Question Number 197595 by universe last updated on 23/Sep/23
Commented by universe last updated on 23/Sep/23
$$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{equilateral}\:\mathrm{triangle}\:? \\ $$
Answered by MM42 last updated on 24/Sep/23
$${AB}=\sqrt{\mathrm{3}}/\mathrm{3}\Rightarrow{AC}=\mathrm{1}+\sqrt{\mathrm{3}}/\mathrm{3} \\ $$$$\Rightarrow{S}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\left(\mathrm{1}+\sqrt{\mathrm{3}}/\mathrm{3}\right)^{\mathrm{2}} =\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}}{\mathrm{6}}\:\:\checkmark \\ $$$$ \\ $$
Commented by MM42 last updated on 23/Sep/23
Commented by a.lgnaoui last updated on 23/Sep/23
$$\:\boldsymbol{\mathrm{S}}\:\:\boldsymbol{\mathrm{doit}}\:\boldsymbol{\mathrm{etre}}\:<\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\left(\boldsymbol{\mathrm{verifier}}\:\boldsymbol{\mathrm{votre}}\:\boldsymbol{\mathrm{calcul}}\right) \\ $$
Answered by a.lgnaoui last updated on 23/Sep/23
$$\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{triangle}}\right)=\frac{\boldsymbol{\mathrm{EN}}×\boldsymbol{\mathrm{AH}}}{\mathrm{2}} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{EN}}=\mathrm{2}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\:\:\boldsymbol{\mathrm{EN}}=\mathrm{1}+\boldsymbol{\mathrm{x}}}\\{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}=\mathrm{1}\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}=\mathrm{1}−\boldsymbol{\mathrm{x}}}\end{cases} \\ $$$$\:\begin{cases}{\boldsymbol{\mathrm{AH}}=\boldsymbol{\mathrm{OA}}+\boldsymbol{\mathrm{OH}}}\\{\boldsymbol{\mathrm{OH}}=\mathrm{1}\:\:\:\:\boldsymbol{\mathrm{OA}}=\boldsymbol{\mathrm{OC}}\mathrm{tan}\:\frac{\boldsymbol{\pi}}{\mathrm{3}}=\frac{\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$$\Rightarrow\boldsymbol{\mathrm{AH}}=\mathrm{1}+\frac{\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{2}+\sqrt{\mathrm{3}}\:−\boldsymbol{\mathrm{x}}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{S}}=\frac{\left(\mathrm{1}+\boldsymbol{\mathrm{x}}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\:−\boldsymbol{\mathrm{x}}\sqrt{\mathrm{3}}\:\right)}{\mathrm{4}}\:\:\:\:\left(\mathrm{i}\right)\: \\ $$$$ \\ $$$$\:\mathrm{tan}\:\frac{\boldsymbol{\pi}}{\mathrm{3}}=\frac{\boldsymbol{\mathrm{CM}}}{\boldsymbol{\mathrm{MN}}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}=\sqrt{\mathrm{3}}\:\Rightarrow\boldsymbol{\mathrm{x}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\boldsymbol{\mathrm{donc}}\:\:\left(\mathrm{i}\right)\Rightarrow \\ $$$$\boldsymbol{{S}}\left(\boldsymbol{\mathrm{Triangle}}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}=\:\:\:\mathrm{1},\mathrm{07735} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by a.lgnaoui last updated on 23/Sep/23
Commented by JDamian last updated on 23/Sep/23
$${you}\:{make}\:{it}\:{so}\:{hard} \\ $$