Find the sides of a right triangle given the area and perimeter

Question

I was sure that this has been asked before, but the title did not match.

So if the area of a right triangle is A and the perimeter is P, express the sides a and b in terms of A and P.

Answer 1

$a+b+\sqrt{a^2+b^2}=a+b+\sqrt{(a+b)^2-4A}=P, {a}{b}=2A$

This leads to $P^2-2P(a+b)=-4A$, hence $a+b=\frac{P^2+4A}{2P}$

So $a$ and $b$ are roots of the quadratic equation $x^2-\frac{P^2+4A}{2P}x+2A=0$