(i) 402
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 2.
Therefore 2 must be subtracted from 402 to get a
perfect square.
∴ 402 – 2 = 400
Hence, the square root of 400 is 20.

(ii) 1989
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 53. Therefore 53 must be
subtracted from 1989 to get a perfect square.
∴ 1989 – 53 = 1936
Hence, the square root of 1936 is 44.

(iii) 3250
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 1. Therefore 1 must be
subtracted from 3250 to get a perfect square.
∴ 3250 – 1 = 3249
Hence, the square root of 3249 is 57.

(iv) 825
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 41. Therefore 41 must be
subtracted from 825 to get a perfect square.
∴ 825 – 41 = 784
Hence, the square root of 784 is 28.

(v) 4000
We know that, if we subtract the remainder from the
number, we get a perfect square.
Here, we get remainder 31. Therefore 31 must be
subtracted from 4000 to get a perfect square.
∴ 4000 – 31 = 3969
Hence, the square root of 3969 is 63.

Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/