Question-226755

Question Number 226755 by Hanuda354 last updated on 13/Dec/25

Answered by TonyCWX last updated on 13/Dec/25

A_(Green) =(1/2)(8^2 )((π/3))+(1/2)(8^2 )((π/3)−((√3)/2))−(1/2)π(4^2 )=((40)/3)π−16(√3) A_(Blue ) = (((12−3(√3)−2π)/(12)))(8^2 )=64−16(√3)−((32)/3)π (A_(Blue) /A_(Green) )=((64−16(√3)−((32)/3)π)/(((40)/3)π−16(√3)))=0.19589756479...

$${A}_{{Green}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}^{\mathrm{2}} \right)\left(\frac{\pi}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}^{\mathrm{2}} \right)\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{4}^{\mathrm{2}} \right)=\frac{\mathrm{40}}{\mathrm{3}}\pi−\mathrm{16}\sqrt{\mathrm{3}} \\ $$$${A}_{{Blue}\:} =\:\left(\frac{\mathrm{12}−\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{2}\pi}{\mathrm{12}}\right)\left(\mathrm{8}^{\mathrm{2}} \right)=\mathrm{64}−\mathrm{16}\sqrt{\mathrm{3}}−\frac{\mathrm{32}}{\mathrm{3}}\pi \\ $$$$ \\ $$$$\frac{{A}_{{Blue}} }{{A}_{{Green}} }=\frac{\mathrm{64}−\mathrm{16}\sqrt{\mathrm{3}}−\frac{\mathrm{32}}{\mathrm{3}}\pi}{\frac{\mathrm{40}}{\mathrm{3}}\pi−\mathrm{16}\sqrt{\mathrm{3}}}=\mathrm{0}.\mathrm{19589756479}… \\ $$

Commented by Hanuda354 last updated on 13/Dec/25

$$\mathrm{Thanks} \\ $$

Answered by mr W last updated on 13/Dec/25

Commented by mr W last updated on 13/Dec/25

A_1 =R×(R/2)−(1/2)×(R/2)×(((√3)R)/2)−(R^2 /2)×(π/6) =(R^2 /2)(1−((√3)/4)−(π/6)) A_2 =(R^2 /2)×(π/3)−(1/2)×(R/2)×(((√3)R)/2)−(π/4)×((R/2))^2 =(R^2 /2)(((5π)/(24))−((√3)/4)) ((blue)/(green))=((2A_1 )/(2A_2 ))=(A_1 /A_2 )=((1−((√3)/4)−(π/6))/(((5π)/(24))−((√3)/4))) =((24−6(√3)−4π)/(5π−6(√3)))≈0.195898

$${A}_{\mathrm{1}} ={R}×\frac{{R}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{{R}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}−\frac{{R}^{\mathrm{2}} }{\mathrm{2}}×\frac{\pi}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:=\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\pi}{\mathrm{6}}\right) \\ $$$${A}_{\mathrm{2}} =\frac{{R}^{\mathrm{2}} }{\mathrm{2}}×\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{{R}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}×\left(\frac{{R}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:=\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\mathrm{5}\pi}{\mathrm{24}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$$$\frac{{blue}}{{green}}=\frac{\mathrm{2}{A}_{\mathrm{1}} }{\mathrm{2}{A}_{\mathrm{2}} }=\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\pi}{\mathrm{6}}}{\frac{\mathrm{5}\pi}{\mathrm{24}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{24}−\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{4}\pi}{\mathrm{5}\pi−\mathrm{6}\sqrt{\mathrm{3}}}\approx\mathrm{0}.\mathrm{195898} \\ $$

Commented by Hanuda354 last updated on 13/Dec/25

$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{Prof}. \\ $$

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