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

## 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 \(6412 + 149 = 6561 \)

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