$$\therefore 390625 =(625)^2$$

The value of 22*22 is 484..

So, the square root of $$1156$$ is $$34$$.

$$\therefore \sqrt {69696}=264$$

$$\therefore \sqrt{7714} = 88$$

The sum of digits of the given number $$5184 (5+1+8+4 = 18)$$ is multiple of $$3$$. Also, since the the number at unit's place is even (i.e, $$4$$), therefore, it is divisible by $$2$$.

The square of the numbers must end with any of the digits $$ 0 $$, $$ 1 $$, $$ 4 $$, $$ 5 $$, $$6$$  or $$ 9  $$. 
Example:-
 $$ { 1 }^{ 2 } = 1 $$
$${ 12 }^{ 2 } = 144$$
$${ 13 }^{ 2 } = 169$$
$${ 5 }^{ 2 } = 25$$ 
Also, a perfect square has even number of zeros at the end. Example :- $$100, 400, 640000$$
Here, the number is $$23453 $$.
It ends with $$ 3 $$. Therefore, it is not a perfect square.

The square of any number must end with any of the digits $$ 0 $$, $$ 1 $$, $$ 4 $$, $$ 5 $$, $$6$$  or $$ 9  $$.
Examples:-
$$ { 1 }^{ 2 } = 1 $$
$${ 12 }^{ 2 } = 144$$
$${ 13 }^{ 2 } = 139$$
$${ 5 }^{ 2 } = 25$$
Here, the number is $$1057$$.
It ends with $$7$$. Therefore, it is not a perfect square.

For example,
 $$ { 1 }^{ 2 } = 1 $$
$${ 12 }^{ 2 } = 144$$
$${ 13 }^{ 2 } = 169$$
$${ 5 }^{ 2 } = 25$$ 

The square of the numbers must end with any of the digits $$ 0 $$, $$ 1 $$, $$ 4 $$, $$ 5 $$, $$6$$  or $$ 9  $$. 
Example:-
 $$ { 1 }^{ 2 }\quad =\quad 1 $$
$${ 12 }^{ 2 }\quad =\quad 144$$
$${ 13 }^{ 2 }\quad =\quad 139$$
$${ 5 }^{ 2 }\quad =\quad 25$$ 

The square of any number ends in the digits $$ 0 $$, $$ 1 $$, $$ 4 $$, $$ 5 $$, $$6$$  or $$ 9  $$. 
Example:-
 $$ { 1 }^{ 2 } = 1 $$
$${ 12 }^{ 2 }= 144$$
$${ 13 }^{ 2 } = 139$$
$${ 5 }^{ 2 } = 25$$ 
Also, a perfect square has even number of zeros at the end. Example :- $$100, 400, 640000 $$.
Here, the number is $$7928$$.
It ends with $$ 8 $$. Therefore it is not a perfect square.

$$1764\, =\, 4\, \times\, 7\, \times\, 7\, \times\, 9\, =\, 2^2\, \times\, 7^2\, \times\,3^2$$
$$\therefore\, \sqrt{1764}\, =\, \sqrt{2^2\, \times\, 7^2\, \times\, 3^2}$$
$$=\, 2\, \times\, 7\, \times\, 3$$
$$=\, 42$$

$$3844=62\times 62$$

Only the square roots of perfect squares will be rational. If it of the form $$\sqrt{a}$$, whereas if $$'a'$$ is not a perfect square then the term will be irrational. $$4, 9$$ are perfect squares.

$$53361\, =\, 11\, \times\, 11\, \times\, 21\, \times\, 21$$
$$=\, 11^2\, \times\, 21^2$$
$$\therefore\, \sqrt{53361}\, =\, \sqrt{11^2\, \times\, 21^2}$$
$$=\, 11\, \times\, 21\, =\, 231$$

The given statement is sometimes true and sometimes it is false.

$$196=2\times 2\times 7\times 7=(2\times 7)^2=(14)^2$$
Hence 196 is perfect square of 14.

So, a number having $$2,3,7$$ or $$  8$$ at unit's place is never a perfect square.

All the perfect square end with any one from $$0,1,4,5,6,9$$

All the perfect square end with $$0,1,4,5,6,9$$

By definition, an odd number is an integer that can be written in the form $$2n + 1$$, for some integer $$n$$, as an odd number leaves remainder 1 when divided by 2.

$$7 \times 7 = 49$$. 

So, $$89722$$ is not a perfect square as the number ends in $$2$$. 

$$222222$$ is not a perfect square as a perfect square never ends in $$2$$.

Square root of $$9 = 3 \times 3 = 3^2$$ is $$3$$

$$1 \times 1 = 1$$

$$2 \times 2 = 4$$

$$3 \times 3 = 9$$

$$4 \times 4 = 16$$

$$5\times 5 = 25$$

All are square numbers, so they are perfect squares.

Therefore, A is the correct answer.