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Question Number 64686 by ajfour last updated on 20/Jul/19

Answered by ajfour last updated on 20/Jul/19

$$LetBbeoriginandxaxisalong$$$$BC.$$$$x^2+y^2=c^2$$$${(x-a)}^2+y^2=b^2$$$$\Rightarrow a(2x-a)=c^2-b^2$$$$x=\frac{a^2+c^2-b^2}{2a}$$$$y=\sqrt{c^2-{\left(\frac{a^2+c^2-b^2}{2a}\right)}^2}$$$${△}^2=\frac{a^2{y}^2}{4}=\frac{a^2}{4}\left[\frac{(2ac-a^2-c^2+b^2)(2ac+a^2+c^2-b^2}{4a^2}\right]$$$$=\frac{[b^2-{(a-c)}^2][{(a+c)}^2-b^2]}{16}$$$$\Rightarrow 16{△}^2=-{(a^2-c^2)}^2-b^4+2b^2(a^2+c^2)$$$$=-a^4-b^4-c^4+2(a^2{c}^2+b^2{c}^2+c^2{a}^2)$$$$=2\mathrm{\Sigma }{a}^2{b}^2-\mathrm{\Sigma }{a}^4$$$$=4\mathrm{\Sigma }{a}^2{b}^2-{\left(\mathrm{\Sigma }{a}^2\right)}^2$$$$=4\mathrm{\Sigma }{a}^2{b}^2-{[{(a+b+c)}^2-2\mathrm{\Sigma }ab]}^2$$$$=4\mathrm{\Sigma }{a}^2{b}^2-[16s^4-16s^2\mathrm{\Sigma }ab+4{\left(\mathrm{\Sigma }ab\right)}^2]$$$$=-16s^4+16s^2\mathrm{\Sigma }ab-16\left(abc\right)s$$$$\Rightarrow △=\sqrt{s^2\mathrm{\Sigma }ab-s^4-\left(abc\right)s}$$$$△=\sqrt{s[s(ab+bc+ca)-s^3-abc]}$$$$=\sqrt{s[s(ab+bc+ca)+s^3-s(a+b+c)-abc}$$$$\Rightarrow$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$△=\sqrt{\mathit{s}(\mathit{s}-\mathit{a})(\mathit{s}-\mathit{b})(\mathit{s}-\mathit{c})}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$

Commented by mr W last updated on 20/Jul/19

$${what}^{\prime }sthequestionsir?$$

Commented by ajfour last updated on 20/Jul/19

$$NothingSir,justwhilingaway$$$$mytime.({}^{\underset{\bullet }{⌢}}{\underset{⌣}{\mid }}^{\underset{\bullet }{⌢}}\partial$$

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