GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2

   

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?

  1. 4, 7p
  2. -4p, 7p
  3. -4p, 7pq
  4. 4p3, -3p
  5. 4p, 0

Solution:
1. 4 and 7p
4 × 7p = (4 × 7) × p = 28p

2. -4p and 7p
-4p and 7p = {(-4 × 7) × p × p = -28p2

3. -4p and 7pq
-4p × 7pq = (-4 × 7) × p × pq = -28 × p2q = -28p2q

4. 4p3 and -3p
4p3 × (-3p) = (4 × (-3)}p3 × p = -12 × p4
= -12p4

5. 4p and 0 ⇒ 4p × 0 = 0

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2

Question 2.
Find the areas of rectangles with the following pairs of monomlaLc as their lengths and breadths respectively?
(p, q): (10m, 5n); (20x2, 5y2); (4x, 3x2); (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 = 20x2
Breadth = 5y2
∴ Area = 20x2 × 5y2
= 20 × 5 × x2 × y2
= 100x2y2

(iv) Length = 4x
Breadth = 3x2
∴ Area = 4x × 3 × x2
= 4 × 3 × x × x2 = 12x3

(v) Length = 3mn
Breadth = 4np
∴ Area = 3mn × 4np
= (3 × 4) × m × n × n × p = 12 mn2p

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2

Question 3.
Complete the table of products?
GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 img 1Solution:
2x × (-5y) = [2 × (-5)] × x × y = -10xy
2x × 3x2 = (2 × 3) × x × x2 = 6x3
2x × (-4xy) = [2 × (-4)] × x × xy = -8x2y
2x × 7x2y = (2 × 7) × x × x2y = 14x3y
2x × (-9x2y2) = [2 × (-9)] × x × x2y2 = -18x3y2
-5y × 2x = [-5 × 2] × y × x = -10xy
-5y × (-5y) = [-5 × 5)] × y × y = 25y2
-5y × 3x2 = (-5 × 3) × y × x2 = -15x2y
-5y × (-4xy) = [-5 × (-4)] × y × xy = 20 x2y
-5y × 7x2y = [-5 × 7] × y × x2y = -35x2y2
-5y × (-9x2y2) = [-5 × (-9)] × y × x2y2 = 45x2y3
3x2 × 2x = [3 × 2] × x2 × x = 6x3
3x2 × (-5y) = [3 × (-5)] × x2 × y = -15x2y
3x2 × 3x2 = [3 × 3] × x2 × x2 = 9x4
3x2 × (-4xy) = [3 × (-4)] × x2 × xy = -12x3y
3x2 × 7x2y = [3 × 7] × x2 × x2y
3x2 × (-9x2y2) = [3 × (-9)] × x2 × x2y2 = -27x4y2
– 4xy × 2x = [-4 × 2] × xy × x = -8x2y
– 4xy × (-5y) = [-4 × (-5)] × xy × y = 20xy2
– 4xy × 3x2 = [-4 × 3] × xy × x2 = -12x3y
– 4xy × 7x2y = [-4 × 7] × xy × x2y = -28x3y2
– 4xy × (-9x2y2) = [-4 × (-9)] × xy × x2y2 = 36x3y3
7x2y × 2x = [7 × 2] × x2y × x = 14x3y
7x2y × (-5y) = [7 × (-5)] × x2y × x = 14x3y
7x2y × (-5y) = [7 × (-5)] × x2y × y = -35x2y2
7x2y × 3x2 = [7 × 3] × x2y × x2 = 21x4y
7x2y × (-4xy) = [7 × (-4)] × x2y × xy = -28x3y2
7x2y × 7x2y = [7 × 7] × x2y × x2y = 49x4y2
7x2y × -9x2y2 = [7 × (-9)] × x2y × x2y = -63x4y3
– 9x2y2 × 2x = [-9 × 2] × x2y2 × x = -18x3y2
– 9x2y2 × (-5y) = [-9 × (-5)] × x2y2 × y = 45x2y3
– 9x2y2 × 3x2 = [-9 × 3] × x2y2 × x2 = -27x4y2
– 9x2y2 × (-4xy) = [-9 × (-4)] × x2y2 × xy = 36x3y3
– 9x2y2 × 7x2y = [-9 × 7] × x2y2 × x2y = -63x4y3
– 9x2y2 × (- 9x2y2) = [-9 × (-9)] × x2y2 × x2y2 = 81x4y4

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 img 3

Question 4.
Obtain the volume of rectangular boxes with the following length, breadth, height respectively?

  1. 5a, 3a2, 7a4
  2. 2p, 4q, 8r
  3. xy, 2x2y, 2xy2
  4. a, 2b, 3c

Solution:
Volume of the rectangular box
= Length × Breadth × Height

1. Length = 5a; Breadth = 3a2, Height
= 5a × 3a2 × 7a4
= (5 × 3 × 7) × a × a2 × a4
= 105 × a7 = 105a7

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 = 2x2y, Height = 2xy2
∴ Volume = Length × Breadth × Height
= xy + 2x2y × 2xy2
= (1 × 2 × 2) × xy × x2y × x2y
= 4 × x4y4 = 4x4y4

4. Length = a, Breadth = 2b, Height = 3c
∴ Volume = Length × Breadth × Height
= a × 2b × 3c
= (1 × 2 × 3) × a × b × c
= 6 × abc = 6abc

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2

Question 5.
obtain the product of

  1. xy, yz, zx
  2. a, a -a2, a2
  3. 2, 4y, 8y2, 16y3
  4. a, 2b, 3c, 6abc
  5. m, -mn, mnp

Solution:
1. xy × yz × zx = (1 × 1 × 1) × x × y × y × z × z × x = 1 × (x2 × y2 × z2) = x2y2z2

2. a × (-a)2 × a3 = [1 × (-1) × 1] × a × a2 × a3
= (-1) × a6 = -a6

3. 2 × 4y × 8y2 × 16y3
= (2 × 4 × 8 × 16) × y × y2 × y3
= 1024 × y6 = 1024y6

4. a × 2b × 3c × 6abc
= (1 × 2 × 3 × 6) × a × b × c × abc
= 36 × a2b2c2 = 36a2b2c2

5. m × (-mn) × mnp = [1 × (-1) × 1] × m × mn × mnp = (-1)m2n2p = -m3n2p

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