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

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 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
Answer:
(i) 4 × 7p = (4 × 7) × p = 28p

(ii) - 4p × 7p = (- 4 × 7) × (p × p)
= - 28p^{1 + 1}
= - 28p²

(iii) - 4p × 7pq = (- 4 × 7) × (p × p × q)
= - 28p¹⁺¹q
= - 28p²q

(iv) 4p³ × - 3p = (4 × - 3) × (p³ × p)
= - 12p³ + 1
= - 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)
Answer:
We know that the area of rectangle = l × b, where l = length and b = breadth
p × q = pq Ans.
10m × 5n = (10 × 5) × (m × n)
= 50 mn
20x² × 5y² = (20 × 5) × (x² × y²)
= 100x²y²
4x × 3x² = (4 × 3) × (x × x²)
= 12x³
and 3mn × 4np = (3 × 4) × (m × n × n × p)
= 12mn²p

Question 3.
Complete the table of products:

Answer:
Completed table is as under:

Question 4.
Obtain the volume of rectangular boxes with the following length, breadth and height respectively:
(i) 5a, 3a², 7a⁴
Answer:
Volume = 5a × 3a2 × 7 a4
= (5 × 3 × 7) × (a × a² × a⁴)
= 105a^{1 + 2 + 4}
= 105a⁷

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

(iii) xy, 2x²y, 2xy²
Answer:
Volume = xy × 2x2y × 2xy2 _ y2y2^2 Ans
= (1 × 2 × 2) × (x × x² × x × y × y × y²)
= 4x^{1 + 2 + 1}y^{1 + 1 + 2}
= 4x⁴y⁴

(iv) a, 2b, 3c
Answer:
Volume = a × 2b × 3c
= (1 × 2 × 3) × (a × b × c)
= 6abc

Question 5.
Obtain the product of .
(i) xy, yz, zx
Answer:
xy × yz × zx = x × x × y × y × z × z
= x^{1 + 1} × y^{1 + 1} × z^{1 + 1}
= x²y²z²

(ii) a, - a², a³
Answer:
a × -a² × a³ = (1 × - 1 × 1) × (a × a² × a³)
= - 1 × a^{1 + 2 + 3}
= - a⁶

(iii) 2, 4y, 8y², 16y³
Answer:
2 × 4y × 8y² × 16y³ = (2 × 4 × 8 × 16) × (y × y² × y³)
= 1024 × y^{1 + 2 + 3}
= 1024y⁶

(iv) a, 2b, 3c, 6abc
Answer:
a × 2b × 3c × 6abc = (1 × 2 × 3 × 6) × (a × a × b × b × c × c)
= 36a² b² c²

(v) m, - mn, mnp
Answer:
m × -mn × mnp = (1 × - 1 × 1) × (m × m × m × n × n × p)
= - 1 × m³ × n² × p
= - m³ × n²p