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Question Number 191930 by a.lgnaoui last updated on 05/May/23

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{inside}\mathrm{Area}\mathrm{of}$$$$\left(\mathrm{ABCDEF}\right)?$$$$\mathrm{Such}\mathrm{that}:\measuredangle \mathrm{AOB}=120\measuredangle \mathrm{ANB}=60;°\mathrm{R}=\mathrm{ON}$$$$(\mathrm{OA}=\mathrm{OB}=32\mathrm{cm})\mathrm{ArcAE}=\mathrm{ArcBF}(\mathrm{r}=12\mathrm{cm})$$$$\mathrm{BASE}\mathrm{is}\mathrm{circulare}$$$$(Aiderletailleurasavoirlasurface$$$$dutissunecessairepourcouvrir$$$$l{}^{\prime }\mathrm{e}spaceindiquedanslafigure?)$$

Commented by a.lgnaoui last updated on 04/May/23

$$\lambda =120$$

Commented by Skabetix last updated on 04/May/23

$$Quelesttonraisonnement?$$$$javoueavoirdesdifficultesaposerleprobleme$$

Answered by a.lgnaoui last updated on 05/May/23

$$$$$$\mathbf{AB}=2\mathbf{OA}\cos 60=64\times \frac{1}{2}=32\mathbf{cm}$$$$\mathbf{AE}=2\mathbf{r}=24\mathbf{cm}\Rightarrow {24}^2={\mathbf{AC}}^2+{\mathbf{CE}}^2$$$${\mathbf{d}}^{\prime }\mathbf{apres}\mathbf{la}\mathbf{figure}\mathbf{AC}=\mathbf{CE}(\measuredangle \mathrm{CEA}=45°)$$$$24=\mathbf{AC}\sqrt{2}\mathbf{AC}=\mathbf{CE}=12\sqrt{2}$$$$\mathbf{alors}\mathbf{la}\mathbf{surface}\mathbf{des}\mathbf{Rectangles}\left(\mathbf{ABCD}\right)$$$$\mathbf{et}\left(\mathbf{CDEF}\right)=2\times \left(\mathbf{AB}\times \mathbf{AC}\right)=2\times 32\times 12\sqrt{2}$$$$=64\times 12\sqrt{2}=768\sqrt{2}=1086,{11\mathrm{cm}}^2$$$$\mathbf{surface}\mathbf{des}\mathbf{ailes}$$$$\mathbf{s}=2({\mathbf{AC}}^2-\frac{\mathit{\pi }}{4}{\mathrm{r}}^2)=2(228-36\mathit{\pi })$$$$=456-72\mathit{\pi }=229,80$$$$$$$$\mathbf{Surface}\mathbf{totale}=1086,11+229,80$$$$=1315,91{\mathbf{cm}}^2$$$$$$

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