Question-63985

Question Number 63985 by Tawa1 last updated on 11/Jul/19

Answered by mr W last updated on 12/Jul/19

Commented by mr W last updated on 12/Jul/19

tan α=(1/2) θ=(π/4)−α radiys R=3 BC=(√2)R=3(√2) CE=2R cos α=2×3×(2/( (√5)))=((12)/( (√5))) A_(ΔABC) =((3×3)/2)=(9/2) A_(ΔBCE) =((BC×EC sin θ)/2)=((3(√2)×12)/(2(√5)))×((√2)/2)((2/( (√5)))−(1/( (√5))))=((18)/5) A_(cap BE) =(3^2 /2)(2θ−sin 2θ)=(9/2)((π/2)−2α−cos 2α) A_(cap BE) =(9/2)((π/2)−2α−1+2×(1/5)) A_(cap BE) =(9/2)((π/2)−2tan^(−1) (1/2)−(3/5)) A_(green) =A_(ΔABC) −A_(ΔBCE) −A_(cap BE) =(9/2)−((18)/5)−(9/2)((π/2)−2tan^(−1) (1/2)−(3/5)) =(9/2)((4/5)−(π/2)+2tan^(−1) (1/2)) A_(blue) =6×12−2×π×3^2 −2×(9/2)((4/5)−(π/2)+2tan^(−1) (1/2)) A_(blue) =((324)/5)−((27π)/2)−18 tan^(−1) (1/2)=14.04

$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\theta=\frac{\pi}{\mathrm{4}}−\alpha \\ $$$${radiys}\:{R}=\mathrm{3} \\ $$$${BC}=\sqrt{\mathrm{2}}{R}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${CE}=\mathrm{2}{R}\:\mathrm{cos}\:\alpha=\mathrm{2}×\mathrm{3}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{12}}{\:\sqrt{\mathrm{5}}} \\ $$$${A}_{\Delta{ABC}} =\frac{\mathrm{3}×\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${A}_{\Delta{BCE}} =\frac{{BC}×{EC}\:\mathrm{sin}\:\theta}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{2}}×\mathrm{12}}{\mathrm{2}\sqrt{\mathrm{5}}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)=\frac{\mathrm{18}}{\mathrm{5}} \\ $$$${A}_{{cap}\:{BE}} =\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}\theta−\mathrm{sin}\:\mathrm{2}\theta\right)=\frac{\mathrm{9}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha−\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$$${A}_{{cap}\:{BE}} =\frac{\mathrm{9}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha−\mathrm{1}+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$${A}_{{cap}\:{BE}} =\frac{\mathrm{9}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{5}}\right) \\ $$$${A}_{{green}} ={A}_{\Delta{ABC}} −{A}_{\Delta{BCE}} −{A}_{{cap}\:{BE}} \\ $$$$=\frac{\mathrm{9}}{\mathrm{2}}−\frac{\mathrm{18}}{\mathrm{5}}−\frac{\mathrm{9}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{9}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{5}}−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${A}_{{blue}} =\mathrm{6}×\mathrm{12}−\mathrm{2}×\pi×\mathrm{3}^{\mathrm{2}} −\mathrm{2}×\frac{\mathrm{9}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{5}}−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${A}_{{blue}} =\frac{\mathrm{324}}{\mathrm{5}}−\frac{\mathrm{27}\pi}{\mathrm{2}}−\mathrm{18}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}=\mathrm{14}.\mathrm{04} \\ $$

Commented by Tawa1 last updated on 12/Jul/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 12/Jul/19

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply