Finding the third side of a triangle given the area

By cosine rule, begin{align} c^2&=a^2+b^2-2abcosgamma tag{1}label{1} , end{align}
And from the formula for the area begin{align} 2S&=absingamma ,\ singamma&=frac {2S}{ab} , end{align}

so, with begin{align} cosgamma&=pmsqrt{1-sin^2gamma} = pmsqrt{1-frac {4S^2}{a^2b^2}} =pmfrac{sqrt{a^2b^2-4S^2}}{ab} end{align}

eqref{1} becomes

begin{align} c^2&=a^2+b^2pm2sqrt{a^2b^2-4S^2} . end{align}

Suppose we know $$K = |triangle ABC|, quad a, b,$$ and we wish to determine $c$. Then $$begin{align*} 16K^2 &= (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \ &= left((a+b)^2 - c^2 right)left(c^2 - (a-b)^2 right) \ &= -c^4 + left( (a-b)^2 + (a+b)^2 right) c^2 - (a+b)^2 (a-b)^2 \ &= -c^4 + 2(a^2 + b^2) c^2 - (a^2 - b^2)^2. end{align*}$$ This of course is a quadratic in $c^2$. We can complete the square to get $$begin{align*} 0 &= (c^2 - (a^2+b^2))^2 + (a^2-b^2)^2 - (a^2+b^2)^2 + 16K^2 \ &= (c^2 - (a^2+b^2))^2 - 4a^2 b^2 + 16K^2, end{align*}$$ from which we obtain $$c^2 = a^2 + b^2 pm 2sqrt{a^2 b^2 - 4K^2}.$$ Note that we must always have $ab ge 2K$, as the area of the triangle is maximized when the angle between sides of fixed length $a$ and $b$ is right; thus the square root is always well-defined. Furthermore, the AM-GM inequality guarantees that when $a, b > 0$, $a^2 + b^2 ge 2 sqrt{a^2 b^2}$, hence $c^2$ admits exactly two positive solutions whenever the strict inequality $ab > 2K$ is satisfied, and one unique solution when $ab = 2K$ and the triangle is right.

$s;$ is the conventional way of representing the semiperimeter $frac{a+b+c}2$ of the triangle with sides $a,,b,,c$.

$sqrt{s(s-a)(s-b)(s-c)}= frac{sqrt{-a^4-b^4-c^4+2b^2c^2+2c^2a^2+2a^2b^2}}4=Delta$

Solving for $c$, $$c=sqrt{a^2+b^2pm 2sqrt{a^2b^2-4Delta^2}}$$

where area $=Delta$