Class 8 Maths Ganita Prakash Chapter 6 Solutions (NCERT 2026–27) – We Distribute, Yet Things Multiply

These Class 8 Maths Ganita Prakash Chapter 6 solutions cover We Distribute, Yet Things Multiply from the new NCF-2023 textbook (Part I, 2026–27). Use the worked exercise answers, original area diagrams and growing-pattern illustrations to understand distributivity and algebraic identities. The explanations also address missing cases and common sign errors.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 6 Exercises: Figure it Out (3 sets) Session: 2026–27
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Chapter 6 Overview

Chapter 6 of Ganita Prakash, We Distribute, Yet Things Multiply, uses the distributive property of multiplication over addition to explain a wide range of multiplication patterns through algebra. It begins with how a product changes when its factors are increased or decreased, builds the general identity (a + m)(b + n) = ab + mb + an + mn, and then derives the three special identities — (a + b)2, (a − b)2 and (a + b)(a − b). Along the way it shows quick methods for multiplying by 11, 101, 1001…, links the ideas to Brahmagupta and Sridharacharya, and explores number patterns geometrically. The Class 8 Maths Ganita Prakash Chapter 6 solutions below work through every exercise question in exact book order.

Key Concepts & Definitions

Distributive property: multiplying a number by a sum equals the sum of the separate products — a(b + c) = ab + ac. It also extends to more terms, e.g. (3a/2)(a − b + 1/5) = (3/2)a2 − (3/2)ab + (3/10)a.

Expanding: writing a product of bracketed expressions as a sum of terms by multiplying every term in one bracket by every term in the other.

Identity: a statement of equality between two algebraic expressions that is true for every value of the letter-numbers (e.g. (a + 1)(b − 1) = ab + b − a − 1).

Like terms: terms with exactly the same letter-numbers (same variables and powers); only like terms can be added into a single term.

Increase / decrease of a product: if a and b become a + m and b + n, the product changes by an + bm + mn (a negative change means a decrease); signs follow the rules of integer multiplication when m or n is negative.

Important Formulas & Identities (Chapter 6)

Distributive property: a(b + c) = ab + ac  •  (a + b)c = ac + bc

General product of two binomials (Identity 1): (a + m)(b + n) = ab + mb + an + mn, and in general (a + b)(c + d) = ac + ad + bc + bd

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

Identity 1B: (a − b)2 = a2 + b2 − 2ab

Identity 1C: (a + b)(a − b) = a2 − b2

Useful pattern results: 2(a2 + b2) = (a + b)2 + (a − b)2  •  a2 = (a + b)(a − b) + b2 (Sridharacharya’s method)

A rectangle of height a plus m and width b plus n is divided into areas ab, an, mb and mn.
The area picture uses positive lengths; the algebraic identity also holds for zero and negative values.
A square of side a plus b contains squares a squared and b squared and two rectangles of area ab.
Page 145: for a = 60 and b = 5, the four areas are 3600, 300, 300 and 25. Diagram is schematic.
In a square of side a, removing a bottom strip and a right strip each of width b leaves a square of side a minus b. Their overlapping corner b squared must be added back once.
Page 146–147: assumes a ≥ b ≥ 0 for the picture. For 55², calculate 3600 − 300 − 300 + 25 = 3025.
Remove a b by b corner from an a by a square, then rotate the remaining lower strip to form a rectangle of sides a plus b and a minus b.
Page 148: the two coloured pieces have unchanged total area. This geometric model assumes a > b > 0.

Figure it Out (Page 142–143)

1. Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.

SOLUTION The middle cell is the product pq, so its row factor is p and its column factor is q. Each neighbour changes p or q by 1.
(p − 1)(q − 1)(p − 1)q(p − 1)(q + 1)
p(q − 1)pqp(q + 1)
(p + 1)(q − 1)(p + 1)q(p + 1)(q + 1)
A three by three product grid has row factors p minus one, p, p plus one and column factors q minus one, q, q plus one. Its centre is pq.
Page 142, Question 1: multiply the row label by the column label to fill each cell.

2. Expand the following products. (i) (3 + u)(v − 3)    (ii) (2/3)(15 + 6a) (iii) (10a + b)(10c + d)    (iv) (3 − x)(x − 6) (v) (−5a + b)(c + d)    (vi) (5 + z)(y + 9)

SOLUTION (i) (3 + u)(v − 3) = 3v − 9 + uv − 3u = 3v − 3u + uv − 9. (ii) (2/3)(15 + 6a) = (2/3)×15 + (2/3)×6a = 10 + 4a. (iii) (10a + b)(10c + d) = 100ac + 10ad + 10bc + bd. (iv) (3 − x)(x − 6) = 3x − 18 − x2 + 6x = −x2 + 9x − 18. (v) (−5a + b)(c + d) = −5ac − 5ad + bc + bd. (vi) (5 + z)(y + 9) = 5y + 45 + yz + 9z = 5y + yz + 9z + 45.

3. Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.

SOLUTION Let the numbers be a and b. We need (a + 2)(b − 4) = ab. Expand: ab − 4a + 2b − 8 = ab ⇒ −4a + 2b − 8 = 0 ⇒ b = 2a + 4. So any a with b = 2a + 4 works. Three examples: (a = 1, b = 6), (a = 2, b = 8), (a = 3, b = 10). Check (a = 1, b = 6): ab = 6 and (1 + 2)(6 − 4) = 3 × 2 = 6. ✓

4. Expand (i) (a + ab − 3b2)(4 + b), and (ii) (4y + 7)(y + 11z − 3).

SOLUTION (i) = 4(a + ab − 3b2) + b(a + ab − 3b2) = 4a + 4ab − 12b2 + ab + ab2 − 3b3 = 4a + 5ab + ab2 − 12b2 − 3b3 (combining the like terms 4ab + ab = 5ab). (ii) = 4y(y + 11z − 3) + 7(y + 11z − 3) = 4y2 + 44yz − 12y + 7y + 77z − 21 = 4y2 + 44yz + 77z − 5y − 21 (combining −12y + 7y = −5y).

5. Expand (i) (a − b)(a + b), (ii) (a − b)(a2 + ab + b2) and (iii) (a − b)(a3 + a2b + ab2 + b3). Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?

SOLUTION (i) (a − b)(a + b) = a2 + ab − ab − b2 = a2 − b2. (ii) (a − b)(a2 + ab + b2) = a3 + a2b + ab2 − a2b − ab2 − b3 = a3 − b3. (iii) Similarly all middle terms cancel, giving a4 − b4. Pattern: (a − b) multiplied by such a chain of n terms gives an − bn. So the next identity is (a − b)(a4 + a3b + a2b2 + ab3 + b4) = a5 − b5, which can be checked by expanding (all intermediate terms cancel in pairs).

Math Talk & Try This (Sections 6.1–6.2)

These are the in-text reflective questions from the chapter. Each is solved with verified working.

[Page 138] Will the product always increase? Find 3 examples where the product decreases.

SOLUTION No. When a is increased by 1 and b is decreased by 1, the change is (a + 1)(b − 1) − ab = b − a − 1, which is negative exactly when b < a + 1 (equivalently a ≥ b for integers). Examples where the product decreases: (a = 10, b = 3): ab = 30 but 11 × 2 = 22; (a = 8, b = 5): 40 vs 9 × 4 = 36; (a = 6, b = 2): 12 vs 7 × 1 = 7. In each, the new product is smaller.

[Page 140] Use Identity 1 to find how the product changes when (i) one number is decreased by 2 and the other increased by 3; (ii) both numbers are decreased, one by 3 and the other by 4.

SOLUTION (i) (a − 2)(b + 3) = ab + 3a − 2b − 6, so the change is 3a − 2b − 6. (ii) (a − 3)(b − 4) = ab − 4a − 3b + 12, so the change is −4a − 3b + 12.

[Pages 143–144] Explain multiplication by 11, 101, 1001 and similar numbers. Evaluate the textbook examples and all six practice products.

SOLUTIONRule: N × (10k + 1) = N × 10k + N. Shift a copy of the number k places to the left and add the original. Add right to left with carries. For 11, this is the adjacent-digit sum rule, with a zero imagined at each end.3874 × 11 = 38740 + 3874 = 42614; 5678 × 11 = 56780 + 5678 = 62458.Page 144, ×11: (i) 94 × 11 = 1034; (ii) 495 × 11 = 5445; (iii) 3279 × 11 = 36069; (iv) 4791256 × 11 = 52703816.For ×101, shift two places: 387400 + 3874 = 391274. For ×1001 shift three places; for ×10001 shift four. Use subtraction for ×99 or ×999.(i) 89 × 101 = 8900 + 89 = 8989.(ii) 949 × 101 = 94900 + 949 = 95849.(iii) 265831 × 1001 = 265831000 + 265831 = 266096831.(iv) 1111 × 1001 = 1111000 + 1111 = 1112111.(v) 9734 × 99 = 973400 − 9734 = 963666.(vi) 23478 × 999 = 23478000 − 23478 = 23454522.
Examples multiply 3874 by 11, 101 and 999 using a shifted copy and addition or subtraction.
Pages 143–144: N(10ᵏ + 1) = N·10ᵏ + N, while N(10ᵏ − 1) = N·10ᵏ − N.

[Pages 145–147] Square identities: additional in-text questions.

SOLUTIONDoes (a + b)2 always exceed a2 + b2? Their difference is 2ab. It is positive when a and b have the same nonzero sign, zero if either is zero, and negative when they have opposite signs. Example: a = 2, b = −1 gives 1 < 5.1042 = (100 + 4)2 = 10000 + 800 + 16 = 10816; 372 = (40 − 3)2 = 1600 − 240 + 9 = 1369.(m + 3)2 = m2 + 6m + 9; (6 + p)2 = p2 + 12p + 36.(3j + 2k)2 = (3j)2 + 2(3j)(2k) + (2k)2 = 9j2 + 12jk + 4k2. Distributing directly gives 9j2 + 6jk + 6jk + 4k2, the same result.992 = (100 − 1)2 = 9801; 582 = (60 − 2)2 = 3364.For the two alternative decompositions of 65: 302 + 2×30×35 + 352 = 900 + 2100 + 1225 = 4225; 522 + 2×52×13 + 132 = 2704 + 1352 + 169 = 4225. Each uses the four-part square diagram above.

[Page 147, Try This] Expand the following using both Identity 1B and the distributive property: (i) (b − 6)2, (ii) (−2a + 3)2, (iii) (7y − (3/4)z)2.

SOLUTION(i) Identity: (b − 6)2 = b2 − 2(b)(6) + 36 = b2 − 12b + 36. Distribute: (b − 6)(b − 6) = b2 − 6b − 6b + 36.(ii) Identity 1B: (−2a + 3)2 = (3 − 2a)2 = 9 − 12a + 4a2. Distribute: (−2a + 3)(−2a + 3) = 4a2 − 6a − 6a + 9 = 4a2 − 12a + 9.(iii) Identity: (7y − 3z/4)2 = 49y2 − 2(7y)(3z/4) + 9z2/16. Distribute: 49y2 − 21yz/4 − 21yz/4 + 9z2/16 = 49y2 − 21yz/2 + 9z2/16.

[Page 148] Use Identity 1C to calculate 98 × 102 and 45 × 55.

SOLUTION 98 × 102 = (100 − 2)(100 + 2) = 1002 − 22 = 10000 − 4 = 9996. 45 × 55 = (50 − 5)(50 + 5) = 502 − 52 = 2500 − 25 = 2475.

Figure it Out (Page 149)

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

SOLUTION Since b − a = −(a − b), squaring removes the sign: (b − a)2 = (a − b)2. Both expand to a2 − 2ab + b2, so they are always equal (neither is greater).

2. Express 100 as the difference of two squares.

SOLUTIONUse a2 − b2 = (a + b)(a − b). Choose a + b = 50 and a − b = 2. Adding gives 2a = 52, so a = 26 and b = 24.100 = 262 − 242 = 676 − 576. If zero is allowed, 100 = 102 − 02 is another representation.

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

SOLUTION 4062 = (400 + 6)2 = 160000 + 4800 + 36 = 164836. 722 = (70 + 2)2 = 4900 + 280 + 4 = 5184. 1452 = (140 + 5)2 = 19600 + 1400 + 25 = 21025. 10972 = (1100 − 3)2 = 1210000 − 6600 + 9 = 1203409. 1242 = (120 + 4)2 = 14400 + 960 + 16 = 15376.

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

SOLUTION The patterns come from the identities 2(a2 + b2) = (a + b)2 + (a − b)2 and a2 − b2 = (a + b)(a − b), which were proved using the distributive property. The distributive property holds for all integers and all rational numbers, so the identities — and hence both patterns — hold for negative integers and fractions too. For example, a = −3, b = 2: 2(9 + 4) = 26 = (−1)2 + (−5)2 = 1 + 25 = 26. ✓

Mind the Mistake, Mend the Mistake (Page 150)

Below, each simplification is checked; if there is a mistake the reason is explained and the correct expression is given.

SOLUTIONS 1. −3p(−5p + 2q) → given “= −3p + 5p − 2q = p − 2q”. Wrong — the −3p was not multiplied into each term. Correct: −3p(−5p + 2q) = 15p2 − 6pq. 2. 2(x − 1) + 3(x + 4) → given “= 2x − 1 + 3x + 4 = 5x + 3”. Wrong — only the first terms were multiplied. Correct: 2x − 2 + 3x + 12 = 5x + 10. 3. y + 2(y + 2) → given “= (y + 2)2 = y2 + 4y + 4”. Wrong — y + 2(y + 2) is a sum, not a square. Correct: y + 2y + 4 = 3y + 4. 4. (5m + 6n)2 → given “= 25m2 + 36n2”. Wrong — the middle term 2ab is missing. Correct: 25m2 + 60mn + 36n2. 5. (−q + 2)2 → given “= q2 − 4q + 4”. Correct — (−q)2 + 2(−q)(2) + 4 = q2 − 4q + 4. ✓ 6. 3a(2b × 3c) → given “= 6ab × 9ac = 54a2bc”. Wrong — the bracket is a single product 6bc, not a sum. Correct: 3a × 6bc = 18abc. 7. (1/2)(10s − 6) + 3 → given “= 5s − 3 + 3 = 5s”. Correct. ✓ 8. 5w2 + 6w → given “= 11w2”. Wrong — 5w2 and 6w are not like terms. Correct: it stays 5w2 + 6w. 9. 2a3 + 3a3 + 6a2b + 6ab2 → given “= 5a3 + 12a2b2”. Wrong — 6a2b and 6ab2 are not like terms and cannot be merged. Correct: 5a3 + 6a2b + 6ab2. 10. (x + 2)(x + 5) → given working ends “= x2 + 7x + 10”. Correct. ✓ 11. (a + 2)(b + 4) → given “= ab + 8”. Wrong — cross terms are missing. Correct: ab + 4a + 2b + 8. 12. ab2 + a2b + a2b2 → given “= ab(a + b + ab)”. Correct — factor out ab: ab(b + a + ab). ✓

Patterns and shaded areas (Pages 150–154)

[Section 6.4] Use the formula k2 + 2k to find the number of circles in Step 15. Also write the number of square tiles in Step n.

SOLUTIONCircles: Step k has a k × k square plus two arms of k circles: k2 + 2k. Step 4 has 24, Step 10 has 120, and Step 15 has 255. Equivalent counts are (k + 1)2 − 1, k(k + 1) + k, and k(k + 2).Square-tile border: Steps 1, 2, 3 and 4 have 8, 12, 16 and 20 tiles. Step n = (n + 2)2 − n2 = 4n + 4. Alternatively, count the two full rows and the two remaining side strips: 2(n + 2) + 2n. Step 10 has 44 tiles.
Steps one to four have 3, 8, 15 and 24 circles, formed from a k by k square and two arms of k circles.
Pages 151–152: the orange circles are the extra arms. The missing bottom-right corner gives another way to count.
Square frames in steps one to four contain 8, 12, 16 and 20 tiles around holes of side one, two, three and four.
Pages 152–153: Step 10 has 4 × 10 + 4 = 44 tiles.

[Page 153–154] Verify (m + n)2 − 4mn = (n − m)2; and write the area of the dashed region (Substitute p = 6, r = 3.5, s = 9).

SOLUTIONInterior square: the outer area is (m + n)2. Removing four rectangles of area mn gives (m + n)2 − 4mn = m2 − 2mn + n2 = (n − m)2. This equals the inner side squared.
Four rectangles of sides m and n surround an inner square of side n minus m within an outer square of side m plus n.
Page 153: subtract four areas mn from the outer square, or square the inner side directly.
Dashed region beside the L: the required region is a rectangle with sides s − r and p − r. Its area is (s − r)(p − r). Another method is outer area minus two full strips plus their overlap: ps − pr − sr + r2.For p = 6, r = 3.5 and s = 9: (9 − 3.5)(6 − 3.5) = 5.5 × 2.5 = 13.75 square units. The orange L-shaped part is not the requested region.
An outer nine by six rectangle contains an L-shaped border of width 3.5. The remaining top-left rectangle has width 5.5 and height 2.5.
Page 154: green represents the book’s hatched region. The orange L-shaped part is excluded.

[Pages 153–154] Three equal rectangles: compare the three shaded-area methods and substitute x = 8, y = 3.

SOLUTIONEach bar has dimensions x by y. Method 1: start with the middle x × x square and subtract the vertical bar: x2 − xy.Method 2: subtract all three bars from the outer rectangle: x(x + 2y) − 3xy = x2 − xy.Method 3: add two equal gaps of width (x − y)/2 and height x: 2 × x(x − y)/2 = x(x − y).All three equal x2 − xy. At x = 8 and y = 3, area = 64 − 24 = 40 square units.
Two horizontal x by y bars and one vertical x by y bar form an I shape. The two shaded gaps have total area x times x minus y.
Pages 153–154: each gap is x units high and (x − y)/2 wide. Diagram uses x > y > 0.

Figure it Out (Page 154–156)

1. 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)

SOLUTION (i) 462 = (40 + 6)2 = 1600 + 480 + 36 = 2116. (ii) 397 × 403 = (400 − 3)(400 + 3) = 4002 − 32 = 160000 − 9 = 159991. (iii) 912 = (100 − 9)2 = 10000 − 1800 + 81 = 8281. (iv) 43 × 45 = (44 − 1)(44 + 1) = 442 − 12 = 1936 − 1 = 1935.

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

SOLUTION (i) (p − 1)(p + 11) = p2 + 11p − p − 11 = p2 + 10p − 11. (ii) (3a − 9b)(3a + 9b) = (3a)2 − (9b)2 = 9a2 − 81b2. (iii) −(2y + 5)(3y + 4) = −(6y2 + 8y + 15y + 20) = −(6y2 + 23y + 20) = −6y2 − 23y − 20. (iv) (6x + 5y)2 = (6x)2 + 2(6x)(5y) + (5y)2 = 36x2 + 60xy + 25y2. (v) (2x − 1/2)2 = (2x)2 − 2(2x)(1/2) + (1/2)2 = 4x2 − 2x + 1/4. (vi) (7p)(3r)(p + 2) = 21pr(p + 2) = 21p2r + 42pr.

3. For each statement identify the appropriate algebraic expression(s). (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

SOLUTION(i) The intended expression is s2 + 2: square an integer, then add 2.(ii) Three listed expressions qualify: m2 + (m + 1)2; m2 + (m − 1)2; and (2m)2 + (2m + 1)2. In each case the two numbers differ by 1. The last describes consecutive pairs starting with an even number.m2 + n2 does not guarantee consecutive numbers unless an extra relation between m and n is given. The square of a sum is different from the sum of two squares.

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

SOLUTION In a calendar, a 2×2 block is a, a + 1 (top) and a + 7, a + 8 (bottom). The two diagonal products are a(a + 8) = a2 + 8a and (a + 1)(a + 7) = a2 + 8a + 7. Their difference is (a + 1)(a + 7) − a(a + 8) = 7, always. The two diagonal products always differ by 7 (e.g. 5 × 11 = 55 and 4 × 12 = 48; 55 − 48 = 7), because the rows are 7 days apart.
A two by two calendar block has 4 and 5 above 11 and 12. A matching algebraic block has a, a plus one, a plus seven and a plus eight.
Page 155, Question 4: the upward diagonal product exceeds the downward diagonal product by 7.

5. 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.

SOLUTION (i) False. (k + 1)(k + 2) − (k + 3) = k2 + 3k + 2 − k − 3 = k2 + 2k − 1, which depends on k (for example, k = 2 gives 7). It equals 2 when k = 1 or k = −3. (ii) False. (2q + 1)(2q − 3) = 4q2 − 4q − 3; the −3 means it is not a multiple of 4 (e.g. q = 1 gives −3). (iii) True. Even number 2k: (2k)2 = 4k2, a multiple of 4. Odd number 2k + 1: (2k + 1)2 = 4k(k + 1) + 1; k(k + 1) is even, so 4k(k + 1) is a multiple of 8, making the square 1 more than a multiple of 8. (iv) False. (6n + 2)2 − (4n + 3)2 = (36n2 + 24n + 4) − (16n2 + 24n + 9) = 20n2 − 5 = 5(4n2 − 1); for n = 1 this is 15, and 15 + 5 = 20 is not an integer square. That counterexample disproves the claim about square numbers.

6. 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?

SOLUTIONLet A = 7a + 3 and B = 7b + 5.Sum: A + B = 7(a + b + 1) + 1, so remainder 1.Difference depends on order: A − B = 7(a − b) − 2 = 7(a − b − 1) + 5, so the first minus the second leaves 5. Reversing the order gives B − A = 7(b − a) + 2, remainder 2. A remainder is nonnegative; −2 is replaced by 5. If “difference” means the larger minus the smaller, either 2 or 5 is possible without knowing which number is larger.Product: AB = 49ab + 35a + 21b + 15 = 7(7ab + 5a + 3b + 2) + 1, so remainder 1.

7. 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? How do we write this as an algebraic equation? Expand both sides to check it is a true identity.

SOLUTION Let the three consecutive numbers be (n − 1), n, (n + 1). n2 − (n − 1)(n + 1) = n2 − (n2 − 1) = 1. So the result is always 1. As an identity: n2 − (n − 1)(n + 1) = 1, true for every n (e.g. 52 − 4 × 6 = 25 − 24 = 1).

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

SOLUTION Let the two numbers be a and b. Their sum is (a + b); half the sum is (1/2)(a + b). Result = (a + b) × (1/2)(a + b) = (1/2)(a + b)2. This is exactly half of the square of the sum, (1/2)(a + b)2. Proved.

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

SOLUTION (i) Write 14 × 26 = (16 − 2)(24 + 2) = 16 × 24 + 16(2) − 2(24) − 4 = (16 × 24) + (32 − 48 − 4) = (16 × 24) − 20. So 16 × 24 is larger (by 20). (ii) Write 25 × 75 = (26 − 1)(74 + 1) = 26 × 74 + 26 − 74 − 1 = (26 × 74) − 49. So 26 × 74 is larger (by 49).

10. A tiny park is coming up in Dhauli. The 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.

SOLUTION In the textbook layout, each outer margin is w and the middle gap contains two adjacent w-wide paths, giving width 2w. Length = w + g + 2w + g + w = (2g + 4w) ft. Breadth of the park = w + g + w = (g + 2w) ft. Total area = (2g + 4w)(g + 2w) = 2g2 + 8wg + 8w2 sq. ft. The two green squares cover 2g2 sq. ft. Area to be tiled = (2g + 4w)(g + 2w) − 2g2 = 8wg + 8w2 = 8w(w + g) sq. ft.
Two green squares of side g sit inside a rectangular walking path. Outer margins have width w and the central strip has width 2w.
Page 156, Question 10: this follows the pictured layout and NCERT solution, with two w-wide paths meeting in the middle. Schematic, not to scale.

11. For each pattern shown, (i) draw the next figure, (ii) find the number of basic units in Step 10, (iii) write an expression for the number of basic units in Step y.

SOLUTION Figure (a): Step 1 = (2 + 1)2 = 9, Step 2 = (2 + 2)2 = 16, Step 3 = (2 + 3)2 = 25; the next (Step 4) figure keeps the bent shape: a 4-by-6 middle block with a 6-tile arm above the right side and another below the left. Total = 4 × 6 + 6 + 6 = (2 + 4)2 = 36 units. In general, y(y + 2) + 2(y + 2) = (y + 2)2. (ii) Step 10 = (2 + 10)2 = 122 = 144. (iii) Step y = (y + 2)2. Figure (b): Step 1 = 22 + 1 = 5, Step 2 = 32 + 2 = 11, Step 3 = 42 + 3 = 19; the next (Step 4) = 52 + 4 = 29 units. (ii) Step 10 = 112 + 10 = 121 + 10 = 131. (iii) Step y = (y + 1)2 + y.
Four bent tile patterns have a middle rectangle y rows by y plus two columns, with an arm of y plus two tiles above its right side and below its left side. Counts are 9, 16, 25 and 36.
Page 156, Question 11(a): Step 4 is a bent 36-tile figure, not a filled 6 by 6 square.
Steps one to four contain blue squares with sides two, three, four and five units, plus orange bottom rows of one, two, three and four units.
Page 156, Question 11(b): count the square and the extra row separately.

Common Mistakes to Avoid

Watch out for these

  • Forgetting the middle term 2ab: writing (a + b)2 = a2 + b2 is wrong; it equals a2 + 2ab + b2.
  • Not distributing to every term — e.g. 2(x − 1) is 2x − 2, not 2x − 1.
  • Adding unlike terms: 5w2 + 6w is not 11w2; only like terms combine.
  • Mixing up a sum with a product inside brackets: 3a(2b × 3c) = 3a × 6bc = 18abc, not 6ab × 9ac.
  • Sign errors with negatives — treat (a − b) as (a + (−b)) and follow integer-multiplication rules.
  • In (a − b)2, the middle term is −2ab; do not make it +2ab.

Practice MCQs & Assertion–Reason

1. The expansion of (a + b)2 is:

(a) a2 + b2    (b) a2 + 2ab + b2    (c) a2 − 2ab + b2    (d) a2 − b2

2. (a + b)(a − b) equals:

(a) a2 + b2    (b) a2 − b2    (c) a2 + 2ab + b2    (d) 2ab

3. Using the identity, 1022 equals:

(a) 10404    (b) 10400    (c) 10204    (d) 10004

4. The product 99 × 101 is:

(a) 9999    (b) 10000    (c) 9899    (d) 10001

5. Which of these are like terms?

(a) 5w2 and 6w    (b) 2a2b and 3ab2    (c) 4xy and −7xy    (d) p2 and p3

6. The expansion of 3a(2a − 5) is:

(a) 6a2 − 5    (b) 6a2 − 15a    (c) 6a − 15a    (d) 5a2 − 15a

7. (x − 5)2 is equal to:

(a) x2 − 25    (b) x2 + 25    (c) x2 − 10x + 25    (d) x2 + 10x + 25

8. The value of 2(a2 + b2) using the chapter’s pattern equals:

(a) (a + b)2    (b) (a − b)2    (c) (a + b)2 + (a − b)2    (d) (a + b)(a − b)

9. To multiply 6453 by 11 quickly, you add:

(a) 6453 to itself    (b) 64530 + 6453    (c) 6453 × 10 only    (d) 6453 + 11

10. The number of circles at Step k in the pattern 3, 8, 15, … is:

(a) k2    (b) k2 + 2k    (c) 2k + 1    (d) k2 − 1

Answer key: 1-(b), 2-(b), 3-(a), 4-(a), 5-(c), 6-(b), 7-(c), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: (a + b)2 = a2 + 2ab + b2.

Reason: (a + b)2 = (a + b)(a + b), expanded using the distributive property.

A-R 2. Assertion: 45 × 55 = 2475.

Reason: 45 × 55 = (50 − 5)(50 + 5) = 502 − 52.

A-R 3. Assertion: 5w2 + 6w can be simplified to 11w2.

Reason: Only like terms (same letter-numbers) can be added into a single term.

A-R 4. Assertion: (a − b)2 and (b − a)2 are always equal.

Reason: Squaring a number and its negative gives the same value.

A-R 5. Assertion: n2 − (n − 1)(n + 1) is always 1.

Reason: (n − 1)(n + 1) = n2 − 1 by the identity (a + b)(a − b) = a2 − b2.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • Distributive property: a(b + c) = ab + ac; it extends to any number of terms and to all integers and fractions.
  • General product: (a + b)(c + d) = ac + ad + bc + bd; the product is the sum of each term of one bracket times each term of the other.
  • Identity 1A: (a + b)2 = a2 + 2ab + b2.
  • Identity 1B: (a − b)2 = a2 + b2 − 2ab.
  • Identity 1C: (a + b)(a − b) = a2 − b2.
  • Fast multiplication: ×11 (shift one place + add), ×101 (shift two places + add), ×1001 (shift three places + add), and ×99, ×999 via (100 − 1), (1000 − 1).
  • Patterns: 2(a2 + b2) = (a + b)2 + (a − b)2; a2 = (a + b)(a − b) + b2 (Sridharacharya).
  • Only like terms can be combined; always distribute to every term and keep the middle term in squares.

How to score full marks in this chapter

Memorise the three identities (1A, 1B, 1C) and recognise when a product fits one of them — this saves time on numbers like 1022, 98 × 102 or 397 × 403. Show one line of working even when using an identity, distribute carefully to every term, and combine only like terms. For “Mind the Mistake”-style questions, always state the error before writing the corrected expression.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 6 about?

Chapter 6, We Distribute, Yet Things Multiply, uses the distributive property to expand products of algebraic expressions and to derive the identities (a + b)2, (a − b)2 and (a + b)(a − b). It also covers fast multiplication tricks and number patterns explained through algebra.

What are the three key identities in this chapter?

(a + b)2 = a2 + 2ab + b2; (a − b)2 = a2 − 2ab + b2; and (a + b)(a − b) = a2 − b2. These are labelled Identity 1A, 1B and 1C in the textbook.

How many exercises does Ganita Prakash Chapter 6 have?

It has three “Figure it Out” exercise sets (on pages 142–143, 149, and 154–156), plus the “Mind the Mistake, Mend the Mistake” activity and several Math Talk / Try This tasks — with worked explanations below.

Are these Class 8 Maths Ganita Prakash Chapter 6 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for the 2026–27 session.

Textbook reference: NCERT Ganita Prakash, Chapter 6 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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