Class 8 Maths NCERT Solutions for Chapter – 9 Algebraic Expressions and Identities Ex – 9.2

Algebraic Expressions and Identities

Question 1.
Find the product of the following pairs of monomials :
(i) 4, 7p
(ii) -4p, 7p
(iii) – 4p, 7pq
(iv) 4p³ , – 3p
(v) 4p, 0

Solution:
(i) 4 x 7p = (4 x 7) x p = 28p
(ii) – 4 p x 7p = (- 4 x 7) x (px P)
= -28p¹ + ¹ = – 28p²
(iii) -4px 7pq = (- 4 x 7) x (p x p x q)
= -28 p¹⁺¹q = – 28p²q
(iv) 4p³ x – 3p = (4 x – 3) x (p³ x p)
= – 12p³⁺¹ = =-12p⁴
(v) 4p x 0 = (4 x 0) x p = 0 x p = 0

Question 2.
Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively :
(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)

Solution:
We know that the area of a rectangle = l x b, where l = length and b = breadth.
Therefore, the areas of rectangles with pair of monomials (p, q); (10m, 5n); (20x², 5y²); (4x, 3x²) and (3mn, 4np) as their lengths and breadths are given by
pxq=pq
10 m x 5n = (10 x 5) x (m x n) = 50 mn
20x² x 5y² = (20 x 5) x (x² x y²) = 100x²y²
4x x 3x² = (4 x 3) x (x x x²)
= 12x²
and, 3 mn x 4np = (3×4 )x(mxnxnxp)
=12 mn²p

Question 3.
Complete the table of products :

Solution:
Completed table is as under :

Question 4.
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:
(i) Required volume = 5a x 3a² x 7a⁴
= (5 x 3 x 7) x (a x a² x a⁴)
= 105a¹⁺²⁺⁴ =105a⁷

(ii) Required volume = 2p x 4q x 8r
= (2 x 4 x 8 )x p x q x r = 64 pqr

(iii) Required volume =xy x 2x²y x 2xy²
= (1 x 2 x 2) x (x x x² x x x y x y x y²)
= 4x¹⁺²⁺¹ y¹⁺¹⁺² = 4x⁴y⁴

(iv) Required volume =a x 2b x 3c
= (1 x 2 x 3) x (a x b x c)
= 6abc

Question 5.
Obtain the product of
(i) xy, yz, zx
(ii) a, – a², a³
(iii) 2, 4y, 8y², 16y³
(iv) a, 26, 3c, 6a6c
(v) m, – mn, mnp

Solution: