NCERT Solutions Ganita Prakash (Part 1) Chapter 6 .2 Special Cases of the Distributive Property — In-text Questions

Book page 1466 Updated on2026-09-05

Q1.

Use Identity 1A to write the expressions for the following. (i) (m + 3)² (ii) (6 + p)²

Answer

Identity 1A: (a + b)² = a² + 2ab + b².

(i) (m + 3)² with a = m, b = 3
= m² + 2 × m × 3 + 3²
= m² + 6m + 9

(ii) (6 + p)² with a = 6, b = p
= 6² + 2 × 6 × p + p²
= 36 + 12p + p²

Test (i) at m = 4: 7² = 49 and 16 + 24 + 9 = 49 ✓

Q2.

Expand (6x + 5)².

Answer

Using the distributive property

(6x + 5)² = (6x + 5)(6x + 5)
= (6x × 6x) + (5 × 6x) + (6x × 5) + 5 × 5
= (6x)² + 2(6x × 5) + 5²
= 36x² + 60x + 25

Using the identity

(6x + 5)² = (6x)² + 5² + 2 × (6x × 5)
= 36x² + 25 + 60x

Test at x = 1: 11² = 121 and 36 + 60 + 25 = 121 ✓

Tip: if you cannot remember the identity, just multiply the two brackets out. The identity is a shortcut for the distributive property, never a replacement for it.

Q3.

Expand (3j + 2k)² using both the identity and by applying the distributive property.

Answer

Using Identity 1A with a = 3j and b = 2k:

(3j + 2k)² = (3j)² + 2(3j)(2k) + (2k)²
= 9j² + 12jk + 4k²

Using the distributive property

(3j + 2k)(3j + 2k) = 3j(3j + 2k) + 2k(3j + 2k)
= 9j² + 6jk + 6jk + 4k²
= 9j² + 12jk + 4k²

Test at j = 1, k = 1: 5² = 25 and 9 + 12 + 4 = 25 ✓

Why it happens: in Identity 1A the letters a and b stand for whole terms, not just single letters. Here a = 3j, so a² = (3j)² = 9j², and the cross term is 2 × 3j × 2k = 12jk.

Q4.

Can we use 60² (=3600) and 5² (=25) to find the value of (60 – 5)² or 55²?

Answer

Yes. Draw a square of side 55 sitting inside a square of side 60.

Start with 60² and take away the two 60 × 5 rectangles along two sides. That removes the little 5 × 5 corner twice, so add it back once.

(60 – 5)² = 60² – (60 × 5) – (5 × 60) + 5²
= 3600 – 300 – 300 + 25
= 3025 sq. units

The 5 × 5 corner lies in both removed strips, so it must be added back once.

Why it happens: the two strips overlap. Subtracting them one after the other subtracts the overlap twice, and the “+ b²” in Identity 1B is precisely the repair. Check: 55 × 55 = 3025 ✓

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