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

NCERT exercise solutions are helpful to improve your algebraic expressions concepts and problem-solving skills. Students will learn how to multiply two monomials as well as algebraic expressions identities. All the exercise questions (5 questions along with their sub questions) have been solved by subject experts. CBSE NCERT Solutions for Class 8 Chapter 9 Algebraic Expressions and Identities Exercise 9.2 questions and answers help students to improve their problem solving skills. Download free Maths NCERT Solutions for Chapter 9 and get going with your homework or prepare for exams.

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## Exercise 9.2 Page No: 143

1.Â Find the product of the following pairs of monomials.
(i) 4, 7p
(ii) â€“ 4p, 7p
(iii) â€“ 4p, 7pq
(iv)Â  4p3, â€“ 3p
(v) 4p, 0

Solution:

(i) 4 , 7 pÂ =Â  4 Ã— 7p = 28p

(ii) â€“ 4pÂ Ã—Â 7pÂ =Â (-4 Ã— 7 ) Ã— (p Ã— p )= -28p2

(iii) â€“ 4pÂ Ã—Â 7pqÂ =(-4 Ã— 7 ) (p Ã— pq) = Â -28p2q

(iv) 4p3Â Ã— â€“ 3pÂ = (4 Ã— -3 ) (p3 Ã— p ) = Â -12p4

(v) 4pÂ Ã— Â 0Â =Â 0

2.Â Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q) ; (10m, 5n) ; (20x2Â , 5y2) ; (4x, 3x2) ; (3mn, 4np)

Solution:

Area of rectangle = Length x breadth. So, it is multiplication of two monomials.

The results can be written in square units.

(i) pÂ Ã—Â qÂ =Â pq

(ii)10mÂ Ã—Â Â 5nÂ =Â 50mn

(iii)Â 20x2Â Ã—Â Â 5y2Â =Â Â 100x2y2

(iv)Â 4xÂ Ã—Â 3x2Â =Â 12x3

(v) 3mnÂ Ã—Â Â 4npÂ =Â 12mn2p

3. Complete the following table of products:

Solution:

4.Â Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) 5a, 3a2, 7a4

(ii) 2p, 4q, 8r

(iii) xy, 2x2y, 2xy2

(iv) a, 2b, 3c

Solution:

Volume of rectangle = length x Â breadth x Â height. To evaluate volume of rectangular boxes, multiply all the monomials.

(i) 5aÂ xÂ 3a2Â xÂ 7a4Â =Â (5 Ã— 3 Ã— 7) (a Ã— a^2 Ã— a^4 ) = 105a7

(ii) 2pÂ xÂ 4qÂ xÂ 8rÂ =Â (2 Ã— 4 Ã— 8 ) (p Ã— q Ã— r ) = 64pqr

(iii) yÂ Ã—Â 2x2y Ã—Â 2xy2Â =(1 Ã— 2 Ã— 2 )( x Ã— x^2 Ã— x Ã— y Ã— y Ã— y^2 ) = Â 4x4y4

(iv) a x Â 2b x 3c = (1 Ã— 2 Ã— 3 ) (a Ã— b Ã— c) = 6abc

5.Â Obtain the product of

(i) xy, Â yz, zx

(ii) a, â€“ a2Â , a3

(iii) 2, 4y, 8y2Â , 16y3

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

(v) m, â€“ mn, mnp

Solution:

(i) xy Ã— yz Ã— zx = x2 y2 z2

(ii) a Ã— â€“ a2Â  Ã— a3 = â€“ a6

(iii) 2 Ã— 4y Ã— 8y2Â Ã— 16y3 = 1024 y6

(iv) a Ã— 2b Ã— 3c Ã— 6abc = 36a2 b2 c2

(v) m Ã— â€“ mn Ã— mnp = â€“m3 n2 p

### Access Other Exercise Solutions of Class 8 Maths Chapter 9 Algebraic Expressions and Identities

Exercise 9.1 Solutions : 4 Questions (Short answers)

Exercise 9.3 Solutions : 5 Questions (Short answers)

Exercise 9.4 Solutions : 3 Questions (Short answers)

Exercise 9.5 Solutions : 8 Questions (Short answers)

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

NCERT Class 8 Chapter 9 exercise 9.2, this exercise is based on concepts like multiplication of algebraic expressions, multiplying a monomial by a monomial, multiplying three or more monomials. Using different situations students are taught how two algebraic expressions have to be multiplied. Problems on the area of rectangular figures, the volume of rectangular boxes are prepared using tables so that students can easily identify the given length, breadth and height of the figures and operate using relevant formulas.

#### 1 Comment

1. wow this is great thanks for all the question and answer