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Question-201979

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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
Q155657 reposted for alternative solutions.
$${Q}\mathrm{155657}\:{reposted}\:{for}\:{alternative} \\ $$$${solutions}. \\ $$
Answered by a.lgnaoui last updated on 18/Dec/23
Stotale=Σ_(i=1) ^(i=8) S_i =𝛑R^2 S1= ((𝛑R^2 )/4) S2+S3=R^2 sin (𝛑/4)=((R^2 (√2))/2) S7=((𝛑R^2 )/8)−(R^2 /2)sin (𝛑/4)=R^2 (((𝛑−2(√2))/8)) S8=𝛑(R^2 /(16))−(R^2 /2)sin (𝛑/8)=R^2 (((𝛑−4(√(2−(√2))))/(16))) S4+S5=((3𝛑)/8)R^2 −(R^2 /2)sin (𝛑/4)−S7=((𝛑R^2 )/4) S3+S6=((3𝛑)/8)R^2 −(R^2 /2)sin (𝛑/4)−S8 =(R^2 /(16))(6𝛑−4(√2) −𝛑+4(√(2−(√2) )) ) =(R^2 /(16))[(5𝛑+4((√(2−(√2) )) −(√2) )] alors S(Rose)=S3+S4+S5+S6 =((𝛑R^2 )/8)+((5𝛑R^2 )/(16))+((((√(2−(√2) )) −(√2))/4))R^2 =[((7𝛑)/(16))+((((√(2 −(√2)))−(√2)))/4)]R^2 S(Vert)=𝛑R^2 −((7𝛑R^2 )/(16))+((((√2))/4)−(((√(2−(√2) )) )R^2 )/4) =[((9𝛑)/(16))+((((√2) −(√(2−(√2))) ))/4)]R^2 ((S(Rose))/(S(Vert))) = ((7𝛑−4((√2) −(√(2−(√2) ))))/(9𝛑+4((√2) −(√(2−(√2))))))
$$\boldsymbol{\mathrm{S}}\mathrm{totale}=\sum_{\boldsymbol{\mathrm{i}}=\mathrm{1}} ^{\boldsymbol{\mathrm{i}}=\mathrm{8}} \boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{i}}} =\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{S}}\mathrm{1}=\:\frac{\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} }{\mathrm{4}}\:\:\:\:\:\boldsymbol{\mathrm{S}}\mathrm{2}+\boldsymbol{\mathrm{S}}\mathrm{3}=\boldsymbol{\mathrm{R}}^{\mathrm{2}} \mathrm{sin}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}=\frac{\boldsymbol{\mathrm{R}}^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{S}}\mathrm{7}=\frac{\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} }{\mathrm{8}}−\frac{\boldsymbol{\mathrm{R}}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}=\boldsymbol{\mathrm{R}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{8}}\right) \\ $$$$\boldsymbol{\mathrm{S}}\mathrm{8}=\boldsymbol{\pi}\frac{\boldsymbol{\mathrm{R}}^{\mathrm{2}} }{\mathrm{16}}−\frac{\boldsymbol{\mathrm{R}}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}\:\frac{\boldsymbol{\pi}}{\mathrm{8}}=\boldsymbol{\mathrm{R}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}−\mathrm{4}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{16}}\right) \\ $$$$\boldsymbol{\mathrm{S}}\mathrm{4}+\boldsymbol{\mathrm{S}}\mathrm{5}=\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{8}}\boldsymbol{\mathrm{R}}^{\mathrm{2}} −\frac{\boldsymbol{\mathrm{R}}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}−\boldsymbol{\mathrm{S}}\mathrm{7}=\frac{\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\boldsymbol{\mathrm{S}}\mathrm{3}+\boldsymbol{\mathrm{S}}\mathrm{6}=\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{8}}\boldsymbol{\mathrm{R}}^{\mathrm{2}} −\frac{\mathrm{R}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}−\boldsymbol{\mathrm{S}}\mathrm{8} \\ $$$$\:\:\:\:\:\:\:=\frac{\boldsymbol{\mathrm{R}}^{\mathrm{2}} }{\mathrm{16}}\left(\mathrm{6}\boldsymbol{\pi}−\mathrm{4}\sqrt{\mathrm{2}}\:−\boldsymbol{\pi}+\mathrm{4}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}\:}\:\right) \\ $$$$\:\:\:\:\:\:=\frac{\boldsymbol{\mathrm{R}}^{\mathrm{2}} }{\mathrm{16}}\left[\left(\mathrm{5}\boldsymbol{\pi}+\mathrm{4}\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}\:}\:−\sqrt{\mathrm{2}}\:\right)\right]\right. \\ $$$$\mathrm{alors} \\ $$$$\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{Rose}}\right)=\boldsymbol{\mathrm{S}}\mathrm{3}+\boldsymbol{\mathrm{S}}\mathrm{4}+\boldsymbol{\mathrm{S}}\mathrm{5}+\boldsymbol{\mathrm{S}}\mathrm{6} \\ $$$$\:\:\:\:\:\:\:=\frac{\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{5}\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} }{\mathrm{16}}+\left(\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}\:}\:−\sqrt{\mathrm{2}}}{\mathrm{4}}\right)\boldsymbol{\mathrm{R}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\left[\frac{\mathrm{7}\boldsymbol{\pi}}{\mathrm{16}}+\frac{\left(\sqrt{\mathrm{2}\:−\sqrt{\mathrm{2}}}−\sqrt{\mathrm{2}}\right)}{\mathrm{4}}\right]\boldsymbol{\mathrm{R}}^{\mathrm{2}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{Vert}}\right)=\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} −\frac{\mathrm{7}\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} }{\mathrm{16}}+\frac{\left(\sqrt{\mathrm{2}}\right.}{\mathrm{4}}−\frac{\left.\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}\:}\:\right)\boldsymbol{\mathrm{R}}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:=\left[\frac{\mathrm{9}\boldsymbol{\pi}}{\mathrm{16}}+\frac{\left(\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\:\right)}{\mathrm{4}}\right]\boldsymbol{\mathrm{R}}^{\mathrm{2}} \\ $$$$\:\: \\ $$$$ \\ $$$$\:\:\frac{\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{Rose}}\right)}{\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{Vert}}\right)}\:=\:\:\frac{\mathrm{7}\boldsymbol{\pi}−\mathrm{4}\left(\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}\:}\right)}{\mathrm{9}\boldsymbol{\pi}+\mathrm{4}\left(\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right)} \\ $$
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) =((πR^2 )/2)
$${wrong}! \\ $$$${S}_{{yellow}} ={S}_{{green}} =\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$
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
α+β=45° CB=CA=(√2)R CE=2R cos α CD=2R cos β Δ_(CBE) =(((√2)R×2R cos α sin β)/2)=(√2)R^2 cos α sin β Δ_(CBD) =(((√2)R×2R cos β sin α)/2)=(√2)R^2 sin α cos β Δ_(CBE) +Δ_(CBD) =(√2)R^2 (cos α sin β+sin α cos β) =(√2) R^2 sin (α+β)=R^2 =Δ_(ABC) S_(yellow) =half of circle=((πR^2 )/2) S_(green) =πR^2 −S_(yellow) =((πR^2 )/2) (S_(yellow) /S_(green) )=1
$$\alpha+\beta=\mathrm{45}° \\ $$$${CB}={CA}=\sqrt{\mathrm{2}}{R} \\ $$$${CE}=\mathrm{2}{R}\:\mathrm{cos}\:\alpha \\ $$$${CD}=\mathrm{2}{R}\:\mathrm{cos}\:\beta \\ $$$$\Delta_{{CBE}} =\frac{\sqrt{\mathrm{2}}{R}×\mathrm{2}{R}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta}{\mathrm{2}}=\sqrt{\mathrm{2}}{R}^{\mathrm{2}} \mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$\Delta_{{CBD}} =\frac{\sqrt{\mathrm{2}}{R}×\mathrm{2}{R}\:\mathrm{cos}\:\beta\:\mathrm{sin}\:\alpha}{\mathrm{2}}=\sqrt{\mathrm{2}}{R}^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta \\ $$$$\Delta_{{CBE}} +\Delta_{{CBD}} =\sqrt{\mathrm{2}}{R}^{\mathrm{2}} \left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta+\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\right) \\ $$$$\:\:\:\:\:\:\:\:=\sqrt{\mathrm{2}}\:{R}^{\mathrm{2}} \:\mathrm{sin}\:\left(\alpha+\beta\right)={R}^{\mathrm{2}} =\Delta_{{ABC}} \\ $$$${S}_{{yellow}} ={half}\:{of}\:{circle}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${S}_{{green}} =\pi{R}^{\mathrm{2}} −{S}_{{yellow}} =\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{S}_{{yellow}} }{{S}_{{green}} }=\mathrm{1} \\ $$

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