We Distribute yet things Multiply - Chapter 6 Class 8 (Ganita Prakash)

Master We Distribute yet things Multiply - Chapter 6 Class 8 (Ganita Prakash) with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.

NCERT Solutions

We Distribute yet things Multiply - Chapter 6 Class 8 (Ganita Prakash) – NCERT Solutions

Each question below opens its complete step-by-step Teachoo solution.

Figure it out - Page 142, 143

5 questions

Question 1

Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 Γ— 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.We follow these rules
First number is same in same row
Second number is same in same column

View solution

Question 2

(i) Expand the following products. (3 + u) (v – 3)Solving
(3+u)(π‘£βˆ’3)
=3(π‘£βˆ’3)+𝑒(π‘£βˆ’3)
=3 Γ— 𝑣+3 Γ— (βˆ’3) +𝑒 Γ— 𝑣+𝑒 Γ— (βˆ’3)
=3π‘£βˆ’9+π‘’π‘£βˆ’3𝑒
=πŸ‘π’—βˆ’πŸ‘π’–+π’–π’—βˆ’πŸ—
Question 2 (ii) Expand the following products. 2/3 "(15 + 6a)" Solving
2/3(15+6a)
=2/3 Γ— 15+2/3 Γ— 6π‘Ž
=2 Γ— 5+2 Γ— 2π‘Ž
=𝟏𝟎+πŸ’πš
Question 2 (iii) Expand the following products. (10a + b) (10c + d)Solving
(10a+b)(10𝑐+𝑑)
=10π‘Ž(10𝑐+𝑑)+𝑏(10𝑐+𝑑)
=10π‘Ž Γ— 10𝑐+10π‘Ž Γ— 𝑑+𝑏 Γ— 10𝑐+𝑏 Γ— 𝑑
=10 Γ— 10 Γ— π‘Žπ‘+10π‘Žπ‘‘+10𝑏𝑐+𝑏𝑑
=πŸπŸŽπŸŽπ’‚π’„+πŸπŸŽπ’‚π’…+πŸπŸŽπ’ƒπ’„+𝒃𝒅

View solution

Question 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. We need two numbers.
Let's call them π‘₯ and 𝑦.

View solution

Question 4

(i) Expand (i) (a + ab – 3b2) (4 + b)Solving
(π‘Ž+π‘Žπ‘βˆ’3𝑏^2 )(4+𝑏)
=π‘Ž(4+𝑏)+π‘Žπ‘(4+𝑏)βˆ’3𝑏^2 (4+𝑏)
=πŸ’π’‚+𝒂𝒃+πŸ’π’‚π’ƒ+𝒂𝒃 Γ— 𝒃 βˆ’πŸπŸπ’ƒ^πŸβˆ’πŸ‘π’ƒ^𝟐 Γ— 𝒃
=4π‘Ž+π‘Žπ‘+4π‘Žπ‘+π‘Žπ‘^2 βˆ’12𝑏^2βˆ’3𝑏^3
=4π‘Ž+(π‘Žπ‘+4π‘Žπ‘)+π‘Žπ‘^2 βˆ’12𝑏^2βˆ’3𝑏^3
=πŸ’π’‚+πŸ“π’‚π’ƒ+𝒂𝒃^𝟐 βˆ’πŸπŸπ’ƒ^πŸβˆ’πŸ‘π’ƒ^πŸ‘

View solution

Question 5

Expand (i) (a – b) (a + b), (ii) (a – b) (a2 + ab + b2 ), (iii) (a – b)(a3 + a2 b + 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?Let’s expand one-by-one

View solution

Figure it out - Page 149

4 questions

Question 1

Which is greater: (a – b)2 or (b – a)2? Justify your answer.Let’s take an example
For a = 5, b = 2
(a – b)2 = (5 – 2)2 = 32
(b – a)2 = (2 – 5)2 = (–3)2 = –3 Γ— –3 = 3 Γ— 3 = 32

View solution

Question 2

Express 100 as the difference of two squares.Difference of two squares means a2 – b2

View solution

Question 3

Find 4062, 722, 1452, 10972, and 1242 using the identities you have learnt so far.To find these, we use Sridharacharya’s identity
a2 = (a + b) Γ— (a – b) + b2

View solution

Question 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.Pattern 1 was
2 Γ— (a2 + b2) = (a + b)2 + (a – b)2
And, Pattern 2 was
a2 – b2 = (a + b) Γ— (a – b)

View solution

Figure it out - Page 154-156

12 questions

Question 1

Compute these products using the suggested identity. (i) 462 using Identity 1A for (a + b)2We can write
46 = (40 + 6)

View solution

Question 2

Use either a suitable identity or the distributive property to find each of the following products. (i) (p – 1) (p + 11)We will use distributive property for this
(π’‘βˆ’πŸ) (𝒑+𝟏𝟏)=𝑝(𝑝+11)βˆ’1(𝑝+11)
=𝑝^2+11π‘βˆ’π‘βˆ’11
=𝒑^𝟐+πŸπŸŽπ’‘βˆ’πŸπŸ

View solution

Question 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 22sNow,
A square number is s2
"Two more" means add 2

View solution

Question 4

Consider any 2 by 2 square of numbers in a calendar, as shown in the figure. Find products of numbers lying along each diagonal β€” 4 Γ— 12 = 48, 5 Γ— 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens. Hint: Label the numbers in each 2 by 2 square asOur hint shows a general 2 Γ— 2 grid

View solution

Question 5

Verify which of the following statements are true. (i) (k + 1) (k + 2) – (k + 3) is always 2.Solving
(π‘˜+1)(π‘˜+2)βˆ’(π‘˜+3)
= [π’Œ(π’Œ+𝟐)+𝟏(π’Œ+𝟐)] βˆ’(π‘˜+3)
= [π‘˜^2+2π‘˜+π‘˜+2] βˆ’(π‘˜+3)
= [π’Œ^𝟐+πŸ‘π’Œ+𝟐] βˆ’(π’Œ+πŸ‘)
= π‘˜^2+3π‘˜+2βˆ’π‘˜βˆ’3
= π‘˜^2+(3π‘˜βˆ’π‘˜)+(2βˆ’3)
= π’Œ^𝟐+πŸπ’Œβˆ’πŸ

View solution

Question 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?Let number A have remainder 3 when divided by 7
A = 7x + 3
Let number B have remainder 5 when divided by 7
B = 7y + 5

View solution

Question 7

Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.Three consecutive numbers are
𝑛, 𝑛+1, 𝑛+2

View solution

Question 8

What is the algebraic expression describing the following steps β€” 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.Let the two numbers be π‘Ž & 𝑏
Now, we are asked to
Add any two numbers: 𝒂+𝒃
And, Multiply this by half of the sum of the two numbers
So,
(𝒂+𝒃)Γ— ((𝒂 + 𝒃)/𝟐)=((π‘Ž + 𝑏) Γ— (π‘Ž + 𝑏))/2
=𝟏/𝟐 Γ—(𝒂 + 𝒃)^𝟐
This is half of the square of the sum of the two numbers.
Hence proved

View solution

Question 9

Which is larger? Find out without fully computing the product. (i) 14 Γ— 26 or 16 Γ— 24Now,
14 Γ— 26 = (20 – 6) Γ— (20 + 6)
Using (a – b) (a + b) = a2 – b2
= 202 – 62

View solution

Question 10

A tiny park is coming up in Dhauli. The plan is shown in the figure. 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.Now,
Area to be tiled
= Area of bigger rectangle – 2 Γ— Area of square plots

View solution

Question 11

– Pattern 1 For each pattern shown below, (i) Draw the next figure in the sequence. (ii) How many basic units are there in Step 10? (iii) Write an expression to describe the number of basic units in Step y.Analysing the pattern
In Step 1
Two vertical strips of 3 units on sides
One horizontal strip of 3 units
Total = 9 squares

View solution

Question 11

– Pattern 2 For each pattern shown below, (i) Draw the next figure in the sequence. (ii) How many basic units are there in Step 10? (iii) Write an expression to describe the number of basic units in Step y.Analysing the pattern
Step 1:
It is a square of 2 Γ— 2
With one additional unit squares
Total squares = 2 Γ— 2 + 1 = 5

View solution

Why Learn This With Teachoo?

We Distribute yet Things Multiply is Chapter 6 of NCERT Class 8 Ganita Prakash Part 1. It develops the distributive property from arithmetic into algebra, uses it for fast multiplication and derives identities for the square of a sum or difference. Students investigate patterns, correct common mistakes and apply algebraic expressions to shaded-area problems. Teachoo explains every transformation and provides step-by-step solutions for the chapter’s Figure it out questions.

Understanding the distributive property

The distributive property connects multiplication with addition and subtraction. It states that a(b + c) = ab + ac and a(b − c) = ab − ac. Each term inside the bracket must be multiplied by the outside factor. The reverse process, ab + ac = a(b + c), extracts a common factor.

This property supports fast mental multiplication. A difficult factor can be split into convenient parts: 37 × 99 may be viewed as 37(100 − 1). The result follows through distribution without a long multiplication algorithm. Students compare methods and decide which decomposition is efficient.

Squares of sums and differences

Multiplying (a + b)(a + b) and distributing carefully produces (a + b)² = a² + 2ab + b². Similarly, (a − b)² = a² − 2ab + b². Area models make the terms visible: a large square can be decomposed into smaller squares and rectangles.

Investigating Patterns encourages students to derive and test identities rather than accept them as unexplained formulas. Mind the Mistake, Mend the Mistake focuses on errors such as writing (a + b)² as a² + b² and forgetting the middle term.

“This Way or That Way, All Ways Lead to the Bay” compares equivalent routes to the same expression. Area of Shaded Region questions translate a geometric diagram into algebra and may be solved either through subtraction of areas or decomposition into parts. Equivalent correct methods should give the same simplified result.

Topics covered on Teachoo

Teachoo covers:

  • distributive property over addition and subtraction;

  • Figure it out solutions for pages 142–143;

  • fast multiplication using distribution;

  • square of a sum and square of a difference;

  • investigating algebraic patterns;

  • Figure it out solutions for page 149;

  • Mind the Mistake, Mend the Mistake;

  • comparing equivalent solution methods;

  • areas of shaded regions; and

  • Figure it out solutions for pages 154–156.

Learning outcomes

Students should be able to expand and factor simple expressions using distribution, choose a convenient split for mental multiplication and derive the square identities. They should detect a missing or incorrect term, show that two methods are equivalent and form an algebraic expression for a shaded region. They should also check an identity numerically without mistaking numerical evidence for a proof.

Why is this chapter important?

Distribution is central to algebraic simplification, equation solving, factorisation and polynomial multiplication. Identities later support quadratic expressions, coordinate geometry and numerical approximation. Area models connect symbolic algebra with geometric meaning, reducing dependence on memorised rules.

How Teachoo helps you study

Teachoo divides the chapter into property, identity, error-analysis and application sections. When expanding, draw arrows from the outside factor to every term. When squaring a binomial, first write it as multiplication of two identical brackets; then distribute. This prevents the common loss of the 2ab term.

Attempt shaded-region questions in two ways whenever possible. Label dimensions, write the area of each relevant shape and simplify only after the geometric relationship is clear. Use Teachoo’s solution to check whether your expression and final simplification are both valid.

Common mistakes to avoid

Never distribute to only the first term. Keep subtraction signs attached to the following term. Remember that (a − b)² contains +b², while the middle term is −2ab. Do not combine unlike terms, and do not assume expressions are equivalent merely because they match for one chosen value.

Quick revision checklist

Expand and factor several expressions in both directions. Derive the two square identities from bracket multiplication and also from an area diagram. Use one identity for a fast numerical square, then verify it conventionally. Complete an error-analysis example in which a middle term is missing. Finally, solve a shaded-area question by two decompositions and confirm that the simplified expressions agree.

Deeper reasoning and concept connections

The strongest way to learn We Distribute yet Things Multiply (Ganita Prakash) is to separate three layers: the object being studied, the rule that describes it and the reason the rule works. A correct numerical result is useful, but a complete mathematical answer also explains the relationship used. Students should compare examples and non-examples, change one condition at a time and observe whether the conclusion still holds.

This chapter is part of a longer progression. Its vocabulary and representations will appear again in algebra, geometry, data, measurement or higher problem-solving. Build links deliberately: translate pictures into statements, statements into operations and operations back into a sensible interpretation. If the final result cannot be explained in ordinary language, the method has probably been followed mechanically rather than understood.

How to solve unfamiliar and competency-based questions

When a question looks new, do not search memory for an identical example. Classify it. Decide whether it asks for recognition, calculation, representation, comparison, explanation or proof. Write the relevant definition or property first. Next, organise the data and select the shortest valid method. This converts an unfamiliar surface story into a familiar mathematical structure.

Use estimation and special cases as quality checks. Test zero, one, equal values, endpoints or a simple symmetric figure whenever they are permitted. A result that violates the diagram, scale, sign, unit or expected range is a signal to recheck the setup. In multi-part cases, carry forward only verified results so one early error does not silently contaminate every later answer.

What complete mastery looks like

For We Distribute yet Things Multiply (Ganita Prakash), a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.

Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.

Additional frequently asked questions

What should a student know before starting We Distribute yet Things Multiply (Ganita Prakash)?

Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.

How can a student check an answer in We Distribute yet Things Multiply (Ganita Prakash)?

Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.

How many questions are enough for strong preparation?

There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.

How should Teachoo solutions be used without becoming dependent on them?

Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.

Frequently asked questions

What is the distributive property?

It is the rule a(b + c) = ab + ac, with a corresponding form for subtraction.

Why is (a + b)² not equal to a² + b²?

Because multiplying (a + b)(a + b) also creates two ab terms, giving a² + 2ab + b².

How does the chapter connect algebra and geometry?

Area decompositions illustrate identities, while shaded-region dimensions are represented and simplified algebraically.

Understand distribution as one idea working in arithmetic, algebra and area. That connection makes later algebra far easier.