# Ex.6.3 Q6 Squares and Square Roots Solutions - NCERT Maths Class 8

## Question

For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

(i) \(252\)

(ii) \(2925\)

(iii) \(396\)

(iv) \(2645\)

(v) \(2800\)

(vi)\(1620\)

## Text Solution

**What is known?**

Numbers

**What is unknown?**

The smallest whole number by which it should be divided so as to get a perfect square number.

**Reasoning:**

To get a perfect square, each factor of the given number must be paired.

**Steps:**

(i) \(252\)

Hence, prime factor \(7\) does not have its pair. If number is divided by \(7\) then rest of the prime factor will be in pairs.

Therefore, \(252\) has to be divided by \(7\) to get a perfect square.

\[252 \div 7 = 36\]

\(36 \) is perfect square

\[\begin{align}36 &= 2 \times 2 \times 3 \times 3\\ &= {2^2} \times {3^2}\\ &= {(2 \times 3)^2}\\\sqrt {36} &= 2 \times 3 = 6\end{align}\]

(ii) \(2925\)

Hence, prime factor \(13\) does not have its pair. If number is divided by \(13\) then rest of the prime factor will be in pairs.

Therefore, \(2925\) has to be divided by \(13\) to get a perfect square.

\[2925 \div 13 = 225\]

\(225\) is a perfect square

\[\begin{align}225 &= 5 \times 5 \times 3 \times 3\\&= {5^2} \times {3^2}\\&= {(5 \times 3)^2}\\\sqrt {225}&= 15\end{align}\]

(iii) \(396\)

Hence, prime factor \(11\) does not have its pair. If number is divided by \(11\) then rest of the prime factor will be in pairs. Therefore, \(396\) has to be divided by \(11\) to get a perfect square.

\[\begin{align} 396 \div 11 &= 36\\ 36 &= 3 \times 3 \times 2 \times 2\\&= {3^2} \times {2^2}\\&= {(3 \times 2)^2}\\\sqrt {36}&= 3 \times 2 = 6\end{align}\]

\(36\) isa perfect square

(iv) \(2645\)

Hence, prime factor \(5\) does not have its pair. If number is divided by \(5\) then rest of the prime factor will be in pairs. Therefore, \(2645\) has to be divided by \(5\) to get a perfect square.

\[2645 \div 5 = 529\]

\(592\) is a perfect square

\[\begin{align}529 &= 23 \times 23\\&= {23^2}\\\sqrt {529}&= 23\end{align}\]

(v) \(2800\)

Hence, prime factor \(7\) does not have its pair. If number is divided by \(7\) then rest of the prime factor will be in pairs. Therefore, \(2800\) has to be divided by \(7\) to get a perfect square

\[2800 \div 7 = 400\]

\(400\) is a perfect square

\[\begin{align}400 &= 2 \times 2 \times 2 \times 2 \times 5 \times 5\\ &= {2^2} \times {2^2} \times {5^2}\\

&= {(2 \times 3 \times 5)^2}\\\sqrt {400} &= 2 \times 2 \times 5\\ &= 20\end{align}\]

(vi) \(1620\)

Hence, prime factor \(5\) does not have its pair. If number is divided by \(5\) then rest of the prime factor will be in pairs. Therefore, \(1620\) has to be divided by \(5\) to get a perfect square.

\[1620 \div 5 = 324\]

\(324\) is a perfect square

\[\begin{align}324 &= 2 \times 2 \times 3 \times 3 \times 3 \times 3\\&= {2^2} \times {3^2} \times {3^2}\\

&= {(2 \times 3 \times 3)^2}\\\sqrt {324}&= 2 \times 3 \times 3\\&= 18\end{align}\]