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

NCERT Solutions Class 8 Mathematics 
Chapter – 9 (Algebraic Expressions and Identities) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 9 Algebraic Expressions and Identities Exercise 9.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter 9: Algebraic Expressions and Identities

• NCERT Solution Class 8 Maths Ex – 9.1
• NCERT Solution Class 8 Maths Ex – 9.3
• NCERT Solution Class 8 Maths Ex – 9.4
• NCERT Solution Class 8 Maths Ex – 9.5

Exercise – 9.2 

1. Find the product of the following pairs of monomials.
(i) 4, 7p
(ii) – 4p, 7p
(iii) – 4p, 7pq
(iv) 4p³, – 3p
(v) 4p, 0

Solution –

(i) 4 , 7 p
=  4 × 7 × p = 28p

(ii) – 4p × 7p
= (-4 × 7 ) × (p × p )
= – 28p²

(iii) – 4p × 7pq
= (-4 × 7 ) (p × pq)
=  -28p²q

(iv) 4p³ × – 3p
= (4 × -3 ) (p³ × p )
=  -12p⁴

(v) 4p × 0
= 0

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.
(p, q) ; (10m, 5n) ; (20x² , 5y²) ; (4x, 3x²) ; (3mn, 4np)

Solution –
Area of rectangle = Length x breadth.
So, it is multiplication of two monomials.
The results can be written in square units.

(i) (p, q)
Length = p
Breadth = q
∴ Area of the rectangle = Length × Breadth
= p × q
= pq

(ii) (10m, 5n)
Length = 10 m
Breadth = 5 n
∴ Area of the rectangle = Length × Breadth
= (10m) × (5n)
= (10 × 5) × (m × n)
= 50 × (mn)
= 50 mn

(iii) (20x², 5y²)
Length = 20x²
Breadth = 5y²
∴ Area of the rectangle = Length × Breadth
= (20 × 2) × (5y²)
= (20 × 5) × (x²×y²)
= 100 × (x²y²)
= 100x²y²

(iv) (4x, 3x²)
Length = 4x
Breadth = 3x²
∴ Area of the rectangle = Length × Breadth
= (4x) × (3x²)
= (4 × 3) × (x × x²)
= 12 × x³
= 12x³

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

3. Complete the following table of products:

Solution –

4. Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.
(i) 5a, 3a², 7a⁴
(ii) 2p, 4q, 8r
(iii) xy, 2x²y, 2xy²
(iv) a, 2b, 3c

Solution – Volume of a rectangular box = length × breadth × height

(i) 5a, 3a², 7a⁴
Volume = 5a × 3a² × 7a⁴
= 5 × 3 × 7 × a × a² × a⁴
= 105 a⁷

(ii) 2p, 4q, 8r
Volume = 2p × 4q × 8r
= 2 × 4 × 8 × p × q × r
= 64 pqr

(iii) xy, 2x²y, 2xy²
Volume = xy × 2x² y × 2 xy²
= 2 × 2 × xy × x² y × xy²
= 4x⁴ y⁴

(iv) a, 2b, 3c
Volume = a × 2b × 3c
= 2 × 3 × a × b × c
= 6abc

5. Obtain the product of
(i) xy,  yz, zx
(ii) a, – a² , a³
(iii) 2, 4y, 8y² , 16y³
(iv) a, 2b, 3c, 6abc
(v) m, – mn, mnp

Solution –

(i) xy,  yz, zx
= xy × yz × zx
= x² y² z²

(ii) a, – a² , a³
= a × – a²  × a³
= – a⁶

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

(iv) a, 2b, 3c, 6abc
= a × 2b × 3c × 6abc
= 36a² b² c²

(v) m, – mn, mnp
= m × – mn × mnp
= –m³ n² p