NCERT Solutions Class 8 Maths Chapter 5 Squares and Square Root Exercise 5.2

Exercise 5.2 in Chapter 5 of Class 8 Maths provides you with the definition of a square root. It is important to understand how to find the square root of a number by methods such as repeated subtraction, prime factorization, and the division method. Finding square roots will be extremely important for students in many other chapters in higher classes, particularly algebra and geometry.

This exercise is designed as per the latest NCERT syllabus, according to the official textbook. The questions in this exercise reinforce your understanding of square roots, which is a common topic in the CBSE exams and other competitive exams like the Olympiads.

With ALLEN’s NCERT Solutions, students will be well equipped to properly think through each question and will understand the correct steps to arrive at the solution. The step-by-step explanations are designed to make the student’s learning easy and mistake-free. 

1.0Download NCERT Solutions Class 8 Maths Chapter 5 Squares and Square Root Exercise 5.2: Free PDF

Find solutions to square root questions from NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.2. Download the free PDF from below:

2.0Key Concepts in Exercise 5.2 of Class 8 Maths Chapter 5

Exercise 5.2 explains how to find square roots of numbers using different simple methods. The key concepts here include :

• Finding square roots through repeated subtraction
• Prime factorization method to find square roots
• Division method for square roots
• Understanding perfect and non-perfect squares
• Square roots of large numbers

3.0NCERT Class 8 Maths Chapter 5: Other Exercises

4.0NCERT Class 8 Maths Chapter 5 Exercise 5.2: Detailed Solutions

• Find the square of the following numbers. (i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

Sol. (i) $32^2=(30+2{)}^2$

$\begin{array}{rl}& =30(30+2)+2(30+2)\\ & =30^2+30\times 2+2\times 30+2^2\\ & =900+60+60+4=1024\end{array}$

(ii) $35^2=(30+5{)}^2=30(30+5)+5(30+5)$ $=30^2+30\times 5+5\times 30+5^2$ $=900+150+150+25=1225$ (iii) $86^2=(80+6{)}^2$ $=80(80+6)+6(80+6)$ $=80^2+80\times 6+6\times 80+6^2$ $=6400+480+480+36=7396$ (iv) $93^3=(90+3{)}^2$ $=90(90+3)+3(90+3)$ $=90^2+90\times 3+3\times 90+3^2$ $=8100+270+270+9=8649$ (v) $71^2=(70+1{)}^2$ $=70(70+1)+1(70+1)$ $=70^2+70\times 1+1\times 70+1^2$ $=4900+70+70+1=5041$ (vi) $46^2=(40+6{)}^2$ $=40(40+6)+6(40+6)$ $=40^2+40\times 6+6\times 40+6^2$ $=1600+240+240+36=2116$

2. Write a Pythagorean triplet whose one member is (i) 6 (ii) 14 (iii) 16 (iv) 18

Sol. (i) $2\mathrm{m}=6\Rightarrow \mathrm{m}=3$ Put $m=3$ in $m^2-1,m^2+1$ $\therefore 2\mathrm{m}=6,{\mathrm{m}}^2-1=3^2-1=9-1=8$ and ${\mathrm{m}}^2+1=9+1=10$ Thus, 6, 8 and 10 is Pythagorean triplets. (ii) $2\mathrm{m}=14\Rightarrow \mathrm{m}=7$

Put $m=7$ in $m^2-1,m^2+1$ $\therefore 2\mathrm{m}=14,{\mathrm{m}}^2-1=7^2-1=49-1=48$ and $m^2+1=7^2+1=49+1=50$ Thus, 14, 48 and 50 is Pythagorean triplets. (iii) $2\mathrm{m}=16\Rightarrow \mathrm{m}=8$

Put $\mathrm{m}=8$ in ${\mathrm{m}}^2-1\&{\mathrm{m}}^2+1$ $\therefore 2\mathrm{m}=16,{\mathrm{m}}^2-1=8^2-1=64-1=63$ and $m^2+1=8^2+1=64+1=65$ Thus 16, 63, and 65 is Pythagorean triplets. (iv) $2\mathrm{m}=18\Rightarrow \mathrm{m}=9$

Put $m=9$ in $m^2-1$ and $m^2+1$ $\therefore 2\mathrm{m}=18,{\mathrm{m}}^2-1=9^2-1=81-1=80$ And $m^2+1=9^2+1=81+1=82$ Thus 18, 80 and 82 is Pythagorean triplets.

5.0Key Features and Benefits of Class 8 Maths Chapter 5 Exercise 5.2

• This exercise outlines three different approaches to work out square roots in a  step-by-step manner.
• The Solutions are developed fully in accordance with the current NCERT syllabus for Class 8.
• This exercise has a mixture of conceptual and application-based problems seen in CBSE examinations.
• Regular practice prepares students to work quickly while solving tricky square root problems.
• Develops logical thinking skills needed to take Olympiad and other competitive examinations.
• Builds number sense, helping the preparation for advanced Maths concepts.