Question Number 226829 by Estevao last updated on 16/Dec/25
Answered by Estevao last updated on 17/Dec/25
$${Good} \\ $$
Answered by mr W last updated on 17/Dec/25
Commented by mr W last updated on 17/Dec/25
$${R}^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} +\left(\mathrm{2}{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{R}=\mathrm{5}{a} \\ $$$$\Rightarrow\frac{{a}}{{R}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\frac{{green}}{{yellow}}=\left(\frac{{a}}{{R}}\right)^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{25}} \\ $$$${X}=\frac{\mathrm{4}}{\mathrm{25}}×\mathrm{25}=\mathrm{4}\:{u}^{\mathrm{2}} \\ $$
Answered by fantastic2 last updated on 18/Dec/25
Commented by fantastic2 last updated on 18/Dec/25
$$\mathrm{2}\left({R}−\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)={R}\sqrt{\mathrm{2}} \\ $$$${R}−\frac{{a}}{\:\sqrt{\mathrm{2}}}=\frac{{R}}{\:\sqrt{\mathrm{2}}} \\ $$$${R}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\frac{{a}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{{a}}{{R}}=\sqrt{\mathrm{2}}−\mathrm{1}\:?? \\ $$$${R}−\frac{{a}}{\:\sqrt{\mathrm{2}}}+\frac{{a}}{\:\sqrt{\mathrm{2}}}+{a}\sqrt{\mathrm{2}}={R}\sqrt{\mathrm{2}} \\ $$$${a}\sqrt{\mathrm{2}}={R}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\frac{{a}}{{R}}=\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:?? \\ $$$$\mathrm{2}\left({R}−\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)={R}−\frac{{a}}{\:\sqrt{\mathrm{2}}}+\frac{{a}}{\:\sqrt{\mathrm{2}}}+{a}\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}{R}−{a}\sqrt{\mathrm{2}}={R}+{a}\sqrt{\mathrm{2}} \\ $$$$\frac{{a}}{{R}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}?? \\ $$
Commented by fantastic2 last updated on 18/Dec/25
$${where}\:{am}\:{i}\:{wrong}? \\ $$
Commented by som(math1967) last updated on 18/Dec/25
$${how}\:{yellow}\:{triangle}\:{is}\:{rt}\angle\:?? \\ $$