Question Number 189790 by normans last updated on 21/Mar/23
Answered by mr W last updated on 22/Mar/23
Commented by mr W last updated on 22/Mar/23
$${BE}={a}\:\mathrm{tan}\:\alpha \\ $$$${BD}=\sqrt{\mathrm{3}}{a} \\ $$$$\Rightarrow{ED}=\left(\sqrt{\mathrm{3}}−\mathrm{tan}\:\alpha\right){a} \\ $$$$\frac{{FD}}{{AD}}=\frac{\mathrm{sin}\:\left(\mathrm{60}°−\alpha\right)}{\mathrm{sin}\:\left(\mathrm{60}°+\alpha\right)}=\frac{\sqrt{\mathrm{3}}−\mathrm{tan}\:\alpha}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\alpha} \\ $$$$\Rightarrow{FD}=\frac{\mathrm{2}{a}\left(\sqrt{\mathrm{3}}−\mathrm{tan}\:\alpha\right)}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\alpha} \\ $$$${A}_{\mathrm{1}} =\frac{{a}^{\mathrm{2}} \:\mathrm{tan}\:\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{2}{A}_{\mathrm{1}} }{{a}^{\mathrm{2}} } \\ $$$${A}_{\mathrm{2}} =\frac{{FD}×{ED}\:\mathrm{sin}\:\mathrm{30}°}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}−\mathrm{tan}\:\alpha\right)^{\mathrm{2}} }{\:\mathrm{2}\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\alpha\right)} \\ $$$${A}_{\mathrm{2}} =\frac{{a}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}−\frac{\mathrm{2}{A}_{\mathrm{1}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} }{\:\mathrm{2}\left(\sqrt{\mathrm{3}}+\frac{\mathrm{2}{A}_{\mathrm{1}} }{{a}^{\mathrm{2}} }\right)} \\ $$$$\mathrm{3}{a}^{\mathrm{4}} −\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{2}{A}_{\mathrm{1}} +{A}_{\mathrm{2}} \right){a}^{\mathrm{2}} +\mathrm{4}{A}_{\mathrm{1}} \left({A}_{\mathrm{1}} −{A}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{2}{A}_{\mathrm{1}} +{A}_{\mathrm{2}} +\sqrt{{A}_{\mathrm{2}} \left(\mathrm{8}{A}_{\mathrm{1}} +{A}_{\mathrm{2}} \right)}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{2}×\mathrm{15}+\mathrm{8}+\sqrt{\mathrm{8}\left(\mathrm{8}×\mathrm{15}+\mathrm{8}\right)}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{70}}{\:\sqrt{\mathrm{3}}} \\ $$$${area}\:{of}\:{hexagon}: \\ $$$$\mathrm{6}×\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{\mathrm{70}}{\:\sqrt{\mathrm{3}}}=\mathrm{105}\:\checkmark \\ $$