Physics Wallah

CBSE Class 9 Maths Notes Chapter 8 Quadrilaterals PDF Download

Here, we have provided CBSE Class 9 Maths Notes Chapter 8 Quadrilaterals. Students can view these CBSE Class 9 Maths Notes Chapter 8 before exams for better understanding of the chapter.
authorImageAnanya Gupta22 May, 2024
Share

Share

CBSE Class 9 Maths Notes Chapter 8

CBSE Class 9 Maths Notes Chapter 8: In Class 9 Math, Chapter 8 talks about Quadrilaterals, which are shapes with four straight sides. This chapter helps you understand different kinds of quadrilaterals, like parallelograms, rectangles, squares, rhombuses, and trapeziums.

You'll learn about their special features and how to recognize them. We'll also explore things like opposite sides and angles, diagonals, and special properties like symmetry and congruence. By understanding this chapter well, you'll be better prepared for more advanced math topics later on.

CBSE Class 9 Maths Notes Chapter 8 Quadrilaterals Overview

These notes on Chapter 8, Quadrilaterals, are written by subject experts of Physics Wallah in simple language to help students understand the concepts easily. In this chapter, you will learn about different types of quadrilaterals, such as parallelograms, rectangles, squares, rhombuses, and trapeziums. The notes explain their properties, like the relationships between sides and angles, and how to identify each type of quadrilateral. By studying these notes, students can grasp the fundamentals of quadrilaterals, which will be useful for more advanced math topics in the future.

CBSE Class 9 Maths Notes Chapter 8 Quadrilaterals PDF

You can access the CBSE Class 9 Maths Notes for Chapter 8 Quadrilaterals in PDF format using the provided link. These notes provide comprehensive explanations and examples to help you grasp the concepts of triangles effectively.

CBSE Class 9 Maths Notes Chapter 8 Quadrilaterals PDF

CBSE Class 9 Maths Notes Chapter 8 Quadrilaterals

Quadrilaterals

Quadrilaterals are a specific type of polygon characterized by having exactly four sides. These geometric shapes are formed by the union of four line segments. Common examples of quadrilaterals include squares, rectangles, parallelograms, and trapezoids. Each quadrilateral has four vertices (corners) and four angles. The sum of the interior angles of a quadrilateral is always 360 degrees. Understanding the properties and types of quadrilaterals is fundamental in geometry, as they form the basis for more complex shapes and are widely used in various mathematical applications.

Examples of Quadrilaterals

Quadrilaterals

Parallelograms

A parallelogram is a specific type of quadrilateral defined by its unique properties. One defining characteristic of a parallelogram is that its opposite sides are both parallel and equal in length. This means that if you extend the opposite sides indefinitely, they will never intersect. Parallelograms encompass various other quadrilaterals, including rectangles, rhombuses, and squares, each possessing additional properties beyond those of a standard parallelogram. In contrast, a trapezium is another type of quadrilateral, but it differs from a parallelogram in that only one pair of its opposite sides are parallel. Therefore, it does not qualify as a parallelogram. Despite this distinction, trapeziums share similarities with parallelograms, such as having opposite sides of equal length. Understanding the properties and distinctions of parallelograms and trapeziums is crucial in geometry, as they form the basis for many geometric concepts and mathematical calculations.
Parallelogram In the diagram,

Opposite Sides

A B βˆ₯ D C 𝐴𝐡βˆ₯𝐷𝐢 and A D βˆ₯ B C 𝐴𝐷βˆ₯𝐡𝐢

A B = D C 𝐴𝐡=𝐷𝐢 and A D = B C 𝐴𝐷=𝐡𝐢

  • Opposite Angles are Equal.

From figure,

A Λ† = C Λ† 𝐴^=𝐢^ and B Λ† = D Λ† 𝐡^=𝐷^

Diagonals of a Parallelogram Bisect Each Other

Parallelogram Diagonals
In the diagram,

O D = O B 𝑂𝐷=𝑂𝐡 and O A = O C 𝑂𝐴=𝑂𝐢

Each diagonal divides the parallelogram into two congruent triangles

Congruent Triagles of Parallelogram
In the diagram,

β–³ A B C β‰… β–³ C D A △𝐴𝐡𝐢≅△𝐢𝐷𝐴

β–³ A B D β‰… β–³ C D B △𝐴𝐡𝐷≅△𝐢𝐷𝐡

Opposite Sides of a Quadrilateral

Opposite Sides of Quadrilateral

Two sides of a quadrilateral, which have no common point, are called opposite sides.

  • In the diagram, A B 𝐴𝐡 and D C 𝐷𝐢 is one pair of opposite sides.

  • D C 𝐷𝐢 and B C 𝐡𝐢 is the other pair of opposite side

Understanding Quadrilateral Properties

Quadrilaterals are four-sided polygons with distinct properties that define their angles and sides. One fundamental aspect of quadrilaterals is the concept of consecutive sides, opposite angles, and consecutive angles. Consecutive sides are two sides of a quadrilateral that share a common endpoint. For example, in a quadrilateral ABCD, AB and BC form one pair of consecutive sides, while BC and CD, CD and DA, and DA and AB represent the other three pairs of consecutive sides. Opposite angles in a quadrilateral are a pair of angles that do not share a side in their intersection. In the quadrilateral ABCD, angles A and C form one pair of opposite angles, while angles B and D constitute another pair of opposite angles. Consecutive angles, on the other hand, are two angles of a quadrilateral that include a side in their intersection. For instance, angles A and B represent one pair of consecutive angles, while angles B and C, C and D, and D and A form the other three pairs of consecutive angles. Understanding these properties helps in identifying and analyzing quadrilaterals, facilitating geometric calculations and problem-solving in various mathematical contexts.

Theorem 1 Statement

  • The diagonals of a parallelogram bisect each other.

  • If two sides of a triangle are unequal, the longer side has the greater angle opposite to it.

  • A B C D 𝐴𝐡𝐢𝐷 is a parallelogram in which diagonals A C 𝐴𝐢 and

    B D 𝐡𝐷

    intersect each other at O 𝑂 .

Consecutive angles of Parallelogram

To prove:

The diagonals A C 𝐴𝐢 and

B D 𝐡𝐷

bisect eac B Λ† 𝐡^ and D Λ† 𝐷^ areh other that is,

A O = O C 𝐴𝑂=𝑂𝐢 and

B O = D O 𝐡𝑂=𝐷𝑂 .

Proof:

A B βˆ₯ C D 𝐴𝐡βˆ₯𝐢𝐷 (By definition of parallelogram)

A C 𝐴𝐢 is a transversal.

∴ O A Λ† B = O C Λ† D . . . . . ( i ) βˆ΄π‘‚π΄^𝐡=𝑂𝐢^𝐷 .....Β (𝑖) (Alternate angles are equal in a parallelogram)

Also,

A B = D C 𝐴𝐡=𝐷𝐢 (Opposite sides are equal in a parallelogram)

Now in Ξ” A O B Δ𝐴𝑂𝐡 and $\Delta COD

A B = D C 𝐴𝐡=𝐷𝐢 (Opposite sides of parallelogram are equal)

O A Λ† B = O C Λ† D 𝑂𝐴^𝐡=𝑂𝐢^𝐷 (Proved by ( i ) (𝑖) )

A O Λ† B = C O Λ† D 𝐴𝑂^𝐡=𝐢𝑂^𝐷 (Vertically opposite angles are equal)

Therefore,

β–³ A O B β‰… β–³ C O D △𝐴𝑂𝐡 ≅ △𝐢𝑂𝐷

( A A S 𝐴𝐴𝑆 Congruency condition)

Therefore,

A O = O C 𝐴𝑂=𝑂𝐢 and B O = O D 𝐡𝑂=𝑂𝐷 (corresponding parts of congruent triangles are congruent) that is the diagonals of a parallelogram bisect each other.

Sufficient Conditions for a Quadrilateral to be a Parallelogram

Identifying a parallelogram involves understanding its defining properties. These properties provide sufficient conditions for determining whether a quadrilateral is a parallelogram. One key property is that if a quadrilateral is a parallelogram, then its opposite sides are equal. Conversely, if both pairs of opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram. Another condition is that if the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram. This means that the point where the diagonals intersect divides each diagonal into two equal parts. Additionally, if either pair of opposite sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram. This condition emphasizes the significance of parallelism in identifying parallelograms. Understanding these conditions enables us to efficiently recognize and classify parallelograms, aiding in geometric analysis and problem-solving.

Theorem 2 2

Statement:

If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Consecutive angles of Parallelogram

Given:

A B C D 𝐴𝐡𝐢𝐷 is a quadrilateral in which diagonals A C 𝐴𝐢 and B D 𝐡𝐷 intersect at O 𝑂 such that A O = O C 𝐴𝑂=𝑂𝐢 and B O = O D 𝐡𝑂=𝑂𝐷 .

To prove:

A B C D 𝐴𝐡𝐢𝐷 is a parallelogram.

Proof:

In triangles A O B 𝐴𝑂𝐡 and C O D 𝐢𝑂𝐷 ,

A O = C O 𝐴𝑂=𝐢𝑂 (Given)

B O = O D 𝐡𝑂=𝑂𝐷 (Given)

A O Λ† B = C O Λ† D 𝐴𝑂^𝐡=𝐢𝑂^𝐷 (Vertically opposite angles are equal)

Therefore,

Ξ” A O B β‰… Ξ” C O D Δ𝐴𝑂𝐡 ≅ Δ𝐢𝑂𝐷

( S A S 𝑆𝐴𝑆 Congruency condition)

Therefore,

O A Λ† B = O C Λ† D 𝑂𝐴^𝐡=𝑂𝐢^𝐷 ( c p c t 𝑐𝑝𝑐𝑑 )

Since these are alternate angles made by the transversal A C 𝐴𝐢 intersecting A B 𝐴𝐡 and C D 𝐢𝐷

Therefore,

A B βˆ₯ C D 𝐴𝐡βˆ₯𝐢𝐷

Similarly,

A D βˆ₯ B C 𝐴𝐷βˆ₯𝐡𝐢

Hence, A B C D 𝐴𝐡𝐢𝐷 is a parallelogram.

Theorem 3 3 :

Statement:

A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel.

Diagonal of a Parallelogram

Given:

A B C D 𝐴𝐡𝐢𝐷 is a quadrilateral in which A B βˆ₯ C D 𝐴𝐡βˆ₯𝐢𝐷 and A B = C D 𝐴𝐡=𝐢𝐷 .

To prove:

A B C D 𝐴𝐡𝐢𝐷 is a parallelogram.

Construction:

Join A C 𝐴𝐢 .

Proof:

In triangles A B C 𝐴𝐡𝐢 and A D C 𝐴𝐷𝐢 ,

A B = C D 𝐴𝐡=𝐢𝐷 (Given)

B A Λ† C = A C Λ† D 𝐡𝐴^𝐢=𝐴𝐢^𝐷 (Alternate angles are equal)

A C = A C 𝐴𝐢=𝐴𝐢 (Common side)

Therefore,

Ξ” A B C β‰… Ξ” C D A Δ𝐴𝐡𝐢 ≅ Δ𝐢𝐷𝐴

( S A S 𝑆𝐴𝑆 Congruency condition)

B C Λ† A = D A Λ† C 𝐡𝐢^𝐴=𝐷𝐴^𝐢 (Corresponding parts of corresponding triangles)

Since these are alternate angles,

A B βˆ₯ C D 𝐴𝐡βˆ₯𝐢𝐷

Thus, in the quadrilateral A B C D 𝐴𝐡𝐢𝐷 , A B βˆ₯ C D 𝐴𝐡βˆ₯𝐢𝐷 and A D βˆ₯ B C 𝐴𝐷βˆ₯𝐡𝐢

Therefore, A B C D 𝐴𝐡𝐢𝐷 is a parallelogram.

Special Parallelograms

Parallelograms encompass a diverse set of quadrilaterals, including rectangles, rhombuses, and squares. Each of these special parallelograms has distinct properties and characteristics.

Rectangle:

A rectangle is a parallelogram with all interior angles measuring 90 degrees, making it a right angle. Consequently, opposite sides of a rectangle are equal in length.

Rhombus:

A rhombus is a parallelogram with all sides of equal length. This means that opposite sides are equal and parallel. However, the angles of a rhombus are not necessarily 90 degrees, except in the case of a square.

Square:

A square is a special case of both a rectangle and a rhombus. It possesses all the properties of a rectangle, including right angles, and all the sides are equal in length like a rhombus.

Relationships Between Special Parallelograms:

In terms of relationships, every rectangle and rhombus is inherently a parallelogram. Therefore, they are depicted as subsets of a parallelogram. Furthermore, because a square possesses the characteristics of both a rectangle and a rhombus, it is represented by the overlapping shaded region in the diagram. Understanding the distinctions and relationships between these special parallelograms is crucial for geometry and problem-solving applications.

Rectangle

A rectangle is a parallelogram with one of its angles as a right angle.

Rectangle

In the above figure,

Let, A Λ† = 90 ∘ 𝐴^=90∘

Since,

A D βˆ₯ B C 𝐴𝐷βˆ₯𝐡𝐢 ,

A Λ† + B Λ† = 180 ∘ 𝐴^+𝐡^=180∘

(Sum of interior angles on the same side of transversal A B 𝐴𝐡 )

Therefore,

B Λ† = 90 ∘ 𝐡^=90∘

Here,

A B βˆ₯ C D 𝐴𝐡βˆ₯𝐢𝐷 and A Λ† = 90 ∘ 𝐴^=90∘ (Given)

Therefore,

A Λ† + D Λ† = 180 ∘ 𝐴^+𝐷^=180∘

∴ D Λ† = 90 ∘ ∴𝐷^=90∘

∴ C Λ† = 90 ∘ ∴𝐢^=90∘

Corollary: Each of the four angles of a rectangle is a right angle.

Rhombus

A rhombus is a parallelogram with a pair of its consecutive sides equal.

Rhombus

A B C D 𝐴𝐡𝐢𝐷 is a rhombus in which A B = B C 𝐴𝐡=𝐡𝐢 .

Since a rhombus is a parallelogram,

A B = D C 𝐴𝐡=𝐷𝐢 and B C = A D 𝐡𝐢=𝐴𝐷

Thus, A B = B C = C D = A D 𝐴𝐡=𝐡𝐢=𝐢𝐷=𝐴𝐷

Corollary: All the four sides of a rhombus are equal (congruent).

Square

A square is a rectangle with a pair of its consecutive sides equal.

Square

Since square is a rectangle, each angle of a rectangle is a right angle and A B = D C 𝐴𝐡=𝐷𝐢 , B C = C D 𝐡𝐢=𝐢𝐷 .

Thus,

A B = B C = C D = A D 𝐴𝐡=𝐡𝐢=𝐢𝐷=𝐴𝐷

Each of the four angles of a square is a right angle and each of the four sides is of the same length.

Theorem 4 4

Statement:

The diagonals of a rectangle are equal in length.

Diagonals of a Rectangle

Given:

A B C D 𝐴𝐡𝐢𝐷 is a rectangle.

A C 𝐴𝐢 and B D 𝐡𝐷 are diagonals.

To prove:

A C = B D 𝐴𝐢=𝐡𝐷

Proof:

Let, A Λ† = 90 ∘ 𝐴^=90∘ (By definition of rectangle)

A Λ† + B Λ† = 180 ∘ 𝐴^+𝐡^=180∘ (Consecutive interior angle)

A Λ† = B Λ† = 90 ∘ 𝐴^=𝐡^=90∘

Now in triangles, A B D 𝐴𝐡𝐷 and A B C 𝐴𝐡𝐢 ,

A B = A B 𝐴𝐡=𝐴𝐡 (Common side)

A Λ† = B Λ† = 90 ∘ 𝐴^=𝐡^=90∘ (Each angle is a right angle)

A D = B C 𝐴𝐷=𝐡𝐢 (Opposite sides of parallelogram)

Therefore,

Ξ” A B D β‰… Ξ” B A C Δ𝐴𝐡𝐷 ≅ Δ𝐡𝐴𝐢

Therefore,

B D = A C 𝐡𝐷=𝐴𝐢 (Corresponding parts of corresponding triangles)

Hence the theorem is proved.

Converse of Theorem 4 4 :

Statement:

If two diagonals of a parallelogram are equal, it is a rectangle.

Diagonals of Rectangle

Given:

A B C D 𝐴𝐡𝐢𝐷 is a parallelogram in which A C = B D 𝐴𝐢=𝐡𝐷 .

To prove:

Parallelogram A B C D 𝐴𝐡𝐢𝐷 is a rectangle.

Proof:

In triangles A B C 𝐴𝐡𝐢 and D B C 𝐷𝐡𝐢 ,

A B = D C 𝐴𝐡=𝐷𝐢 (Opposite sides of parallelogram)

B C = B C 𝐡𝐢=𝐡𝐢 (Common side)

A C = B D 𝐴𝐢=𝐡𝐷 (Given)

Therefore,

Ξ” A B C β‰… Ξ” D C B Δ𝐴𝐡𝐢 ≅ Δ𝐷𝐢𝐡

(

S S S 𝑆𝑆𝑆

congruency condition)

Therefore,

A B Λ† C = D C Λ† B 𝐴𝐡^𝐢=𝐷𝐢^𝐡 (Corresponding parts of corresponding triangles)

But these angles are consecutive interior angles on the same side of transversal B C 𝐡𝐢 and A B βˆ₯ D C 𝐴𝐡βˆ₯𝐷𝐢 .

Therefore,

A B Λ† C + D C Λ† B = 180 ∘ 𝐴𝐡^𝐢+𝐷𝐢^𝐡=180∘

But,

A B Λ† C = D C Λ† B 𝐴𝐡^𝐢=𝐷𝐢^𝐡

Therefore,

A B Λ† C = D C Λ† B = 90 ∘ 𝐴𝐡^𝐢=𝐷𝐢^𝐡=90∘

Therefore, by definition of rectangle, parallelogram A B C D 𝐴𝐡𝐢𝐷 is a rectangle.

Hence the theorem is proved.

Theorem 5 5 :

Statement:

The diagonals of a rhombus are perpendicular to each other.

Diagonals of Rhombus

Given:

A B C D 𝐴𝐡𝐢𝐷 is a rhombus.

Diagonal A C 𝐴𝐢 and B D 𝐡𝐷 intersect at O 𝑂 .

To prove:

A C 𝐴𝐢 and B D 𝐡𝐷 bisect each other at right angles.

Proof:

A rhombus is a parallelogram such that

AB = DC = AD = BC . . . . . . ( i ) AB = DC = AD = BCΒ ......(i)

Also the diagonals of a parallelogram bisect each other.

Hence,

B O = D O 𝐡𝑂=𝐷𝑂 and A O = OC . . . . . . ( ii ) 𝐴𝑂=OCΒ ......(ii)

Now, compare triangles A O B 𝐴𝑂𝐡 and A O D 𝐴𝑂𝐷 ,

A B = A D 𝐴𝐡=𝐴𝐷 (From ( i ) (i) above)

B O = D O 𝐡𝑂=𝐷𝑂 (From ( ii ) (ii) above)

AO = AO AO = AO (Common side)

Therefore,

Ξ” A O B β‰… Ξ” A O D Δ𝐴𝑂𝐡 ≅ Δ𝐴𝑂𝐷

(

S S S 𝑆𝑆𝑆

congruency condition)

Therefore,

A O Λ† B = A O Λ† D 𝐴𝑂^𝐡=𝐴𝑂^𝐷

(Corresponding parts of corresponding parts)

B D 𝐡𝐷

is a straight line segment.

Therefore,

A O Λ† B + A O Λ† D = 180 ∘ 𝐴𝑂^𝐡+𝐴𝑂^𝐷=180∘

But,

A O Λ† B = A O Λ† D 𝐴𝑂^𝐡=𝐴𝑂^𝐷

(Proved)

Therefore,

A O Λ† B = A O Λ† D = 180 ∘ 2 𝐴𝑂^𝐡=𝐴𝑂^𝐷=180∘2
A O Λ† B = A O Λ† D = 90 ∘ 𝐴𝑂^𝐡=𝐴𝑂^𝐷=90∘

That is, the diagonals bisect at right angles.

Hence the theorem is proved.

Converse of Theorem 5 5 :

Statement:

If the diagonals of a parallelogram are perpendicular then it is a rhombus.

Given:

A B C D 𝐴𝐡𝐢𝐷 is a parallelogram in which A C 𝐴𝐢 and B D 𝐡𝐷 are perpendicular to each other.

Diagonals of Rhombus

To prove:

A B C D 𝐴𝐡𝐢𝐷 is a rhombus.

Proof:

Let A C 𝐴𝐢 and B D 𝐡𝐷 intersect at right angles at O 𝑂 .

A O Λ† B = 90 ∘ 𝐴𝑂^𝐡=90∘

In triangles A O D 𝐴𝑂𝐷 and C O D 𝐢𝑂𝐷 ,

A O = O C 𝐴𝑂=𝑂𝐢 (Diagonals bisect each other)

O D = O D 𝑂𝐷=𝑂𝐷 (Common side)

A O Λ† D = C O Λ† D = 90 ∘ 𝐴𝑂^𝐷=𝐢𝑂^𝐷=90∘

(Given)

Therefore,

Ξ” A O D β‰… Ξ” C O D Δ𝐴𝑂𝐷 ≅ Δ𝐢𝑂𝐷

(

S A S 𝑆𝐴𝑆

congruency condition)

A D = D C 𝐴𝐷=𝐷𝐢

That is, the adjacent sides are equal.

Therefore, by definition, A B C D 𝐴𝐡𝐢𝐷 is a rhombus.

Hence the theorem is proved.

Benefits of CBSE Class 9 Maths Notes Chapter 8 Quadrilaterals

  • Conceptual Understanding: These notes provide a comprehensive explanation of the properties and characteristics of quadrilaterals, helping students build a strong conceptual foundation in geometry.
  • Clarity in Definitions: By clearly defining terms such as parallelograms, rectangles, rhombuses, and squares, the notes ensure that students understand the distinctions between different types of quadrilaterals.
  • Problem-Solving Skills: Through worked examples and exercises, students can enhance their problem-solving abilities in geometry. Practice questions included in the notes enable students to apply the concepts they've learned to solve a variety of problems.
  • Preparation for Exams: CBSE Class 9 Maths exams often include questions related to quadrilaterals. These notes serve as a valuable resource for exam preparation, ensuring that students are well-equipped to tackle questions on this topic.
CBSE Maths Notes For Class 9
Chapter 1 – Number System
Chapter 2 – Polynomials
Chapter 3 – Coordinate Geometry
Chapter 4 – Linear Equations in Two Variables
Chapter 5 – Introduction to Euclid’s Geometry
Chapter 6 – Lines and Angles
Chapter 7 – Triangles
Chapter 8 – Quadrilaterals
Chapter 9 - Areas of Parallelograms and Triangles
CBSE Class 9 Maths Notes Chapter 10 - Circles Notes
CBSE Class 9 Maths Notes Chapter 11 - Constructions Notes
CBSE Class 9 Maths Notes Chapter 12 - Herons Formula Notes
CBSE Class 9 Maths Notes Chapter 13 - Surface Areas and Volumes Notes
CBSE Class 9 Maths Notes Chapter 14 - Statistics Notes
CBSE Class 9 Maths Notes Chapter 15 - Probability Notes

CBSE Class 9 Maths Notes Chapter 8 FAQs

What are quadrilaterals?

Quadrilaterals are polygons with four sides. They come in various shapes and sizes, such as squares, rectangles, parallelograms, rhombuses, trapezoids, and kites.

What are the properties of a parallelogram?

Parallelograms have opposite sides that are equal and parallel. Their opposite angles are also equal.

What are the conditions for a quadrilateral to be a parallelogram?

A quadrilateral is a parallelogram if its opposite sides are equal and parallel, or if its diagonals bisect each other, or if one pair of opposite sides are equal and parallel.

How can we calculate the area and perimeter of quadrilaterals?

The area of a quadrilateral can be calculated using various formulas depending on the type of quadrilateral. Perimeter is calculated by adding the lengths of all sides. These calculations are essential in fields such as construction, architecture, and engineering.
Join 15 Million students on the app today!
Point IconLive & recorded classes available at ease
Point IconDashboard for progress tracking
Point IconMillions of practice questions at your fingertips
Download ButtonDownload Button
Banner Image
Banner Image
Free Learning Resources
Know about Physics Wallah
Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others

Copyright Β© 2025 Physicswallah Limited All rights reserved.