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Question Number 176913 by a.lgnaoui last updated on 28/Sep/22

$$Determinerlahauteur\mathrm{D}E(r+x)enfonctionder$$$$r=\mathrm{OOC}=\mathrm{BH}\mathrm{BF}=20$$$$\mathrm{pour}\mathrm{que}\mathrm{distance}(\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DE}+\mathrm{EF}\mathrm{soit}$$$$\mathrm{sgale}\mathrm{AC}+\mathrm{arcCDF}$$

Commented by a.lgnaoui last updated on 28/Sep/22

Answered by a.lgnaoui last updated on 29/Sep/22

$$\tan \left(30\right)=\frac{\sqrt{3}}{3}=\frac{\mathrm{BH}}{\mathrm{BC}}=\frac{\mathrm{r}}{\mathrm{BC}}\Rightarrow \mathrm{BC}=r\sqrt{3}$$$$\measuredangle B=\measuredangle C\mathrm{AB}=\mathrm{BC}$$$$\mathrm{CD}=\mathrm{r}\sqrt{2}\mathrm{DE}=\mathrm{r}+\mathrm{x}\mathrm{EF}=\sqrt{{\mathrm{x}}^2+{\mathrm{r}}^2}$$$$\mathrm{A}BzCDEF=2r\sqrt{3}+r\sqrt{2}+r+x+\sqrt{x^2+r^2}$$$$=(2\sqrt{3}+\sqrt{2}+1)r+x+\sqrt{x^2+r^2}$$$$AC+arc\left(CDF\right)=2\mathrm{BC}\times \cos \left(30\right)+\pi r=2r\sqrt{3}\times \frac{\sqrt{3}}{2}+\pi r=(3+\pi )r$$$$egalitedes2trajets$$$$[2\sqrt{3}+\sqrt{2}+1)r+x+\sqrt{x^2+r^2}]=(3+\pi )r$$$$\sqrt{x^2+r^2}=(2+\pi +2\sqrt{3}-\sqrt{2})r-x$$$$x^2+r^2={(2+\pi +2\sqrt{3}-\sqrt{2})}^2{r}^2+x^2-2(2+\pi +2\sqrt{3}-\sqrt{2})rx$$$$r={(2+\pi +2\sqrt{3}-\sqrt{2})}^{2}r-2x(2+\pi -2\sqrt{3}-\sqrt{2})$$$$x=\frac{1}{2}[\frac{{(\pi +1+2\sqrt{3}-\sqrt{2})}^2-1}{(\pi +2+2\sqrt{3}-\sqrt{2})}r=3,526r$$$$$$$$\mathrm{BF}=20=\mathrm{BC}+\mathrm{CF}=\mathrm{r}\sqrt{3}+2\mathrm{r}=(\sqrt{3}+2)\mathrm{r}$$$$\mathrm{r}=\frac{20}{\sqrt{3}+2}=5,358\Rightarrow x=18,89$$$$lmahauteurh(\mathrm{DE}=\mathrm{r}+\mathrm{x})=5,358+18,89=24,25$$

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