Question Number 190306 by cortano12 last updated on 31/Mar/23
Answered by Frix last updated on 31/Mar/23
$$\mathrm{Area}\:\mathrm{of}\:\mathrm{triangle}\:{u},\:{v},\:{w}\:\mathrm{is} \\ $$$$\frac{\sqrt{\left({u}+{v}+{w}\right)\left({u}+{v}−{w}\right)\left({u}+{w}−{v}\right)\left({v}+{w}−{u}\right)}}{\mathrm{4}}= \\ $$$$=\frac{\sqrt{\mathrm{2}\left({u}^{\mathrm{2}} {v}^{\mathrm{2}} +{u}^{\mathrm{2}} {w}^{\mathrm{2}} +{v}^{\mathrm{2}} {w}^{\mathrm{2}} \right)−\left({u}^{\mathrm{4}} +{v}^{\mathrm{4}} +{w}^{\mathrm{4}} \right)}}{\mathrm{4}} \\ $$$$\mathrm{Now}\:\mathrm{simply}\:\mathrm{insert} \\ $$
Commented by som(math1967) last updated on 31/Mar/23
$$=\frac{\mathrm{1}}{\mathrm{4}}×\sqrt{\mathrm{6}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)+\mathrm{2}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)−\mathrm{2}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)−\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\sqrt{\mathrm{4}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)} \\ $$