Question-192325

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

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

surface S=(1/2)(AB×BF)=(1/2)CD×(BD−DF) S=2CD Calcul CD △ACE ∡EAC=∡DEF=𝛉 CE=AEsin θ cos θ=(5/(AE))⇒AE=(5/(cos θ)) CE=5tan 𝛉 EFsin θ=1 ⇒ ED=EFcos θ=(1/(tan θ)) ⇒ { ((CE=5tan θ)),((ED=(1/(tan θ)))) :} CD=5tan θ+(1/(tan θ)) △ABF tan 2θ=((HF)/4) =((CD)/4) ⇒((2tan θ)/(1−tan^2 θ))=(5/4)tan θ+(1/(4tan θ)) posons t=tan 𝛉 ((2t)/(1−t^2 ))=((5t)/4)+(1/(4t))⇒ t^4 +((4t^2 )/5)−(1/5)=0 z^2 +((4z)/5)−(1/5)=0(z=t^2 ) z=((−(4/5)±(6/5))/2) z={((−1)/5),((+1)/5)} z>0⇒ z=(1/5) ⇒ t=((√5)/5) ⇒ CD=(√5)+(5/( (√5)))=2(√5) Surface sombre(S)=4(√5)

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

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

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