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

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

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

1. Find the product of the following pairs of monomials:

Question (i)
4, 7p
Solution:
= 4 × 7p
= 4 × 7 × p
= 28p

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

Question (ii)
– 4p, 7p
Solution:
= – 4p × 7p
= (-4 × 7) × p × p
= – 28p2

Question (iii)
– 4p, 7pq
Solution:
= (- 4p) × 7pq
= – 4 × 7 × p × pq
= – 28p2q

Question (iv)
4p3, – 3p
Solution:
= 4p3 × (- 3p)
= 4 × (- 3) × p3 × p
= – 12p4

Question (v)
4p, 0
Solution:
= 4p × 0
= 0

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

2. Find the areas of rectangles with the following pairs of monomials as their s lengths and breadths respectively:
(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)
Solution:
Area of a rectangle = length × breadth
(i) length = p, breadth = q
∴ Area = length × breadth
= p × q = pq unit2

(ii) length = 10m, breadth = 5n
∴ Area = length × breadth
= 10m × 5n
= 50mn unit2

(iii) length = 20x2, breadth = 5y2
∴ Area = length × breadth
= 20x2 × 5y2
= 100x2y2 unit2

(iv) length = 4x, breadth 3x2
∴ Area = length × breadth
= 4x × 3x2
= 12x3 unit2

(v) length = 3mn, breadth = 4np
∴ Area = length × breadth
= 3mn × 4np
= 12 mn2p unit2

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

3. Complete the table of products:
Solution:

First monomial → Second monomial ↓ 2x -5 y 3x2 -4xy 7x2y – 9x2y2
2x 4x2 -10xy 6x3 -8x2y 14x3y – 18x3y2
– 5 y – 10xy 25y2 – 15x2y 20xy2 – 35x2y2 45x2y3
3x2 6x3 – 15x2y 9x4 – 12x3y 21x4y – 27x4y2
– 4xy – 8x2y 20xy2 -12x3y 16x2y2 – 28x3y2 36x3y3
7x2y 14x3y – 35x2y2 21x4y – 28x3y2 49x4y2 – 63x4y3
– 9x2y2 – 18x3y2 45x2y3 – 27x4y2 36x3y3 – 63x4y3 81x4y4

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

Question (i)
5a, 3a2, 7a4
Solution:
Volume of rectangular box = length × breadth × height
length = 5a, breadth = 3a2, height = 7a4
∴ Volume = length × breadth × height
= 5a × 3a2 × 7a4
= (5 × 3 × 7) × a × a2 × a4
= 105a7 cubic unit

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

Question (ii)
2p, 4q, 8r
Solution:
length = 2p, breadth = 4q, height = 8r
∴ Volume = length × breadth × height
= 2p × 4q × 8r
= (2 × 4 × 8) × p × q × r
= 64pqr cubic unit

Question (iii)
xy, 2x2y, 2xy2
Solution:
length = xy, breadth = 2x2y, height = 2 xy2
∴ Volume = length × breadth × height
= xy × 2x2y × 2xy2
= (1 × 2 × 2) × xy × x2y × xy2
= 4x4y4 cubic unit

Question (iv)
a, 2b, 3c
Solution:
length = a, breadth = 2b, height = 3c
∴ Volume = length × breadth × height
= a × 2b × 3c
= (1 × 2 × 3) × a × b × c
= 6abc cubic unit

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

5. Obtain the product of:

Question (i)
xy, yz, zx
Solution:
= xy × yz × zx
= [x × y) × (y × z) × (z × x)
= x × x × y × y × z × z
= x2y2z2

Question (ii)
a, -a2, a3
Solution:
= (a) × (- a2) × (a3)
= – a × a2 × a3
= – a6

Question (iii)
2, 4y, 8y2, 16y3
Solution:
= 2 × 4y × 8y2 × 16y3
= 2 × 4 × 8 × 16 × y × y2 × y3
= 1024y6

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

Question (iv)
a, 2b, 3c, 6abc
Solution:
= a × 2b × 3c × 6abc
= 1 × 2 × 3 × 6 × a × b × c × abc
= 36 a2b2c2

Question (v)
m, – mn, mnp
Solution:
= (m) × (- mn) × (mnp)
= – (m × mn × mnp)
= – m3n2p

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