Let side of two squares be a₁ and a₂·
Then, perimeter of first square = 4a₁
According to the question,
 Perimeter = 40
\(\therefore\) 4a₁ = 40 \(\Rightarrow\) a₁ = 10
Similarly, second square's perimeter = 4a₂
But according to the question,
Perimeter = 96
\(\therefore\) 4a₂ = 96 \(\Rightarrow\) a₂ = 24
Let a be the side of another square.
Now, sum of the areas of first and second square area of
= Area of (first square) + (Area of second square)
=\({ a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }\)
Area =(10)² + (24)² =100 + 576
\(\therefore\) Area = 676
\(\Rightarrow\)a² = 676 \(\Rightarrow\) a = \(\sqrt { 676 } \)
Then, a=26 m
Another square's perimeter = 4a = 4 x 26 = 104 m
So, perimeter of another square is 104 m.

\(\frac { 401 }{ 607 } \)

If a number contains 3 zeros at the end, then its square will have 3\(\times\)2 = 6 zeros at the end
The number of zeros at the end of a number
=\(\frac { 1 }{ 2 } \times \) The number of zeros at the end of its square
Yes ; we can say that square numbers can only have even number of zeros at the end.

Since 6 x 6 = 36
\(\therefore\) The unit digit of (12796) will be 6.

We have, \(\sqrt { 16+\sqrt { 81 } } =\sqrt { 16+9 } =\sqrt { 25 } =5\)

9604

Given number is 390625.
Here, number of digits, n = 6             [even]
\(\therefore\) Number of digits in the square root of 390625
\(=\frac { n }{ 2 } =\frac { 6 }{ 2 } =3\)
Hence, the number of digits in the square root of 390625 is 3 digits.

Given number is 36864.
Here, number of digits, n = 5       [odd]
\(\therefore\) Number of digits in the square root of 36864
\(\frac { n+1 }{ 2 } =\frac { 5+1 }{ 2 } =\frac { 6 }{ 2 } =3\)
Hence, the number of digits in the square root of 36864 is 3 digits.

The possible one's digits of the square root of the numbers.
In a given number 657666025, one's digit is 5.
We know that, square of5 gives out 5 in one's digit.
Hence, possible one's digit of the square root digit is 5.

We have, 11² = 121
Therefore, the square root of 121 is 11.

Let 2m = 18
⇒ m  =\(\frac { 18 }{ 2 } \) =9
∴ m² - 1 = 9² - 1 = 81 - 1 = 80
and m² +1 = 9² +1 = 81 + 1 = 80
So, a Pythagorean triplet, whose one number is 18, is 18, 80, 82.

Let 2m = 16
⇒ m =\(\frac { 16 }{ 2 } \) = 8
∴  m²- 1 = 8² - 1 = 64 - 1 = 63
and m²+ 1 = 8² + 1 = 64 +1 = 65
So, a Pythagorean triplet, whose one number is 16, is 16, 63, 65

15² = 1\(\times\)(1+1)\(\times\)hundreds + 25
=(1\(\times\)2)\(\times\)100 + 25 = 200 + 25 = 225

The given number 222000 ends with 0, so it cannot be a perfect square due to odd number of zeroes.

Given number is 272.
Unit's digit of 272 = 2
So, the square of unit's digit = 2² = 4
Hence, unit's digit of the square of unit's digit of the given number is 4.

Given, pair of numbers 100² and 101².
Non-square numbers between 100² and 101² = 2n
= 2 \(\times\) 100 = 200 [\(\because\)n=100]

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.

Since, number 26 ends with digit 6, so the number getting from the square of 26, will be the end of the digit 6.

We know that, if a number has 1 or 9 in the unit's place, then its square ends with digit 1. Therefore,161² and 109² would end with digit 1.

We know that,
7 X 7 = 49 and 8 X 8 = 64
Hence, there is no perfect square number lying between 50 and 60.