NCERT Class 8 Mathematics Solutions: Chapter 6 – Squares and Square Roots Exercise 6.3 Part 4
Get top class preparation for NSTSE right from your home: fully solved questions with step-by-step explanation- practice your way to success.
Question: 5 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
Answer:
(i) 252
2 | 252 |
2 | 126 |
3 | 63 |
3 | 21 |
7 | 7 |
1 |
Here, prime factor 7 has no pair.
Therefore 252 must be multiplied by 7 to make it a perfect square.
(ii) 180
2 | 180 |
2 | 90 |
3 | 45 |
3 | 15 |
5 | 5 |
1 |
Here, prime factor 5 has no pair.
Therefore 180 must be multiplied by 5 to make it a perfect square.
(iii) 1008
2 | 1008 |
2 | 504 |
2 | 252 |
2 | 126 |
3 | 63 |
3 | 21 |
7 | 7 |
1 |
Here, prime factor 7 has no pair.
Therefore 1008 must be multiplied by 7 to make it a perfect square.
(iv) 2028
2 | 2028 |
2 | 1014 |
3 | 507 |
13 | 169 |
13 | 13 |
1 |
Here, prime factor 3 has no pair.
Therefore 2028 must be multiplied by 3 to make it a perfect square.
(v) 1458
2 | 1458 |
3 | 729 |
3 | 243 |
3 | 81 |
3 | 27 |
3 | 9 |
3 | 3 |
1 |
Here, prime factor 2 has no pair.
Therefore 1458 must be multiplied by 3 to make it a perfect square.
(vi) 768
2 | 768 |
2 | 384 |
2 | 192 |
2 | 96 |
2 | 48 |
2 | 24 |
2 | 12 |
2 | 6 |
3 | 3 |
1 |
Here, prime factor 3 has no pair.
Therefore 768 must be multiplied by 3 to make it a perfect square.