 ## Book: NCERT - Mathematics

### Chapter: 6. Squares and Square Roots

#### Subject: Maths - Class 8th

##### Q. No. 4 of Exercises 6.4

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##### 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

(i) The square root of 402 can be calculated by long division method as follows:

 2 22 4̅ 0̅2̅4 4 002

The remainder is 2

This represents that the square of 20 is less than 402 by 2.

Hence, a perfect square will be obtained by subtracting 2 from the given number 402

Therefore, required perfect square = 402 - 2

= 400 = 20

(ii) The square root of 1989 can be calculated by long division method as follows:

 44 198916 8 389

The remainder is 53

This represents that the square of 44 is less than 1989 by 53

Hence, a perfect square will be obtained by subtracting 53 from the given number 1989.

Therefore, required perfect square = 1989 - 53

= 1936 = 44

(iii) The square root of 3250 can be calculated by long division method as follows:

 55 325025 10 750

The remainder is 1

This represents that the square of 57 is less than 3250 by 1

Hence, a perfect square can be obtained by subtracting 1 from the given number 3250.

Therefore, required perfect square = 3250 - 1

= 3249 = 57

(iv) The square root of 825can be calculated by long division method as follows:

 22 8254 4 425

The remainder is 41

This represents that the square of 28 is less than 825 by 41

Hence, a perfect square can be calculated by subtracting 41 from the given number 825

Therefore, required perfect square = 825 – 41

= 784 = 28

(v) The square root of 4000 can be calculated by long division method as follows:

 66 400036 12 400

The remainder is 31

This represents that the square of 63 is less than 4000 by 31

Hence, a perfect square can be calculated by subtracting 31 from the given number 4000

Therefore, required perfect square = 4000 – 31

= 3969 = 63

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