Author: Tinku Tara

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}}…

Question Number 212720 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 $${AB}=\sqrt{\mathrm{2}\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{6};\:{Let}\:\angle{OAB}=\theta \\ $$$$\frac{{sin}\theta}{{r}}=\frac{{sin}\left(\mathrm{180}−\mathrm{2}\theta\right)}{{AB}}\Rightarrow{cos}\theta=\frac{\mathrm{3}}{{r}}\Rightarrow{sin}\mathrm{2}\theta=\frac{\mathrm{6}\sqrt{{r}^{\mathrm{2}} −\mathrm{9}}}{{r}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}}…

Question Number 212723 by ajfour last updated on 22/Oct/24 Commented by ajfour last updated on 22/Oct/24 $${Find}\:\:{R}_{{min}} \\ $$ Answered by mr W last updated…

Question Number 212719 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 Commented by A5T last updated on 22/Oct/24 $${a}=\mathrm{2}\left(\frac{\pi}{\mathrm{6}}−\frac{{sin}\mathrm{60}°}{\mathrm{2}}\right)=\mathrm{2}\left(\frac{\pi}{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\…