What is Mensuration?
Mensuration is the branch of mathematics that deals with measuring various geometric figures and shapes. It entails calculating the areas, volumes, and other properties of shapes. Geometry covers various Mensuration formulas that are useful for the exam point. To excel in this topic, you must be familiar with the formulas and concepts used to solve the questions.
What are the 2D and 3D shapes?
2D Shapes- A two-dimensional shape in geometry is a flat plane figure or a shape with only two dimensions, namely length and width. Two-dimensional or 2-D shapes have no thickness and can only be measured on two sides. Only the area and perimeter of two-dimensional shapes can be computed.
3D Shapes- A three-dimensional shape is one with three dimensions: length, width, and thickness. Volume, Curved Surface Area, and Total Surface Area of 3D shapes are calculated.
Mensuration in Mathematics – Important Terminologies
|
Terms |
Abbreviation |
Unit |
Definition |
|
Area |
A |
m2 or cm2 |
The surface covered by the closed shape is known as Area. |
|
Perimeter |
P |
m or cm |
The measure of the continuous lines along the boundary of a given figure is known as the Perimeter. |
|
Volume |
V |
m3 or cm3 |
The capacity of a 3D shaped space is known as its Volume. |
|
Curved Surface Area |
CSA |
m2 or cm2 |
If there is a curved surface, then the total area of that shape is known as its Curved Surface Area. |
|
Lateral Surface Area |
LSA |
m2 or cm2 |
The total area of all the lateral surfaces a figure surrounds is known as the Lateral Surface Area. |
|
Total Surface Area |
TSA |
m2 or cm2 |
If there are several surfaces, then the sum of all the areas of all these surfaces is known as the Total Surface Area. |
|
Square Unit |
- |
m2 or cm2 |
The area covered by a square of side one unit is known as a square unit. |
|
Cube Unit |
- |
m3 or cm3 |
The volume occupied by a cube having a side of one unit is known as a cube unit. |
Mensuration Formula
1. Area of Square Formula
|
A = a2 |
Where,
A = Area
a = Side of the Square
2. Perimeter of Square Formula
|
P = 4a |
Where,
P = Perimeter of Square
a = Side of the Square
3. Perimeter of Rectangle Formula
|
P = 2 * (L + B) |
Where,
P = Perimeter of Rectangle
L = Length of Rectangle
B = Breadth of Rectangle
4. Area of Rectangle Formula
|
A = L * B |
Where,
A = Area of Rectangle
L = Length of Rectangle
B = Breadth of Rectangle
5. Surface Area of a Cube
|
S = 6 * A2 |
Where,
S = Surface Area of Cube
A = Length of Side of Cube
6. Surface Area of Cuboid
|
S = 2 * (LB + BH + LH) |
Where,
S = Surface Area of Cuboid
L = Length of Cuboid
B = Breadth of Cuboid
H = Height of Cuboid
7. Surface Area of a Cylinder
|
S = 2 * π * R * (R + H) |
Where,
S = Surface Area of Cylinder
R = Radius of Circular Base
H = Height of Cylinder
8. Surface Area of a Sphere
|
S = 4 * π * R2 |
Where,
S = Surface Area of Cylinder
R = Radius of Circular Base
H = Height of Cylinder
9. Surface Area of a Right Circular Cone
|
S = π * R * (L + R) |
Where,
S = Surface Area of Cone
R = Radius of Circular Base
L = Slant Height of Cone
10. Volume of Cube
|
V = a3 |
Where,
V = Volume of Cube
a = Side of Cube
11. Volume of Cuboid
|
V = L * B * H |
Where,
V = Volume of Cuboid
L = Length of Cuboid
B = Breadth of Cuboid
H = Height of Cuboid
12. Volume of Cylinder
|
V = π * R2 * H |
Where,
V = Volume of Cylinder
R = Radius of Circular Base
H = Height of Cylinder
13. Volume of a Right Circular Cone
|
V = 1/3 * π * R2 * H |
Where,
V = Volume of Cone
R = Radius of Circular Base
H = Height of Cone
14. Volume of a Sphere
|
V = 4/3 * π * R3 |
Where,
V = Volume of Sphere
R = Radius of Sphere
15. Volume of a Right Circular Cone
|
V = 1/3 * π * R2 * H |
Where,
V = Volume of Cone
R = Radius of Circular Base
H = Height of Cone
Example 1. Find out the height of a cylinder with a circular base of radius 70 cm and volume 154000 cubic cm.
Solution: A given here,
R = 70 cm
V= 154000 cubic cm
Since formula is,
V = π * R2 * H
i.e. H =V * π * R²
=15400015400
= 10 cm
Therefore, height of the cylinder will be 10 cm.
Example 2. The radius of a cylinder is 10 cm, and the height is 4 cm. The number of centimetres that may be added either to the radius or to the height to get the same increase in the volume of the cylinder is:
Solution: Let ‘a' cm is added in radius and height
π (10+a)² 4 = π (10)² (4 +a)
(10+a)² 4 = 10² (4 +a)
⇒ a = 5 cm
