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:

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