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

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

Question 3.
Complete the table of products?
Solution:
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

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

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