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Question Number 194967 by Rupesh123 last updated on 21/Jul/23

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

$$4\mathrm{cercles}\mathrm{disposes}\mathrm{comme}\mathrm{suit}$$$$\bullet 1)\mathrm{cercle}\mathbf{R}\mathrm{ouge}\mathbf{C}1\left(\mathrm{Rayon}\mathbf{R}1\right)$$$$2\mathbf{R}1=\mathbf{AB}=12\mathbf{R}1=6$$$$\bullet 2)\mathrm{cercle}\mathbf{V}\mathrm{ert}\mathbf{C}2\left(\mathbf{R}\mathrm{ayon}\mathbf{R}2\right)/$$$$\mathrm{demi}-\mathrm{cercle}\mathbf{C}(\mathbf{A},\mathbf{B},\mathbf{C})$$$$2\mathbf{R}2=\mathbf{AC}=\sqrt{{12}^2-5^2}=13\mathbf{R}2=6,5$$$$\bullet 3)\mathrm{cercle}\mathbf{Noir}:\mathbf{C}\left(\mathbf{Rayon}\mathbf{R}\mathrm{a}\mathrm{determiner}\right)$$$$$$$$\mathbf{Espace}\mathbf{entre}\mathbf{C}1\mathbf{et}\mathbf{C}2(\mathbf{Aire}=\mathbf{b})$$$$\mathbf{Espace}\mathbf{entre}\mathbf{C}2\mathbf{et}\mathbf{C}(\mathbf{Aire}=\mathbf{c})$$$$\bullet 3)\mathbf{Portion}\mathbf{de}\mathbf{cercle}\mathbf{C}2\mathbf{delimite}\mathbf{par}$$$$\mathrm{arc}\left(\mathrm{BC}\right)\mathbf{diametre}\mathbf{BC}=5(\mathbf{R}3=2,5)$$$$$$$$\mathbf{Calcul}\mathbf{de};\mathbf{a};\mathbf{b};\mathbf{c}$$$$\ast \mathrm{Calcul}\mathrm{de}\left(\mathrm{a}\right)$$$$\mathrm{Aire}\left(\mathrm{arc}\mathbf{MBC}\mathrm{forme}\mathrm{par}\mathrm{angle}\mathit{\theta }\right)$$$$(\sin \frac{\theta }{2}=\frac{\mathrm{BC}/2}{\mathrm{AM}}=\frac{2,5}{6,5}=\frac{5}{13})\theta =45,25$$$$\mathrm{Aire}\left(\mathrm{arcMBC}\right)={\mathrm{a}}_1=\frac{\pi \times {(6,5)}^2\times 45,25}{360}$$$${\mathrm{a}}_1=5,3\mathit{\pi }$$$$\mathrm{portion}\left(\mathrm{BHC}\right)={\mathrm{a}}_1-\left(\mathrm{aire}\mathrm{triangle}\mathrm{MBC}\right)$$$${\mathrm{a}}_1-\frac{\mathrm{BC}}{2}\times \mathrm{MCcos}\frac{\theta }{2}=2,5\times 6,5\times 0,923=15$$$$\mathrm{portion}\mathrm{BHC}=5,3\mathit{\pi }-15$$$$\Rightarrow \mathbf{a}=\mathit{\pi }\times {(2,5)}^2/2+15-5,3\mathit{\pi }$$$$$$$$\mathrm{Aire}\mathbf{a}=15-2,175\mathit{\pi }$$$$$$$$\ast \mathrm{calcul}\mathrm{de}\mathrm{l}\mathrm{aire}\left(\mathbf{b}\right)$$$$\mathbf{b}=\mathit{\pi }\times [{(6,5)}^2-6^2]=6,25\mathit{\pi }$$$$\ast \mathrm{Calcul}\mathrm{de}\mathbf{c}$$$$\mathbf{c}=\mathit{\pi }{\mathbf{R}}^2-\mathit{\pi }\times {(6,5)}^2=\mathit{\pi }({\mathbf{R}}^2-42,25)$$$$$$$$\mathrm{Aire}\mathbf{c}=\mathit{\pi }{\mathbf{R}}^2-42,25\mathit{\pi }$$$$$$$$\mathbf{c}=\mathbf{a}+\mathbf{b}\Rightarrow$$$$\mathit{\pi }{\mathrm{R}}^2-42,25\mathit{\pi }=15-2,175\mathit{\pi }+6,25\mathit{\pi }$$$$\mathit{\pi }{\mathbf{R}}^2-(42,25+6,25-2,175)\mathit{\pi }=15$$$$\mathit{\pi }{\mathbf{R}}^2-46,32\mathit{\pi }=15$$$$\Rightarrow {\mathbf{R}}^2=46,25+\frac{15}{\mathit{\pi }}$$$$$$$$\mathrm{le}\mathrm{Rayon}:\mathbf{R}=\sqrt{46,25+\frac{15}{\mathit{\pi }}}$$$$$$$$$$

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

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