Figure it Out — Squares and Differences of Squares — We Distribute Yet Things Multiply

1. 12 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 149

   Which is greater: (a − b)² or (b − a)²? Justify your answer.

   Hint. Expand both, or think about what squaring does to a sign.

   Neither — they are always exactly equal.

   Justification 1 — by expanding. (a − b)² = a² − 2ab + b² (b − a)² = b² − 2ab + a² These are the same three terms written in a different order, so the two expressions are identical.

   Justification 2 — by the sign argument, which is quicker. The two quantities a − b and b − a are negatives of each other: b − a = −(a − b). Squaring removes the sign, since (−x)² = x². So the two squares must agree whatever a and b are.

   Check with numbers. Take a = 7, b = 3: (7 − 3)² = 16 and (3 − 7)² = (−4)² = 16 ✓ Take a = 2, b = 9: (2 − 9)² = 49 and (9 − 2)² = 49 ✓

   Why it matters. This is the reason (a − b)² is the natural way to measure how far apart two numbers are — it never depends on which one you subtract from which. The same idea underlies the modulus and, later, the distance formula in coordinate geometry.

   Contrast: (a − b)³ and (b − a)³ are not equal — they are negatives of each other, since an odd power keeps the sign.

   ✦ They are equal, because b − a = −(a − b) and squaring removes the sign; both expand to a² − 2ab + b².

2. 23 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 149

   Express 100 as the difference of two squares.

   Hint. Use Identity 1C backwards: a² − b² = (a + b)(a − b), so you need two factors of 100.

   Step 1 — Turn it into a factor problem. We want a² − b² = 100. By Identity 1C read backwards, a² − b² = (a + b)(a − b), so we need two factors of 100 to serve as (a + b) and (a − b).

   Step 2 — Note the parity restriction. Adding the two equations a + b = m and a − b = n gives 2a = m + n, so m and n must have the same parity for a to be a whole number. Since 100 is even, both factors must be even.

   Step 3 — Try the even factor pairs of 100.

   a + b	a − b	a	b	Check
   50	2	26	24	26² − 24² = 676 − 576 = 100 ✓
   10	10	10	0	10² − 0² = 100 ✓ (trivial)

   The pair (50, 2) gives the genuine answer: 100 = 26² − 24²

   Step 4 — Why other pairs fail. The factor pairs (100, 1), (20, 5) and (4, 25) each mix an even with an odd number, so m + n is odd and a would be a fraction. That is why only 50 × 2 (and the trivial 10 × 10) work.

   The general rule this reveals: a number can be written as a difference of two squares exactly when it is odd or a multiple of 4. Numbers that are twice an odd number — 2, 6, 10, 14 — never can.

   ✦ 100 = 26² − 24², found from 100 = (a+b)(a−b) with a + b = 50 and a − b = 2.

3. 34 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 149

   Find 406², 72², 145², 1097² and 124² using the identities you have learnt so far.

   Hint. Split each number as a round number plus or minus a small one, then use Identity 1A or 1B.

   The strategy: write each number as a round value ± a small one, then apply (a + b)² = a² + 2ab + b² or (a − b)² = a² − 2ab + b². Choosing a round a makes a² and 2ab easy.

   406² — use (a + b)² with a = 400, b = 6 = 400² + 2(400)(6) + 6² = 160000 + 4800 + 36 = 164836

   72² — use (a + b)² with a = 70, b = 2 = 4900 + 2(70)(2) + 4 = 4900 + 280 + 4 = 5184

   145² — use (a + b)² with a = 140, b = 5 = 19600 + 2(140)(5) + 25 = 19600 + 1400 + 25 = 21025 (Faster still: any number ending in 5 squares to [n(n+1)] followed by 25, so 14 × 15 = 210 gives 21025 ✓)

   1097² — use (a − b)² with a = 1100, b = 3 = 1210000 − 2(1100)(3) + 9 = 1210000 − 6600 + 9 = 1203409 (Subtracting from 1100 is far easier than adding onto 1000, since 1097 is close to 1100.)

   124² — use (a + b)² with a = 120, b = 4 = 14400 + 2(120)(4) + 16 = 14400 + 960 + 16 = 15376

   How to choose a and b. Pick the nearest convenient round number, so b stays small and b² is trivial. For 1097 that means going down from 1100 rather than up from 1000 — a difference of 3 instead of 97.

   ✦ 406² = 164836, 72² = 5184, 145² = 21025, 1097² = 1203409, 124² = 15376.

4. 43 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 149

   Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.

   Hint. Look at how the identities were derived — did the derivation ever assume the numbers were whole or positive?

   Yes — the identities hold for all numbers: counting numbers, negative integers, fractions and decimals alike.

   Why. The identities were derived using only the distributive property, together with the ordinary rules for multiplying signs. Nowhere in the derivation was it assumed that a and b are whole or positive. Since the distributive property holds for every kind of number, so must anything derived from it alone.

   Verification with negative integers. Take a = −3, b = −5. (a + b)² = (−8)² = 64 a² + 2ab + b² = 9 + 2(−3)(−5) + 25 = 9 + 30 + 25 = 64 ✓

   Verification with fractions. Take a = ½, b = ⅓. (a + b)² = (5/6)² = 25/36 a² + 2ab + b² = ¼ + 2(½)(⅓) + ⅑ = ¼ + ⅓ + ⅑ With denominator 36: 9/36 + 12/36 + 4/36 = 25/36 ✓

   Verification with a mix. Take a = 2.5, b = −1.5. (a + b)(a − b) = (1)(4) = 4 a² − b² = 6.25 − 2.25 = 4 ✓

   Why this matters. It is exactly what makes the identities useful later. In Class 9 and 10 you will apply them to surds like (√3 + √2)(√3 − √2) = 3 − 2 = 1, and in algebra to whole expressions such as (x² + y)(x² − y) = x⁴ − y². None of that would be allowed if the identities only worked for counting numbers.

   ✦ Both patterns hold for every kind of number, because they follow from the distributive property alone, which never assumed the numbers were whole or positive — verified above with negatives, fractions and decimals.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp106.pdf). The chapter develops the distributive property into the three standard identities — 1A (a+b)², 1B (a−b)², 1C (a+b)(a−b) — via a multiplication-grid model, then applies them to fast mental multiplication and to area/tile patterns. Every expansion here was independently re-expanded term by term and every numeric answer recomputed before comparison with the book's printed answer key. TWO NOTES: (1) the twelve 'Mind the Mistake, Mend the Mistake' items on page 150 have NO answers in the printed key — each has been worked out from first principles here, including identifying which four of the twelve are in fact already correct, and this is stated openly in the solution; (2) the circle-pattern activity in §6.4 ('This Way or That Way') is omitted because the circle counts cannot be recovered from the text without the printed figure.. Questions are referenced from the NCERT textbook for identification.