NCERT Solutions Ganita Prakash (Part 1) Chapter 6 Investigating Patterns — In-text Questions

Book page 148 Updated on2026-09-05

Q1.

Take a pair of natural numbers. Calculate the sum of their squares. Can you write twice this sum as a sum of two squares? Try this with other pairs of numbers. Have you figured out a pattern?

Answer

Yes — and the two new squares are built from the sum and the difference of the pair.

Pair a, b	a² + b²	2(a² + b²)	(a + b)² + (a – b)²
2, 1	5	10	3² + 1² = 10 ✓
3, 1	10	20	4² + 2² = 20 ✓
5, 3	34	68	8² + 2² = 68 ✓
6, 5	61	122	11² + 1² = 122 ✓
7, 4	65	130	11² + 3² = 130 ✓

Pattern: 2(a² + b²) = (a + b)² + (a – b)².

Q2.

Do the identities below help in explaining the observed pattern? [(a + b)² = a² + 2ab + b² and (a – b)² = a² – 2ab + b²]

Answer

Yes — adding them proves the pattern in one line.

(a + b)² + (a – b)² = (a² + 2ab + b²) + (a² – 2ab + b²)

Collect like terms: a² + a² = 2a², b² + b² = 2b², and 2ab – 2ab = 0.

2(a² + b²) = (a + b)² + (a – b)²

Test at a = 6, b = 5: 2(36 + 25) = 122 and 11² + 1² = 122 ✓

Why it happens: the cross terms are equal in size but opposite in sign, so they destroy each other. Adding the two identities is like adding the two square pictures: the strips of one fill exactly the gaps of the other.

Did you know? Because the proof uses only the two identities, it holds for negative numbers and fractions too — try a = 1⁄2, b = 1⁄3: 2(1⁄4 + 1⁄9) = 13⁄18, and (5⁄6)² + (1⁄6)² = 25⁄36 + 1⁄36 = 26⁄36 = 13⁄18 ✓

Q3.

Here is a related pattern. Try to describe the pattern using algebra to determine if the pattern always holds. [9 × 9 – 1 × 1 = 10 × 8; 8 × 8 – 6 × 6 = 14 × 2; 7 × 7 – 2 × 2 = 9 × 5; 10 × 10 – 4 × 4 = 14 × 6]

Answer

Read each line as a difference of two squares and look at the numbers on the right.

Line	a, b	a + b	a – b
9² – 1² = 10 × 8	9, 1	10	8
8² – 6² = 14 × 2	8, 6	14	2
7² – 2² = 9 × 5	7, 2	9	5
10² – 4² = 14 × 6	10, 4	14	6

So the guess is a² – b² = (a + b)(a – b). Expanding settles it:

(a + b)(a – b) = a² – ab + ba – b²
ab + (–ab) = 0, so
Identity 1C: (a + b)(a – b) = a² – b²

It is a true identity, so the pattern always holds. You had already met it in Figure it Out 5 (i).

Why it happens: as with Pattern 1, the two middle products are the same size with opposite signs and cancel. This is why a difference of squares can always be turned into a product — and that is what makes 397 × 403 or 45 × 55 so quick.

Q4.

Use Identity 1C to calculate 98 × 102, and 45 × 55.

Answer

Look for the number midway between the two factors.

98 × 102 = (100 – 2)(100 + 2)
= 100² – 2²
= 10000 – 4 = 9996

45 × 55 = (50 – 5)(50 + 5)
= 50² – 5²
= 2500 – 25 = 2475

Tip: the trick works whenever the two numbers are equally far from a round number. Their midpoint is a, and half their difference is b.

Q5.

Show that (a + b) × (a – b) = a² – b² geometrically.

Answer

Start with a square of side a and cut out a square of side b from one corner. The area left is a² – b², and it is an L-shape.

Cut the L into two rectangles and slide one against the other:

Rectangle 1: a wide, (a – b) tall
Rectangle 2: b wide, (a – b) tall

Placing Rectangle 2 beside Rectangle 1 makes a single rectangle of width a + b and height a – b.

The L-shaped region of area a² – b² is re-cut into one rectangle measuring (a + b) by (a – b).

a² – b² = area of the L = (a + b)(a – b)

Why it happens: cutting and sliding never changes area. Both rectangles have the same height a – b, so they fit side by side exactly, and their widths a and b add to a + b.

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