triangle dimensions
Isosceles Triangle½ × b × h2a + b
Shapes-7
Equilateral triangle dimensions
Equilateral Triangle(√3/4) × a23a
Shapes-7
Equilateral triangle dimensions
Right Angle Triangle½ × b × hb + hypotenuse + h
Shapes-5
Right Angle Triangle dimensions
Rhombus½ × d1 × d24 × side
Shapes-10
Rhombus Diagonals
Parallelogramsb × h2(l + b)
Shapes-9
Parallelogram dimensions
Trapezium½ h(a + c)a + b + c + d
Shapes-8
Trapezium dimensions

Learn More:

Mensuration Formula for 3D Shapes

The following table provides a list of all mensuration formulas for 3D shapes:

ShapeVolume Curved Surface Area or Lateral Surface AreaTotal Surface AreaFigure
Cubea3LSA = 4 a26a2
Shapes-11
cube dimensions
Cuboidsl × b × hLSA = 2h(l + b)2(lb +bh +hl)
Shapes-12
cuboid dimensions
Sphere(4/3)πr34πr24πr2
Shapes-13
sphere dimensions
Hemisphere(⅔)πr32πr23πr2
Shapes-14
hemisphere dimensions
Cylinderπr2h2πrh2πrh + 2πr2
Shapes-15
cylinder dimensions
Cone(⅓)πr2hπrlπr(r + l)
Shapes-16
cone dimensions

Learn More :

Solved Problems on Mensuration

Let's solve some example problems on mensuration.

Problem 1: Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.

Solution:

Radius of the cone, r = 1.5 cm
Height of the cone, h = 5 cm

∴ Volume of the cone, V = 13πr2h=13×227×(1.5)2×5= 11.79 cm3

Thus, the volume of the cone is 11.79 cm3

Problem 2: The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.

Solution:

The dimensions of the cuboid are 44 cm, 21 cm and 12 cm.

Let the radius of the cone be r cm.

Height of the cone, h = 24 cm

It is given that cuboid is melted to form a cone.

∴ Volume of metal in cone = Volume of metal in cuboid
⇒(1/3)πr2h = 44 × 21 × 12             

 (Volume of cuboid=Length×Breadth×Height)
⇒(1/3) × (22/7) × r2 × 24 = 44 × 21 × 12
⇒r = √(44 × 21 × 12 × 21) / (22 × 24)
=21 cm

Thus, the radius of the base of cone is 21 cm. 

Problem 3: The radii of two circular ends of frustum shape bucket are 14 cm and 7 cm. The height of the bucket is 30 cm. How many liters of water can it hold? (1 litre = 1000 cm3).

Solution:

Radius of one circular end, r1 = 14 cm
Radius of other circular end, r2 = 7 cm
Height of the bucket, h = 30 cm

∴ Volume of water in the bucket = Volume of frustum of cone

=(1/3)πh(r12 + r1r2 + r22)
=13×22/7 × 30 × (142 + 14 × 7 + 72)
=13×22/7 × 30 × 343 = 10780 cm3
=107801000 = 10.780 L

Thus, the bucket can hold 10.780 litres of water.

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