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Question-189790

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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 tan α BD=(√3)a ⇒ED=((√3)−tan α)a ((FD)/(AD))=((sin (60°−α))/(sin (60°+α)))=(((√3)−tan α)/( (√3)+tan α)) ⇒FD=((2a((√3)−tan α))/( (√3)+tan α)) A_1 =((a^2 tan α)/2) ⇒tan α=((2A_1 )/a^2 ) A_2 =((FD×ED sin 30°)/2)=((a^2 ((√3)−tan α)^2 )/( 2((√3)+tan α))) A_2 =((a^2 ((√3)−((2A_1 )/a^2 ))^2 )/( 2((√3)+((2A_1 )/a^2 )))) 3a^4 −2(√3)(2A_1 +A_2 )a^2 +4A_1 (A_1 −A_2 )=0 ⇒a^2 =((2A_1 +A_2 +(√(A_2 (8A_1 +A_2 ))))/( (√3))) ⇒a^2 =((2×15+8+(√(8(8×15+8))))/( (√3)))=((70)/( (√3))) area of hexagon: 6×(((√3)a^2 )/4)=((6(√3))/4)×((70)/( (√3)))=105 ✓
$${BE}={a}\:\mathrm{tan}\:\alpha \\ $$$${BD}=\sqrt{\mathrm{3}}{a} \\ $$$$\Rightarrow{ED}=\left(\sqrt{\mathrm{3}}−\mathrm{tan}\:\alpha\right){a} \\ $$$$\frac{{FD}}{{AD}}=\frac{\mathrm{sin}\:\left(\mathrm{60}°−\alpha\right)}{\mathrm{sin}\:\left(\mathrm{60}°+\alpha\right)}=\frac{\sqrt{\mathrm{3}}−\mathrm{tan}\:\alpha}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\alpha} \\ $$$$\Rightarrow{FD}=\frac{\mathrm{2}{a}\left(\sqrt{\mathrm{3}}−\mathrm{tan}\:\alpha\right)}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\alpha} \\ $$$${A}_{\mathrm{1}} =\frac{{a}^{\mathrm{2}} \:\mathrm{tan}\:\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{2}{A}_{\mathrm{1}} }{{a}^{\mathrm{2}} } \\ $$$${A}_{\mathrm{2}} =\frac{{FD}×{ED}\:\mathrm{sin}\:\mathrm{30}°}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}−\mathrm{tan}\:\alpha\right)^{\mathrm{2}} }{\:\mathrm{2}\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\alpha\right)} \\ $$$${A}_{\mathrm{2}} =\frac{{a}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}−\frac{\mathrm{2}{A}_{\mathrm{1}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} }{\:\mathrm{2}\left(\sqrt{\mathrm{3}}+\frac{\mathrm{2}{A}_{\mathrm{1}} }{{a}^{\mathrm{2}} }\right)} \\ $$$$\mathrm{3}{a}^{\mathrm{4}} −\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{2}{A}_{\mathrm{1}} +{A}_{\mathrm{2}} \right){a}^{\mathrm{2}} +\mathrm{4}{A}_{\mathrm{1}} \left({A}_{\mathrm{1}} −{A}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{2}{A}_{\mathrm{1}} +{A}_{\mathrm{2}} +\sqrt{{A}_{\mathrm{2}} \left(\mathrm{8}{A}_{\mathrm{1}} +{A}_{\mathrm{2}} \right)}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{2}×\mathrm{15}+\mathrm{8}+\sqrt{\mathrm{8}\left(\mathrm{8}×\mathrm{15}+\mathrm{8}\right)}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{70}}{\:\sqrt{\mathrm{3}}} \\ $$$${area}\:{of}\:{hexagon}: \\ $$$$\mathrm{6}×\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{\mathrm{70}}{\:\sqrt{\mathrm{3}}}=\mathrm{105}\:\checkmark \\ $$

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