Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000
Solution:
The given numbers that are not perfect squares
If we subtract the remainder from the number, we get a perfect square.
(i) The square root of 402 is calculated as
The square of 20 is less than 402 by 2. If we subtract the remainder from the number, we get a perfect square.
Therefore, required perfect square = 402 - 2 = 400
√400 = 20
(ii) The square root of 1989 is calculated as
The remainder obtained is 53. The square of 44 is less than the given number 1989 by 53.
Therefore, required perfect square = 1989 - 53 = 1936
√1936 = 44
(iii) The square root of 3250 is calculated as
The remainder obtained is 1. The square of 57 is less than 3250 by 1. Therefore, the required perfect square = 3250 - 1 = 3249
√3249 = 57
(iv) The square root of 825 is calculated as
The remainder is 41.it shows that the square of 28 is less than 825 by 41. Therefore, required perfect square = 825 - 41 = 784
√784 = 28
(v) The square root of 4000 is calculated as
The remainder is 31, it represents that the square of 63 is less than 4000 by 31. Therefore, the square perfect square = 4000 - 31 = 3969.
√3969 = 63
☛ Check: NCERT Solutions for Class 8 Maths Chapter 6
Video Solution:
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000
NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.4 Question 4
Summary:
The least numbers that need to be subtracted from the given numbers (i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000 to get a perfect square are (i) 2, (ii) 52, (iii) 1, (iv) 41, and (v) 31 and the square roots are as follows, 20, 44, 57, 28 and 63.
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