So, we know that:
[tex]A=\dfrac{1}{2}ab\\P=a+b+c\\a^2+b^2=c^2[/tex]
We also know that [tex](a+b)^2=a^2+2ab+b^2[/tex]. We will use that fact to find [tex]c[/tex] in terms of [tex]A[/tex] and [tex]P[/tex].
[tex]A=\dfrac{1}{2}ab\Rightarrow 2ab=4A\\P=a+b+c\Rightarrow a+b=P-c[/tex]
Therefore:
[tex](P-c)^2=c^2+4A\\P^2-2cP+c^2=c^2+4a\\2cP=P^2-4A\\c=\dfrac{P^2-4A}{2P}[/tex]
