Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1 Solutions: Exploring Algebraic Identities

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1 Solutions explain the identity (a + b)² = a² + 2ab + b² through binomial square expansion, algebraic expressions and quick numerical square calculations.

Algebra becomes easier when students recognise patterns instead of expanding every expression from the beginning. Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1 Solutions introduce the identity a plus b whole square, written as (a + b)² = a² + 2ab + b². This identity helps students expand binomials, simplify calculations and understand the first major idea in algebraic identities Class 9.

In Ganita Manjari Class 9 Chapter 4 Exercise 4.1, students apply this identity to expressions such as (7x + 4y)², (2.5p + 1.5q)² and binomials with fractional terms. The same method is also used to calculate squares like 64², 105² and 205² without direct multiplication. These Class 9 Maths Chapter 4 Exercise 4.1 Solutions from Exploring Algebraic Identities Exercise 4.1 give step-by-step Class 9 Maths algebraic identities answers for binomial square expansion Class 9 and expanding algebraic expressions Class 9.

Key Takeaways

Algebraic Identity: An identity is true for all values of its variables.
Binomial Square: (a + b)² expands as a² + 2ab + b².
Middle Term: The 2ab term must be included in every binomial square expansion.
Quick Squares: The same identity helps calculate values like 64² and 105².

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1 Solutions

Exercise 4.1 is the starting point for Class 9 Ganita Manjari algebraic identities solutions. It focuses on the identity (a + b)² = a² + 2ab + b², so students first learn to identify the two terms a and b, then square them correctly and include the middle term 2ab.

This concept is important for later parts of Class 9 Maths Ganita Manjari Chapter 4 Solutions, where the same identity is used in factorisation, numerical shortcuts and more advanced algebraic expressions.

Exercise 4.1 Question 1

Using the identity (a + b)² = a² + 2ab + b², expand the following:

(i) (7x + 4y)²

Solution:

Use the identity:

(a + b)² = a² + 2ab + b²

Here:

a = 7x
b = 4y

So,

(7x + 4y)² = (7x)² + 2(7x)(4y) + (4y)²

= 49x² + 56xy + 16y²

Answer: 49x² + 56xy + 16y²

(ii) ((7/5)x + (3/2)y)²

Solution:

Here:

a = (7/5)x
b = (3/2)y

Using the identity:

((7/5)x + (3/2)y)² = ((7/5)x)² + 2((7/5)x)((3/2)y) + ((3/2)y)²

Now simplify each term.

((7/5)x)² = 49x²/25

2((7/5)x)((3/2)y) = 21xy/5

((3/2)y)² = 9y²/4

Therefore:

((7/5)x + (3/2)y)² = 49x²/25 + 21xy/5 + 9y²/4

Answer: 49x²/25 + 21xy/5 + 9y²/4

(iii) (2.5p + 1.5q)²

Solution:

Here:

a = 2.5p
b = 1.5q

Using the identity:

(2.5p + 1.5q)² = (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²

Now simplify.

(2.5p)² = 6.25p²

2(2.5p)(1.5q) = 7.5pq

(1.5q)² = 2.25q²

Therefore:

(2.5p + 1.5q)² = 6.25p² + 7.5pq + 2.25q²

Answer: 6.25p² + 7.5pq + 2.25q²

(iv) ((3/4)s + 8t)²

Solution:

Here:

a = (3/4)s
b = 8t

Using the identity:

((3/4)s + 8t)² = ((3/4)s)² + 2((3/4)s)(8t) + (8t)²

Now simplify.

((3/4)s)² = 9s²/16

2((3/4)s)(8t) = 12st

(8t)² = 64t²

Therefore:

((3/4)s + 8t)² = 9s²/16 + 12st + 64t²

Answer: 9s²/16 + 12st + 64t²

(v) (x + 1/2y)²

Solution:

Here, the expression is:

(x + 1/(2y))²

Take:

a = x
b = 1/(2y)

Using the identity:

(a + b)² = a² + 2ab + b²

we get:

(x + 1/(2y))² = x² + 2(x)(1/(2y)) + (1/(2y))²

Now simplify.

2(x)(1/(2y)) = x/y

(1/(2y))² = 1/(4y²)

Therefore:

(x + 1/(2y))² = x² + x/y + 1/(4y²)

Answer: x² + x/y + 1/(4y²)

(vi) (1/x + 1/y)²

Solution:

Here:

a = 1/x
b = 1/y

Using the identity:

(1/x + 1/y)² = (1/x)² + 2(1/x)(1/y) + (1/y)²

Now simplify.

(1/x)² = 1/x²

2(1/x)(1/y) = 2/(xy)

(1/y)² = 1/y²

Therefore:

(1/x + 1/y)² = 1/x² + 2/(xy) + 1/y²

Answer: 1/x² + 2/(xy) + 1/y²

Exercise 4.1 Question 2

Using the same identity, find the values of the following:

(i) (64)²

Solution:

Write 64 as:

64 = 60 + 4

Using:

(a + b)² = a² + 2ab + b²

we get:

64² = (60 + 4)²

= 60² + 2(60)(4) + 4²

= 3600 + 480 + 16

= 4096

Answer: 64² = 4096

(ii) (105)²

Solution:

Write 105 as:

105 = 100 + 5

Using the identity:

105² = (100 + 5)²

= 100² + 2(100)(5) + 5²

= 10000 + 1000 + 25

= 11025

Answer: 105² = 11025

(iii) (205)²

Solution:

Write 205 as:

205 = 200 + 5

Using the identity:

205² = (200 + 5)²

= 200² + 2(200)(5) + 5²

= 40000 + 2000 + 25

= 42025

Answer: 205² = 42025

Final Answers for Exercise 4.1

Concept Used in Exploring Algebraic Identities Exercise 4.1

Exploring Algebraic Identities Exercise 4.1 is based on the identity:

(a + b)² = a² + 2ab + b²

This identity is used for binomial square expansion Class 9 questions. A binomial has two terms, and when its square is expanded, the answer has three main parts:

For example:

(7x + 4y)² = (7x)² + 2(7x)(4y) + (4y)²

= 49x² + 56xy + 16y²

The same identity also helps in numerical calculations. Instead of multiplying 105 × 105 directly, students can write 105 = 100 + 5 and use the identity to get 11025.

About Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1

Ganita Manjari Class 9 Chapter 4 Exercise 4.1 appears in the chapter Exploring Algebraic Identities. It comes after students are introduced to the visual model of (a + b)² using squares and rectangles.

These Class 9 Maths Chapter 4 Exercise 4.1 Solutions help students revise:

• algebraic identities,
• binomial square expansion,
• expanding algebraic expressions,
• squaring decimal and fractional terms,
• using identities for quick numerical calculations,
• identifying a and b in a binomial.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4