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

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

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

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2.

### Ex 9.2 Class 8 Maths Question 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 × 7p = (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 x 0 = (4 × 0) × p = 0 × p = 0

### Ex 9.2 Class 8 Maths Question 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:
(i)
Length = p units and breadth = q units
Area of a rectangle = length × breadth = p × q = pq sq. units
(ii) Length = 10m units and breadth = 5n units
Area of a rectangle = length × breadth = 10m × 5n = (10 × 5) × m × n = 50mn sq. units
(iii) Length = 20x2 units and breadth = 5y2 units
Area of a rectangle = length × breadth = 20x2 × 5y2 = (20 × 5) × x2 × y2 = 100x2y2 sq. units
(iv) Length = 4x units and breadth = 3x2 units
Area of a rectangle = length × breadth = 4x × 3x2 = (4 × 3) × x × x2 = 12x3 sq. units
(v) Length = 3mn units and breadth = 4np units
Area of a rectangle = length × breadth = 3mn × 4np = (3 × 4) × mn × np = 12mn2p sq. units

### Ex 9.2 Class 8 Maths Question 3.

Solution:
The completed table is given below.

### Ex 9.2 Class 8 Maths Question 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:
(i) Length = 5a, breadth = 3a2, and height = 7a4

Volume of a rectangular box = length × breadth × height

= 5a × 3a2 × 7a4
= (5 × 3 × 7) × (a × a2 × a4)
= 105a1+2+4 = 105a7

(ii) Length = 2p, breadth = 4q, and height = 8r

Volume of a rectangular box = length × breadth × height

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

(iii) Length = xy, breadth = 2x2y, and height = 2xy2

Volume of a rectangular box = length × breadth × height

= xy × 2x2y × 2xy2
= (1 × 2 × 2) × (x × x2 × x × y × y × y2)
= 4x1+2+1 y1+1+2 = 4x4y4

(iv) Length = a, breadth = 2b, and height = 3c

Volume of a rectangular box = length × breadth × height

= a × 2b × 3c
= (1 × 2 × 3) × (a × b × c)
= 6abc

### Ex 9.2 Class 8 Maths Question 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:

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