Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.2 Textbook Exercise Questions and Answers.

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

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

Question (i)

4, 7p

Solution:

= 4 × 7p

= 4 × 7 × p

= 28p

Question (ii)

– 4p, 7p

Solution:

= – 4p × 7p

= (-4 × 7) × p × p

= – 28p^{2}

Question (iii)

– 4p, 7pq

Solution:

= (- 4p) × 7pq

= – 4 × 7 × p × pq

= – 28p^{2}q

Question (iv)

4p^{3}, – 3p

Solution:

= 4p^{3} × (- 3p)

= 4 × (- 3) × p^{3} × p

= – 12p^{4}

Question (v)

4p, 0

Solution:

= 4p × 0

= 0

2. Find the areas of rectangles with the following pairs of monomials as their s lengths and breadths respectively:

(p, q); (10m, 5n); (20x^{2}, 5y^{2}); (4x, 3x^{2}); (3mn, 4np)

Solution:

Area of a rectangle = length × breadth

(i) length = p, breadth = q

∴ Area = length × breadth

= p × q = pq unit^{2}

(ii) length = 10m, breadth = 5n

∴ Area = length × breadth

= 10m × 5n

= 50mn unit^{2}

(iii) length = 20x^{2}, breadth = 5y^{2}

∴ Area = length × breadth

= 20x^{2} × 5y^{2}

= 100x^{2}y^{2} unit^{2}

(iv) length = 4x, breadth 3x^{2}

∴ Area = length × breadth

= 4x × 3x^{2}

= 12x^{3} unit^{2}

(v) length = 3mn, breadth = 4np

∴ Area = length × breadth

= 3mn × 4np

= 12 mn^{2}p unit^{2}

3. Complete the table of products:

Solution:

First monomial → Second monomial ↓ | 2x | -5 y | 3x^{2} |
-4xy | 7x^{2}y |
– 9x^{2}y^{2} |

2x | 4x^{2} |
-10xy | 6x^{3} |
-8x^{2}y |
14x^{3}y |
– 18x^{3}y^{2} |

– 5 y | – 10xy | 25y^{2} |
– 15x^{2}y |
20xy^{2} |
– 35x^{2}y^{2} |
45x^{2}y^{3} |

3x^{2} |
6x^{3} |
– 15x^{2}y |
9x^{4} |
– 12x^{3}y |
21x^{4}y |
– 27x^{4}y^{2} |

– 4xy | – 8x^{2}y |
20xy^{2} |
-12x^{3}y |
16x^{2}y^{2} |
– 28x^{3}y^{2} |
36x^{3}y^{3} |

7x^{2}y |
14x^{3}y |
– 35x^{2}y^{2} |
21x^{4}y |
– 28x^{3}y^{2} |
49x^{4}y^{2} |
– 63x^{4}y^{3} |

– 9x^{2}y^{2} |
– 18x^{3}y^{2} |
45x^{2}y^{3} |
– 27x^{4}y^{2} |
36x^{3}y^{3} |
– 63x^{4}y^{3} |
81x^{4}y^{4} |

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

Question (i)

5a, 3a^{2}, 7a^{4}

Solution:

Volume of rectangular box = length × breadth × height

length = 5a, breadth = 3a^{2}, height = 7a^{4}

∴ Volume = length × breadth × height

= 5a × 3a^{2} × 7a^{4}

= (5 × 3 × 7) × a × a^{2} × a^{4}

= 105a^{7} cubic unit

Question (ii)

2p, 4q, 8r

Solution:

length = 2p, breadth = 4q, height = 8r

∴ Volume = length × breadth × height

= 2p × 4q × 8r

= (2 × 4 × 8) × p × q × r

= 64pqr cubic unit

Question (iii)

xy, 2x^{2}y, 2xy^{2}

Solution:

length = xy, breadth = 2x^{2}y, height = 2 xy^{2}

∴ Volume = length × breadth × height

= xy × 2x^{2}y × 2xy^{2}

= (1 × 2 × 2) × xy × x^{2}y × xy^{2}

= 4x^{4}y^{4} cubic unit

Question (iv)

a, 2b, 3c

Solution:

length = a, breadth = 2b, height = 3c

∴ Volume = length × breadth × height

= a × 2b × 3c

= (1 × 2 × 3) × a × b × c

= 6abc cubic unit

5. Obtain the product of:

Question (i)

xy, yz, zx

Solution:

= xy × yz × zx

= [x × y) × (y × z) × (z × x)

= x × x × y × y × z × z

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

Question (ii)

a, -a^{2}, a^{3}

Solution:

= (a) × (- a^{2}) × (a^{3})

= – a × a^{2} × a^{3}

= – a^{6}

Question (iii)

2, 4y, 8y^{2}, 16y^{3}

Solution:

= 2 × 4y × 8y^{2} × 16y^{3}

= 2 × 4 × 8 × 16 × y × y^{2} × y^{3}

= 1024y^{6}

Question (iv)

a, 2b, 3c, 6abc

Solution:

= a × 2b × 3c × 6abc

= 1 × 2 × 3 × 6 × a × b × c × abc

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

Question (v)

m, – mn, mnp

Solution:

= (m) × (- mn) × (mnp)

= – (m × mn × mnp)

= – m^{3}n^{2}p