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NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.2

Mastering the expansion of algebraic terms is simple with our NCERT Solutions for Class 8 Maths Ch 6: We Distribute, Yet Things Multiply – Exercise 6.2. This exercise focuses on the multiplication of monomials. You will learn how to combine numerical coefficients and apply the laws of exponents to variables.

Class 8 Math Lessons (Ch 6): Learn to Multiply Algebraic Terms with Ease! The solutions to Exercise 6.2 will be presented in a step-by-step approach to help students find the product of two or more monomials and calculate the area of rectangles using algebraic dimensions. These solutions follow CBSE Guidelines to ensure accuracy in fundamental algebraic operations.

1.0Download NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply Ex 6.2 : Free PDF

Our easy-to-follow NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2 are available for download as a PDF. Perfect for offline study and quick reference.

NCERT Solutions Class 8 Maths Chapter 6 Ex 6.2

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2.0Detailed NCERT Class 8 Maths Chapter 6 Solutions of Exercise 6.1

1.Which is greater: $(a - b)^{2}$ or $(b - a)^{2}$ ? Justify your answer.

Sol. Here, $(a - b)^{2} = a^{2} + b^{2} - 2 ab$ and $(b - a)^{2} = b^{2} + a^{2} - 2 ab$ $b^{2} + a^{2} = a^{2} + b^{2}$ and $ba = ab$ $(b - a)^{2} = a^{2} + b^{2} - 2 ab$

Comparing (1) and (2), we get $(a - b)^{2} = (b - a)^{2}$

2. Express 100 as the difference of two squares.

Sol. $a^{2} - b^{2} = 100$ $(a + b) (a - b) = 100$ [100= $1 \times 100, 2 \times 50, 4 \times 25, 5 \times 20, 10 \times 10$ ] We can take anyone Let us take $50 \times 2 = 100$ Hence, $(a + b) (a - b) = 50 \times 2$ $a + b = 50$ $a - b = 2$

Adding (1) and (2) 2a = 52 $\Rightarrow a = 26$ Substituting a = 26 in (1) $26 + b = 50$ $\Rightarrow b = 50 - 26 = 24$ Let us check $2 6^{2} - 2 4^{2} = 676 - 576 = 100$ Hence, $2 6^{2} - 2 4^{2} = 100$

3. Find $40 6^{2}, 7 2^{2}, 14 5^{2}, 109 7^{2}$ , and $12 4^{2}$ using the identities you have learnt so far.

Sol. (i) $40 6^{2} = (400 + 6)^{2}$

$= 40 0^{2} + 2 \times 400 \times 6 + 6^{2} = 160000 + 4800 + 36 = 164836$

(ii) $7 2^{2} = (50 + 22)^{2}$

$= 5 0^{2} + 2 \times 50 \times 22 + 2 2^{2} = 2500 + 2200 + 484 = 5184$

(iii) $14 5^{2} = (150 - 5)^{2}$

$= 15 0^{2} - 2 \times 150 \times 5 + 5^{2} = 22500 - 1500 + 25 = 21025$

(iv) $109 7^{2} = (1100 - 3)^{2}$

$= 110 0^{2} - 2 \times 1100 \times 3 + 3^{2} = 1210000 - 6600 + 9 = 1203409$

(v) $12 4^{2} = (100 + 24)^{2}$

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.

Sol. $2 (a^{2} + b^{2}) = (a + b)^{2} + (a - b)^{2}$

Case-I

Let $a = 4, b = 2$ LHS $= 2 (4^{2} + 2^{2}) = 2 \times (16 + 4) = 40$ RHS $= (4 + 2)^{2} + (4 - 2)^{2} = 6^{2} + 2^{2} = 36 + 4 = 40$ $∴$ Pattern 1 holds for counting numbers.

Case-II

Let $a = - 4, b = - 2$ LHS $= 2 ((- 4)^{2} + (- 2)^{2})$ $= 2 \times (16 + 4)$ $= 2 \times 20$ $= 40$ RHS $= (- 4 + (- 2))^{2} + (- 4 - (- 2))^{2}$ $= (- 4 - 2)^{2} + (- 4 + 2)^{2}$ $= (- 6)^{2} + (- 2)^{2}$ $= 36 + 4$ $= 40$ LHS $=$ RHS $∴$ Pattern 1 holds for negative integers also.

Case-III

Let $a = \frac{1}{2}, b = \frac{1}{3}$

$LHS = 2 ((\frac{1}{2})^{2} + (\frac{1}{3})^{2}) = 2 (\frac{1}{4} + \frac{1}{9}) = 2 \times \frac{13}{36} = \frac{13}{18}$ $RHS = (\frac{1}{2} + \frac{1}{3})^{2} + (\frac{1}{2} - \frac{1}{3})^{2} = (\frac{5}{6})^{2} + (\frac{1}{6})^{2} = \frac{25}{36} + \frac{1}{36} = \frac{26}{36} = \frac{13}{18}$

The pattern holds for fractions also.

Pattern 2

$a^{2} - b^{2} = (a + b) (a - b)$

Case I

Let $a = 5, b = 3$ LHS $= 5^{2} - 3^{2} = 25 - 9 = 16$ RHS $= (5 + 3) (5 - 3) = 8 \times 2 = 16$ ∴ LHS $=$ RHS ∴ Pattern 2 holds for counting numbers

Case II

Let $a = - 5, b = - 3$ Now, LHS $= (- 5)^{2} - (- 3)^{2} = 25 - 9 = 16$ And RHS $= [(- 5) + (- 3)] [(- 5) - (- 3)]$ $= (- 5 - 3) (- 5 + 3)$ $= (- 8) (- 2)$ = 16 ∴ LHS $=$ RHS ∴ Pattern 2 hold for negative integers also.

Case III

Let $a = \frac{1}{2}, b = \frac{1}{3}$ LHS $= (\frac{1}{2})^{2} - (\frac{1}{3})^{2}$ $= \frac{1}{4} - \frac{1}{9}$ $= \frac{9 - 4}{36}$ $= \frac{5}{36}$ And RHS $= (\frac{1}{2} + \frac{1}{3}) (\frac{1}{2} - \frac{1}{3}) = (\frac{3 + 2}{6}) (\frac{3 - 2}{6}) = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ ∴ LHS $=$ RHS ∴ Pattern 2 holds for fractions also.

3.0Key Concepts of Chapter 6 We Distribute, Yet Things Multiply Exercise 6.2

Multiplying Two Monomials: To find the product, multiply the numerical coefficients together and the algebraic parts (variables) together.
Laws of Exponents: Remember that when multiplying variables with the same base, you add their powers: $x^{a} \times x^{b} = x^{a + b}$ .
Area as a Product: The area of a rectangle is Length x Breadth. If the dimensions are given as monomials, the area is their algebraic product.
Volume of a Rectangular Box: Volume is calculated as length x breadth x height. This involves finding the product of three monomials.

4.0NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply : All Exercises

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.1

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.2

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.3

5.0Benefits of NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2

Conceptual Clarity: Helps students understand that the order of multiplication (commutative property) does not change the result.
Pattern Recognition: Reinforces the use of exponents when the same variable appears multiple times in a product.
Application-Based: Connects algebra to geometry by solving for area and volume.

NCERT Class 8 Maths Ch. 6 We Distribute, Yet Things Multiply Other Exercises:-

Exercise 6.1

Exercise 6.2

Exercise 6.3

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 - A Square and a Cube

Chapter 2 - Power play

Chapter 3 - A story of Numbers

Chapter 4 - Quadrilaterals

Chapter 5 - Number Play

Chapter 6 - We distribute, yet things multiply

Chapter 7 - Proportional Reasoning

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NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.2

Mastering the expansion of algebraic terms is simple with our NCERT Solutions for Class 8 Maths Ch 6: We Distribute, Yet Things Multiply – Exercise 6.2. This exercise focuses on the multiplication of monomials. You will learn how to combine numerical coefficients and apply the laws of exponents to variables.

Class 8 Math Lessons (Ch 6): Learn to Multiply Algebraic Terms with Ease! The solutions to Exercise 6.2 will be presented in a step-by-step approach to help students find the product of two or more monomials and calculate the area of rectangles using algebraic dimensions. These solutions follow CBSE Guidelines to ensure accuracy in fundamental algebraic operations.

1.0Download NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply Ex 6.2 : Free PDF

Our easy-to-follow NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2 are available for download as a PDF. Perfect for offline study and quick reference.

NCERT Solutions Class 8 Maths Chapter 6 Ex 6.2

Download PDF

2.0Detailed NCERT Class 8 Maths Chapter 6 Solutions of Exercise 6.1

1.Which is greater: $(a - b)^{2}$ or $(b - a)^{2}$ ? Justify your answer.

Sol. Here, $(a - b)^{2} = a^{2} + b^{2} - 2 ab$ and $(b - a)^{2} = b^{2} + a^{2} - 2 ab$ $b^{2} + a^{2} = a^{2} + b^{2}$ and $ba = ab$ $(b - a)^{2} = a^{2} + b^{2} - 2 ab$

Comparing (1) and (2), we get $(a - b)^{2} = (b - a)^{2}$

2. Express 100 as the difference of two squares.

Sol. $a^{2} - b^{2} = 100$ $(a + b) (a - b) = 100$ [100= $1 \times 100, 2 \times 50, 4 \times 25, 5 \times 20, 10 \times 10$ ] We can take anyone Let us take $50 \times 2 = 100$ Hence, $(a + b) (a - b) = 50 \times 2$ $a + b = 50$ $a - b = 2$

Adding (1) and (2) 2a = 52 $\Rightarrow a = 26$ Substituting a = 26 in (1) $26 + b = 50$ $\Rightarrow b = 50 - 26 = 24$ Let us check $2 6^{2} - 2 4^{2} = 676 - 576 = 100$ Hence, $2 6^{2} - 2 4^{2} = 100$

3. Find $40 6^{2}, 7 2^{2}, 14 5^{2}, 109 7^{2}$ , and $12 4^{2}$ using the identities you have learnt so far.

Sol. (i) $40 6^{2} = (400 + 6)^{2}$

$= 40 0^{2} + 2 \times 400 \times 6 + 6^{2} = 160000 + 4800 + 36 = 164836$

(ii) $7 2^{2} = (50 + 22)^{2}$

$= 5 0^{2} + 2 \times 50 \times 22 + 2 2^{2} = 2500 + 2200 + 484 = 5184$

(iii) $14 5^{2} = (150 - 5)^{2}$

$= 15 0^{2} - 2 \times 150 \times 5 + 5^{2} = 22500 - 1500 + 25 = 21025$

(iv) $109 7^{2} = (1100 - 3)^{2}$

$= 110 0^{2} - 2 \times 1100 \times 3 + 3^{2} = 1210000 - 6600 + 9 = 1203409$

(v) $12 4^{2} = (100 + 24)^{2}$

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.

Sol. $2 (a^{2} + b^{2}) = (a + b)^{2} + (a - b)^{2}$

Case-I

Let $a = 4, b = 2$ LHS $= 2 (4^{2} + 2^{2}) = 2 \times (16 + 4) = 40$ RHS $= (4 + 2)^{2} + (4 - 2)^{2} = 6^{2} + 2^{2} = 36 + 4 = 40$ $∴$ Pattern 1 holds for counting numbers.

Case-II

Let $a = - 4, b = - 2$ LHS $= 2 ((- 4)^{2} + (- 2)^{2})$ $= 2 \times (16 + 4)$ $= 2 \times 20$ $= 40$ RHS $= (- 4 + (- 2))^{2} + (- 4 - (- 2))^{2}$ $= (- 4 - 2)^{2} + (- 4 + 2)^{2}$ $= (- 6)^{2} + (- 2)^{2}$ $= 36 + 4$ $= 40$ LHS $=$ RHS $∴$ Pattern 1 holds for negative integers also.

Case-III

Let $a = \frac{1}{2}, b = \frac{1}{3}$

$LHS = 2 ((\frac{1}{2})^{2} + (\frac{1}{3})^{2}) = 2 (\frac{1}{4} + \frac{1}{9}) = 2 \times \frac{13}{36} = \frac{13}{18}$ $RHS = (\frac{1}{2} + \frac{1}{3})^{2} + (\frac{1}{2} - \frac{1}{3})^{2} = (\frac{5}{6})^{2} + (\frac{1}{6})^{2} = \frac{25}{36} + \frac{1}{36} = \frac{26}{36} = \frac{13}{18}$

The pattern holds for fractions also.

Pattern 2

$a^{2} - b^{2} = (a + b) (a - b)$

Case I

Let $a = 5, b = 3$ LHS $= 5^{2} - 3^{2} = 25 - 9 = 16$ RHS $= (5 + 3) (5 - 3) = 8 \times 2 = 16$ ∴ LHS $=$ RHS ∴ Pattern 2 holds for counting numbers

Case II

Let $a = - 5, b = - 3$ Now, LHS $= (- 5)^{2} - (- 3)^{2} = 25 - 9 = 16$ And RHS $= [(- 5) + (- 3)] [(- 5) - (- 3)]$ $= (- 5 - 3) (- 5 + 3)$ $= (- 8) (- 2)$ = 16 ∴ LHS $=$ RHS ∴ Pattern 2 hold for negative integers also.

Case III

Let $a = \frac{1}{2}, b = \frac{1}{3}$ LHS $= (\frac{1}{2})^{2} - (\frac{1}{3})^{2}$ $= \frac{1}{4} - \frac{1}{9}$ $= \frac{9 - 4}{36}$ $= \frac{5}{36}$ And RHS $= (\frac{1}{2} + \frac{1}{3}) (\frac{1}{2} - \frac{1}{3}) = (\frac{3 + 2}{6}) (\frac{3 - 2}{6}) = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ ∴ LHS $=$ RHS ∴ Pattern 2 holds for fractions also.

3.0Key Concepts of Chapter 6 We Distribute, Yet Things Multiply Exercise 6.2

Multiplying Two Monomials: To find the product, multiply the numerical coefficients together and the algebraic parts (variables) together.
Laws of Exponents: Remember that when multiplying variables with the same base, you add their powers: $x^{a} \times x^{b} = x^{a + b}$ .
Area as a Product: The area of a rectangle is Length x Breadth. If the dimensions are given as monomials, the area is their algebraic product.
Volume of a Rectangular Box: Volume is calculated as length x breadth x height. This involves finding the product of three monomials.

4.0NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply : All Exercises

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.1

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.2

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.3

5.0Benefits of NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2

Conceptual Clarity: Helps students understand that the order of multiplication (commutative property) does not change the result.
Pattern Recognition: Reinforces the use of exponents when the same variable appears multiple times in a product.
Application-Based: Connects algebra to geometry by solving for area and volume.

Frequently Asked Questions

What is a variable?

How do I identify a coefficient if no number is written?

Why can we only add "Like Terms"?

Is x/2 a term with a coefficient?

Join ALLEN!

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.2

1.0Download NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply Ex 6.2 : Free PDF

2.0Detailed NCERT Class 8 Maths Chapter 6 Solutions of Exercise 6.1

Case-I

Case-II

Case-III

Pattern 2

Case I

Case II

Case III

3.0Key Concepts of Chapter 6 We Distribute, Yet Things Multiply Exercise 6.2

4.0NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply : All Exercises

5.0Benefits of NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2

Frequently Asked Questions

What is a variable?

How do I identify a coefficient if no number is written?

Why can we only add "Like Terms"?

Is x/2 a term with a coefficient?

Join ALLEN!

NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply - Exercise 6.2

1.0Download NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply Ex 6.2 : Free PDF

2.0Detailed NCERT Class 8 Maths Chapter 6 Solutions of Exercise 6.1

Case-I

Case-II

Case-III

Pattern 2

Case I

Case II

Case III

3.0Key Concepts of Chapter 6 We Distribute, Yet Things Multiply Exercise 6.2

4.0NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply : All Exercises

5.0Benefits of NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2