Define a $1$-$1$ onto function with domain $A$ onto the set $\{1, 2, ... n\}$ Let $A = \{x^2 : x \in \mathbb{N} \text{ and } 0 \leq x^2 \leq 90\}$.
Define a 1-1 onto function with domain $A$ onto a set of the form $\{1, 2, \ldots, n\}$ to show the cardinality of $A$ is $n$.

I understand that if a bijection exists between two sets, then the cardinality of the sets must be equal. 
What I do not understand, is that it seems the domain $A$ has a fixed cardinality of $9$. This is the size of the set of the squares of the natural numbers that fit $0 \leq x^2 \leq 90$.
However, if we pick a value of $n \neq 9$, how can a bijection exist?
 A: $A = \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81\}$
In this case, you need $n = 10:$ there are 10 values $x$ including $0$ such that $0\leq x^2 \leq 90$. 
So the range is $B = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
Try $f(a) = \sqrt{a} + 1$, where $a = x^2 \in A$.
Now just show that $f: A \to B$ is one-to-one and onto. 
Also note, you can show that for any $n > 10$, any function $g: A \to B_n$ would fail to be onto $B_n$, where $B_n$ is the set $\{1, 2, 3, ... , n\}, n>10$. And show that any function $h: A \to B_n$ would fail to be injective if $B_n$ is the set $\{1, 2, ... n\},$ $1 \leq n\lt 10$.

If $A$ cannot contain $0^2 = 0$, then use the following:
Simply use $A = \{1, 4, 9, 16, 25, 36, 49, 64, 81\}$, 
$B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, (so $n = 9$), 
$f(x) = \sqrt a, \;a\in A$, for your one-to-one function $f:A\to B$, 
Then show that $f$ is one-to-one and onto. In this case, you can show why, for any $n > 9,\,$ any function $\,g: A \to B_n\,$ would fail to be onto $\,B_n$, where $\,B_n\,$ is the set $\{1, 2, 3, ... , n\}, n>9$. And finally, show why any function $\,h: A \to B_n\,$ would fail to be injective if $\,B_n\,$ is the set $\{1, 2, ... n\},$ $1 \leq n\lt 9$.
