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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

2.0Detailed NCERT Class 8 Maths Chapter 6 Solutions of Exercise 6.2

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

3.0Case-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.

4.0Case-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.

5.0Case-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.

6.0Pattern 2

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

7.0Case 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

8.0Case 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.

9.0Case 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.

Very short answer type questions

1.Prove by expansion:

$(2 x + 3)^{2} - (2 x - 3)^{2} = 24 x$

2. Simplify: $(3 a + 4 b) (3 a - 4 b) + (a + b)^{2}$

3. If $(x + 5)^{2} - (x - 5)^{2} = k$ , find $k$ in terms of $x$ .

4. Expand and simplify: $(x + 1) (x + 2) (x + 3)$

5. Show that $(a + 3)^{2} - (a - 3)^{2} = 12 a$ .

Short answer type questions

6.Simplify: $(x + 7)^{2} - (x - 3)^{2} + (x + 2)^{2}$ .

7. Factorize and simplify: $4 x^{2} - 9 - (2 x - 3)^{2}$ .

8. Verify the identity: $(x + y)^{2} + (x - y)^{2} = 2 (x^{2} + y^{2})$ by substituting $x = 4, y = 5$ .

9. Given $a = 3, b = 4$ , verify $(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2})$ . 10. Expand and simplify: $(2 x + 3) (x - 4) + (x + 5)^{2}$

11. A square's side length increases by 8 meters and its area increases by $240 m^{2}$ . Find the original side length using identities.

12. Simplify the expression: $(3 x + 2)^{2} - (x - 4)^{2}$

13. Show that $(a + b) (a - b) + (b + c) (b - c) + (c + a) (c - a) = 0$ .

14. Simplfy: $(1.5 x - 4 y) (1.5 x + 4 y + 3) - 4.5 x + 12 y$

15. A rectangular lawn measures $(x + 5)$ meters by ( $x - 5$ ) meters. Find its area and express it in simplified form.

Long answer type questions

16. A square garden has a side of $x$ meters. $A$ path of uniform width 3 meters is built outside the garden. Express the area of the path using identities.

17. Using the distributive property, simplify and factorise the expression: $(2 x + 3) (x + 4) - (x + 3) (2 x + 5)$

18. Prove that $(a + b)^{2} - (a - b)^{2} = 4 ab$ using the distributive property.

19. Find the product of $(x + 4) (x^{2} - 4 x + 16)$ and simplify it using identities.

20. The length and breadth of a rectangle are $x + 3$ and $x - 3$ respectively. Find the difference between the squares of length and breadth.

21. Simplify: $(2 x + 1)^{2} - (2 x - 1)^{2} + (x + 1)^{2}$

22. Expand and simplify $(3 x - 2)^{2} - (x - 4)^{2}$ .

23. Using identities, prove that $(x + y)^{2} - (x - y)^{2} = 4 x y$

24. If $(x + 2)^{2} = 49$ , find the possible values of $x$ .

25. Factorize: $9 x^{2} - 6 x + 1$ using identities.

26. Find the value of k such that $(x + k)^{2} - (x - k)^{2} = 60$

27. Prove that the difference of squares $(a + b)^{2} - (a - b)^{2}$ is always divisible by 4 .

28. Expand and simplify $(x + 2) (x + 3) (x + 4)$ using distributive property stepwise.

29. If $(x + 1)^{2} + (x - 1)^{2} = 50$ , find the value of $x$ .

30. Simplify $(x + 5)^{2} - (x + 3)^{2} + (x + 1)^{2}$ and write the answer in terms of x .

10.0Direction (Q. 31 to Q.34):

A school is designing a rectangular playground. They want to tile two square sections inside the ground: The first section will have side length of x meters.

The second section will be 4 meters longer on each side than the first. To calculate the total number of tiles needed, they assume 1 tile per square meter.

Answer the following: 31. Write an expression for the total area of both square sections.

32. Simplify the expression using identities.

33. If the first square has side $x = 6$ meters, find the total number of tiles needed.

34. Show that the increase in area from the first to second section matches the identity $(a + b)^{2} - a^{2}$ .

35. Without actual multiplication, write the products of the following number in one line. (i) $132 \times 11$ (ii) $4376 \times 11$ (iii) $7315 \times 101$ (iv) $666 \times 99$ (v) $531 \times 999$ (vi) $637513 \times 1001$

ANSWER KEY

Very short answer type questions

$10 a^{2} + 2 ab - 15 b^{2}$
20 x
$x^{3} + 6 x^{2} + 11 x + 6$

Short answer type questions

$x^{2} + 24 x + 44$

7. $6 (2 x - 3)$

$3 x^{2} + 5 x + 13$
11 meters

12. $8 x^{2} + 20 x - 12$ $2.25 x^{2} - 16 y^{2}$

15. $(x^{2} - 25) m^{2}$

Long answer type questions

16. $12 x + 36$ 17. -3 19. $x^{3} + 4^{3}$ 20. 12x. 21. $x^{2} + 10 x + 1$ 22. $8 x^{2} - 4 x - 12$ 23. $4 x y$ 24. $x = 5$ or $x = - 9$ 25. $(3 x - 1)^{2}$ 26. $\frac{15}{x}$ 28. $x^{3} + 9 x^{2} + 26 x + 24$ 29. $\pm 26$ 30. $x^{2} + 6 x + 17$ 31. $x^{2} + (x + 4)^{2}$ 32. $2 x^{2} + 8 x + 16$ 33. 136 tiles 35. (i) 1452 (ii) 48136 (iii) 738815 (iv) 65934 (v) 530469 (vi) 638150513

11.0Key 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.

12.0NCERT 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

13.0Benefits 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

2.0Detailed NCERT Class 8 Maths Chapter 6 Solutions of Exercise 6.2

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

3.0Case-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.

4.0Case-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.

5.0Case-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.

6.0Pattern 2

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

7.0Case 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

8.0Case 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.

9.0Case 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.

Very short answer type questions

1.Prove by expansion:

$(2 x + 3)^{2} - (2 x - 3)^{2} = 24 x$

2. Simplify: $(3 a + 4 b) (3 a - 4 b) + (a + b)^{2}$

3. If $(x + 5)^{2} - (x - 5)^{2} = k$ , find $k$ in terms of $x$ .

4. Expand and simplify: $(x + 1) (x + 2) (x + 3)$

5. Show that $(a + 3)^{2} - (a - 3)^{2} = 12 a$ .

Short answer type questions

6.Simplify: $(x + 7)^{2} - (x - 3)^{2} + (x + 2)^{2}$ .

7. Factorize and simplify: $4 x^{2} - 9 - (2 x - 3)^{2}$ .

8. Verify the identity: $(x + y)^{2} + (x - y)^{2} = 2 (x^{2} + y^{2})$ by substituting $x = 4, y = 5$ .

9. Given $a = 3, b = 4$ , verify $(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2})$ . 10. Expand and simplify: $(2 x + 3) (x - 4) + (x + 5)^{2}$

11. A square's side length increases by 8 meters and its area increases by $240 m^{2}$ . Find the original side length using identities.

12. Simplify the expression: $(3 x + 2)^{2} - (x - 4)^{2}$

13. Show that $(a + b) (a - b) + (b + c) (b - c) + (c + a) (c - a) = 0$ .

14. Simplfy: $(1.5 x - 4 y) (1.5 x + 4 y + 3) - 4.5 x + 12 y$

15. A rectangular lawn measures $(x + 5)$ meters by ( $x - 5$ ) meters. Find its area and express it in simplified form.

Long answer type questions

16. A square garden has a side of $x$ meters. $A$ path of uniform width 3 meters is built outside the garden. Express the area of the path using identities.

17. Using the distributive property, simplify and factorise the expression: $(2 x + 3) (x + 4) - (x + 3) (2 x + 5)$

18. Prove that $(a + b)^{2} - (a - b)^{2} = 4 ab$ using the distributive property.

19. Find the product of $(x + 4) (x^{2} - 4 x + 16)$ and simplify it using identities.

20. The length and breadth of a rectangle are $x + 3$ and $x - 3$ respectively. Find the difference between the squares of length and breadth.

21. Simplify: $(2 x + 1)^{2} - (2 x - 1)^{2} + (x + 1)^{2}$

22. Expand and simplify $(3 x - 2)^{2} - (x - 4)^{2}$ .

23. Using identities, prove that $(x + y)^{2} - (x - y)^{2} = 4 x y$

24. If $(x + 2)^{2} = 49$ , find the possible values of $x$ .

25. Factorize: $9 x^{2} - 6 x + 1$ using identities.

26. Find the value of k such that $(x + k)^{2} - (x - k)^{2} = 60$

27. Prove that the difference of squares $(a + b)^{2} - (a - b)^{2}$ is always divisible by 4 .

28. Expand and simplify $(x + 2) (x + 3) (x + 4)$ using distributive property stepwise.

29. If $(x + 1)^{2} + (x - 1)^{2} = 50$ , find the value of $x$ .

30. Simplify $(x + 5)^{2} - (x + 3)^{2} + (x + 1)^{2}$ and write the answer in terms of x .

10.0Direction (Q. 31 to Q.34):

A school is designing a rectangular playground. They want to tile two square sections inside the ground: The first section will have side length of x meters.

The second section will be 4 meters longer on each side than the first. To calculate the total number of tiles needed, they assume 1 tile per square meter.

Answer the following: 31. Write an expression for the total area of both square sections.

32. Simplify the expression using identities.

33. If the first square has side $x = 6$ meters, find the total number of tiles needed.

34. Show that the increase in area from the first to second section matches the identity $(a + b)^{2} - a^{2}$ .

35. Without actual multiplication, write the products of the following number in one line. (i) $132 \times 11$ (ii) $4376 \times 11$ (iii) $7315 \times 101$ (iv) $666 \times 99$ (v) $531 \times 999$ (vi) $637513 \times 1001$

ANSWER KEY

Very short answer type questions

$10 a^{2} + 2 ab - 15 b^{2}$
20 x
$x^{3} + 6 x^{2} + 11 x + 6$

Short answer type questions

$x^{2} + 24 x + 44$

7. $6 (2 x - 3)$

$3 x^{2} + 5 x + 13$
11 meters

12. $8 x^{2} + 20 x - 12$ $2.25 x^{2} - 16 y^{2}$

15. $(x^{2} - 25) m^{2}$

Long answer type questions

16. $12 x + 36$ 17. -3 19. $x^{3} + 4^{3}$ 20. 12x. 21. $x^{2} + 10 x + 1$ 22. $8 x^{2} - 4 x - 12$ 23. $4 x y$ 24. $x = 5$ or $x = - 9$ 25. $(3 x - 1)^{2}$ 26. $\frac{15}{x}$ 28. $x^{3} + 9 x^{2} + 26 x + 24$ 29. $\pm 26$ 30. $x^{2} + 6 x + 17$ 31. $x^{2} + (x + 4)^{2}$ 32. $2 x^{2} + 8 x + 16$ 33. 136 tiles 35. (i) 1452 (ii) 48136 (iii) 738815 (iv) 65934 (v) 530469 (vi) 638150513

11.0Key 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.

12.0NCERT 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

13.0Benefits 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

How do I handle negative signs in multiplication?

Can I multiply a monomial with a number?

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 6.2

3.0Case-I

4.0Case-II

5.0Case-III

6.0Pattern 2

7.0Case I

8.0Case II

9.0Case III

Very short answer type questions

Short answer type questions

Long answer type questions

10.0Direction (Q. 31 to Q.34):

ANSWER KEY

Very short answer type questions

Short answer type questions

Long answer type questions

11.0Key Concepts of Chapter 6 We Distribute, Yet Things Multiply Exercise 6.2

12.0NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply : All Exercises

13.0Benefits of NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2

Frequently Asked Questions

How do I handle negative signs in multiplication?

Can I multiply a monomial with a number?

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 6.2

3.0Case-I

4.0Case-II

5.0Case-III

6.0Pattern 2

7.0Case I

8.0Case II

9.0Case III

Very short answer type questions

Short answer type questions

Long answer type questions

10.0Direction (Q. 31 to Q.34):

ANSWER KEY

Very short answer type questions

Short answer type questions

Long answer type questions

11.0Key Concepts of Chapter 6 We Distribute, Yet Things Multiply Exercise 6.2

12.0NCERT Solutions for Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply : All Exercises

13.0Benefits of NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2