Mathematics – Class 8 – Chapter 9 – Algebraic Expression and Identities – Exercise 9.2 -NCERT Exercise Solution

1. Find the product of the following pairs of monomials.

(i) 4, 7p

Solution:

= 4 × 7 × p

= 28p

(ii) – 4p, 7p

Solution:

= (-4 × 7) × (p × p)

= -28p2

(iii) – 4p, 7pq

Solution:

= (-4 × 7) (p × pq)

= -28p2 q

(iv)  4p3, – 3p

Solution:

= (4 × -3) (p3 × p)

= -12p4

(v) 4p, 0

Solution:

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:

We know that the area of rectangle = Length x breadth.

(i) p × q = pq sq units

(ii)10m × 5n = 50mn sq units

(iii) 20x2 ×  5y2 =  100x2 y2 sq units

(iv) 4x × 3×2 = 12×3 sq units

(v) 3mn ×  4np = 12mn2 p sq units

3. Complete the following table of products:

Mathematics - Class 8 - Chapter 9 - Exercise 9.2 - Algebraic Expression and Identities - NCERT Exercise Solution

Solution:

Mathematics - Class 8 - Chapter 9 - Exercise 9.2 - Algebraic Expression and Identities - NCERT Exercise 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, 2x2 y, 2xy2

(iv) a, 2b, 3c

Solution:

We know that volume of rectangle = length x  breadth x  height.

Now,

(i) 5a, 3a2, 7a4

=5a x 3a2 x 7a4

= (5 × 3 × 7) (a × a2 × a4)

= 105a7

(ii) 2p, 4q, 8r

= 2p x 4q x 8r

= (2 × 4 × 8) (p × q × r)

= 64pqr

(iii) xy, 2x2 y, 2xy2

= y × 2x2 y × 2xy2

= (1 × 2 × 2) (x × x2 × x × y × y × y2 )

= 4x4 y4

(iv) a, 2b, 3c

= a x 2b x 3c

= (1 × 2 × 3) (a × b × c)

= 6abc

5. Obtain the product of

(i) xy,  yz, zx

Solution:

= xy × yz × zx

= x2 y2 z2

(ii) a, – a2 , a3

Solution:

= a × – a2 × a3

= – a6

(iii) 2, 4y, 8y2 , 16y3

Solution:

= 2 × 4y × 8y2 × 16y3

= 1024 y6

(iv) a, 2b, 3c, 6abc

Solution:

= a × 2b × 3c × 6abc

= 36a2 b2c2

(v) m, – mn, mnp

Solution:

= m × – mn × mnp

= –m3 n2 p

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