Question Number 15623 by tawa tawa last updated on 12/Jun/17
Answered by ajfour last updated on 12/Jun/17
Commented by ajfour last updated on 12/Jun/17
$$\:\:{BD}^{\mathrm{2}} ={BC}^{\:\mathrm{2}} −{CD}^{\mathrm{2}} \\ $$$$\:\:\:{AC}={BC}=\:{r}\:=\mathrm{7} \\ $$$$\:\:\:\:\:\:\:{h}^{\mathrm{2}} =\:\mathrm{49}−\mathrm{16}=\mathrm{33} \\ $$$$\:\:\:\:\:\:\:{h}=\sqrt{\mathrm{33}}\:. \\ $$$$\:{Area}\:{of}\:\Delta{BCD}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\sqrt{\mathrm{33}} \\ $$$$\:{Area}\:{of}\:{sector}\:{ABC}\:=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}\:\theta=\frac{\mathrm{4}}{\mathrm{7}}\:\Rightarrow\:\:\theta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{7}}\right) \\ $$$$\:{Shaded}\:{area}\:=\:{Area}\:{of}\:{sector}\:{ABC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{Area}\:{of}\:\Delta{BCD} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{7}\right)^{\mathrm{2}} \mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{7}}\right)−\mathrm{2}\sqrt{\mathrm{33}} \\ $$$${Shaded}\:{area}\:=\frac{\mathrm{49}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{7}}\right)−\mathrm{2}\sqrt{\mathrm{33}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\:\mathrm{12}.\mathrm{093}\:. \\ $$
Commented by tawa tawa last updated on 12/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$