Question Number 212733 by cherokeesay last updated on 22/Oct/24
Answered by A5T last updated on 22/Oct/24
Commented by A5T last updated on 22/Oct/24
$$\left({R}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{4}\Rightarrow{R}=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$${G}+{B}=\frac{\sqrt{\mathrm{2}}\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}=\sqrt{\mathrm{2}}+\mathrm{2} \\ $$$${Y}+{B}=\frac{\sqrt{\mathrm{2}}\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\right)}{\:\mathrm{2}}=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{G}={Y}\Rightarrow\frac{{Green}_{{area}} }{{Yellow}_{{area}} }=\mathrm{1} \\ $$
Commented by cherokeesay last updated on 22/Oct/24
$${nice}\:!\:{thank}\:{you}\:{sir}\:! \\ $$