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Question Number 192325 by cherokeesay last updated on 14/May/23

Answered by a.lgnaoui last updated on 14/May/23

$$\mathbf{s}\mathrm{urface}\mathbf{S}=\frac{1}{2}\left(\mathbf{AB}\times \mathbf{BF}\right)=\frac{1}{2}\mathbf{CD}\times (\mathbf{BD}-\mathbf{DF})$$$$\mathbf{S}=2\mathbf{CD}$$$$\mathbf{C}\mathrm{alcul}\mathbf{CD}$$$$△\mathrm{ACE}\measuredangle \mathrm{EAC}=\measuredangle \mathrm{DEF}=\mathit{\theta }$$$$\mathbf{C}\mathrm{E}=\mathrm{AEsin}\theta \cos \theta =\frac{5}{\mathrm{AE}}\Rightarrow \mathrm{AE}=\frac{5}{\cos \theta }$$$$\mathrm{CE}=5\tan \mathit{\theta }$$$$\mathrm{EFsin}\theta =1\Rightarrow \mathrm{ED}=\mathrm{EFcos}\theta =\frac{1}{\tan \theta }$$$$\Rightarrow \{\begin{array}{l}\mathrm{CE}=5\tan \theta \\ \mathrm{ED}=\frac{1}{\tan \theta }\end{array}$$$$\mathrm{CD}=5\tan \theta +\frac{1}{\tan \theta }$$$$△\mathrm{ABF}\tan 2\theta =\frac{\mathrm{HF}}{4}=\frac{\mathrm{CD}}{4}$$$$\Rightarrow \frac{2\tan \theta }{1-\tan {}^2\theta }=\frac{5}{4}\tan \theta +\frac{1}{4\tan \theta }$$$$\mathrm{posons}\mathbf{t}=\tan \mathit{\theta }$$$$\frac{2\mathbf{t}}{1-{\mathbf{t}}^2}=\frac{5\mathbf{t}}{4}+\frac{1}{4\mathbf{t}}\Rightarrow$$$${\mathbf{t}}^4+\frac{4{\mathbf{t}}^2}{5}-\frac{1}{5}=0{\mathbf{z}}^2+\frac{4\mathbf{z}}{5}-\frac{1}{5}=0(\mathbf{z}={\mathbf{t}}^2)$$$$\mathbf{z}=\frac{-\frac{4}{5}\pm \frac{6}{5}}{2}\mathrm{z}=\{\frac{-1}{5},\frac{+1}{5}\}\mathrm{z}>0\Rightarrow$$$$\mathrm{z}=\frac{1}{5}\Rightarrow \mathbf{t}=\frac{\sqrt{5}}{5}$$$$\Rightarrow \mathbf{CD}=\sqrt{5}+\frac{5}{\sqrt{5}}=2\sqrt{5}$$$$\mathbf{Surface}\mathbf{sombre}\left(\mathbf{S}\right)=4\sqrt{5}$$$$$$

Commented by a.lgnaoui last updated on 14/May/23

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