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) \(768\)

 Video Solution
Squares And Square Roots
Ex 6.3 | Question 5

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 \(2 \! \times \! 2 \! \times \! 2 \! \times \! 2 \! \times \! 3 \! = \! 48\)

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