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Question Number 190306 by cortano12 last updated on 31/Mar/23

Answered by Frix last updated on 31/Mar/23

$$\mathrm{Area}\mathrm{of}\mathrm{triangle}u,v,w\mathrm{is}$$$$\frac{\sqrt{(u+v+w)(u+v-w)(u+w-v)(v+w-u)}}{4}=$$$$=\frac{\sqrt{2(u^2{v}^2+u^2{w}^2+v^2{w}^2)-(u^4+v^4+w^4)}}{4}$$$$\mathrm{Now}\mathrm{simply}\mathrm{insert}$$

Commented by som(math1967) last updated on 31/Mar/23

$$=\frac{1}{4}\times \sqrt{6(a^2{b}^2+b^2{c}^2+a^2{b}^2)+2(a^4+b^4+c^4)-2(a^4+b^4+c^4)-2(a^2{b}^2+b^2{c}^2+a^2{b}^2)}$$$$=\frac{1}{4}\times \sqrt{4(a^2{b}^2+b^2{c}^2+c^2{a}^2)}$$$$=\frac{1}{2}\sqrt{(a^2{b}^2+b^2{c}^2+c^2{a}^2)}$$

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