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Question Number 77057 by Maclaurin Stickker last updated on 03/Jan/20

Commented by Maclaurin Stickker last updated on 03/Jan/20

I wouldn′t like an answer. I would like a tip on how to solve this problem.

$${I}\:{wouldn}'{t}\:{like}\:{an}\:{answer}.\:{I}\:{would}\:{like} \\ $$$${a}\:{tip}\:{on}\:{how}\:{to}\:{solve}\:{this}\:{problem}. \\ $$

Commented by mr W last updated on 03/Jan/20

is the answer in Q76898 no help for you? also here we should try to determine the length of the diagonal.

$${is}\:{the}\:{answer}\:{in}\:{Q}\mathrm{76898}\:{no}\:{help} \\ $$$${for}\:{you}? \\ $$$${also}\:{here}\:{we}\:{should}\:{try}\:{to}\:{determine} \\ $$$${the}\:{length}\:{of}\:{the}\:{diagonal}. \\ $$

Answered by mr W last updated on 03/Jan/20

Commented by mr W last updated on 03/Jan/20

α=(((n−2)×180°)/n)=(((9−2)×180)/9)=140° θ=((360°)/n)=((360)/9)=40° 2β+3α=3×180° ⇒β=((540−420)/2)=60° γ=β−θ=60−40=20° AB=((BD)/(tan (θ/4)))=(a/(2 tan 10°)) EB=CG×sin θ=a sin 40° GE=(a/2)+a cos θ=(a/2)(1+2 cos 40°) FE=GE×tan γ=(a/2)(1+2 cos 40°)tan 20° AF=AB−FE−EB AF=(a/(2 tan 10°))−(a/2)(1+2 cos 40°)tan 20°−a sin 40° ((AF)/a)=(1/(2 tan 10°))−((tan 20°)/2)−cos 40° tan 20°−sin 40° ((AF)/a)=(1/(2 tan 10°))−((tan 20°)/2)−(2 cos^2 20°−1) tan 20°−sin 40° ((AF)/a)=(1/(2 tan 10°))+((tan 20°)/2)−2 sin 40° ((AF)/a)=(√3) ⇒AF=(√3)a area of square =(((AF)/(√2)))^2 =((AF^2 )/2) =((3a^2 )/2)=((3×2^2 )/2)=6

$$\alpha=\frac{\left({n}−\mathrm{2}\right)×\mathrm{180}°}{{n}}=\frac{\left(\mathrm{9}−\mathrm{2}\right)×\mathrm{180}}{\mathrm{9}}=\mathrm{140}° \\ $$$$\theta=\frac{\mathrm{360}°}{{n}}=\frac{\mathrm{360}}{\mathrm{9}}=\mathrm{40}° \\ $$$$\mathrm{2}\beta+\mathrm{3}\alpha=\mathrm{3}×\mathrm{180}° \\ $$$$\Rightarrow\beta=\frac{\mathrm{540}−\mathrm{420}}{\mathrm{2}}=\mathrm{60}° \\ $$$$\gamma=\beta−\theta=\mathrm{60}−\mathrm{40}=\mathrm{20}° \\ $$$${AB}=\frac{{BD}}{\mathrm{tan}\:\frac{\theta}{\mathrm{4}}}=\frac{{a}}{\mathrm{2}\:\mathrm{tan}\:\mathrm{10}°} \\ $$$${EB}={CG}×\mathrm{sin}\:\theta={a}\:\mathrm{sin}\:\mathrm{40}° \\ $$$${GE}=\frac{{a}}{\mathrm{2}}+{a}\:\mathrm{cos}\:\theta=\frac{{a}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\mathrm{40}°\right) \\ $$$${FE}={GE}×\mathrm{tan}\:\gamma=\frac{{a}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\mathrm{40}°\right)\mathrm{tan}\:\mathrm{20}° \\ $$$${AF}={AB}−{FE}−{EB} \\ $$$${AF}=\frac{{a}}{\mathrm{2}\:\mathrm{tan}\:\mathrm{10}°}−\frac{{a}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\mathrm{40}°\right)\mathrm{tan}\:\mathrm{20}°−{a}\:\mathrm{sin}\:\mathrm{40}° \\ $$$$\frac{{AF}}{{a}}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:\mathrm{10}°}−\frac{\mathrm{tan}\:\mathrm{20}°}{\mathrm{2}}−\mathrm{cos}\:\mathrm{40}°\:\mathrm{tan}\:\mathrm{20}°−\mathrm{sin}\:\mathrm{40}° \\ $$$$\frac{{AF}}{{a}}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:\mathrm{10}°}−\frac{\mathrm{tan}\:\mathrm{20}°}{\mathrm{2}}−\left(\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}°−\mathrm{1}\right)\:\mathrm{tan}\:\mathrm{20}°−\mathrm{sin}\:\mathrm{40}° \\ $$$$\frac{{AF}}{{a}}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:\mathrm{10}°}+\frac{\mathrm{tan}\:\mathrm{20}°}{\mathrm{2}}−\mathrm{2}\:\mathrm{sin}\:\mathrm{40}° \\ $$$$\frac{{AF}}{{a}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{AF}=\sqrt{\mathrm{3}}{a} \\ $$$${area}\:{of}\:{square}\:=\left(\frac{{AF}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\frac{{AF}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}×\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{6} \\ $$

Commented by Tawa11 last updated on 29/Dec/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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