Don't be so negative

According to the given conditions, we have to work on numbers $A ≤ 6$ .

It's clear that A is not a positive number because of $A^2 = 6^2 = 36,$ which is not greater than 36.

So A must be negative. $e.g. ~ A^2 = (-7)^2 = 49,$ which is greater than 36.

Hence, it is $\color{#D61F06} \boxed{\text{no}}.$

The problem states that $A$ is a number , and it doesn't specify what kind. This means that it could be any kind of number. It also states $A ≤ 6$ . Since we are not limited to just positive numbers, we can also assume $A$ is a negative number too.

If the letter A is equal to -7, $(-7)^2$ = 49

Therefore, $A^2$ is not necessarily less than 36.

If $0\le A<6\\ 0\le { A }^{ 2 }<36$

Negative numbers if squared, the results are always positive.