Question Number 193093 by Mingma last updated on 04/Jun/23
Answered by Subhi last updated on 04/Jun/23
$$\frac{\mathrm{1}}{\mathrm{2}}.\pi.{r}^{\mathrm{2}} \:=\:\mathrm{3}\pi \\ $$$${r}\:=\:\sqrt{\mathrm{6}} \\ $$$${AB}\:=\:{AD}\:=\:{BE}\:=\:\:\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\left({CD}\right)^{\mathrm{2}} \:=\:{CE}.{BC}={CE}.\left({CE}+\mathrm{2}\sqrt{\mathrm{6}}\right) \\ $$$$\left({CD}\right)^{\mathrm{2}} =\left({CE}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{6}}.{CE}\:\:\rightarrow\left(\mathrm{1}\right) \\ $$$${or} \\ $$$$\left\{\left({CD}\right)^{\mathrm{2}} +\mathrm{6}=\left({CE}+\sqrt{\mathrm{6}}\right)^{\mathrm{2}} =\left({CE}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{6}}.{CE}+\mathrm{6}\right\} \\ $$$$\frac{{CD}}{{sin}\left(\mathrm{90}−{x}\right)}=\frac{\sqrt{\mathrm{6}}}{{sin}\left({x}\right)}\:\:\:\:\:\left({sin}\:{law}\right) \\ $$$$\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)}=\frac{\sqrt{\mathrm{6}}}{{CD}}\:\:\Rrightarrow\:{tan}\left({x}\right)=\frac{\sqrt{\mathrm{6}}}{{CD}}\:\:\left({i}\right) \\ $$$${in}\:{the}\:\Delta\:{ABC} \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{6}}}{{sin}\left({x}\right)}=\frac{\mathrm{2}\sqrt{\mathrm{6}}+{CE}}{{sin}\left(\mathrm{90}−{x}\right)} \\ $$$${tan}\left({x}\right)\:=\:\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{2}\sqrt{\mathrm{6}}+{CE}}\:\:\:\left({ii}\right) \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{2}\sqrt{\mathrm{6}}+{CE}}=\frac{\sqrt{\mathrm{6}}}{{CD}} \\ $$$${CD}\:=\:\sqrt{\mathrm{6}}+\frac{{CE}}{\mathrm{2}}\:\:\rightarrow\:\left(\mathrm{2}\right) \\ $$$${substitute}\:{in}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{6}+\sqrt{\mathrm{6}}.{CE}+\frac{\left({CE}\right)^{\mathrm{2}} }{\mathrm{4}}=\left({CE}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{6}}.{CE} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\left({CE}\right)^{\mathrm{2}} +\sqrt{\mathrm{6}}.{CE}−\mathrm{6}\:=\:\mathrm{0} \\ $$$${CE}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}\:\:\:\:\:,\:\:{CD}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$$\left[\Delta{ABC}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\sqrt{\mathrm{6}}×\left(\mathrm{2}\sqrt{\mathrm{6}}+\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}\right)=\mathrm{16} \\ $$$${the}\:{shaded}\:{area}\:=\:\mathrm{16}−\mathrm{3}\pi \\ $$
Commented by Subhi last updated on 04/Jun/23
$${You}\:{can}\:{also}\:{use}\:{the}\:{similarity} \\ $$$$\Delta{DCO}\:{is}\:{similar}\:{to}\:\Delta{BCA} \\ $$$$\frac{{DC}}{{CE}+\mathrm{2}\sqrt{\mathrm{6}}}=\frac{{DO}}{{BA}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}\sqrt{\mathrm{6}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{DC}\:=\:{CE}+\mathrm{2}\sqrt{\mathrm{6}} \\ $$
Commented by Subhi last updated on 04/Jun/23
Commented by Mingma last updated on 04/Jun/23
Perfect