Latest Updated : October 2023

If you are in search of **NCERT solutions for Class 8 Maths Chapter** **5 Squares and Square Roots**, 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** of all the chapters, **exercise wise** with **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 bookmark **ncertforclass8.com** to boost your preparation.

## NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Ex 5.2

**Class 8 Maths Chapter** **5 Squares and Square Roots** **Exercise 5.**2

**Exercise 5.2**

**1. Find the square of the following numbers.(i) 32 **

**(ii) 35**

(iii) 86

(iii) 86

**(iv) 93**

(v) 71

(v) 71

**(vi) 46**

Solution:

`\[\left(i\right)\ \ \left(32\right)^{2\ }=\ \left(30+2\right)^2\]`

`\[=\left(30\right)^{2\ }+\ \left(2\right)^{2\ }+\ 2\ X\ 30\ X\ 2\ \ \ \ \ \ \ \ \ \ \ \ \ \left[\sin ce\ \left(a+b\right)^{2\ }=\ a^2\ +\ b^2\ +\ 2ab\right]\]`

`\[=\ 900\ +\ 4\ +\ 120\]`

`\[=1024\]`

`\[\left(ii\right)\ \left(35\right)^{2\ }=\ \left(30+5\right)^2\]`

`\[=\left(30\right)^{2\ }+\ \left(5\right)^{2\ }+\ 2\ X\ 30\ X\ 5\ \ \ \ \ \ \ \ \ \ \ \ \left[\sin ce\ \left(a+b\right)^{2\ }=\ a^2\ +\ b^2\ +\ 2ab\right]\]`

`\[=900\ +\ 25\ +\ 300\]`

`\[=1225\]`

`\[\left(iii\right)\ \left(86\right)^{2\ }=\ \left(80+6\right)^2\]`

`\[=\left(80\right)^{2\ }+\ \left(6\right)^{2\ }+\ 2\ X\ 80\ X\ 6\ \ \ \ \ \ \ \ \ \ \ \ \left[\sin ce\ \left(a+b\right)^{2\ }=\ a^2\ +\ b^2\ +\ 2ab\right]\]`

`\[=1600\ +\ 36\ +\ 960\ \]`

`\[=\ 7386\]`

`\[\left(iv\right)\ \left(93\right)^{2\ }=\ \left(90+3\right)^2\]`

`\[=\left(90\right)^{2\ }+\ \left(3\right)^{2\ }+\ 2\ X\ 90\ X\ 3\ \ \ \ \ \ \ \ \ \ \left[\sin ce\ \left(a+b\right)^{2\ }=\ a^2\ +\ b^2\ +\ 2ab\right]\]`

`\[=8100\ +\ 9\ +\ 540\]`

`\[=\ 8649\]`

`\[\left(v\right)\ \ \left(71\right)^{2\ }=\ \left(70+1\right)^2\]`

`\[=\left(70\right)^{2\ }+\ \left(1\right)^{2\ }+\ 2\ X\ 70\ X\ 1\ \ \ \ \ \ \ \ \ \ \left[\sin ce\ \left(a+b\right)^{2\ }=\ a^2\ +\ b^2\ +\ 2ab\right]\]`

`\[=4900\ +\ 1\ +\ 140\]`

`\[=5041\]`

`\[\left(vi\right)\ \left(46\right)^{2\ }=\ \left(40+6\right)^2\]`

`\[=\left(40\right)^{2\ }+\ \left(1\right)^{2\ }+\ 2\ X\ 40\ X\ 6\ \ \ \ \ \ \ \ \ \ \left[\sin ce\ \left(a+b\right)^{2\ }=\ a^2\ +\ b^2\ +\ 2ab\right]\]`

`\[=1600\ +\ 36\ +\ 480\]`

`\[=2116\]`

**2. Write a Pythagorean triplet whose one member is.(i) 6 **

**(ii) 14**

**(iii) 16**

**(iv) 18**

Solution:

`\[\left(i\right)\ There\ are\ three\ numbers\ 2m,\ m^2-1\ and\ m^2+1\ in\ a\ Pythagorean\ Triplet.\]`

`\[Here\ 2m=6\]`

`\[So,\ m=\ \frac{6}{2}\ =\ 3\]`

`\[Second\ number\ =\ \left(m^{2\ }-\ 1\right)\ =\ 3^{2\ }-\ 1\ =\ 9-1\ =\ 8\]`

`\[Third\ number\ =\ \left(m^{2\ }+\ 1\right)\ =\ 3^{2\ }+\ 1\ =\ 9\ +\ 1\ =\ 10\]`

`\[Hence,\ Pythagorean\ triplet\ is\ \left(6,8,10\right).\]`

`\[\left(ii\right)\ There\ are\ three\ numbers\ 2m,\ m^2-1\ and\ m^2+1\ in\ a\ Pythagorean\ Triplet.\]`

`\[Here\ 2m=14\]`

`\[m=\frac{14}{2\ }=\ 7\]`

`\[Second\ number\ =\ \left(m^{2\ }-\ 1\right)\ =\ 7^2\ -\ 1\ =\ 49-1\ =\ 48\]`

`\[Third\ number\ =\ \left(m^{2\ }+\ 1\right)\ =\ 7^{2\ }+\ 1\ =\ 49\ +\ 1\ =\ 50\]`

`\[Hence,\ Pythagorean\ triplet\ is\ \left(14,48,50\right).\]`

`\[\left(iii\right)\ There\ are\ three\ numbers\ 2m,\ m^2-1\ and\ m^2+1\ in\ a\ Pythagorean\ Triplet.\]`

`\[Here,\ 2m\ =\ 16\]`

`\[m=\ \frac{16\ }{2}=\ 8\]`

`\[Second\ number\ =\ \left(m^{2\ }-\ 1\right)\ =\ 8^2\ -\ 1\ =\ 64-1\ =\ 63\]`

`\[Third\ number\ =\ \left(m^{2\ }+\ 1\right)\ =\ 8^{2\ }+\ 1\ =\ 64\ +\ 1\ =\ 65\]`

`\[Hence,\ Pythagorean\ triplet\ is\ \left(16,63,65\right).\]`

`\[\left(iv\right)\ There\ are\ three\ numbers\ 2m,\ m^2-1\ and\ m^2+1\ in\ a\ Pythagorean\ Triplet.\]`

`\[Here,\ 2m\ =\ 18\]`

`\[m=\ \frac{18\ }{2\ }=\ 9\]`

`\[Second\ number\ =\ \left(m^{2\ }-\ 1\right)\ =\ 9^2\ -\ 1\ =\ 81-1\ =\ 80\]`

`\[Third\ number\ =\ \left(m^{2\ }+\ 1\right)\ =\ 8^{2\ }+\ 1\ =\ 81\ +\ 1\ =\ 82\]`

`\[Hence,\ Pythagorean\ triplet\ is\ \left(18,80,82\right).\]`

**Using Class 8 NCERT solutions offers several key advantages for students:**

**Accurate Answers:**NCERT solutions for Class 8 Maths are prepared by expert educators and thoroughly reviewed to ensure quality and accurate answers.**Aligned with Curriculum:**NCERT Solutions for Class 8 Maths are specifically designed to go along with the CBSE curriculum, making them the ideal resource for students studying in CBSE affiliated schools.**Crisp an Clear Explanation**: The NCERT solutions for Class 8 Maths provide step by step explanations, making complex concepts easier to understand. This helps students to grasp the concepts easily and build a better foundation.**Time-Saver:**Instead of spending extra time searching for answers, students can quickly access these NCERT solutions for Class 8 Maths , saving time and allowing them to focus on other aspects of their studies.**Extra Questions:**NCERT Solutions for Class 8 Maths also has extra questions that offer additional opportunities for students to build better problem solving skills.**Free Forever:**We provide answers which will always be free to access and share.**Self-Study Support:**These are beneficial for students who prefer self study. They can independently work through the solutions and improve their understanding of the subject at their own pace.

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