You can Download KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Questions and Answers to help you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2

Question 1.

Find the product of the following pairs of monomials.

(i) 4, 7p

(ii) -4, 7p

(iii) -4p, 7pq

(iv) 4p^{3}, -3p

(v) 4p, 0

Solution:

(i) 4, 7p = 28p

(ii) -4, 7p = -28p^{2}

(iii) -4p, 7pq = -28p^{2}q

(iv) 4p^{3}, -3p = -12p^{2}

(v) 4p, 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); (20x^{2}, 5y^{2}); (4x, 3x^{2}); (3mn, 4np)

Solution:

Since the area of rectangle is the product of length × Breadth

(i) p × q = pq

(ii) 10m × 5n = 50mn

(iii) 20x^{2} × 5y^{2} = 100x^{2}y^{2}

(iv) 4x × 3x^{2} = 12x^{3}

(v) 3mn × 4np = 12mn^{2}p

Question 3.

Complete the table of products.

Solution:

Question 4.

Obtain the volume of rectangular boxes with the following length, breadth, and height respectively.

(i) 5a, 3a^{2}, 7a4

(ii) 2p, 4q, 8r

(iii) xy, 2x^{2}y, 2xy^{2}

(iv) a, 2b, 3c

Solution:

Volume of rectangular boxes is length × breadth × height

(i) 5a, 3a^{2}, 7a^{4}

= 5 × 3 × 7 × a^{1+2+4}

= 105a^{7}

(ii) 2p, 4q, 8r

= 2p, 4q, 8r

= 2 × 4 × 8pqr

= 64pqr

(iii) xy, 2x^{2}y, 2xy^{2}

= 1 × 2 × 2 × xy × x^{2}y × xy^{2}

= 4x^{4}y^{4}

(iv) a × 2b × 3c

= 1 × 2 × 3 × abc

= 6abc

Question 5.

Obtain the product of

(i) xy, yz, zx

(ii) a, -a^{2}, a^{3}

(iii) 2, 4y, 8y^{2}, 16y^{3}

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

(v) m, -mn, mnp

Solution:

(i) xy × yz × zx

= x^{2}y^{2}z^{2}

(ii) ax – a^{2} × a^{3}

= 1 × 1 × 1 × a × a^{2} × a^{3}

= 1 × a^{6}

= -a^{6}

(iii) 2 × 4y × 8y^{2} × 16y^{3}

= 2 × 4 × 8 × 16 × y × y^{2} × y^{3}

= 102y^{6}

(iv) a × 2b × 3c × 6abc

= 1 × 2 × 3 × 6 × a × b × c × abc

= 36a^{2}b^{2}c^{2}

(v) m × -mn × mnp

= 1 × -1 × 1 × m × -mn × mnp

= -1m^{2}n^{2}p