Question Number 219067 by Spillover last updated on 19/Apr/25
Commented by Spillover last updated on 19/Apr/25
Quarter circle area = Semicircle area.
Green area/Orange area = ?
Answered by mr W last updated on 20/Apr/25
Commented by mr W last updated on 20/Apr/25
$$\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{R}=\sqrt{\mathrm{2}}{r} \\ $$$$\mathrm{2}{r}\:\mathrm{tan}\:\alpha+{R}\:\mathrm{tan}\:\alpha=\frac{{R}}{\mathrm{tan}\:\alpha} \\ $$$$\Rightarrow\mathrm{tan}^{\mathrm{2}} \:\alpha=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$${A}_{{green}} =\frac{\mathrm{2}{R}\:\mathrm{sin}\:\alpha×\mathrm{2}{r}\:\mathrm{sin}\:\alpha\:\mathrm{tan}\:\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\:\mathrm{tan}\:\alpha\:{r}^{\mathrm{2}} \\ $$$${A}_{{orange}} =\frac{\left(\mathrm{2}{r}\right)^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha\:{r}^{\mathrm{2}} \\ $$$$\frac{{A}_{{green}} }{{A}_{{orange}} }=\frac{\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\:\mathrm{tan}\:\alpha}{\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha} \\ $$$$\:\:\:\:\:\:=\sqrt{\mathrm{2}}\:\mathrm{tan}^{\mathrm{2}} \:\alpha=\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\mathrm{2}−\sqrt{\mathrm{2}}\:\checkmark \\ $$
Commented by Spillover last updated on 20/Apr/25
$${thank}\:{you}\:{Mr}\:{W} \\ $$
Answered by Spillover last updated on 20/Apr/25
