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Question-145573

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Question Number 145573 by mnjuly1970 last updated on 06/Jul/21
Answered by Olaf_Thorendsen last updated on 06/Jul/21
p = half perimeter p = (1/2)(1+2r+2+2r+3+2r) = 3(r+1) p−a = 3(r+1)−(1+2r) = r+2 p−b = 3(r+1)−(2+2r) = r+1 p−c = 3(r+1)−(3+2r) = r S = (√(p(p−a)(p−b)(p−c))) S = (√(3(r+1)(r+2)(r+1)r)) S = (r+1)(√(3r(r+2))) S_1 = (1/2)θ_1 r^2 S_2 = (1/2)θ_2 r^2 S_3 = (1/2)θ_3 r^2 S_1 +S_2 +S_3 = (1/2)(θ_1 +θ_2 +θ_3 )r^2 = (1/2)πr^2 ((S_1 +S_2 +S_3 )/S) = (((1/2)πr^2 )/( (r+1)(√(3r(r+2))))) = (π/6) (r+1)(√(3r(r+2))) = 3r^2 (r+1)^2 3r(r+2) = 9r^4 (r+1)^2 (r+2) = 3r^3 (r^2 +2r+1)(r+2) = 3r^3 −2r^3 +4r^2 +5r+2 = 0 r ≈ 2,959
$${p}\:=\:{half}\:{perimeter} \\ $$$${p}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}{r}+\mathrm{2}+\mathrm{2}{r}+\mathrm{3}+\mathrm{2}{r}\right)\:=\:\mathrm{3}\left({r}+\mathrm{1}\right) \\ $$$${p}−{a}\:=\:\mathrm{3}\left({r}+\mathrm{1}\right)−\left(\mathrm{1}+\mathrm{2}{r}\right)\:=\:{r}+\mathrm{2} \\ $$$${p}−{b}\:=\:\mathrm{3}\left({r}+\mathrm{1}\right)−\left(\mathrm{2}+\mathrm{2}{r}\right)\:=\:{r}+\mathrm{1} \\ $$$${p}−{c}\:=\:\mathrm{3}\left({r}+\mathrm{1}\right)−\left(\mathrm{3}+\mathrm{2}{r}\right)\:=\:{r} \\ $$$${S}\:=\:\sqrt{{p}\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)} \\ $$$${S}\:=\:\sqrt{\mathrm{3}\left({r}+\mathrm{1}\right)\left({r}+\mathrm{2}\right)\left({r}+\mathrm{1}\right){r}} \\ $$$${S}\:=\:\left({r}+\mathrm{1}\right)\sqrt{\mathrm{3}{r}\left({r}+\mathrm{2}\right)} \\ $$$$ \\ $$$${S}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\theta_{\mathrm{1}} {r}^{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\theta_{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$${S}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\theta_{\mathrm{3}} {r}^{\mathrm{2}} \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} +\theta_{\mathrm{3}} \right){r}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}} \\ $$$$\frac{{S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} }{{S}}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}} }{\:\left({r}+\mathrm{1}\right)\sqrt{\mathrm{3}{r}\left({r}+\mathrm{2}\right)}}\:=\:\frac{\pi}{\mathrm{6}} \\ $$$$\left({r}+\mathrm{1}\right)\sqrt{\mathrm{3}{r}\left({r}+\mathrm{2}\right)}\:=\:\mathrm{3}{r}^{\mathrm{2}} \\ $$$$\left({r}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{3}{r}\left({r}+\mathrm{2}\right)\:=\:\mathrm{9}{r}^{\mathrm{4}} \\ $$$$\left({r}+\mathrm{1}\right)^{\mathrm{2}} \left({r}+\mathrm{2}\right)\:=\:\mathrm{3}{r}^{\mathrm{3}} \\ $$$$\left({r}^{\mathrm{2}} +\mathrm{2}{r}+\mathrm{1}\right)\left({r}+\mathrm{2}\right)\:=\:\mathrm{3}{r}^{\mathrm{3}} \\ $$$$−\mathrm{2}{r}^{\mathrm{3}} +\mathrm{4}{r}^{\mathrm{2}} +\mathrm{5}{r}+\mathrm{2}\:=\:\mathrm{0} \\ $$$${r}\:\approx\:\mathrm{2},\mathrm{959} \\ $$
Commented by mnjuly1970 last updated on 06/Jul/21
thanks alot mr olaf
$${thanks}\:{alot}\:{mr}\:{olaf} \\ $$

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