Question Number 65128 by Tawa1 last updated on 25/Jul/19
Answered by ajfour last updated on 25/Jul/19
Commented by ajfour last updated on 25/Jul/19
$${OP}\:=\mathrm{8} \\ $$$${with}\:{O}\:{as}\:{origin} \\ $$$${P}\:\left(\mathrm{4},\:\sqrt{\mathrm{64}−\mathrm{16}}\right)\:\equiv\left(\mathrm{4},\:\mathrm{4}\sqrt{\mathrm{3}}\right) \\ $$$${OQ}=\mathrm{8}\sqrt{\mathrm{2}} \\ $$$${eq}.\:{of}\:{OQ}\:{is}\:\:{y}−{x}=\mathrm{0} \\ $$$$\bot\:{distance}\:{of}\:{P}\:{from}\:{line}\:{OQ} \\ $$$$\:\:\:{p}=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{4}}{\:\sqrt{\mathrm{2}}} \\ $$$${shaded}\:{area}=\:{Area}\left(\bigtriangleup{OPQ}\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}\sqrt{\mathrm{2}}×\frac{\mathrm{4}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{16}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:. \\ $$
Commented by Tawa1 last updated on 25/Jul/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 25/Jul/19
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 25/Jul/19
Commented by Tawa1 last updated on 25/Jul/19
$$\mathrm{This}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 25/Jul/19
Commented by mr W last updated on 25/Jul/19
$${DC}={DB}={DE}={r} \\ $$$${CF}=\frac{{CB}}{\mathrm{2}}=\mathrm{2} \\ $$$$\frac{{CF}}{{CD}}=\frac{{CD}}{{AC}} \\ $$$${CD}^{\mathrm{2}} ={CF}×{AC} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\mathrm{2}×\mathrm{9}=\mathrm{18} \\ $$$$\Rightarrow{r}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$
Answered by mr W last updated on 25/Jul/19
Commented by mr W last updated on 25/Jul/19
$${OC}={OP}={CP}={r}=\mathrm{8} \\ $$$$\Rightarrow\angle{OCP}=\mathrm{60}°=\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\angle{PCQ}=\mathrm{30}°=\frac{\pi}{\mathrm{6}} \\ $$$${A}_{{shade}} ={Cap}_{{OQ}} −{Cap}_{{OP}} −{Cap}_{{PQ}} \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\right)−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{3}}−\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right)−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{2}}+\mathrm{sin}\:\frac{\pi}{\mathrm{3}}+\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$$=\frac{\mathrm{64}}{\mathrm{2}}\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$=\mathrm{16}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$
Commented by mr W last updated on 25/Jul/19
$${An}\:{other}\:{way}: \\ $$$${A}_{{shade}} =\frac{\mathrm{1}}{\mathrm{2}}×{OP}×{OQ}×\mathrm{sin}\:\angle{POQ} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}×\mathrm{8}\sqrt{\mathrm{2}}×\mathrm{sin}\:\left(\mathrm{60}°−\mathrm{45}°\right) \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\mathrm{16}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$