Question Number 82659 by naka3546 last updated on 23/Feb/20
Commented by naka3546 last updated on 23/Feb/20
$${area}\:\:{of}\:\:{red}\:\:{region}\:\:{is}\:\:\:… \\ $$
Commented by john santu last updated on 23/Feb/20
$${area}\:=\:\mathrm{16}\left(\mathrm{0}.\mathrm{4}−\mathrm{0}.\mathrm{5tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right) \\ $$
Commented by naka3546 last updated on 23/Feb/20
$${show}\:\:{your}\:\:{workings}\:,\:\:{please}\: \\ $$
Answered by mr W last updated on 24/Feb/20
Commented by mr W last updated on 24/Feb/20
$${r}=\mathrm{4} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{4}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\beta=\pi−\mathrm{2}\alpha=\pi−\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\beta=\mathrm{sin}\:\mathrm{2}\alpha=\mathrm{2}×\frac{\mathrm{1}×\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${A}_{{gray}\:{shaded}} =\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\beta−\mathrm{sin}\:\beta\right)=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$${A}_{{blue}\:{shaded}} ={r}^{\mathrm{2}} −\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}={r}^{\mathrm{2}} \left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$$${A}_{{red}} ={A}_{\Delta{ABC}} −{A}_{{gray}\:{shaded}} −{A}_{{blue}\:{shaded}} \\ $$$$=\frac{\mathrm{4}×\mathrm{8}}{\mathrm{2}}−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{4}}{\mathrm{5}}\right)−{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\mathrm{16}−\mathrm{8}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{6}}{\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{32}}{\mathrm{5}}−\mathrm{4}\pi+\mathrm{16}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\mathrm{1}.\mathrm{251991} \\ $$