# NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.2

## Introduction:

In this exercise/article we will learn about multiplication of algebraic expression. Multiplying a monomials by a monomials and multiply two monomials and multiplying three or more monomials find out factor of monomials. [ using identity = volume of rectangle = Length × Breadth × Height ] only on using formula and multiply do in this exercise .

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities :

- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.1
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.2
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.3
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.4
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.5

**Class 8 Maths Exercise 9.2 (Page-143)**

**Q1. **Find the product of the following pairs of monomials.

**(i) **4 , 7p

**Solution :**

We have 4 , 7p

we know very well

= 4 × 7p

= 28p

So, answer is 28p

**(ii) **- 4p , 7p

**Solution :**

We have - 4p , 7p

we know very well

= - 4p × 7p

= - 28p^{2}

So, answer is - 28p^{2}

**(iii) **- 4p , 7pq

**Solution :**

We have ** **- 4p , 7pq

we know very well

= ** **- 4p × 7pq

= - 28p^{2}q

So, answer is - 28pq

**(iv)** 4p^{3} , - 3p

**Solution :**

We have 4p^{3} , - 3p

we know very well

= 4p^{3} × - 3p

= - 12p^{4}

So, answer is - 12p^{4}

**(v) **4p , o

**Solution :**

We have 4p , o

we know very well

= 4p × o

= 0

So, answer is 0

**Q2. **Find the area of rectangles with the following of monomials as their lengths and breadths respectively.

( p , q ) ; ( 10m , 5n ) ; ( 20x^{2} , 5y^{2} ) ; ( 4x , 3x^{2} ) ; ( 3mn , 4np )

**Solution :**

We have given ( p , q )

here, Length = p, breadth = q

Area of rectangle = length × breadth

= p × q

= pq

We have given ( 10m , 5n )

here, Length = 10m, breadth = 5n

Area of rectangle = length × breadth

= 10m × 5n

= 50mn

We have given ( 20x^{2} , 5y^{2} )

here, Length = 20x^{2}, breadth = 5y^{2}

Area of rectangle = length × breadth

= 20x^{2} × 5y^{2}

= 100x^{2}y^{2}

We have ( 4x , 3x^{2} )

here, Length = 4x, breadth = 3x^{2}

Area of rectangle = length × breadth

= 4x × 3x^{2}

= 12x^{3}

We have ( 3mn , 4np )

here, Length = 3mn, breadth = 4np

Area of rectangle = length × breadth

= 3mn × 4np

= 12mn^{2}p

## Class 8 Maths Chapter 9 Exercise 9.2

**Q3.** Complete the table products.

**Solution :**

Multiply the first monomial by the second monomials

**Q4. **Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

**(i) **5a, 3a^{2} , 7a^{4}

**Solution :**

We have 5a, 3a^{2} , 7a^{4}

here, Length = 5a, breadth = 3a^{2} and height = 7a^{4}

Volume of rectangle = length × breadth × height

= 5a × 3a^{2 } × 7a^{4}

= 105a^{7}

So, answer is 105a^{7}

**(ii) **2p, 4q, 8r

**Solution :**

We have 2p, 4q, 8r

here, Length = 2p, breadth = 4q and height = 8r

Volume of rectangle = length × breadth × height

= 2p × 4q × 8r

= 64pqr

So, answer is 64pqr

**(iii) **xy, 2x^{2}y, 2xy^{2}

**Solution :**

We have xy, 2x^{2}y, 2xy^{2}

here, Length = xy, breadth = 2x^{2}y and height = 2xy^{2}

Volume of rectangle = length × breadth × height

= xy × 2x^{2}y × 2xy^{2}

= 4x^{4}y^{4}

So, answer is 4x^{4}y^{4}

**(iv) **a, 2b, 3c

**Solution :**

We have a, 2b, 3c

here, Length = a, breadth = 2b and height = 3c

Volume of rectangle = length × breadth × height

= a × 2b × 3c

= 6abc

So, answer is 6abc

**Q5. **Obtain the product of

**(i) **xy, yz, zx

**Solution :**

We have xy, yz, zx

multiply all these

= xy × yz × zx

= x^{2}y^{2}z^{2}

So, answer is x^{2}y^{2}z^{2}

**(ii) **a, - a^{2} , a^{3}

**Solution :**

We have a, - a^{2} , a^{3}

multiply all these

= a × - a^{2} × a^{3}

= - a^{6}

So, answer is - a^{6}

**(iii) **2, 4y, 8y^{2 }, 16y^{3}

**Solution :**

We have 2, 4y, 8y^{2 }, 16y^{3}

multiply all these

= 2 × 4y × 8y^{2} × 16y^{3}

= 1024y^{6}

So, answer is 1024y^{6}

**(iv) **a, 2b, 3c, 6abc

We have a, 2b, 3c, 6abc

multiply all these

= a × 2b × 3c × 6abc

= 36a^{2}b^{2}c^{2}

= So, answer is 36a^{2}b^{2}c^{2}

**(v) **m, - mn, mnp

**Solution :**

We have m, - mn, mnp

multiply all these

= m × - mn × mnp

= - m^{3}n^{2}p

So, answer is - m^{3}n^{2}p .

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities :

- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.1
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.2
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.3
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.4
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.5