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

Ex 9.2 Class 8 Maths Question 1.

Find the product of the following pairs of monomials.

(i) 4, 7p

(ii) -4p, 7p

(iii) -4p, 7pq

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

(v) 4p, 0

Solution:

(i) 4 × 7p = (4 × 7) × p = 28p

(ii) -4p × 7p = (-4 × 7) × p × p = -28p^{2}

(iii) -4p × 7pq = (-4 × 7) × p × pq = -28p^{2}q

(iv) 4p^{3} × -3p = (4 × -3) × p^{3} × p = -12p^{4}

(v) 4p x 0 = (4 × 0) × p = 0 × p = 0

Ex 9.2 Class 8 Maths Question 2.

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

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

Solution:

(i) Length = p units and breadth = q units

Area of the rectangle = length × breadth = p × q = pq sq units

(ii) Length = 10 m units, breadth = 5n units

Area of the rectangle = length × breadth = 10 m × 5 n = (10 × 5) × m × n = 50 mn sq units

(iii) Length = 20x^{2} units, breadth = 5y^{2} units

Area of the rectangle = length × breadth = 20x^{2} × 5y^{2} = (20 × 5) × x^{2} × y^{2} = 100x^{2}y^{2} sq units

(iv) Length = 4x units, breadth = 3x^{2} units

Area of the rectangle = length × breadth = 4x × 3x^{2} = (4 × 3) × x × x^{2} = 12x^{3} sq units

(v) Length = 3mn units, breadth = 4np units

Area of the rectangle = length × breadth = 3mn × 4np = (3 × 4) × mn × np = 12mn^{2}p sq units

Ex 9.2 Class 8 Maths Question 3.

Complete the table of Products.

Solution:

Completed Table

Ex 9.2 Class 8 Maths Question 4.

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

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

(ii) 2p, 4q, 8r

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

(iv) a, 2b, 3c

Solution:

(i) Here, length = 5a, breadth = 3a^{2}, height = 7a^{4}

Volume of the box = l × b × h = 5a × 3a^{2} × 7a^{4} = 105 a^{7} cu. units

(ii) Here, length = 2p, breadth = 4q, height = 8r

Volume of the box = l × b × h = 2p × 4q × 8r = 64pqr cu. units

(iii) Here, length = xy, breadth = 2x^{2}y, height = 2xy^{2}

Volume of the box = l × b × h = xy × 2x^{2}y × 2xy^{2} = (1 × 2 × 2) × xy × x^{2}y × xy^{2} = 4x^{4}y^{4} cu. units

(iv) Here, length = a, breadth = 2b, height = 3c

Volume of the box = length × breadth × height = a × 2b × 3c = (1 × 2 × 3)abc = 6 abc cu. units

Ex 9.2 Class 8 Maths Question 5.

Obtain the product of

(i) xy, yz, zx

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

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

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

(v) m, -mn, mnp

Solution:

(i) xy × yz × zx = x^{2}y^{2}z^{2}

(ii) a × (-a^{2}) × a^{3} = -a^{6}

(iii) 2 × 4y × 8y^{2} × 16y^{3} = (2 × 4 × 8 × 16) × y × y^{2} × y^{3} = 1024y^{6}

(iv) a × 2b × 3c × 6abc = (1 × 2 × 3 × 6) × a × b × c × abc = 36 a^{2}b^{2}c^{2}

(v) m × (-mn) × mnp = [1 × (-1) × 1 ]m × mn × mnp = -m^{3}n^{2}p

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