5.Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.

(i) 525

(ii) 1750

(iii) 252

(iv) 1825

(v) 6412

(i) 525

(ii) 1750

(iii) 252

(iv) 1825

(v) 6412

(i)

\(\qquad \begin{array}{c|lcr}
&22\\
\hline
2 &\;\; \;\bar{5}\bar{25}\\
&-4\\
\hline
42 &\;\;\; 125\\
&\;\;\; 84\\
\hline
&\;\;\;41\\
\hline
\end{array}
\)

So we got remainder as 41.It represents that the square of 22 is less than 525.So, the next number is 23 and \(23^2 = 529\).

Hence, the number to be added = 529 – 525 = 4

∴Required perfect square = 529

Thus we have, \(\sqrt{529}=23\)

(ii)

\(\qquad \begin{array}{c|lcr}
&41\\
\hline
4 &\;\; \;\bar{17}\bar{50}\\
&-16\\
\hline
81 &\;\;\; 150\\
&\;\;\; 81\\
\hline
&\;\;\;69\\
\hline
\end{array}
\)

So we got remainder as 69.It represents that the square of 41 is less than 1750.So, the next number is 42 and \(42^2 = 1764\).

Hence, the number to be added = 1764– 1750 =14

∴Required perfect square = 1764

Thus we have, \(\sqrt{1764}=42\)

(iii)

\(\qquad \begin{array}{c|lcr}
&15\\
\hline
1 &\;\; \;\bar{2}\bar{52}\\
&-16\\
\hline
25 &\;\;\; 152\\
&\;\;\; 125\\
\hline
&\;\;\;27\\
\hline
\end{array}
\)

So we got remainder as 27.It represents that the square of 15 is less than 252.So, the next number is 16 and \(16^2 = 256\).

Hence, the number to be added = 256 – 252 = 4

∴Required perfect square = 256

Thus we have, \(\sqrt{256}=16\)

(iv)

\(\qquad \begin{array}{c|lcr}
&42\\
\hline
4 &\;\; \;\bar{18}\bar{25}\\
&-16\\
\hline
82 &\;\;\; 225\\
&\;\;\; 164\\
\hline
&\;\;\;61\\
\hline
\end{array}
\)

So we got remainder as 61.It represents that the square of 42 is less than 1825.So, the next number is 43 and \(43^2 = 1849\).

Hence, the number to be added = 1849 – 1825 = 24

∴New number = 1849

Thus we have,\(\sqrt{1849}=43\)

(v)

\(\qquad \begin{array}{c|lcr}
&80\\
\hline
8 &\;\; \;\bar{64}\bar{12}\\
&-64\\
\hline
160 &\;\;\; 12\\
&\;\;\; 0\\
\hline
&\;\;\;12\\
\hline
\end{array}
\)

So we got remainder as 12.It represents that the square of 80 is less than 6412.So, the next number is 81 and \(81^2 = 6561\).

Hence, the number to be added = 6561 – 6412 = 149

∴New number = 6561

Thus, we have \(\sqrt{6561}=81\)