NCERT Solutions
Class 8 Maths
Squares and Square Roots

Ex.6.4 Q.5
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:
1. 525
2. 1750
3. 252
4. 1825
5. 6412
1. 525
Since remainder is 41. Therefore 222 < 525
Next perfect square number 232 = 529
Hence, number to be added = 529 – 525 = 4
So, 525 + 4 = 529
Hence, the square root of 529 is 23.
2. 1750
Since remainder is 69. Therefore 412 < 1750
Next perfect square number 422 = 1764
Hence, number to be added = 1764 – 1750 = 14
So, 1750 + 14 = 1764
Hence, the square root of 1764 is 42.
3. 252
Since remainder is 27. Therefore 152 < 252
Next perfect square number 162 = 256
Hence, number to be added = 256 – 252 = 4
So, 252 + 4 = 256
Hence, the square root of 256 is 16.
4. 1825
Since remainder is 61. Therefore 422 < 1825
Next perfect square number 432 = 1849
Hence, number to be added = 1849 – 1825 = 24
So, 1825 + 24 = 1849
Hence, the square root of 1849 is 43.
5. 6412
Since remainder is 12. Therefore 802 < 6412
Next perfect square number 812 = 6561
Hence, number to be added = 6561 – 6412 = 149
So, 6412 + 149 = 6561
Hence, the square root of 6561 is 81.