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Question Number 200636 by a.lgnaoui last updated on 21/Nov/23

1−Determiner la valeur de EF 2−Laire du triangle ADE

$$\mathrm{1}−\mathrm{Determiner}\:\mathrm{la}\:\mathrm{valeur}\:\mathrm{de}\:\:\boldsymbol{\mathrm{EF}} \\ $$$$\mathrm{2}−\mathrm{Laire}\:\mathrm{du}\:\mathrm{triangle}\:\:\boldsymbol{\mathrm{ADE}} \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 21/Nov/23

Commented by witcher3 last updated on 21/Nov/23

(ED)⊥(BD) ..? and (AB)⊥(AC) or AE...

$$\left(\mathrm{ED}\right)\bot\left(\mathrm{BD}\right)\:..? \\ $$$$\mathrm{and}\:\left(\mathrm{AB}\right)\bot\left(\mathrm{AC}\right)\:\mathrm{or}\:\mathrm{AE}... \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 21/Nov/23

BD⊥DE and AB⊥AC

$$\boldsymbol{\mathrm{BD}}\bot\mathrm{D}\boldsymbol{\mathrm{E}}\:\:\:\:\boldsymbol{\mathrm{and}}\:\:\:\mathrm{AB}\bot\boldsymbol{\mathrm{A}}\mathrm{C} \\ $$

Commented by a.lgnaoui last updated on 21/Nov/23

∡C=30

$$\measuredangle\boldsymbol{\mathrm{C}}=\mathrm{30} \\ $$

Commented by witcher3 last updated on 21/Nov/23

AC=(6/(tg(30)))=6(√3) AD=6(√3)−8 BD=(√(6^2 +(6(√3)−8)^2 )) ∠EDB=60−tan^(−1) (((6(√3)−8)/6))=θ BE=(√(BD^2 +DE^2 )),DE=BDtan(θ) EA^2 =BE^2 +AB^2 −2BE.ABcos(60) we know evrey length..a,b,c side of triangle S=(√((p/2)((p/2)−a)((p/2)−b)((p/2)−c))) P=a+b+c,

$$\mathrm{AC}=\frac{\mathrm{6}}{\mathrm{tg}\left(\mathrm{30}\right)}=\mathrm{6}\sqrt{\mathrm{3}} \\ $$$$\mathrm{AD}=\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8} \\ $$$$\mathrm{BD}=\sqrt{\mathrm{6}^{\mathrm{2}} +\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}\right)^{\mathrm{2}} } \\ $$$$\angle\mathrm{EDB}=\mathrm{60}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}}{\mathrm{6}}\right)=\theta \\ $$$$\mathrm{BE}=\sqrt{\mathrm{BD}^{\mathrm{2}} +\mathrm{DE}^{\mathrm{2}} },\mathrm{DE}=\mathrm{BDtan}\left(\theta\right) \\ $$$$\mathrm{EA}^{\mathrm{2}} =\mathrm{BE}^{\mathrm{2}} +\mathrm{AB}^{\mathrm{2}} −\mathrm{2BE}.\mathrm{ABcos}\left(\mathrm{60}\right) \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{evrey}\:\mathrm{length}..\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{side}\:\mathrm{of}\:\mathrm{triangle} \\ $$$$\mathrm{S}=\sqrt{\frac{\mathrm{p}}{\mathrm{2}}\left(\frac{\mathrm{p}}{\mathrm{2}}−\mathrm{a}\right)\left(\frac{\mathrm{p}}{\mathrm{2}}−\mathrm{b}\right)\left(\frac{\mathrm{p}}{\mathrm{2}}−\mathrm{c}\right)} \\ $$$$\mathrm{P}=\mathrm{a}+\mathrm{b}+\mathrm{c}, \\ $$

Answered by mr W last updated on 21/Nov/23

AC=6(√3) AD=6(√3)−8 BD=(√(6^2 +(6(√3)−8)^2 ))=4(√(13−6(√3))) ∠ABD=tan^(−1) ((6(√3)−8)/6) DE=BD tan (60°−∠ABD) =4(√(13−6(√3)))×(((√3)−((6(√3)−8)/6))/(1+(√3)×((6(√3)−8)/6))) =((4(√(13−6(√3))))/(3−(√3))) ∠CDE=∠ABD=tan^(−1) ((6(√3)−8)/6) Δ_(ADE) =((AD×DE×sin ∠CDE)/2) =((6(√3)−8)/2)×((4(√(13−6(√3))))/(3−(√3)))×((6(√3)−8)/( (√((6(√3)−8)^2 +6^2 )))) =((57−29(√3))/( 3))≈2.256842 BE=((BD)/(cos (60°−∠ABD)))=((42−10(√3))/3) EC=12−((42−10(√3))/3)=((10(√3)−6)/3) tan ∠CAE=((((10(√3)−6)/3)×(1/2))/(6(√3)−((10(√3)−6)/3)×((√3)/2)))=((45+2(√3))/( 183)) x=FE=((DE)/(cos (∠CDE−∠CAE))) =((DE)/(cos (∠ABD−∠CAE))) =((4(√(13−6(√3))))/(3−(√3)))×(1/((6/( (√(6^2 +(6(√3)−8)^2 ))))×((183)/( (√(183^2 +(45+2(√3))^2 ))))+((6(√3)−8)/( (√(6^2 +(6(√3)−8)^2 ))))×((45+2(√3))/( (√(183^2 +(45+2(√3))^2 )))))) =((4(√(13−6(√3))))/(3−(√3)))×((√(458598−210816(√3)))/(774+254(√3))) =((244(√(2622−1476(√3))))/(390−3(√3)))≈5.131542

$${AC}=\mathrm{6}\sqrt{\mathrm{3}} \\ $$$${AD}=\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8} \\ $$$${BD}=\sqrt{\mathrm{6}^{\mathrm{2}} +\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}\right)^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{13}−\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$$\angle{ABD}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}}{\mathrm{6}} \\ $$$${DE}={BD}\:\mathrm{tan}\:\left(\mathrm{60}°−\angle{ABD}\right) \\ $$$$\:\:\:\:\:\:=\mathrm{4}\sqrt{\mathrm{13}−\mathrm{6}\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}−\frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}}{\mathrm{6}}}{\mathrm{1}+\sqrt{\mathrm{3}}×\frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}}{\mathrm{6}}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{4}\sqrt{\mathrm{13}−\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{3}−\sqrt{\mathrm{3}}} \\ $$$$\angle{CDE}=\angle{ABD}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}}{\mathrm{6}} \\ $$$$\Delta_{{ADE}} =\frac{{AD}×{DE}×\mathrm{sin}\:\angle{CDE}}{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}}{\mathrm{2}}×\frac{\mathrm{4}\sqrt{\mathrm{13}−\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{3}−\sqrt{\mathrm{3}}}×\frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}}{\:\sqrt{\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }} \\ $$$$\:\:\:=\frac{\mathrm{57}−\mathrm{29}\sqrt{\mathrm{3}}}{\:\mathrm{3}}\approx\mathrm{2}.\mathrm{256842} \\ $$$${BE}=\frac{{BD}}{\mathrm{cos}\:\left(\mathrm{60}°−\angle{ABD}\right)}=\frac{\mathrm{42}−\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${EC}=\mathrm{12}−\frac{\mathrm{42}−\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{3}}=\frac{\mathrm{10}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\angle{CAE}=\frac{\frac{\mathrm{10}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{6}\sqrt{\mathrm{3}}−\frac{\mathrm{10}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{45}+\mathrm{2}\sqrt{\mathrm{3}}}{\:\mathrm{183}} \\ $$$${x}={FE}=\frac{{DE}}{\mathrm{cos}\:\left(\angle{CDE}−\angle{CAE}\right)} \\ $$$$\:\:\:=\frac{{DE}}{\mathrm{cos}\:\left(\angle{ABD}−\angle{CAE}\right)} \\ $$$$\:\:\:=\frac{\mathrm{4}\sqrt{\mathrm{13}−\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{3}−\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\frac{\mathrm{6}}{\:\sqrt{\mathrm{6}^{\mathrm{2}} +\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}\right)^{\mathrm{2}} }}×\frac{\mathrm{183}}{\:\sqrt{\mathrm{183}^{\mathrm{2}} +\left(\mathrm{45}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }}+\frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}}{\:\sqrt{\mathrm{6}^{\mathrm{2}} +\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{8}\right)^{\mathrm{2}} }}×\frac{\mathrm{45}+\mathrm{2}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{183}^{\mathrm{2}} +\left(\mathrm{45}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }}} \\ $$$$\:\:\:=\frac{\mathrm{4}\sqrt{\mathrm{13}−\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{3}−\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{458598}−\mathrm{210816}\sqrt{\mathrm{3}}}}{\mathrm{774}+\mathrm{254}\sqrt{\mathrm{3}}} \\ $$$$\:\:\:=\frac{\mathrm{244}\sqrt{\mathrm{2622}−\mathrm{1476}\sqrt{\mathrm{3}}}}{\mathrm{390}−\mathrm{3}\sqrt{\mathrm{3}}}\approx\mathrm{5}.\mathrm{131542} \\ $$

Commented by mr W last updated on 21/Nov/23

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