NCERT Solution Class 8th Maths Chapter – 8 Algebraic Expressions and Identities Exercise 8.2

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 = 28 p
(ii) -4p × 7 p = (-4 × 7) × (p × p) = -28 p²
(iii) -4p × 7 pq = (-4 × 7) (p × pq) = -28p²q
(iv) 4p³ × – 3p = (4 × -3) (p³ × p) = -12p⁴
(v) 4p × 0 = 0

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: Area of rectangle = Length x breadth. So, it is the multiplication of two monomials.
The results can be written in square units.
(i) p × q = pq
(ii) 10m × 5n = 50mn
(iii) 20x² × 5y² = 100x²y²
(iv) 4x × 3x² = 12x³
(v) 3mn × 4np = 12mn²p

solution:

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, 2y²
(iv) a, 2b, 3c

Solution: Volume of rectangle = length × breadth × height. To evaluate volume of rectangular boxes, multiply all the monomials.
(i) 5a × 3a² × 7a⁴ = (5 × 3 × 7) (a × a² × a⁴) = 105a⁷
(ii) 2p × 4q × 8r = (2 × 4 × 8 ) (p × q × r) = 64pqr
(iii) xy × 2x²y × 2xy² = (1 × 2 × 2) (x × y × x² × y × x × y² ) = 4x⁴y⁴
(iv) a × 2b × 3c = (1 × 2 × 3) (a × b × c) = 6abc

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 = x²y²z²
(ii) a × – a² × a³ = – a⁶
(iii) 2 × 4y × 8y² × 16y³ = 1024y⁶
(iv) a × 2b × 3c × 6abc = 36a²b²c²
(v) m × – mn × mnp = –m³n²p