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Question Number 191854 by Rupesh123 last updated on 01/May/23

Answered by a.lgnaoui last updated on 02/May/23

$$\mathrm{l}\mathrm{image}\mathrm{comporte}3\mathrm{principale}\mathbf{S}\mathrm{s}\mathrm{surfaces}$$$$\bullet \mathbf{S}1:\mathbf{S}\mathrm{urface}\mathrm{du}\mathrm{quart}\mathrm{ceecle}\mathrm{C}1(\mathrm{rayon}{\mathbf{r}}_1=2)$$$$\bullet \mathrm{S}2:\mathbf{S}\mathrm{urface}\mathrm{petit}\mathrm{cercleC}2(\mathrm{ryon}:{\mathbf{r}}_2)$$$$\bullet \mathbf{S}3:\mathrm{Surface}\mathrm{portion}\mathrm{cercle}\left(\mathrm{rayon}{\mathbf{r}}_3\right)$$$$\mathbf{a}\mathrm{vec}{\mathrm{r}}_3=\mathbf{AB}=\mathbf{BC}$$$$$$$${\mathbf{S}}_1=\frac{\mathit{\pi }\times 4}{4}=\mathit{\pi };$$$$\mathbf{OF}=\mathbf{OE}+\mathbf{EF}=\mathbf{OE}+{\mathbf{r}}_2\Rightarrow 2=\mathbf{OE}+{\mathbf{r}}_2$$$${\mathbf{OE}}^2=2{\mathbf{r}}_2^2\Rightarrow \mathbf{OE}={\mathbf{r}}_2\sqrt{2}$$$$\Rightarrow {\mathbf{r}}_2=2\sqrt{2}-2$$$${\mathbf{S}}_2=\mathit{\pi }{\mathbf{r}}_2^2=4\mathit{\pi }{(\sqrt{2}-1)}^2$$$$$$$$$$$$$$$$\frac{{\mathbf{S}}_1-{\mathbf{S}}_2}{2}={\mathbf{A}}_1+\frac{{\mathbf{A}}_2}{2}$$$$\mathrm{Calcul}{\mathbf{A}}_2$$$${\mathbf{A}}_2={\mathbf{OH}}^2-\frac{{\mathbf{S}}_2}{4}={\mathbf{r}}_2^2-\frac{\mathit{\pi }{\mathbf{r}}_2^2}{4}=\frac{3}{4}\mathit{\pi }{\mathbf{r}}_2^2$$$$\mathrm{donc}{\mathbf{A}}_1+{\mathbf{A}}_2=\frac{{\mathbf{S}}_1-{\mathbf{S}}_2}{2}+\frac{3}{8}\mathit{\pi }{\mathbf{r}}_2^2$$$$=\frac{\mathit{\pi }-4\mathit{\pi }{(\sqrt{2}-1)}^2}{2}+\frac{3}{8}\mathit{\pi }\times 4{(\sqrt{2}-1)}^2$$$$$$$${\mathrm{A}}_1+{\mathrm{A}}_2=(\sqrt{2}-1)\mathit{\pi }$$$$\bullet \mathrm{calcul}{\mathbf{A}}_3$$$$\mathbf{S}\mathrm{urface}\mathbf{Porti}\mathrm{o}\mathbf{n}\mathrm{arc}\mathrm{de}\mathrm{cercle}\mathbf{OAC}$$$$\mathbf{S}\mathrm{urface}\left(\mathrm{Arc}\mathrm{de}\mathrm{cercle}\mathrm{BAC}\right)-\left(\mathrm{Surface}\mathrm{triangle}\mathrm{OAB}\right)$$$$\Rightarrow \mathrm{Surface}\mathrm{arc}=\mathit{\pi }{\mathbf{AB}}^2\left(\frac{\mathit{\theta }}{2\pi }\right)\left(\mathit{\theta }\mathrm{en}\mathrm{radian}\right)$$$${\mathbf{r}}_3=\mathbf{AB}=2\sqrt{2}\left(\mathbf{B}\mathrm{centre}\mathrm{cercle}\mathrm{C}3\right)$$$$\sin \theta =\cos \theta =\frac{\sqrt{2}}{2}\Rightarrow \theta =\frac{\pi }{4}$$$${\mathrm{S}}_3=\frac{\mathit{\pi }\times 8}{8}=\mathit{\pi }$$$$\mathrm{surface}\mathrm{triangle}=\frac{{\mathrm{AB}}^2}{2}=2$$$$\Rightarrow {\mathbf{A}}_3=({\mathbf{S}}_3-4)\mathbf{A}3=\mathit{\pi }-2$$$$\mathbf{Conclusion}:$$$$\mathbf{S}={\mathbf{A}}_1+{\mathbf{A}}_2+{\mathbf{A}}_3$$$$(\sqrt{2}-1)\mathit{\pi }+\mathit{\pi }-2$$$$$$$$\mathbf{S}=\mathit{\pi }\sqrt{2}-2$$$$$$

Commented by a.lgnaoui last updated on 03/May/23

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