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Question Number 201979 by mr W last updated on 17/Dec/23

Commented by mr W last updated on 19/Dec/23

$$Q155657repostedforalternative$$$$solutions.$$

Answered by a.lgnaoui last updated on 18/Dec/23

$$\mathbf{S}\mathrm{totale}=\sum _{\mathbf{i}=1}^{\mathbf{i}=8}{\mathbf{S}}_{\mathbf{i}}=\mathit{\pi }{\mathbf{R}}^2$$$$\mathbf{S}1=\frac{\mathit{\pi }{\mathbf{R}}^2}{4}\mathbf{S}2+\mathbf{S}3={\mathbf{R}}^2\sin \frac{\mathit{\pi }}{4}=\frac{{\mathbf{R}}^2\sqrt{2}}{2}$$$$\mathbf{S}7=\frac{\mathit{\pi }{\mathbf{R}}^2}{8}-\frac{{\mathbf{R}}^2}{2}\sin \frac{\mathit{\pi }}{4}={\mathbf{R}}^2\left(\frac{\mathit{\pi }-2\sqrt{2}}{8}\right)$$$$\mathbf{S}8=\mathit{\pi }\frac{{\mathbf{R}}^2}{16}-\frac{{\mathbf{R}}^2}{2}\sin \frac{\mathit{\pi }}{8}={\mathbf{R}}^2\left(\frac{\mathit{\pi }-4\sqrt{2-\sqrt{2}}}{16}\right)$$$$\mathbf{S}4+\mathbf{S}5=\frac{3\mathit{\pi }}{8}{\mathbf{R}}^2-\frac{{\mathbf{R}}^2}{2}\sin \frac{\mathit{\pi }}{4}-\mathbf{S}7=\frac{\mathit{\pi }{\mathbf{R}}^2}{4}$$$$$$$$\mathbf{S}3+\mathbf{S}6=\frac{3\mathit{\pi }}{8}{\mathbf{R}}^2-\frac{{\mathrm{R}}^2}{2}\sin \frac{\mathit{\pi }}{4}-\mathbf{S}8$$$$=\frac{{\mathbf{R}}^2}{16}(6\mathit{\pi }-4\sqrt{2}-\mathit{\pi }+4\sqrt{2-\sqrt{2}})$$$$=\frac{{\mathbf{R}}^2}{16}[(5\mathit{\pi }+4(\sqrt{2-\sqrt{2}}-\sqrt{2})]$$$$\mathrm{alors}$$$$\mathbf{S}\left(\mathbf{Rose}\right)=\mathbf{S}3+\mathbf{S}4+\mathbf{S}5+\mathbf{S}6$$$$=\frac{\mathit{\pi }{\mathbf{R}}^2}{8}+\frac{5\mathit{\pi }{\mathbf{R}}^2}{16}+\left(\frac{\sqrt{2-\sqrt{2}}-\sqrt{2}}{4}\right){\mathbf{R}}^2$$$$=[\frac{7\mathit{\pi }}{16}+\frac{(\sqrt{2-\sqrt{2}}-\sqrt{2})}{4}]{\mathbf{R}}^2$$$$$$$$\mathbf{S}\left(\mathbf{Vert}\right)=\mathit{\pi }{\mathbf{R}}^2-\frac{7\mathit{\pi }{\mathbf{R}}^2}{16}+\frac{(\sqrt{2}}{4}-\frac{\sqrt{2-\sqrt{2}}){\mathbf{R}}^2}{4}$$$$=[\frac{9\mathit{\pi }}{16}+\frac{(\sqrt{2}-\sqrt{2-\sqrt{2}})}{4}]{\mathbf{R}}^2$$$$$$$$$$$$\frac{\mathbf{S}\left(\mathbf{Rose}\right)}{\mathbf{S}\left(\mathbf{Vert}\right)}=\frac{7\mathit{\pi }-4(\sqrt{2}-\sqrt{2-\sqrt{2}})}{9\mathit{\pi }+4(\sqrt{2}-\sqrt{2-\sqrt{2}})}$$

Commented by a.lgnaoui last updated on 18/Dec/23

Commented by mr W last updated on 19/Dec/23

$$wrong!$$$$S_{yellow}=S_{green}=\frac{\pi R^2}{2}$$

Commented by a.lgnaoui last updated on 18/Dec/23

Commented by a.lgnaoui last updated on 19/Dec/23

Answered by mr W last updated on 19/Dec/23

Commented by mr W last updated on 19/Dec/23

$$\alpha +\beta =45°$$$$CB=CA=\sqrt{2}R$$$$CE=2R\cos \alpha$$$$CD=2R\cos \beta$$$${\mathrm{\Delta }}_{CBE}=\frac{\sqrt{2}R\times 2R\cos \alpha \sin \beta }{2}=\sqrt{2}{R}^2\cos \alpha \sin \beta$$$${\mathrm{\Delta }}_{CBD}=\frac{\sqrt{2}R\times 2R\cos \beta \sin \alpha }{2}=\sqrt{2}{R}^2\sin \alpha \cos \beta$$$${\mathrm{\Delta }}_{CBE}+{\mathrm{\Delta }}_{CBD}=\sqrt{2}{R}^2(\cos \alpha \sin \beta +\sin \alpha \cos \beta )$$$$=\sqrt{2}{R}^2\sin (\alpha +\beta )=R^2={\mathrm{\Delta }}_{ABC}$$$$S_{yellow}=halfofcircle=\frac{\pi R^2}{2}$$$$S_{green}=\pi R^2-S_{yellow}=\frac{\pi R^2}{2}$$$$\frac{S_{yellow}}{S_{green}}=1$$

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