Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2

Question 1.

Find the product of the following pairs of monomials?

- 4, 7p
- -4p, 7p
- -4p, 7pq
- 4p
^{3}, -3p - 4p, 0

Solution:

1. 4 and 7p

4 × 7p = (4 × 7) × p = 28p

2. -4p and 7p

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

3. -4p and 7pq

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

4. 4p^{3} and -3p

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

= -12p^{4}

5. 4p and 0 ⇒ 4p × 0 = 0

Question 2.

Find the areas of rectangles with the following pairs of monomlaLc as their lengths and breadths respectively?

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

Solution:

(i) Length = p

Breadth = q

∴ Area of the rectangle = q = p × q = pq

(ii) Length = 10m

Breadth = 5n

∴ Area = 10m × 5n

= 10 × 5 × m × n

= 50 mn

(iii) Length = 20x^{2}

Breadth = 5y^{2}

∴ Area = 20x^{2} × 5y^{2}

= 20 × 5 × x^{2} × y^{2}

= 100x^{2}y^{2}

(iv) Length = 4x

Breadth = 3x^{2}

∴ Area = 4x × 3 × x^{2}

= 4 × 3 × x × x^{2} = 12x^{3}

(v) Length = 3mn

Breadth = 4np

∴ Area = 3mn × 4np

= (3 × 4) × m × n × n × p = 12 mn^{2}p

Question 3.

Complete the table of products?

Solution:

2x × (-5y) = [2 × (-5)] × x × y = -10xy

2x × 3x^{2} = (2 × 3) × x × x^{2} = 6x^{3
}2x × (-4xy) = [2 × (-4)] × x × xy = -8x^{2}y

2x × 7x^{2}y = (2 × 7) × x × x^{2}y = 14x^{3}y

2x × (-9x^{2}y^{2}) = [2 × (-9)] × x × x^{2}y^{2} = -18x^{3}y^{2
}-5y × 2x = [-5 × 2] × y × x = -10xy

-5y × (-5y) = [-5 × 5)] × y × y = 25y^{2}

-5y × 3x^{2} = (-5 × 3) × y × x^{2} = -15x^{2}y

-5y × (-4xy) = [-5 × (-4)] × y × xy = 20 x^{2}y

-5y × 7x^{2}y = [-5 × 7] × y × x^{2}y = -35x^{2}y^{2}

-5y × (-9x^{2}y^{2}) = [-5 × (-9)] × y × x^{2}y^{2} = 45x^{2}y^{3}

3x^{2} × 2x = [3 × 2] × x^{2} × x = 6x^{3}

3x^{2} × (-5y) = [3 × (-5)] × x^{2} × y = -15x^{2}y

3x^{2} × 3x^{2} = [3 × 3] × x^{2} × x^{2} = 9x^{4}

3x^{2} × (-4xy) = [3 × (-4)] × x^{2} × xy = -12x^{3}y

3x^{2} × 7x^{2}y = [3 × 7] × x^{2} × x^{2}y

3x^{2} × (-9x^{2}y^{2}) = [3 × (-9)] × x^{2} × x^{2}y^{2} = -27x^{4}y^{2}

– 4xy × 2x = [-4 × 2] × xy × x = -8x^{2}y

– 4xy × (-5y) = [-4 × (-5)] × xy × y = 20xy^{2}

– 4xy × 3x^{2} = [-4 × 3] × xy × x^{2} = -12x^{3}y

– 4xy × 7x^{2}y = [-4 × 7] × xy × x^{2}y = -28x^{3}y^{2}

– 4xy × (-9x^{2}y^{2}) = [-4 × (-9)] × xy × x^{2}y^{2} = 36x^{3}y^{3
}7x^{2}y × 2x = [7 × 2] × x^{2}y × x = 14x^{3}y

7x^{2}y × (-5y) = [7 × (-5)] × x^{2}y × x = 14x^{3}y

7x^{2}y × (-5y) = [7 × (-5)] × x^{2}y × y = -35x^{2}y^{2}

7x^{2}y × 3x^{2} = [7 × 3] × x^{2}y × x^{2} = 21x^{4}y

7x^{2}y × (-4xy) = [7 × (-4)] × x^{2}y × xy = -28x^{3}y^{2}

7x^{2}y × 7x^{2}y = [7 × 7] × x^{2}y × x^{2}y = 49x^{4}y^{2}

7x^{2}y × -9x^{2}y^{2} = [7 × (-9)] × x^{2}y × x^{2}y = -63x^{4}y^{3}

– 9x^{2}y^{2} × 2x = [-9 × 2] × x^{2}y^{2} × x = -18x^{3}y^{2}

– 9x^{2}y^{2} × (-5y) = [-9 × (-5)] × x^{2}y^{2} × y = 45x^{2}y^{3}

– 9x^{2}y^{2} × 3x^{2} = [-9 × 3] × x^{2}y^{2} × x^{2} = -27x^{4}y^{2}

– 9x^{2}y^{2} × (-4xy) = [-9 × (-4)] × x^{2}y^{2} × xy = 36x^{3}y^{3}

– 9x^{2}y^{2} × 7x^{2}y = [-9 × 7] × x^{2}y^{2} × x^{2}y = -63x^{4}y^{3}

– 9x^{2}y^{2} × (- 9x^{2}y^{2}) = [-9 × (-9)] × x^{2}y^{2} × x^{2}y^{2} = 81x^{4}y^{4}

Question 4.

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

- 5a, 3a
^{2}, 7a^{4} - 2p, 4q, 8r
- xy, 2x
^{2}y, 2xy^{2} - a, 2b, 3c

Solution:

Volume of the rectangular box

= Length × Breadth × Height

1. Length = 5a; Breadth = 3a^{2}, Height

= 5a × 3a^{2} × 7a^{4}

= (5 × 3 × 7) × a × a^{2} × a^{4}

= 105 × a^{7} = 105a^{7}

2. Length = 2p, Breadth = 4q, Height = 8r

∴ Volume = Length × Breadth × Height

= 2p × 4q × 8r

= (2 × 4 × 8) × p × q × r

= 64 × pqr = 64pqr

3. Length = xy, Breadth = 2x^{2}y, Height = 2xy^{2}

∴ Volume = Length × Breadth × Height

= xy + 2x^{2}y × 2xy^{2}

= (1 × 2 × 2) × xy × x^{2}y × x^{2}y

= 4 × x^{4}y^{4} = 4x^{4}y^{4}

4. Length = a, Breadth = 2b, Height = 3c

∴ Volume = Length × Breadth × Height

= a × 2b × 3c

= (1 × 2 × 3) × a × b × c

= 6 × abc = 6abc

Question 5.

obtain the product of

- xy, yz, zx
- a, a -a
^{2}, a^{2} - 2, 4y, 8y
^{2}, 16y^{3} - a, 2b, 3c, 6abc
- m, -mn, mnp

Solution:

1. xy × yz × zx = (1 × 1 × 1) × x × y × y × z × z × x = 1 × (x^{2} × y^{2} × z^{2}) = x^{2}y^{2}z^{2}

2. a × (-a)^{2} × a^{3} = [1 × (-1) × 1] × a × a^{2} × a^{3}

= (-1) × a^{6} = -a^{6}

3. 2 × 4y × 8y^{2} × 16y^{3}

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

= 1024 × y^{6} = 1024y^{6}

4. a × 2b × 3c × 6abc

= (1 × 2 × 3 × 6) × a × b × c × abc

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

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