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Question Number 76898 by Maclaurin Stickker last updated on 31/Dec/19

Answered by mr W last updated on 31/Dec/19

Commented by mr W last updated on 31/Dec/19

a=2 α=140° β=40° AB=BC=CA=a(1+2 cos β) ∠BAC=60° BD=(a/(2 cos β)) AD=AB−BD=a(1+2 cos β−(1/(2 cos β))) AE=2a cos β ED^2 =AD^2 +AE^2 −2×AD×AE×cos 60° ED^2 =a^2 (1+2 cos β−(1/(2 cos β)))^2 +4a^2 cos^2 β−2a^2 (1+2 cos β−(1/(2 cos β))) cos β ED^2 =a^2 (4 cos^2 β+(1/(4 cos^2 β))+2 cos β−(1/(cos β))+1) ED^2 =a^2 (3+2 cos 2β+(1/(4 cos^2 β))+((cos 2β)/(cos β))) ED^2 =a^2 (3+2 cos 80°+(1/(4 cos^2 40°))+((cos 80°)/(cos 40°))) =4a^2 area of square =(((DE)/(√2)))^2 =((DE^2 )/2) =((4a^2 )/2)=2a^2 =2×2^2 =8 side length of square =(√2)a

$${a}=\mathrm{2} \\ $$$$\alpha=\mathrm{140}° \\ $$$$\beta=\mathrm{40}° \\ $$$${AB}={BC}={CA}={a}\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\beta\right) \\ $$$$\angle{BAC}=\mathrm{60}° \\ $$$${BD}=\frac{{a}}{\mathrm{2}\:\mathrm{cos}\:\beta} \\ $$$${AD}={AB}−{BD}={a}\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\beta−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\beta}\right) \\ $$$${AE}=\mathrm{2}{a}\:\mathrm{cos}\:\beta \\ $$$${ED}^{\mathrm{2}} ={AD}^{\mathrm{2}} +{AE}^{\mathrm{2}} −\mathrm{2}×{AD}×{AE}×\mathrm{cos}\:\mathrm{60}° \\ $$$${ED}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\beta−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\beta}\right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\beta−\mathrm{2}{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\beta−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\beta}\right)\:\mathrm{cos}\:\beta \\ $$$${ED}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\beta+\frac{\mathrm{1}}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\beta}+\mathrm{2}\:\mathrm{cos}\:\beta−\frac{\mathrm{1}}{\mathrm{cos}\:\beta}+\mathrm{1}\right) \\ $$$${ED}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\beta+\frac{\mathrm{1}}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\beta}+\frac{\mathrm{cos}\:\mathrm{2}\beta}{\mathrm{cos}\:\beta}\right) \\ $$$${ED}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{2}\:\mathrm{cos}\:\mathrm{80}°+\frac{\mathrm{1}}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}°}+\frac{\mathrm{cos}\:\mathrm{80}°}{\mathrm{cos}\:\mathrm{40}°}\right) \\ $$$$=\mathrm{4}{a}^{\mathrm{2}} \\ $$$${area}\:{of}\:{square}\:=\left(\frac{{DE}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\frac{{DE}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2}{a}^{\mathrm{2}} =\mathrm{2}×\mathrm{2}^{\mathrm{2}} =\mathrm{8} \\ $$$$ \\ $$$${side}\:{length}\:{of}\:{square}\:=\sqrt{\mathrm{2}}{a} \\ $$

Commented by Maclaurin Stickker last updated on 31/Dec/19

Thanks. And Happy new year!

$${Thanks}.\:{And}\:{Happy}\:{new}\:{year}! \\ $$

Commented by mr W last updated on 31/Dec/19

thanks! the same to you!

$${thanks}!\:{the}\:{same}\:{to}\:{you}! \\ $$

Commented by Maclaurin Stickker last updated on 06/Jan/20

Sir, I founf 6 for the area.

$${Sir},\:{I}\:{founf}\:\mathrm{6}\:{for}\:{the}\:{area}. \\ $$

Commented by mr W last updated on 07/Jan/20

6 is correct sir! i made an error somewhere.

$$\mathrm{6}\:{is}\:{correct}\:{sir}! \\ $$$${i}\:{made}\:{an}\:{error}\:{somewhere}. \\ $$

Commented by Tawa11 last updated on 29/Dec/21

Great sir

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

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