Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.2 Solutions: Exploring Algebraic Identities

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.2 Solutions explain factorisation using identities, perfect square trinomials and quick square calculations with (a − b)².

Exercise 4.2 moves from expanding identities to using them in reverse. In Ganita Manjari Class 9 Chapter 4 Exercise 4.2, students identify expressions that match the form a² + 2ab + b² and factor them as (a + b)². This is the first major step in factorisation using identities Class 9.

The second part of the exercise uses identity a minus b whole square, or (a − b)² = a² − 2ab + b², to calculate values such as 79², 193² and 299² quickly. These Class 9 Maths Chapter 4 Exercise 4.2 Solutions from Exploring Algebraic Identities Exercise 4.2 give step-by-step Class 9 Maths algebraic identities answers for both algebraic factorisation and numerical simplification.

Key Takeaways

Perfect Square Trinomial: a² + 2ab + b² can be factorised as (a + b)².
Common Factor: Some expressions need a common factor taken out before factorisation.
Identity Use: (a − b)² helps find squares of numbers near round numbers.
Reverse Identity: Expansion identities can also be used for factorisation.

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.2 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.2 Solutions

Exercise 4.2 is based on recognising a perfect square trinomial Class 9 expression. A trinomial of the form:

a² + 2ab + b²

can be written as:

(a + b)²

Similarly, the identity:

(a − b)² = a² − 2ab + b²

is useful for finding squares of numbers close to 10, 100 or 300. This exercise is an important part of Class 9 Ganita Manjari algebraic identities solutions because it connects expansion, factorisation and mental calculation.

Exercise 4.2 Question 1

Factor completely:

(i) 9x² + 24xy + 16y²

Solution:

Write each term in square-identity form.

9x² = (3x)²

16y² = (4y)²

24xy = 2(3x)(4y)

So,

9x² + 24xy + 16y² = (3x)² + 2(3x)(4y) + (4y)²

Using a² + 2ab + b² = (a + b)²,

9x² + 24xy + 16y² = (3x + 4y)²

Answer: (3x + 4y)²

(ii) 4s² + 20st + 25t²

Solution:

4s² = (2s)²

25t² = (5t)²

20st = 2(2s)(5t)

So,

4s² + 20st + 25t² = (2s)² + 2(2s)(5t) + (5t)²

Therefore:

4s² + 20st + 25t² = (2s + 5t)²

Answer: (2s + 5t)²

(iii) 49x² + 28xy + 4y²

Solution:

49x² = (7x)²

4y² = (2y)²

28xy = 2(7x)(2y)

So,

49x² + 28xy + 4y² = (7x)² + 2(7x)(2y) + (2y)²

Therefore:

49x² + 28xy + 4y² = (7x + 2y)²

Answer: (7x + 2y)²

(iv) 64p² + 32/3 pq + 4/9 q²

Solution:

64p² = (8p)²

4/9 q² = (2q/3)²

Now check the middle term:

2(8p)(2q/3) = 32pq/3

So,

64p² + 32/3 pq + 4/9 q² = (8p)² + 2(8p)(2q/3) + (2q/3)²

Therefore:

64p² + 32/3 pq + 4/9 q² = (8p + 2q/3)²

Answer: (8p + 2q/3)²

(v) 3a² + 4ab + 4/3 b²

Solution:

First take 1/3 as a common factor.

3a² + 4ab + 4/3 b² = 1/3(9a² + 12ab + 4b²)

Now factor the expression inside the bracket.

9a² = (3a)²

4b² = (2b)²

12ab = 2(3a)(2b)

So,

9a² + 12ab + 4b² = (3a + 2b)²

Therefore:

3a² + 4ab + 4/3 b² = 1/3(3a + 2b)²

Answer: 1/3(3a + 2b)²

(vi) 9/5 s² + 6sv + 5v²

Solution:

First take 1/5 as a common factor.

9/5 s² + 6sv + 5v² = 1/5(9s² + 30sv + 25v²)

Now factor the expression inside the bracket.

9s² = (3s)²

25v² = (5v)²

30sv = 2(3s)(5v)

So,

9s² + 30sv + 25v² = (3s + 5v)²

Therefore:

9/5 s² + 6sv + 5v² = 1/5(3s + 5v)²

Answer: 1/5(3s + 5v)²

Exercise 4.2 Question 2

Find the values of the following using the identity (a − b)² = a² − 2ab + b².

(i) (79)²

Solution:

Write 79 as:

79 = 80 − 1

Using:

(a − b)² = a² − 2ab + b²

we get:

79² = (80 − 1)²

= 80² − 2(80)(1) + 1²

= 6400 − 160 + 1

= 6241

Answer: 79² = 6241

(ii) (193)²

Solution:

Write 193 as:

193 = 200 − 7

Using the identity:

193² = (200 − 7)²

= 200² − 2(200)(7) + 7²

= 40000 − 2800 + 49

= 37249

Answer: 193² = 37249

(iii) (299)²

Solution:

Write 299 as:

299 = 300 − 1

Using the identity:

299² = (300 − 1)²

= 300² − 2(300)(1) + 1²

= 90000 − 600 + 1

= 89401

Answer: 299² = 89401

Final Answers for Exercise 4.2

Concept Used in Exploring Algebraic Identities Exercise 4.2

Exploring Algebraic Identities Exercise 4.2 mainly uses two identities:

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

The first identity is used in reverse to factor perfect square trinomials. For example:

9x² + 24xy + 16y² = (3x)² + 2(3x)(4y) + (4y)²

So,

9x² + 24xy + 16y² = (3x + 4y)²

The second identity is used for quick numerical calculation. For example:

299² = (300 − 1)²

= 300² − 2(300)(1) + 1²

= 89401

This makes the exercise useful for both algebraic identities Class 9 and arithmetic shortcuts.

About Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.2

Ganita Manjari Class 9 Chapter 4 Exercise 4.2 appears after students learn the identity (a + b)² and are introduced to (a − b)². The exercise checks whether students can recognise a trinomial as a perfect square and factor it correctly.

These Class 9 Maths Chapter 4 Exercise 4.2 Solutions help students revise:

• perfect square trinomials,
• factorisation using identities,
• common factor method,
• identity a plus b whole square,
• identity a minus b whole square,
• numerical square calculations,
• algebraic identities Class 9.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4