Question
Mathematics Question on Square Roots of Decimals
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.
- 402
- 1989
- 3250
- 825
- 4000
(i) The square root of 402 can be calculated by long division method as follows.
| 20
---|---
2| 4ˉ0ˉ2ˉ
−4
4| 02
00
| 2
The remainder is 2.
It represents that the square of 20 is less than 402 by 2.
Therefore, a perfect square will be obtained by subtracting 2 from the given number 402.
Therefore, required perfect square = 402−2=400
And, 400=20
(ii) The square root of 1989 can be calculated by long division method as follows.
| 44
---|---
4| 1ˉ9ˉ8ˉ9ˉ
−16
84| 389
336
| 53
The remainder is 53.
It represents that the square of 44 is less than 1989 by 53.
Therefore, a perfect square will be obtained by subtracting 53 from the given number 1989.
Therefore, required perfect square = 1989−53=1936
And, 1936=44
(iii) The square root of 3250 can be calculated by long division method as follows.
| 57
---|---
5| 3ˉ2ˉ5ˉ0ˉ
−25
107| 750
749
| 1
The remainder is 1.
It represents that the square of 57 is less than 3250 by 1.
Therefore, a perfect square can be obtained by subtracting 1 from the given number 3250.
Therefore, required perfect square = 3250−1=3249
And, 3249=57
(iv) The square root of 825 can be calculated by long division method as follows.
| 28
---|---
2| 8ˉ2ˉ5ˉ
-4
48| 425
384
| 41
The remainder is 41.
It represents that the square of 28 is less than 825 by 41.
Therefore, a perfect square can be calculated by subtracting.
(v) 4000 We know that, if we subtract the remainder from the number, we get a perfect square.
| 63
---|---
6| 4ˉ0ˉ0ˉ0ˉ
−36
123| 400
-369
| 31
Here, we get remainder 31.
Therefore 31 must be subtracted from 4000 to get a perfect square.
∴ 4000–31=3969