Exercise 9.2

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

Sol :

The product will be as follows.

(i) 4 × 7p = 4 × 7 × p = 28p

(ii) − 4p × 7p = − 4 × p × 7 × p = (− 4 × 7) × (p × p) = − 28 p2

(iii) − 4p × 7pq = − 4 × p × 7 × p × q = (− 4 × 7) × (p × p × q) = − 28p2q

(iv) 4p3 × − 3p = 4 × (− 3) × p × p × p × p = − 12 p4

(v) 4p × 0 = 4 × p × 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)

Sol :

We know that,

Area of rectangle = Length × Breadth

Area of 1st rectangle = p × q = pq

Area of 2nd rectangle = 10m × 5n = 10 × 5 × m × n = 50 mn

Area of 3rd rectangle = 20x2 × 5y2 = 20 × 5 × x2 × y2 = 100 x2y2

Area of 4th rectangle = 4x × 3x2 = 4 × 3 × x × x2 = 12x3

Area of 5th rectangle = 3mn × 4np = 3 × 4 × m × n × n × p = 12mn2p

Question 3

Complete the table of products.

 2x − 5y 3x2 − 4xy 7x2y − 9x2y2 2x 4x2 … … … … … − 5y … … − 15x2y … … … 3x2 … … … … … … − 4xy … … … … … … 7x2y … … … … … … − 9x2y2 … … … … … …

Sol :

The table can be completed as follows.

 2x − 5y 3x2 − 4xy 7x2y − 9x2y2 2x 4x2 − 10xy 6x3 − 8x2y 14x3y − 18x3y2 − 5y − 10xy 25 y2 − 15x2y 20xy2 − 35x2y2 45x2y3 3x2 6x3 − 15x2y 9x4 − 12x3y 21x4y − 27x4y2 − 4xy − 8x2y 20xy2 − 12x3y 16x2y2 − 28x3y2 36x3y3 7x2y 14x3y − 35x2y2 21x4y − 28x3y2 49x4y2 − 63x4y3 − 9x2y2 − 18x3y2 45 x2y3 − 27x4y2 36x3y3 − 63x4y3 81x4y4

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

Sol :

We know that,

Volume = Length × Breadth × Height

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

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

(iii) Volume = xy × 2x2y × 2xy2 = 2 × 2 × xy ×x2y × xy2 = 4x4y4

1 thought on “NCERT solution class 8 chapter 9 Algebraic Expression and Identities”

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