Figure it Out — Identities, Patterns and Applications — We Distribute Yet Things Multiply

1. 13 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 155

   Compute these products using the suggested identity. (i) 46² using Identity 1A for (a + b)² (ii) 397 × 403 using Identity 1C for (a + b)(a − b) (iii) 91² using Identity 1B for (a − b)² (iv) 43 × 45 using Identity 1C for (a + b)(a − b)

   Hint. For (iv), the two numbers are not equidistant from a round number — but they are equidistant from their own midpoint.

   (i) 46² using (a + b)² with a = 40, b = 6 = 40² + 2(40)(6) + 6² = 1600 + 480 + 36 = 2116

   (ii) 397 × 403 using (a + b)(a − b) with a = 400, b = 3 The midpoint of 397 and 403 is 400, and each lies 3 away: = (400 − 3)(400 + 3) = 400² − 3² = 160000 − 9 = 159991

   (iii) 91² using (a − b)² with a = 100, b = 9 = 100² − 2(100)(9) + 9² = 10000 − 1800 + 81 = 8281 (Going down from 100 is much easier than up from 90, since 100² and 2 × 100 × 9 are both immediate.)

   (iv) 43 × 45 using (a + b)(a − b) Here the midpoint is not a round number — it is 44, since 43 and 45 sit either side of it: = (44 − 1)(44 + 1) = 44² − 1² = 1936 − 1 = 1935

   The choice of a in each case. For a square, pick the nearest round number so that b stays small. For a product of two numbers, pick their midpoint — which exists as a whole number whenever the two numbers have the same parity, as in (ii) and (iv).

   ✦ (i) 2116 (ii) 159991 (iii) 8281 (iv) 1935

2. 24 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 155

   Use either a suitable identity or the distributive property to find each product. (i) (p − 1)(p + 11) (ii) (3a − 9b)(3a + 9b) (iii) −(2y + 5)(3y + 4) (iv) (6x + 5y)² (v) (2x − ½)² (vi) (7p) × (3r) × (p + 2)

   Hint. Only (ii) fits an identity exactly; the rest are quicker by plain distribution.

   (i) (p − 1)(p + 11) — no identity fits, so distribute: = p² + 11p − p − 11 = p² + 10p − 11

   (ii) (3a − 9b)(3a + 9b) — this is Identity 1C with a = 3a and b = 9b: = (3a)² − (9b)² = 9a² − 81b² Note both coefficients get squared, so 9² = 81, not 9.

   (iii) −(2y + 5)(3y + 4) — expand the brackets first, then apply the leading minus to every term: (2y + 5)(3y + 4) = 6y² + 8y + 15y + 20 = 6y² + 23y + 20 Negating: = −6y² − 23y − 20

   (iv) (6x + 5y)² — Identity 1A with a = 6x and b = 5y: = (6x)² + 2(6x)(5y) + (5y)² = 36x² + 60xy + 25y²

   (v) (2x − ½)² — Identity 1B with a = 2x and b = ½: = (2x)² − 2(2x)(½) + (½)² = 4x² − 2x + ¼ = 4x² − 2x + ¼

   (vi) (7p) × (3r) × (p + 2) — multiply the two monomials first, then distribute: 7p × 3r = 21pr 21pr(p + 2) = 21p²r + 42pr

   How to choose your route. Use an identity only when the expression genuinely matches its shape — a sum times the same difference for 1C, or a bracket squared for 1A/1B. Forcing an identity onto (i) or (iii) wastes time and invites errors; plain distribution is faster there.

   ✦ (i) p² + 10p − 11 (ii) 9a² − 81b² (iii) −6y² − 23y − 20 (iv) 36x² + 60xy + 25y² (v) 4x² − 2x + ¼ (vi) 21p²r + 42pr

3. 32 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 156

   For each statement identify the appropriate algebraic expression. (i) Two more than a square number: 2 + s, (s + 2)², s² + 2, s² + 4, 2s², 2²s. (ii) The sum of the squares of two consecutive numbers: m² + n², (m + n)², m² + 1, m² + (m + 1)², m² + (m − 1)², (m + (m + 1))², (2m)² + (2m + 1)².

   Hint. Translate the English phrase piece by piece, in the order it is said.

   (i) Two more than a square number → s² + 2

   Read the phrase in order: take a square number, which is s²; then two more than it means add 2. So the answer is s² + 2.

   Why each of the others is wrong: • 2 + s — this is two more than s, not two more than s squared. • (s + 2)² — this squares after adding, so it is 'the square of two more than a number'. A different thing entirely: for s = 3 it gives 25, not 11. • s² + 4 — four more, not two. • 2s² — twice a square number. • 2²s — four times a number.

   (ii) The sum of the squares of two consecutive numbers → m² + (m + 1)²

   Two consecutive numbers are m and m + 1. Squaring each and adding gives m² + (m + 1)².

   Why each of the others is wrong: • m² + n² — a sum of two squares, but nothing says n follows m. • (m + n)² — the square of a sum, not a sum of squares. • m² + 1 — the second term is not squared. • m² + (m − 1)² — these are m and m − 1, which are consecutive, so this is arguably also a sum of the squares of two consecutive numbers. The book's answer takes the numbers as m and the next one, which is the standard reading. • (m + (m + 1))² — the square of the sum, again the wrong order. • (2m)² + (2m + 1)² — these are consecutive, but one is forced to be even, which the statement never said.

   The lesson: the order of operations must match the order of the words. 'The square of a sum' and 'the sum of squares' are different, and reading carefully is what separates them.

   ✦ (i) s² + 2 (ii) m² + (m + 1)²

4. 43 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 156

   Consider any 2 by 2 square of numbers in a calendar. Find the products of the numbers along each diagonal — for example 4 × 12 = 48 and 5 × 11 = 55. Do this for other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.

   Hint. Label the top-left number a. What are the other three in terms of a, given that a calendar week has 7 days?

   Step 1 — Label the square algebraically. A calendar row advances by 1 and a calendar column advances by 7, since a week has seven days. So a 2 × 2 block with top-left entry a is

   a	a + 1
   a + 7	a + 8

   Step 2 — Form the two diagonal products. Main diagonal: a(a + 8) = a² + 8a Other diagonal: (a + 1)(a + 7) = a² + 7a + a + 7 = a² + 8a + 7

   Step 3 — Take the difference. (a² + 8a + 7) − (a² + 8a) = 7

   So the two diagonal products always differ by exactly 7, whatever block you choose. The a² and 8a terms cancel completely, which is why the answer does not depend on a at all.

   Step 4 — Check against examples. • Block 4, 5, 11, 12: 5 × 11 = 55 and 4 × 12 = 48. Difference 7 ✓ • Block 3, 4, 10, 11: 4 × 10 = 40 and 3 × 11 = 33. Difference 7 ✓ • Block 16, 17, 23, 24: 17 × 23 = 391 and 16 × 24 = 384. Difference 7 ✓

   Why 7 and nothing else. The 7 comes directly from the width of the calendar week. On a grid stepping down by k instead, the same working gives (a + 1)(a + k) − a(a + k + 1) = k. So a 5-day-week calendar would give a difference of 5.

   ✦ The two diagonal products always differ by exactly 7, because the block is a, a+1, a+7, a+8, and (a+1)(a+7) − a(a+8) = 7 — the 7 being the number of days in a calendar week.

5. 54 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 157

   Verify which of the following statements are true. (i) (k + 1)(k + 2) − (k + 3) is always 2. (ii) (2q + 1)(2q − 3) is a multiple of 4. (iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8. (iv) (6n + 2)² − (4n + 3)² is 5 less than a square number.

   Hint. Expand each fully and look at what is left. Only one of the four is true.

   (i) FALSE. Expanding: (k + 1)(k + 2) − (k + 3) = (k² + 3k + 2) − (k + 3) = k² + 2k − 1 This depends on k, so it is certainly not always 2. Check: k = 1 gives 1 + 2 − 1 = 2 (which is where the claim comes from), but k = 2 gives 4 + 4 − 1 = 7 ✗

   (ii) FALSE. Expanding: (2q + 1)(2q − 3) = 4q² − 6q + 2q − 3 = 4q² − 4q − 3 The first two terms are multiples of 4, but the −3 never is, so the whole expression is always 3 short of a multiple of 4. Check: q = 1 gives 3 × (−1) = −3 ✗, and q = 2 gives 5 × 1 = 5 ✗

   (iii) TRUE — both halves. Even squares: an even number is 2n, and (2n)² = 4n², which is a multiple of 4 ✓ Odd squares: an odd number is 2n + 1, and (2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1. Now n and n + 1 are consecutive, so one of them is even and n(n + 1) is always even. Writing n(n + 1) = 2m gives = 8m + 1. So every odd square is exactly 1 more than a multiple of 8 ✓ Check: 3² = 9 = 8 + 1 ✓, 5² = 25 = 24 + 1 ✓, 7² = 49 = 48 + 1 ✓, 9² = 81 = 80 + 1 ✓

   (iv) FALSE. Expanding both squares: (6n + 2)² = 36n² + 24n + 4 (4n + 3)² = 16n² + 24n + 9 Difference = 20n² − 5 For this to be 5 less than a square we would need 20n² to be a perfect square. But 20n² = 4 × 5n², whose square root is 2n√5 — irrational unless n = 0. So it fails for every positive n. Check: n = 1 gives 15, and 15 + 5 = 20 is not a square ✗; n = 2 gives 75, and 80 is not a square ✗

   ✦ Only (iii) is true. (i) simplifies to k² + 2k − 1, (ii) to 4q² − 4q − 3 which is always 3 short of a multiple of 4, and (iv) to 20n² − 5, where 20n² is never a perfect square.

6. 63 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 157

   A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference and product are divided by 7?

   Hint. Write both in the form 7k + r, then combine — the identities from this chapter make the product easy.

   Write the two numbers as n₁ = 7a + 3 and n₂ = 7b + 5.

   Sum. n₁ + n₂ = 7a + 7b + 8 = 7(a + b) + 8 Since 8 exceeds the divisor 7, take out one more group of 7: = 7(a + b + 1) + 1 Remainder 1.

   Difference (taking the larger remainder first, n₂ − n₁). n₂ − n₁ = (7b + 5) − (7a + 3) = 7(b − a) + 2 Here 2 is already less than 7, so no regrouping is needed. Remainder 2.

   Product — this is where the chapter's expansion skills pay off. n₁ × n₂ = (7a + 3)(7b + 5) = 49ab + 35a + 21b + 15 Every term except the 15 already carries a factor of 7, and 15 = 14 + 1: = 7(7ab + 5a + 3b + 2) + 1 Remainder 1.

   The shortcut this reveals. You never need the original numbers — just combine the remainders and reduce mod 7: • sum: 3 + 5 = 8 → 8 − 7 = 1 ✓ • difference: 5 − 3 = 2 ✓ • product: 3 × 5 = 15 → 15 − 14 = 1 ✓

   Check with actual numbers: take n₁ = 10 and n₂ = 12. Sum 22 = 7(3) + 1 ✓; difference 2 ✓; product 120 = 7(17) + 1 ✓

   ✦ Sum leaves remainder 1, difference leaves 2, and product leaves 1 — obtained by combining the remainders 3 and 5 directly and reducing modulo 7.

7. 73 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 157

   Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat with other sets. What pattern do you notice? Write it as an algebraic equation and expand both sides to check it is a true identity.

   Hint. Naming the MIDDLE number n rather than the first makes the algebra collapse immediately.

   Step 1 — Try it with numbers first. • 4, 5, 6: 5² − (4 × 6) = 25 − 24 = 1 • 9, 10, 11: 10² − (9 × 11) = 100 − 99 = 1 • 19, 20, 21: 20² − (19 × 21) = 400 − 399 = 1

   The pattern: the answer is always 1.

   Step 2 — Set it up algebraically. The key move is to call the middle number n, so the three are n − 1, n, n + 1. The claim becomes n² − (n − 1)(n + 1) = 1

   Step 3 — Expand and verify. The product (n − 1)(n + 1) is Identity 1C with a = n and b = 1: (n − 1)(n + 1) = n² − 1 So n² − (n² − 1) = n² − n² + 1 = 1 ✓

   The identity holds for every n, so this is a genuine identity and not a coincidence.

   Step 4 — Why it works, in words. The two outer numbers sit equally far either side of the middle one. Identity 1C says that pushing a pair apart symmetrically from n by 1 costs exactly 1² from the product. So the product always falls short of n² by exactly 1.

   The generalisation worth knowing. Spread the outer numbers by k instead of 1: n² − (n − k)(n + k) = n² − (n² − k²) = k² So with 5, 8, 11 (spread 3): 8² − (5 × 11) = 64 − 55 = 9 = 3² ✓

   ✦ The answer is always 1, expressed by the identity n² − (n − 1)(n + 1) = 1, which follows because (n − 1)(n + 1) = n² − 1 by Identity 1C.

8. 82 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 157

   What is the algebraic expression describing the following steps — add any two numbers, then multiply this by half of the sum of the two numbers? Prove that the result is half of the square of the sum of the two numbers.

   Hint. Write the sum once, give it a name if it helps, then follow the instruction literally.

   Step 1 — Translate the steps. Let the two numbers be a and b. • "Add any two numbers" → a + b • "Half of the sum of the two numbers" → ½(a + b) • "Multiply this by" → the expression is (a + b) × ½(a + b)

   Step 2 — Simplify. Since both factors contain (a + b), multiplying gives (a + b) × ½(a + b) = ½ × (a + b) × (a + b) = ½(a + b)²

   Step 3 — Read off the conclusion. This is exactly "half of the square of the sum of the two numbers", which is what was to be proved.

   Step 4 — Expand it if a fully written form is wanted. ½(a + b)² = ½(a² + 2ab + b²) = ½a² + ab + ½b²

   Check with a = 3 and b = 5: Sum = 8, half the sum = 4, product = 8 × 4 = 32. And ½(a + b)² = ½ × 64 = 32 ✓ Also ½(9) + 15 + ½(25) = 4.5 + 15 + 12.5 = 32 ✓

   Where students slip: writing ½(a + b)² as ½a² + ½b², which drops the ab term. The square of a sum always carries a middle term.

   ✦ The expression is (a + b) × ½(a + b) = ½(a + b)², which is by definition half the square of the sum; expanded it is ½a² + ab + ½b².

9. 93 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 158

   Which is larger? Find out without fully computing the product. (i) 14 × 26 or 16 × 24 (ii) 25 × 75 or 26 × 74

   Hint. Write one product in terms of the other using Identity 1, and look only at the sign of the change.

   The trick: express one product as the other plus a correction, then judge by the sign of the correction alone.

   (i) 14 × 26 versus 16 × 24 Write 14 = 16 − 2 and 26 = 24 + 2: 14 × 26 = (16 − 2)(24 + 2) = 16 × 24 + 16(2) − 2(24) − 2(2) = 16 × 24 + 32 − 48 − 4 = 16 × 24 − 20

   The correction is negative, so 16 × 24 is larger, by 20. (Check: 364 versus 384 ✓)

   (ii) 25 × 75 versus 26 × 74 Write 25 = 26 − 1 and 75 = 74 + 1: 25 × 75 = (26 − 1)(74 + 1) = 26 × 74 + 26 − 74 − 1 = 26 × 74 − 49

   Again negative, so 26 × 74 is larger, by 49. (Check: 1875 versus 1924 ✓)

   The principle behind both. In each pair the two numbers have the same sum — 14 + 26 = 40 = 16 + 24, and 25 + 75 = 100 = 26 + 74. Among all pairs with a fixed sum, the product is largest when the numbers are closest together, and it shrinks as they spread apart. In (i) the pair 16, 24 is closer than 14, 26; in (ii) 26, 74 is closer than 25, 75.

   The algebra behind that: if the two numbers are m − d and m + d, their product is m² − d², which falls as d grows.

   ✦ (i) 16 × 24 is larger, by 20. (ii) 26 × 74 is larger, by 49. In each pair the sum is fixed, and the closer-together pair always has the bigger product since the product is m² − d².

10. 104 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 158

    A tiny park is coming up in Dhauli. Two square plots, each of area g² sq ft, will have a green cover. All the remaining area is a walking path w ft wide that needs to be tiled. Write an expression for the area that needs to be tiled.

    Hint. Find the outer dimensions of the whole park first, then subtract the two green squares.

    Step 1 — Find the length of the whole park. Reading across, the widths are: path (w), first green square (g), path between the squares (2w), second green square (g), path (w). Length = w + g + 2w + g + w = 2g + 4w ft

    Step 2 — Find the breadth. Reading down: path (w), green square (g), path (w). Breadth = w + g + w = g + 2w ft

    Step 3 — Find the total area. Total = (2g + 4w)(g + 2w) = 2g² + 4gw + 4gw + 8w² = 2g² + 8gw + 8w² sq ft

    Step 4 — Subtract the green cover. The two squares occupy 2g² sq ft, so Tiled area = (2g² + 8gw + 8w²) − 2g² = 8gw + 8w² = 8w(g + w) sq ft

    Step 5 — Sanity check the result. The answer contains no g² term, which is right — the green squares have been removed exactly. Every remaining term carries a factor of w, which is also right, since a path of zero width would need no tiling at all: putting w = 0 gives 0 ✓

    A numerical check. Take g = 10 and w = 2. Park: length 2(10) + 4(2) = 28, breadth 10 + 4 = 14, total area 392. Green: 2 × 100 = 200. Tiled = 192. Formula: 8(2)(10 + 2) = 16 × 12 = 192 ✓

    ✦ Tiled area = (2g + 4w)(g + 2w) − 2g² = 8w(g + w) sq ft.

11. 114 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 158

    For each pattern shown, (i) draw the next figure in the sequence, (ii) find how many basic units are in Step 10, and (iii) write an expression for the number of basic units in Step y. Pattern (a): 9, 16, 25 units in Steps 1, 2, 3. Pattern (b): 5, 11, 19 units in Steps 1, 2, 3.

    Hint. For (a) the counts are perfect squares. For (b), try subtracting the step number from each count.

    Pattern (a) — 9, 16, 25, …

    Step 1 — Recognise the counts. 9 = 3², 16 = 4², 25 = 5². The bases run 3, 4, 5, which is always 2 more than the step number.

    Step	Base	Units
    1	3	3² = 9
    2	4	4² = 16
    3	5	5² = 25
    4	6	6² = 36

    (i) The next figure is a 6 × 6 arrangement with 36 basic units. (ii) Step 10: (10 + 2)² = 12² = 144 (iii) Step y: (y + 2)²

    Pattern (b) — 5, 11, 19, …

    Step 1 — Look for the structure. The differences are 6 and 8, which are themselves growing, so this is not linear. Try peeling off the step number: 5 − 1 = 4 = 2², 11 − 2 = 9 = 3², 19 − 3 = 16 = 4². So each count is a square plus the step number, with the square's base one more than the step.

    Step	Structure	Units
    1	2² + 1	5
    2	3² + 2	11
    3	4² + 3	19
    4	5² + 4	29

    (i) The next figure has 29 basic units. (ii) Step 10: (10 + 1)² + 10 = 121 + 10 = 131 (iii) Step y: (y + 1)² + y

    Checking the general expressions. Putting y = 1, 2, 3 into (y + 1)² + y gives 4 + 1 = 5 ✓, 9 + 2 = 11 ✓, 16 + 3 = 19 ✓

    How to attack this kind of question. First test whether the counts are squares outright, as in (a). If not, take the differences; if those are still growing, subtract something simple (often the step number) and test again — which is exactly what unlocks (b).

    ✦ (a) Step 4 has 36, Step 10 has 144, and Step y has (y + 2)². (b) Step 4 has 29, Step 10 has 131, and Step y has (y + 1)² + y.

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.