If $\, 5$ & $4\to 100$ , $8$ & $5 \to 400$ , $2$ & $19 \to 361$ , $A$ & $5 \to 625$ , $12$ & $3 \to 324$ , $7$ & $6 \to 441$ ; then $A=?$

Question

Suppose it is given :
$\quad \, 5$ & $4\to 100$ ,
$\quad 8$ & $5 \to 400$ ,
$\quad 2$ & $19 \to 361$ ,
$\quad A$ & $5 \to 625$ ,
$\quad 12$ & $3 \to 324$ ,
$\quad 7$ & $6 \to 441$
$\quad$;then $A=?$
where we are getting $100$ from $5$ and $4$ via some logic or combinations of operations
similarly, $400$ from $8$ and $5$ via the same logic as used in $100$
and so on ...

My Thoughts:
$100=10^2$ , $400= 20^2$ , $361=19^2$ , $625=25^2$ , $324=18^2$ , $441=21^2$
$\implies$ all RHS are perfect square

For $1st$ relation,
i.e, $\, 5$ & $4\to 100$
LOGIC: $\quad 5^2 * 4 =100$
but the same logic don't apply to 2nd relation,
i.e, $8^2 * 5 =320 \neq 400$
so, how to solve this reasoning problem? Any suggestions please...

Answer 1

Note that $100=10^2=(\frac{5\cdot 4}2)^2$, $400=20^2=(\frac{8\cdot 5}2)^2$, etc. Given the information, it seems that $a\&b \to (\frac{ab}2)^2$. So $A$ would be $10$ (if only $A>0$ is allowed.)