Question

find square root usig division​

Answer 1

Question :-

Find the square root of the following by Long Division Method

a) 7225

b) 961

c) 8649

$\large\underline{\sf{Solution-a}}$

$\rm \: \sqrt{7225}$

So, using Long Division Method, we have

$\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:85 \:\:}}}\\ {\underline{\sf{8}}}& {\sf{\:\:7225\:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \: \: \:64 \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{165}}}& {\sf{\:\: 825   \:\:}} \\{\sf{}}& \underline{\sf{\:\:825 \:  \:}} \\ {\underline{\sf{}}}& {\sf{\: \: 0\:\:}} \end{array}\end{gathered}\end{gathered}$

Hence,

$\rm\implies \: \sqrt{7225} = 85$

$\large\underline{\sf{Solution-b}}$

$\rm \: \sqrt{961}$

So, using Long Division Method, we have

$\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:31 \:\:}}}\\ {\underline{\sf{3}}}& {\sf{\:\:961\:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \: \: \:9 \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{61}}}& {\sf{\:\: \: \: 061   \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: 61 \:  \:}} \\ {\underline{\sf{}}}& {\sf{\: \: \: \: \: 0\:\:}} \end{array}\end{gathered}\end{gathered}$

Hence,

$\rm\implies \: \sqrt{961} = 31$

$\large\underline{\sf{Solution-c}}$

$\rm \: \sqrt{8649}$

So, using Long Division Method, we have

$\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:93\:\:}}}\\ {\underline{\sf{9}}}& {\sf{\:\:8649\:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \: \: \:81 \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{183}}}& {\sf{\:\: 549   \:\:}} \\{\sf{}}& \underline{\sf{\:\:549 \:  \:}} \\ {\underline{\sf{}}}& {\sf{\: \: 0\:\:}} \end{array}\end{gathered}\end{gathered}$

Hence,

$\rm\implies \: \sqrt{8649} = 93$

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ADDITIONAL INFORMATION

1. The number of digits in square root of consisting of 2n + 1 number of digits is n + 1.

2. The number of digits in square root of consisting of 2n number of digits is n.

3. The number of natural number lying between squares of two consecutive natural number n and n + 1 is 2n.

4. The sum of first n odd natural numbers is square of number of odd natural numbers.

For example :- 1 + 3 + 5 + 7 = 4² = 16

Answer 2

Answer in attachment

hope it helpful for you