Ex.6.4 Q5 Squares and Square Roots - NCERT Maths Class 8

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Question

Q5. 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) \(6412\)

Text Solution

What is known?

Numbers that are not perfect square

What is unknown?

What must be added to the numbers so as to get perfect square

Steps:

(i)

Square root of \(525\) is calculated by long division method as follows.

It is evident that \({22^2} < 525\)

Next, the perfect square of \(23 \) is

\({23^2} = 529\)

Hence, the number to be added to \(529\)

\[\begin{align}&= {23^2} - 525\\&= 529 - 525\\&= 4\end{align}\]

The required perfect square is \(525 + 4 = 529\)

\(\sqrt {529}= 23\)

(ii)

Square root of \(1750 \) can be calculated by the long division method as follows.

The remainder is \(69\).

This shows that \({41^2} < 1750\)

Next perfect square is \({42^2} = 1764\)

Hence number to be added to \(1750 \)

\[\begin{align}&= {42^2} - 1750\\&= 1764 - 1750\\&= 14\end{align}\]

The required perfect square is \(1750 + 14 = 1764 \)

\(\sqrt {1764}= 42\)

(iii)

Square root of \(252\) is calculated as follows.

 

 

 

 

 

 

The remainder is \(27.\)

This shows that \({15^2} < 252\)

Next perfect square is \({16^2} = 256\)

Hence number to be added to \(252 \)

\[\begin{align}&= {16^2} - 252\\&= 256 - 252\\&= 4\end{align}\]

The required perfect square is \(252 + 4 = 256\)

\({\rm{}}\;\sqrt {256}= 16\)

(iv) 

Square root of \(1825\) is calculated as follows.

The remainder is \(61.\)

This shows that \({42^2} < 1825\)

Next perfect square is \({43^2} = 1849\)

Hence number to be added to \(1825 \)

\[\begin{align}&= {43^2} - 1825\\&= 1849 - 1825\\&= 24\end{align}\]

The required perfect square is \(1825 + 24 = 1849\)

\({\rm{And}}\;\sqrt {1849}= 43\)

(v)

Square root of \(6412 \) is calculated as follows.

\({80^2} < 6412\)The remainder is \(12. \)This shows that

Next perfect square is \({81^2} = 6561\)

Hence number to be added to \(6412\) is

\[\begin{align}&= {16^2} - 252\\&= 256 - 252\\&= 4\end{align}\]

The required perfect square is \(6412 + 149 = 6561 \)

\({\rm{And}}\;\sqrt {6561}= 81\)