Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3 Solutions: Exploring Algebraic Identities

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3 Solutions cover square identities, factorisation using identities and expansion of three-term algebraic expressions.

Exercise 4.3 brings together the identities students have learnt so far in Chapter 4. In this exercise, students use (a + b)², (a − b)² and (a + b + c)² to find numerical squares, factor algebraic expressions and expand expressions with three terms.

In Ganita Manjari Class 9 Chapter 4 Exercise 4.3, the focus is on choosing the right identity for each question. Some questions are easier with (a − b)², such as 198², while others need the identity a plus b plus c whole square, such as (p + 3q + 7r)². These Class 9 Maths Chapter 4 Exercise 4.3 Solutions from Exploring Algebraic Identities Exercise 4.3 give clear steps for Class 9 Maths algebraic identities answers, factorisation and expansion.

Key Takeaways

Square Identities: Questions use (a + b)², (a − b)² and (a + b + c)².
Three-Term Square: (a + b + c)² includes three square terms and three double-product terms.
Identity Selection: Students must choose the identity that fits the expression.
Identity Check: A statement is an identity only if it is true for all values.

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3 Solutions

Exercise 4.3 is an important part of Class 9 Ganita Manjari algebraic identities solutions because it combines numerical calculation and algebraic manipulation. Students use identities to avoid long multiplication, recognise perfect square identities Class 9, and expand expressions with two or three terms.

The main identities used are:

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Exercise 4.3 Question 1

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

(i) 117²

Solution:

Write 117 as:

117 = 100 + 10 + 7

Use:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

So,

117² = (100 + 10 + 7)²

= 100² + 10² + 7² + 2(100)(10) + 2(10)(7) + 2(7)(100)

= 10000 + 100 + 49 + 2000 + 140 + 1400

= 13689

Answer: 117² = 13689

(ii) 78²

Solution:

Write 78 as:

78 = 80 − 2

Use:

(a − b)² = a² − 2ab + b²

So,

78² = (80 − 2)²

= 80² − 2(80)(2) + 2²

= 6400 − 320 + 4

= 6084

Answer: 78² = 6084

(iii) 198²

Solution:

Write 198 as:

198 = 200 − 2

So,

198² = (200 − 2)²

= 200² − 2(200)(2) + 2²

= 40000 − 800 + 4

= 39204

Answer: 198² = 39204

(iv) 214²

Solution:

Write 214 as:

214 = 200 + 10 + 4

Use the identity for the square of three terms.

214² = (200 + 10 + 4)²

= 200² + 10² + 4² + 2(200)(10) + 2(10)(4) + 2(4)(200)

= 40000 + 100 + 16 + 4000 + 80 + 1600

= 45796

Answer: 214² = 45796

(v) 1104²

Solution:

Write 1104 as:

1104 = 1100 + 4

Use:

(a + b)² = a² + 2ab + b²

So,

1104² = (1100 + 4)²

= 1100² + 2(1100)(4) + 4²

= 1210000 + 8800 + 16

= 1218816

Answer: 1104² = 1218816

(vi) 1120²

Solution:

Write 1120 as:

1120 = 1100 + 20

So,

1120² = (1100 + 20)²

= 1100² + 2(1100)(20) + 20²

= 1210000 + 44000 + 400

= 1254400

Answer: 1120² = 1254400

Exercise 4.3 Question 2

Factor using suitable identities:

(i) 16y² − 24y + 9

Solution:

16y² = (4y)²

9 = 3²

−24y = −2(4y)(3)

So,

16y² − 24y + 9 = (4y)² − 2(4y)(3) + 3²

Using:

a² − 2ab + b² = (a − b)²

we get:

16y² − 24y + 9 = (4y − 3)²

Answer: (4y − 3)²

(ii) 9/4 s² + 6st + 4t²

Solution:

9/4 s² = (3s/2)²

4t² = (2t)²

6st = 2(3s/2)(2t)

So,

9/4 s² + 6st + 4t² = (3s/2)² + 2(3s/2)(2t) + (2t)²

Therefore:

9/4 s² + 6st + 4t² = (3s/2 + 2t)²

Answer: (3s/2 + 2t)²

(iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²

Solution:

Write the expression as the square of three terms.

m²/9 = (m/3)²

k²/4 = (k/2)²

9n² = (3n)²

Now check the middle terms.

2(m/3)(k/2) = mk/3

2(k/2)(3n) = 3kn

2(m/3)(3n) = 2mn

So,

m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²

= (m/3)² + (k/2)² + (3n)² + 2(m/3)(k/2) + 2(k/2)(3n) + 2(m/3)(3n)

= (m/3 + k/2 + 3n)²

Answer: (m/3 + k/2 + 3n)²

(iv) p²/16 − 2 + 16/p²

Solution:

p²/16 = (p/4)²

16/p² = (4/p)²

−2 = −2(p/4)(4/p)

So,

p²/16 − 2 + 16/p² = (p/4)² − 2(p/4)(4/p) + (4/p)²

Therefore:

p²/16 − 2 + 16/p² = (p/4 − 4/p)²

Answer: (p/4 − 4/p)²

(v) 9a² + 4b² + c² − 12ab + 6ac − 4bc

Solution:

This expression can be compared with:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here, take the three terms as:

3a, −2b and c

Now:

(3a)² = 9a²

(−2b)² = 4b²

c² = c²

2(3a)(−2b) = −12ab

2(3a)(c) = 6ac

2(−2b)(c) = −4bc

Therefore:

9a² + 4b² + c² − 12ab + 6ac − 4bc = (3a − 2b + c)²

Answer: (3a − 2b + c)²

Exercise 4.3 Question 3

Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:

(i) (p + 3q + 7r)²

Solution:

Here:

a = p
b = 3q
c = 7r

Using:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

we get:

(p + 3q + 7r)²

= p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)

= p² + 9q² + 49r² + 6pq + 42qr + 14pr

Answer: p² + 9q² + 49r² + 6pq + 42qr + 14pr

(ii) (3x − 2y + 4z)²

Solution:

Here:

a = 3x
b = −2y
c = 4z

Using the identity:

(3x − 2y + 4z)²

= (3x)² + (−2y)² + (4z)² + 2(3x)(−2y) + 2(−2y)(4z) + 2(4z)(3x)

= 9x² + 4y² + 16z² − 12xy − 16yz + 24xz

Answer: 9x² + 4y² + 16z² − 12xy − 16yz + 24xz

Exercise 4.3 Question 4

Is this an identity?

(a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c²

Solution:

To check whether it is an identity, test it with simple values.

Take:

a = 1, b = 1, c = 1

Now calculate the left-hand side.

(a + b − c)² = (1 + 1 − 1)² = 1² = 1

(a − b + c)² = (1 − 1 + 1)² = 1² = 1

(a − b − c)² = (1 − 1 − 1)² = (−1)² = 1

So,

LHS = 1 + 1 + 1 = 3

Now calculate the right-hand side.

RHS = 2a² + 2b² + 2c²

= 2(1)² + 2(1)² + 2(1)²

= 2 + 2 + 2

= 6

Since:

3 ≠ 6

the given statement is not true for all values of a, b and c.

Answer: No, this is not an identity.

Final Answers for Exercise 4.3

Concept Used in Exploring Algebraic Identities Exercise 4.3

Exploring Algebraic Identities Exercise 4.3 uses three main identities:

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

The first two identities help in quick square calculations such as 78² and 198². The third identity is used for square of three terms Class 9 questions such as (p + 3q + 7r)² and (3x − 2y + 4z)².

The factorisation questions use the same identities in reverse. For example:

16y² − 24y + 9 = (4y)² − 2(4y)(3) + 3²

So,

16y² − 24y + 9 = (4y − 3)²

This is why Exercise 4.3 is important for both factorisation using identities Class 9 and expanding algebraic expressions Class 9.

About Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3

Ganita Manjari Class 9 Chapter 4 Exercise 4.3 appears in the section on more identities. It comes after students learn to expand and factor expressions using (a + b)² and (a − b)².

These Class 9 Maths Chapter 4 Exercise 4.3 Solutions help students revise:

• algebraic identities Class 9,
• square of two terms,
• square of three terms,
• perfect square identities,
• factorisation using identities,
• quick numerical squares,
• checking whether a statement is an identity.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4