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

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts.

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

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)
= (-28) × p²
= -28p²

(iii) -4p × 7pq
= {(-4) × 7} × {p × (pq)}
= -28 × (p × p × q)
= -28p²q

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

(v) 4p × 0
= (4 × 0) × p
= 0 × p
= 0

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) (p, q)
Area of the rectangle = length × breadth
= p x q
= pq

(ii) (10m, 5n)
Area of the rectangle = length × breadth
= 10m × 5n
= (10 × 5) × (m × n)
= 50 × mn
= 50mn

(iii) 20x², 5y²
Area of the rectangle = length × breadth
= 20x² × 5y²
= (20 × 5) × (x² × y²)
= 100 × x²y²
= 100x²y²

(iv) (4x, 3x²)
Area of the rectangle = length × breadth
= 4x × 3x²
= (4 × 3) × (x × x²)
= 12 × x³
= 12x³

(v) (3mn, 4np)
Area of the rectangle = length × breadth
= 3mn × 4np
= (3 × 4) × (mn × np)
= 12 × m × (n × n) × p
= 12 × m × n² × p
= 12mn²p

Question 3.
Complete the table products.

Solution:

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) 5a, 3a², 7a⁴
Volume of the rectangular box = length × breadth × height
= (5a) × (3a²) × (7a⁴)
= (5 × 3 × 7) × (a × a² × a⁴)
= 105a⁷

(ii) 2p, 4q, 8r
Volume of the rectangular box = Length × Breadth × Height
= 2p × 4q × 8r
= (2 × 4 × 8) × (p × q × r)
= 64pqr

(iii) xy; 2x²y; 2xy²
Volume of the rectangular box = length × breadth × height
= xy × 2x²y × 2xy²
= (1 × 2 × 2) × (x × x² × x) × (y × y × y²)
= 4 × x⁴ × y⁴
= 4x⁴y⁴

(iv) a, 2b, 3c
Volume of the rectangular box = length × breadth × height
= a × 2b × 3c
= (1 × 2 × 3) × (a × b × c)
= 6abc

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:
(i) xy, yz, zx
(xy) × (yz) × (zx)
= (x × x) × (y × y) × (z × z)
= x² × y² × z²
= x²y²z²

(ii) a; -a²; a³
(a) × (-a²) × (a³)
= -(a × a² × a³)
= -a⁶

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

(iv) a, 2b, 3c, 6abc
(a) × (2b) × (3c) × (6abc)
= (2 × 3 × 6) × (a × a) × (b × b) × (c × c)
= 36 × a² × b² × c²
= 36a²b²c²

(v) m, -mn, mnp
(m) × (-mn) × (mnp)
= -1 × (m × m × m) × (n × n) × p
= -1 × m³ × n² × p
= -m³n²p