Michelle Mendoza

2022-07-15

Finding the third side of a triangle given the area

Perimeter =$\frac{(a+b+c)}{2}$

Area = $A=\sqrt{p(p-a)(p-b)(p-c)}$

How do I simplify the above two equations to solve for c?

Perimeter =$\frac{(a+b+c)}{2}$

Area = $A=\sqrt{p(p-a)(p-b)(p-c)}$

How do I simplify the above two equations to solve for c?

treccinair

Beginner2022-07-16Added 18 answers

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

$\begin{array}{rl}\mathrm{cos}\gamma & =\pm \sqrt{1-{\mathrm{sin}}^{2}\gamma}=\pm \sqrt{1-\frac{4{S}^{2}}{{a}^{2}{b}^{2}}}=\pm \frac{\sqrt{{a}^{2}{b}^{2}-4{S}^{2}}}{ab}\end{array}$

(1) becomes

$\begin{array}{rl}{c}^{2}& ={a}^{2}+{b}^{2}\pm 2\sqrt{{a}^{2}{b}^{2}-4{S}^{2}}.\end{array}$

$\begin{array}{}\text{(1)}& {c}^{2}& ={a}^{2}+{b}^{2}-2ab\mathrm{cos}\gamma ,\end{array}$

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

$\begin{array}{rl}\mathrm{cos}\gamma & =\pm \sqrt{1-{\mathrm{sin}}^{2}\gamma}=\pm \sqrt{1-\frac{4{S}^{2}}{{a}^{2}{b}^{2}}}=\pm \frac{\sqrt{{a}^{2}{b}^{2}-4{S}^{2}}}{ab}\end{array}$

(1) becomes

$\begin{array}{rl}{c}^{2}& ={a}^{2}+{b}^{2}\pm 2\sqrt{{a}^{2}{b}^{2}-4{S}^{2}}.\end{array}$

Holetaug

Beginner2022-07-17Added 8 answers

$s\phantom{\rule{thickmathspace}{0ex}}$ is the conventional way of representing the semiperimeter $\frac{a+b+c}{2}$ of the triangle with sides $a\phantom{\rule{thinmathspace}{0ex}},b\phantom{\rule{thinmathspace}{0ex}},c$

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

Solving for $c$,

$c=\sqrt{{a}^{2}+{b}^{2}\pm 2\sqrt{{a}^{2}{b}^{2}-4{\mathrm{\Delta}}^{2}}}$

where area $=\mathrm{\Delta}$

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

Solving for $c$,

$c=\sqrt{{a}^{2}+{b}^{2}\pm 2\sqrt{{a}^{2}{b}^{2}-4{\mathrm{\Delta}}^{2}}}$

where area $=\mathrm{\Delta}$

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