Exercise 9.2 - Chapter 9 Algebraic Expressions & Identities class 8 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

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:

The product will be as follows.

(i) 4 × 7p = 4 × 7 × p = 28p

(ii) − 4p × 7p = − 4 × p × 7 × p = (− 4 × 7) × (p × p) = − 28 p²

(iii) − 4p × 7pq = − 4 × p × 7 × p × q = (− 4 × 7) × (p × p × q) = − 28p²q

(iv) 4p³ × − 3p = 4 × (− 3) × p × p × p × p = − 12 p⁴

(v) 4p × 0 = 4 × p × 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:

We know that,

Area of rectangle = Length × Breadth

Area of 1^st rectangle = p × q = pq

Area of 2^nd rectangle = 10m × 5n = 10 × 5 × m × n = 50 mn

Area of 3^rd rectangle = 20x² × 5y² = 20 × 5 × x² × y² = 100 x²y²

Area of 4^th rectangle = 4x × 3x² = 4 × 3 × x × x² = 12x³

Area of 5^th rectangle = 3mn × 4np = 3 × 4 × m × n × n × p = 12mn²p

Question 3:

Complete the table of products.

Answer:

The table can be completed as follows.

Question 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, 2xy²

(iv) a, 2b, 3c

Answer:

We know that,

Volume = Length × Breadth × Height

(i) Volume = 5a × 3a² × 7a⁴ = 5 × 3 × 7 × a × a² × a⁴ = 105 a⁷

(ii) Volume = 2p × 4q × 8r = 2 × 4 × 8 × p × q × r = 64pqr

(iii) Volume = xy × 2x²y × 2xy² = 2 × 2 × xy ×x²y × xy² = 4x⁴y⁴

(iv) Volume = a × 2b × 3c = 2 × 3 × a × b × c = 6abc

Question 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

Answer:

(i) xy × yz × zx = x²y²z²

(ii) a × (− a²) × a³ = − a⁶

(iii) 2 × 4y × 8y² × 16y³ = 2 × 4 × 8 × 16 × y × y² × y³ = 1024 y⁶

(iv) a × 2b × 3c × 6abc = 2 × 3 × 6 × a × b × c × abc = 36a²b²c²

(v) m × (− mn) × mnp = − m³n²p