Question Number 209937 by Ari last updated on 26/Jul/24
Answered by mr W last updated on 26/Jul/24
Commented by mr W last updated on 26/Jul/24
$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${blue}\:{line}=\sqrt{\left(\mathrm{6}+\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{4}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} \right)}=\mathrm{2}\sqrt{\mathrm{19}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{2}\sqrt{\mathrm{19}}}{\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{2}\sqrt{\mathrm{57}}}{\mathrm{3}} \\ $$
Commented by mr W last updated on 26/Jul/24
$${or} \\ $$$${R}=\sqrt{\frac{\left(\mathrm{6}×\mathrm{10}+\mathrm{4}×\mathrm{4}\right)\left(\mathrm{6}×\mathrm{4}+\mathrm{10}×\mathrm{4}\right)\left(\mathrm{6}×\mathrm{4}+\mathrm{10}×\mathrm{4}\right)}{\left(−\mathrm{4}+\mathrm{6}+\mathrm{4}+\mathrm{10}\right)\left(\mathrm{4}−\mathrm{6}+\mathrm{4}+\mathrm{10}\right)\left(\mathrm{4}+\mathrm{6}−\mathrm{4}+\mathrm{10}\right)\left(\mathrm{4}+\mathrm{6}+\mathrm{4}−\mathrm{10}\right)}} \\ $$$$\:\:\:\:=\frac{\mathrm{2}\sqrt{\mathrm{57}}}{\mathrm{3}} \\ $$