Ncert Solutions for Algebraic Expressions and Identities Class 8 Maths Chapter 9 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
Answer:

1. 4 x 7 p = 28p
2. - 4p x 7p = -28p²
3.  - 4p x 7pq = -28p²q
4. 4p³ x - 3p = -12p⁴
5. 4p x 0 = 0

Question 2
 Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

1. (p, q)
2. (10m, 5n)
3. (20x² 5y²)
4. (4x, 3x²)\
5. (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.

1. 5a, 3a², 7a⁸
2.  (2p, 4q, 8r)
3. xy, 2x²y, 2xy²
4. (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 = 4x⁴y⁴
(iv) a x  2b x 3c = 6abc
Question 5
 Obtain the product of

1. xy,  yz, zx
2. a, – a² a³
3. 2, 4y, 8y² 16y³
4. a, 2b, 3c, 6abc
5. m, – mn, mnp

Answer:
 (i) x²y²z²
(ii) –a⁵
(iii) 1024y⁶
(iv) 36a²b²c²
(v) –m³n²p

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