Question Number 175901 by infinityaction last updated on 09/Sep/22
Commented by HeferH last updated on 09/Sep/22
Answered by HeferH last updated on 09/Sep/22
$$\:{R}\:=\:\frac{\mathrm{15}\:−\:\mathrm{6}\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$$\: \\ $$
Commented by infinityaction last updated on 09/Sep/22
$${solution}\:?? \\ $$
Commented by HeferH last updated on 09/Sep/22
Commented by HeferH last updated on 09/Sep/22
Commented by Tawa11 last updated on 15/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 09/Sep/22
$$\sqrt{\left({R}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}−{R}\right)^{\mathrm{2}} }+\sqrt{\left({R}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}−{R}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\sqrt{\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{2}{R}−\frac{\mathrm{1}}{\mathrm{4}}\right)}+\sqrt{\mathrm{2}{R}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{3}\left(\mathrm{8}{R}−\mathrm{1}\right)=\left(\mathrm{3}−\mathrm{4}\sqrt{\mathrm{2}{R}}\right)^{\mathrm{2}} \\ $$$$\mathrm{0}=\mathrm{3}−\mathrm{6}\sqrt{\mathrm{2}{R}}+\mathrm{2}{R} \\ $$$$\sqrt{{R}}=\frac{\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{15}−\mathrm{6}\sqrt{\mathrm{6}}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{1515} \\ $$
Commented by Tawa11 last updated on 15/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$