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Question Number 134002 by Dwaipayan Shikari last updated on 26/Feb/21

$$Whatwillbetheminimumareaofaheptagoninscribedin$$$$anunitsquare?$$

Commented by Dwaipayan Shikari last updated on 26/Feb/21

Answered by mr W last updated on 26/Feb/21

Commented by mr W last updated on 26/Feb/21

$$sidelengthofheptagon=a$$$$\theta =\frac{(7-2)\times 180°}{7}=\frac{900°}{7}$$$$\alpha =90°-\frac{\theta }{2}=\frac{180°}{7}$$$$\beta =\theta -2\alpha =\frac{540°}{7}$$$$\gamma =180°-45°-\beta =\frac{405°}{7}$$$$DB=b=2a\cos \alpha =2a\cos \frac{180°}{7}$$$$AB=\frac{a}{\sqrt{2}}$$$$BC=b\sin \gamma =2a\cos \frac{180°}{7}\sin \frac{405°}{7}$$$$AC=(\frac{1}{\sqrt{2}}+2\cos \frac{180°}{7}\sin \frac{405°}{7})a=1$$$$\Rightarrow a=\frac{1}{\frac{1}{\sqrt{2}}+2\cos \frac{180°}{7}\sin \frac{405°}{7}}$$$$areaofheptagon$$$$A=\frac{7a^2}{4\tan \frac{180°}{7}}$$$$=\frac{7}{4\tan \frac{180°}{7}{(\frac{1}{\sqrt{2}}+2\cos \frac{180°}{7}\sin \frac{405°}{7})}^2}$$$$\approx 0.728878175$$$$$$$$check:$$$$ED=\frac{b}{\sqrt{2}}$$$$DC=b\cos \gamma$$$$EC=b(\frac{1}{\sqrt{2}}+\cos \gamma )=2a\cos \frac{180°}{7}(\frac{1}{\sqrt{2}}+\cos \frac{405°}{7})$$$$=2\times \frac{1}{\frac{1}{\sqrt{2}}+2\cos \frac{180°}{7}\sin \frac{405°}{7}}\times \cos \frac{180°}{7}(\frac{1}{\sqrt{2}}+\cos \frac{405°}{7})$$$$=1$$

Commented by Dwaipayan Shikari last updated on 26/Feb/21

$$Greatsir!$$

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