**.**

**NCERT Solutions for Class 8 Maths Chapter 8 ****Algebraic Expressions and Identities**** Ex 8.2**

**Ex 8.2 Class 8 Maths Question 1.**

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

**(i)** 4, 7p

**(ii)** -4p, 7p

**(iii)** -4p, 7pq

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

**(v)** 4p, 0

**Solution:****(i) **4 × 7p = (4 × 7) × p = 28p

**(ii)**-4p × 7p = (-4 × 7) × p × p = -28p

^{2}

**(iii)**-4p × 7pq = (-4 × 7) × p × pq = -28p

^{2}q

**(iv)**4p

^{3}× -3p = (4 × -3) × p

^{3}× p = -12p

^{4}

**(v)**4p x 0 = (4 × 0) × p = 0 × p = 0

**Ex 8.2 Class 8 Maths 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:****(i)** Length = p units and breadth = q units

Area of a rectangle = length × breadth = p × q = pq sq. units

**(ii)**Length = 10m units and breadth = 5n units

Area of a rectangle = length × breadth = 10m × 5n = (10 × 5) × m × n = 50mn sq. units

**(iii)**Length = 20x

^{2}units and breadth = 5y

^{2}units

Area of a rectangle = length × breadth = 20x

^{2}× 5y

^{2}= (20 × 5) × x

^{2}× y

^{2}= 100x

^{2}y

^{2}sq. units

**(iv)**Length = 4x units and breadth = 3x

^{2}units

Area of a rectangle = length × breadth = 4x × 3x

^{2}= (4 × 3) × x × x

^{2}= 12x

^{3}sq. units

**(v)**Length = 3mn units and breadth = 4np units

Area of a rectangle = length × breadth = 3mn × 4np = (3 × 4) × mn × np = 12mn

^{2}p sq. units

** **

**Ex 8.2 Class 8 Maths Question 3.**

### Complete the table of products.

**Solution:**

The completed table is given below.

**Ex 8.2 Class 8 Maths Question 4.**

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

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

**(ii)** 2p, 4q, 8r

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

**(iv)** a, 2b, 3c

**Solution:****(i)** Length = 5a, breadth = 3a^{2}, and height = 7a^{4}

Volume of a rectangular box = length × breadth × height

= 5a × 3a^{2} × 7a^{4}

= (5 × 3 × 7) × (a × a^{2} × a^{4})

= 105a^{1+2+4} = 105a^{7}

**(ii)** Length = 2p, breadth = 4q, and height = 8r

Volume of a rectangular box = length × breadth × height

= 2p × 4q × 8r

= (2 × 4 × 8) × p × q × r = 64pqr

**(iii)** Length = xy, breadth = 2x^{2}y, and height = 2xy^{2}

Volume of a rectangular box = length × breadth × height

= xy × 2x^{2}y × 2xy^{2}

= (1 × 2 × 2) × (x × x^{2} × x × y × y × y^{2})

= 4x^{1+2+1} y^{1+1+2} = 4x^{4}y^{4}

**(iv)** Length = a, breadth = 2b, and height = 3c

Volume of a rectangular box = length × breadth × height

= a × 2b × 3c

= (1 × 2 × 3) × (a × b × c)

= 6abc

**Ex 8.2 Class 8 Maths 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:**

**Related Links:**

**NCERT Solutions for Maths Class 9**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11**