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NCERT Solutions for Class 9th Maths Chapter 4 Think and Reflect — More Identities

Book page 78 Updated on2026-09-08

Q1.
Try to evaluate the following using a suitable identity: (i) 35²   (ii) 65²   (iii) 85²   (iv) 105². Do you observe any interesting pattern?
Answer

Every number here ends in 5, so Śhrīdharāchārya’s form a² = (a + b)(a − b) + b² with b = 5 is the right tool.

35² = (35 + 5)(35 − 5) + 5² = 40 × 30 + 25 = 1200 + 25 = 1225
65² = 70 × 60 + 25 = 4200 + 25 = 4225
85² = 90 × 80 + 25 = 7200 + 25 = 7225
105² = 110 × 100 + 25 = 11000 + 25 = 11025

The pattern. Every answer ends in 25, and the digits in front are the tens digit multiplied by the next whole number:

Numbernn(n + 1)Square
3533 × 4 = 121225
6566 × 7 = 424225
8588 × 9 = 727225
1051010 × 11 = 11011025
Why the rule works — a proof, not a coincidence: any number ending in 5 is 10n + 5. Then
(10n + 5)² = (10n)² + 2(10n)(5) + 5²
= 100n² + 100n + 25
= 100·n(n + 1) + 25

The term 100·n(n + 1) simply writes n(n + 1) two places to the left, and the +25 fills the last two places. So “n times n + 1, then write 25” is exactly the identity in disguise.

Try This: 995². Here n = 99, so n(n + 1) = 99 × 100 = 9900, and 995² = 990025.
Q2.
Observe the two rows of figures below (Fig. 4.6). They represent an algebraic identity. Try to identify it.
Answer

Read the labels off the two rows. The top row has four squares, of sides a + b + c, a + b − c, a − b + c and a − b − c. The bottom row has three squares, of sides 2a, 2b and 2c — and the pieces are coloured to show that the top four cut up and reassemble into the bottom three.

a + b + ca + b − ca − b + ca − b − csum of these four areas2a2b2cequals the sum of these three
The four squares of the top row have exactly the same total area as the three squares of the bottom row.
(a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = (2a)² + (2b)² + (2c)²
that is,   = 4a² + 4b² + 4c²

Proof. Expand the four squares and add:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(a + b − c)² = a² + b² + c² + 2ab − 2bc − 2ca
(a − b + c)² = a² + b² + c² − 2ab − 2bc + 2ca
(a − b − c)² = a² + b² + c² − 2ab + 2bc − 2ca
Sum = 4a² + 4b² + 4c² — every cross term cancels
Why every cross term cancels: look down the 2ab column: the signs are +, +, −, −. Down the 2bc column: +, −, −, +. Down the 2ca column: +, −, +, −. In each column two plus signs meet two minus signs, so each cross term appears twice with each sign. The four sign patterns are precisely the four ways of choosing signs for b and c, which is what forces this balance.
Check it yourself: a = 3, b = 2, c = 1. Left: 6² + 4² + 2² + 0² = 36 + 16 + 4 + 0 = 56. Right: 36 + 16 + 4 = 56. ✓
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