Question Number 219120 by Spillover last updated on 19/Apr/25
Answered by mr W last updated on 20/Apr/25
Commented by mr W last updated on 20/Apr/25
$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${eqn}.\:{of}\:{AD}: \\ $$$${y}=\frac{{x}+\mathrm{2}}{\mathrm{3}} \\ $$$${eqn}.\:{of}\:{circle}\:{A}: \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left({x}_{{C}} +\mathrm{2}\right)^{\mathrm{2}} +\left(\frac{{x}_{{C}} +\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left({x}_{{C}} +\mathrm{2}\right)^{\mathrm{2}} =\frac{\mathrm{36}}{\mathrm{5}} \\ $$$$\Rightarrow{x}_{{C}} =\frac{\mathrm{6}}{\:\sqrt{\mathrm{5}}}−\mathrm{2}\:\Rightarrow{y}_{{C}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$${eqn}.\:{of}\:{circle}\:{E}: \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left({x}_{{B}} −\mathrm{2}\right)^{\mathrm{2}} +\left(\frac{{x}_{{B}} +\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}{x}_{{B}} ^{\mathrm{2}} −\mathrm{16}{x}_{{B}} −\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow{x}_{{B}} =−\frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow{y}_{{B}} =\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${A}_{{green}} =\left({x}_{{C}} −{x}_{{B}} \right)\left({y}_{{C}} +{y}_{{B}} \right) \\ $$$$\:\:\:\:=\left(\frac{\mathrm{6}}{\:\sqrt{\mathrm{5}}}−\mathrm{2}+\frac{\mathrm{4}}{\mathrm{5}}\right)\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}+\frac{\mathrm{2}}{\:\mathrm{5}}\right)=\frac{\mathrm{48}}{\mathrm{25}} \\ $$$${area}\:{of}\:{square}\:=\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{8} \\ $$$${fraction}\:{of}\:{green}\:{area}\: \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{48}}{\mathrm{25}×\mathrm{8}}=\frac{\mathrm{6}}{\mathrm{25}}=\mathrm{0}.\mathrm{24} \\ $$
Commented by Spillover last updated on 20/Apr/25
$${thank}\:{you} \\ $$
Answered by Spillover last updated on 20/Apr/25
