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Chapter 9 Algebraic Expressions and Identities Ex 9.2 Solutions

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

Answer - 1 : -

(i) 4 , 7 p =  4 × 7 × p = 28p

(ii) – 4p × 7p = (-4 × 7 ) × (p × p )= -28p2

(iii) – 4p × 7pq =(-4 × 7 ) (p × pq) =  -28p2q

(iv) 4p3 ×– 3p = (4 × -3 ) (p3 ×p ) =  -12p4

(v) 4p ×  0 = 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)

Answer - 2 : -

Area of rectangle = Length x breadth. So, it is multiplication of two monomials.
The results can be written in square units.
(i) p × q = pq
(ii)10m ×  5n = 50mn

(iii) 20x2 ×  5y2 =  100x2y2

(iv) 4x × 3x2 = 12x3

(v) 3mn ×  4np = 12mn2p

Question - 3 : - Complete the following table of products:

Answer - 3 : -



Question - 4 : -

Obtain the volumeof rectangular boxes with the following length, breadth and heightrespectively.

(i) 5a, 3a2, 7a4

(ii) 2p, 4q, 8r

(iii) xy, 2x2y, 2xy2

(iv) a, 2b, 3c

Answer - 4 : -

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

(i) 5a x 3a2 x 7a4 = (5 × 3 × 7) (a × a2 × a4 ) = 105a7

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

(iii) y × 2x2y × 2xy2 =(1 × 2 × 2 )( x × x2 × x × y × y × y2 ) =  4x4y4

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

Question - 5 : -

Obtain the productof

(i) xy,  yz, zx

(ii) a, – a2 , a3

(iii) 2, 4y, 8y2 , 16y3

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

(v) m, – mn, mnp

Answer - 5 : -

(i) xy × yz × zx = x2 y2 z2

(ii) a × – a2  ×a3 = – a6

(iii) 2 × 4y × 8y2 ×16y3 = 1024 y6

(iv) a × 2b × 3c × 6abc = 36a2 b2 c2

(v) m × – mn × mnp = –m3 n2 p

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