Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.2 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.2

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) 4p3, -3p = [(4) × (-3)] × (p3 × p)

= -12 p^{4}.

(v) 4p × 0 = (4 × 0) × p = 0 × 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); (20x^{2}, 5y^{2}); (4x, 3x^{2}); (3mn, 4np)

Solution:

Area of rectangle = l × b sq. unit

(i) If l = p, b = q

Area of rectangle = p × q= pq sq.\unit.

(ii) If l = 10m, b = 5n

Area of rectangle = 10m × bn

= 50mn sq. unit.

(iii) If l = 20x^{2}, b =-5y^{2}

Area of rectangle = 20x^{2} × 5y^{2}

= 100x^{2}y^{2} sq. unit

(iv) If l = 4x, b = 3x^{2}

Area of rectangle = 4x × 3x^{2} = 12x^{3}sq. unit

(v) If l = 3mn, b = 4np

Area of rectangle = 3mn × 4np

= 12mn^{2}p sq. unit.

Question 3.

Complete the table of products.

Solution:

Complete the table of products as shown in Table :

Question 4.

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

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

(ii) 2p, 4p, 8r,

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

(iv) a, 2b, 3c.

Solution:

Volume of cuboid = l × b × h

(i) l = 5a, b = 3a^{2}, h = 7a^{4}

Volume of rectangular box

= l × b × h cubic unit

= 5a × 3a^{2} × 7a^{4}

= (5 × 3 × 7) × (a × a2 × a4)

= 105a^{7} cubic unit.

(ii) l = 2p, b = 4q, h = 8r

v = l × b × h, = 2p × 4q × 8r

= (2 × 4 × 8) × (p × q × r)

= 64pqr.

(iii) l = xy, b = 2x^{2}y, h = 2xy^{2}

v = xy × 2x^{2}y × 2xy^{2}

= (1 × 2 × 2) × (xy × x^{2}y × xy^{2})

= 4x^{4}y^{4}.

(iv) l = a, b = 2b, h = 3c

v = l × b × h = a × 2b × 3c

= (1 × 2 × 3) (a × b × c) = 6abc.

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})

= 1024 y^{6}.

(iv) a × 2b × 3c × 6abc

= (2 × 3 × 6) x (a x b × c × abc)

= 36a^{2}b^{2}c^{2}.

(v) mx (-mn) × mnp = -m^{3}n^{2}p.