Latest Updated : November 2023

If you are in search of **NCERT solutions for Class 8 Maths Chapter** **8 Algebraic Expressions and Identities**, then you are at the right place. When it comes to **Mathematics**, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed **NCERT solutions for Class 8 Maths**Â of all the chapters,Â **exercise wise**Â withÂ **updated syllabus by NCERT**. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmarkÂ **ncertforclass8.com**Â to boost your preparation.

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

**Class 8 Maths Chapter** **8 Algebraic Expressions and Identities** **Exercise **8.2

**Exercise 8.2**

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

`\[(i)\ 4\ ,\ 7p\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)-4p\ \ ,\ \ 7p\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iii)\ -4p\ \ ,\ \ 7pq\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iv)\ 4p^2-3p\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (v)\ \ 4p\ ,\ \ 0\]`

Solution:

`\[(i)\ \ \ 4,7p=4\times7\times p\ =\ 28p\]`

`\[(ii)-4p\times7p=(-4\times7)\times(p\times p)=\ -28p^2\]`

`\[(iii)-4p\times7pq=(-4\times7)\ \times(p\times pq)=\ -28p^2q\]`

`\[(iv)\ \ 4p^3\times-3p=(4\times-3)\ \times\ (p^3\times p)=-12p^4\]`

`\[(v)\ 4p\times0=0\]`

**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:

`\[Area\ of\ rec\tan gle=\ Length\ \times\ breadth\]`

`\[(i)\ p\times q=pq\ square\ units\]`

`\[(ii)\ 10m\times5n=50mn\ \ \ \ square\ units\]`

`\[(iii)\ 20x^2\times5y^2=\ 100x^2y^2\ \ \ square\ units\]`

`\[(iv)\ 4x\times3x^2=12x^3\ \ \ square\ units\]`

`\[(v)\ 3mn\times4np=12mn^2p\ \ square\ units\]`

**3. Complete the table of products**

Solution:

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

`\[(i)\ \ 5a,\ 3a^2\ ,7a^4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)\ \ 2p,\ 4q,\ 8r\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iii)\ xy,\ 2x^2y\ ,\ \ 2xy^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iv)\ a,\ \ 2b,\ \ 3c\]`

Solution:

Volume of rectangle = length x breadth x height.

`\[(i)\ \ 5a\times3a^2\times7a^4=(5\times3\times7)\ \times(a^1\times a^2\times a^4)=105a^7\ \]`

`\[(ii)\ \ 2p\times4q\times8r=(2\times4\times8)\times(p\times q\times r)=64pqr\]`

`\[(iii)\ y\times2x^2y\times2xy^2=(1\times2\times2)\times(x\times x^2\times x\times y\times y\times y^2)=\ 4x^44y^{4\ }\]`

`\[(iv)\ a\ \times2b\times3c=(1\times2\times3)\times(a\times b\times c)=6abc\]`

**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:

`\[(i)\ xy\times yz\times zx=x^2y^2z^2\]`

`\[(ii)\ a\times-a^2\times a^3=-a^6\]`

`\[(iii)\ 2\times4y\times8y^2\times16y^3=1024\ y^6\]`

`\[(iv)\ a\times2b\times3c\times6abc=36\ a^2b^2c^2\]`

`\[(v)\ m\times-mn\times mnp=-m^3n^2p\]`

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