Question

Please solve with explanation (high points)

Answer 1

So, we know that:

$A=\dfrac{1}{2}ab\\P=a+b+c\\a^2+b^2=c^2$

We also know that $(a+b)^2=a^2+2ab+b^2$. We will use that fact to find $c$ in terms of $A$ and $P$.

$A=\dfrac{1}{2}ab\Rightarrow 2ab=4A\\P=a+b+c\Rightarrow a+b=P-c$

Therefore:

$(P-c)^2=c^2+4A\\P^2-2cP+c^2=c^2+4a\\2cP=P^2-4A\\c=\dfrac{P^2-4A}{2P}$