Michelle Mendoza

2022-07-15

Finding the third side of a triangle given the area
Perimeter =$\frac{\left(a+b+c\right)}{2}$
Area = $A=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}$
How do I simplify the above two equations to solve for c?

treccinair

By cosine rule,
$\begin{array}{}\text{(1)}& {c}^{2}& ={a}^{2}+{b}^{2}-2ab\mathrm{cos}\gamma ,\end{array}$
And from the formula for the area
$\begin{array}{rl}2S& =ab\mathrm{sin}\gamma ,\\ \mathrm{sin}\gamma & =\frac{2S}{ab},\end{array}$
so, with
$\begin{array}{rl}\mathrm{cos}\gamma & =±\sqrt{1-{\mathrm{sin}}^{2}\gamma }=±\sqrt{1-\frac{4{S}^{2}}{{a}^{2}{b}^{2}}}=±\frac{\sqrt{{a}^{2}{b}^{2}-4{S}^{2}}}{ab}\end{array}$
(1) becomes
$\begin{array}{rl}{c}^{2}& ={a}^{2}+{b}^{2}±2\sqrt{{a}^{2}{b}^{2}-4{S}^{2}}.\end{array}$

Holetaug

$s\phantom{\rule{thickmathspace}{0ex}}$ is the conventional way of representing the semiperimeter $\frac{a+b+c}{2}$ of the triangle with sides $a\phantom{\rule{thinmathspace}{0ex}},b\phantom{\rule{thinmathspace}{0ex}},c$
$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\frac{\sqrt{-{a}^{4}-{b}^{4}-{c}^{4}+2{b}^{2}{c}^{2}+2{c}^{2}{a}^{2}+2{a}^{2}{b}^{2}}}{4}=\mathrm{\Delta }$
Solving for $c$,
$c=\sqrt{{a}^{2}+{b}^{2}±2\sqrt{{a}^{2}{b}^{2}-4{\mathrm{\Delta }}^{2}}}$
where area $=\mathrm{\Delta }$

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