In this page we have NCERT Solutions for Algebraic Expressions and Identities Class 8 Maths Chapter 9 for
Exercise 9.2 . Hope you like them and do not forget to like , social share
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Question 1
Find the product of the following pairs of monomials.
(i) 4, 7p
(ii) – 4p, 7p
(iii) – 4p, 7pq
(iv) 4p
3, – 3p
(v) 4p, 0
Answer:
- 4 x 7 p = 28p
- - 4p x 7p = -28p2
- - 4p x 7pq = -28p2q
- 4p3 x - 3p = -12p4
- 4p x 0 = 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: Now we know that
Area of rectangle = Length x breadth
So it is multiplication of monomials
(i) p x q = pq
(ii)10m x 5n = 50mn
(iii) 20x2 x 5y2 = 100x2y2
(iv) 4x x 3x2 = 12x3
(v)3mn x 4np = 12mn2p
Question 3.
Complete the following table of products:
Answer
Question 4
Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
- 5a, 3a2, 7a8
- (2p, 4q, 8r)
- xy, 2x2y, 2xy2
- (a, 2b, 3c
Answer: We know that volume of rectangular box is given by
Volume = length x breadth x height
(i)
5a x
3a2 x
7a8 =
105a11
(ii)
2p x
4q x
8r =
64pqr
(iii)
xy x
2x2y x
2xy2 =
4x4y
4
(iv) a x 2b x 3c = 6abc
Question 5
Obtain the product of
- xy, yz, zx
- a, – a2 a3
- 2, 4y, 8y2 16y3
- a, 2b, 3c, 6abc
- m, – mn, mnp
Answer:
(i)
x2y
2z
2
(ii) –a
5
(iii) 1024y
6
(iv) 36a
2b
2c
2
(v) –m
3n
2p
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