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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

$$AB=\sqrt{3}/3\Rightarrow AC=1+\sqrt{3}/3$$$$\Rightarrow S=\frac{\sqrt{3}}{4}\times {(1+\sqrt{3}/3)}^2=\frac{2\sqrt{3}+3}{6}✓$$$$$$

Commented by MM42 last updated on 23/Sep/23

Commented by a.lgnaoui last updated on 23/Sep/23

$$\mathbf{S}\mathbf{doit}\mathbf{etre}<\frac{3}{2}\left(\mathbf{verifier}\mathbf{votre}\mathbf{calcul}\right)$$

Answered by a.lgnaoui last updated on 23/Sep/23

$$\mathbf{S}\left(\mathbf{triangle}\right)=\frac{\mathbf{EN}\times \mathbf{AH}}{2}$$$$\{\begin{array}{l}\mathbf{EN}=2\mathbf{x}+\mathbf{y}\mathbf{EN}=1+\mathbf{x}\\ \mathbf{x}+\mathbf{y}=1\mathbf{y}=1-\mathbf{x}\end{array}$$$$\{\begin{array}{l}\mathbf{AH}=\mathbf{OA}+\mathbf{OH}\\ \mathbf{OH}=1\mathbf{OA}=\mathbf{OC}\tan \frac{\mathit{\pi }}{3}=\frac{(1-\mathbf{x})\sqrt{3}}{2}\end{array}$$$$\Rightarrow \mathbf{AH}=1+\frac{(1-\mathbf{x})\sqrt{3}}{2}=\frac{2+\sqrt{3}-\mathbf{x}\sqrt{3}}{2}$$$$$$$$\mathbf{S}=\frac{(1+\mathbf{x})(2+\sqrt{3}-\mathbf{x}\sqrt{3})}{4}\left(\mathrm{i}\right)$$$$$$$$\tan \frac{\mathit{\pi }}{3}=\frac{\mathbf{CM}}{\mathbf{MN}}=\frac{1}{\mathbf{x}}=\sqrt{3}\Rightarrow \mathbf{x}=\frac{\sqrt{3}}{3}$$$$\mathbf{donc}\left(\mathrm{i}\right)\Rightarrow$$$$\mathit{S}\left(\mathbf{Triangle}\right)=\frac{1}{2}+\frac{\sqrt{3}}{3}=1,07735$$$$$$

Commented by a.lgnaoui last updated on 23/Sep/23

Commented by JDamian last updated on 23/Sep/23

$$youmakeitsohard$$

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