# Ex.6.3 Q5 Squares and Square Roots - NCERT Maths Class 8

## Question

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

(i) \(252 \)

(ii) \(180\)

(iii) \(1008\)

(iv) \(2028 \)

(v) \(1458 \)

(vi) \(768\)

## Text Solution

**What is known?**

Numbers

**What is unknown?**

The smallest whole number by which it should be multiplied 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 \(7\) gets a pair, then the number becomes a perfect square. Therefore, \(252\) has to be multiplied with \(7\) to get a perfect square.

So,** **perfect square is \(252 \times 7 = 1764\)

\(1764 \! = \! 2 \! \times \! 2 \! \times \! 3 \! \times \! 3 \! \times \! 7 \! \times \! 7\)

So, perfect root of \(1764\) is \( 2 \times 3 \times 7 = 42\)

(ii) \(180\)

Hence, prime factor \(5\) does not have its pair. If \(5\) gets a pair, then the number becomes a perfect square. Therefore, \(180\) has to be multiplied with \(5\) to get a perfect square.

So, perfect square is \(180 \times 5 = 900\)

\(900 \! = \! 2 \! \times \! 2 \! \times \! 3 \! \times \! 3 \! \times \! 5 \! \times \! 5\)

So, perfect root of \(900\) is \( 2 \times 3 \times 5 = 30\)

(iii) \(1008\)

Hence, prime factor \(7\) does not have its pair. If \(7\) gets a pair, then the number becomes a perfect square. Therefore, \(1008\) has to be multiplied with \( 7\) to get a perfect square.

So, perfect square is \(1008 \times 7 = 7056\)

\(7056 \! = \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 3 \! \times \! 3 \! \times \! 7 \! \times \! 7\)

So, perfect root of

\(7056\) is \(2 \times 2 \times 3 \times 7= 84\)

(iv) \(2028\)

Hence, prime factor \(3\) does not have its pair. If \(3\) gets a pair, then the number becomes a perfect square. Therefore, \(2028 \) has to be multiplied with \(3\) to get a perfect square.

So, perfect square is \(2028 \times 3 = 6084\)

\(6084 \! = \! 2 \! \times \! 2 \! \times \! 13 \! \times \! 13 \! \times \! 3 \! \times \! 3\)

So, perfect root of \(6084 \) is \( 2 \times 13 \times 3 = 78\)

(v) \(1458\)

Hence, prime factor \(2\) does not have its pair. If \(2\) gets a pair, then the number becomes a perfect square. Therefore, \(1458\) has to be multiplied with \(2\) to get a perfect square.

So, perfect square is \(1458 \times 2 = 2916\)

\(2916 \! = \! 3 \! \times \! 3 \! \times \! 3 \! \times \! 3 \! \times \! 3 \! \times \! 3 \! \times \! 2 \! \times \! 2 \)

So, perfect root of \(2916\) is \(3 \! \times \! 3 \! \times \! 3 \! \times \! 2 \! = \! 54\)

(vi) \(768\)

Hence, prime factor \(3\) does not have its pair. If \(3\) gets a pair, then the number becomes a perfect square. Therefore, \(768\) has to be multiplied with \(3\) to get a perfect square.

So, perfect square is \( 768 \times 3 = 2304\)

\(2304 \! = \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 3 \! \times \! 3\)

So, perfect root of \(2304 \) is \(2 \! \times \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 3 \! = \! 48\)