Question Number 25479 by AdolfHitler last updated on 11/Dec/17
Answered by jota^ last updated on 11/Dec/17
$${r}\:=\:{radius}\:{circle} \\ $$$${l}\:=\:{side}\:{square} \\ $$$${r}={l}\sqrt{\mathrm{2}} \\ $$$${The}\:{triangle}\:\bigtriangleup\mathrm{APQ} \\ $$$${A}\left(−\mathrm{r},\:\mathrm{0}\right) \\ $$$${P}\left(\frac{{l}\mathrm{cos}\:\alpha}{\mathrm{2}},\:\frac{{l}\mathrm{sin}\:\alpha}{\mathrm{2}}\right) \\ $$$${Q}\left({l}\mathrm{cos}\:\alpha−\frac{{l}\mathrm{sin}\:\alpha}{\mathrm{2}},\:{l}\mathrm{sin}\:\alpha+\frac{{l}\mathrm{cos}\:\alpha}{\mathrm{2}}\right) \\ $$$$\mathrm{Area}=\frac{\mathrm{1}}{\mathrm{2}}\mid\overset{\rightarrow} {\mathrm{AP}}×\overset{\rightarrow} {\mathrm{AQ}}\mid \\ $$