When 'n' is a member of a Pythagorean
triplet, then the triplet is
n² - 1, 2n, n² + 1
or (13²- 1), (2 x 13), (13² + 1)
or (169 - 1), (26), (169 + 1)
or 168, 26 and 170

We know that, between n2 and (n + 1)², there are 2n non-square numbers.
\(\therefore\) Between 27 and 28, there are 2 x 27, i.e, 54 numbers.

Here, n = 19
∴ \(\frac { { n }^{ 2 }-1 }{ 2 } =\frac { { 19 }^{ 2 }-1 }{ 2 } =\frac { 361-1 }{ 2 } \)
= \(\frac { 360 }{ 2 } \) = 180
and  \(\frac { { n }^{ 2 }+1 }{ 2 } =\frac { { 19 }^{ 2 }+1 }{ 2 } =\frac { 361+1 }{ 2 } \)
 = \(\frac { 362 }{ 2 } \) = 181
∴ 19² = 180 + 181 = 361

Here , n = 11
∴ \(\frac { { n }^{ 2 }-1 }{ 2 } =\frac { { 11 }^{ 2 }-1 }{ 2 } =\frac { 121-1 }{ 2 } \)
=\(\frac { 120 }{ 2 } \)= 60
and  \(\frac { { n }^{ 2 }+1 }{ 2 } =\frac { { 11 }^{ 2 }+1 }{ 2 } =\frac { 121+1 }{ 2 } \)
=\(\frac { 121+1 }{ 2 } \)=61
∴  11² = 60 + 61 = 121.

\(\because \quad 6.1=\frac { 61 }{ 10 } \)
\(\therefore \quad { (6.1) }^{ 2 }={ \left( \frac { 61 }{ 10 } \right) }^{ 2 }=\frac { 61\times 61 }{ 10\times 10 } =\frac { 3721 }{ 100 } =37.21\)

4, 9,16,25,36 and 49 are six perfect squares, which lie between 1 and 50.

\(\frac { 39 }{ 11 } \)

81

Given decimal number is 2.56.
Firstly, place the bar over the number and then use the division method
\(\therefore \ \sqrt { 2.56 } =1.6\)
Hence, the square root of 2.56 is 1.6.

Given number is 144.
Here, number of digits, n =3                      [odd]
\(\therefore\) Number of digits in the square root of 144
\(=\frac { n+1 }{ 2 } =\frac { 3+1 }{ 2 } =\frac { 4 }{ 2 } =2\)
Hence, number of digits in the square root of 144 is 2 digits.

Therefore,\(\sqrt { 1024 } \)= 32

We know that, the number of zeroes in the end of a number is half of the number of zeroes in the end of its square. The number of zeros in the square of 60 is 2.

Given number is 21222.
Unit's digit of 21222 = 2
So, square of unit's digit = (2)² = 4
Hence, unit's digit of the square of unit's digit of the given number is 4.

A list of five numbers for which we cannot decide whether they are square numbers just by looking at their unit's digits are 1331, 2744, 3375,17576 and 24389.

We know that, a number ends with 2, 3, 7 or 8 is neverapeIfea square.
The number 7928 ends with 8, which is not one of the end digits of 0, 1, 4, 5, 6 or 9, so it is not a perfect square.