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Question Number 181808 by mathocean1 last updated on 30/Nov/22

Commented by CElcedricjunior last updated on 01/Dec/22

$${\mathit{A}}_{\mathit{t}}={\mathit{A}}_{red}+A_{bleu}=>A_{red}=A_{\mathit{t}}-{\mathit{A}}_{bleu}$$$$or{A}_t=\frac{\mathit{B}\mathit{H}}{2}=\frac{8\times \sqrt{8^2-4^2}}{2}=\frac{8\times 4\sqrt{3}}{2}=16\sqrt{3}$$$$A_{\mathit{b}\mathit{l}\mathit{e}\mathit{u}}=3\left(\frac{4^2\mathit{\pi }}{6}\right)=\frac{4^2\mathit{\pi }}{2}=\frac{16\mathit{\pi }}{2}=8\mathit{\pi }$$$${\mathit{A}}_{red}=16\sqrt{3}-8\mathit{\pi }\approx 2.5800um$$$$=======================$$$$......lecelebrecedricjunior............$$$$=====================.=.$$$$$$$$$$

Answered by BaliramKumar last updated on 30/Nov/22

$$red=\frac{\sqrt{3}}{4}\times 8^2-3\times \frac{60}{360}\times \pi \times 4^2=16\sqrt{3}-8\pi$$

Answered by HeferH last updated on 30/Nov/22

$$8(2\sqrt{3}-\pi )$$

Answered by a.lgnaoui last updated on 01/Dec/22

$$airetriagle=8\times 4\times \sin \left(\frac{\pi }{3}\right)=16\sqrt{3}$$$$aireregionRouge=16\sqrt{3}-8\pi$$$$=16\sqrt{3}-8\pi$$$$Aire=2,58$$

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