# What-will-be-the-minimum-area-of-a-heptagon-inscribed-in-an-unit-square-

Question Number 134002 by Dwaipayan Shikari last updated on 26/Feb/21
$${What}\:{will}\:{be}\:{the}\:{minimum}\:{area}\:{of}\:{a}\:{heptagon}\:{inscribed}\:{in} \\$$$${an}\:{unit}\:{square}? \\$$
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
$${side}\:{length}\:{of}\:{heptagon}\:=\:{a} \\$$$$\theta=\frac{\left(\mathrm{7}−\mathrm{2}\right)×\mathrm{180}°}{\mathrm{7}}=\frac{\mathrm{900}°}{\mathrm{7}} \\$$$$\alpha=\mathrm{90}°−\frac{\theta}{\mathrm{2}}=\frac{\mathrm{180}°}{\mathrm{7}} \\$$$$\beta=\theta−\mathrm{2}\alpha=\frac{\mathrm{540}°}{\mathrm{7}} \\$$$$\gamma=\mathrm{180}°−\mathrm{45}°−\beta=\frac{\mathrm{405}°}{\mathrm{7}} \\$$$${DB}={b}=\mathrm{2}{a}\:\mathrm{cos}\:\alpha=\mathrm{2}{a}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}} \\$$$${AB}=\frac{{a}}{\:\sqrt{\mathrm{2}}} \\$$$${BC}={b}\:\mathrm{sin}\:\gamma=\mathrm{2}{a}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}} \\$$$${AC}=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}}\right){a}=\mathrm{1} \\$$$$\Rightarrow{a}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}}} \\$$$${area}\:{of}\:{heptagon} \\$$$${A}=\frac{\mathrm{7}{a}^{\mathrm{2}} }{\mathrm{4}\:\mathrm{tan}\:\frac{\mathrm{180}°}{\mathrm{7}}} \\$$$$=\frac{\mathrm{7}}{\mathrm{4}\:\mathrm{tan}\:\frac{\mathrm{180}°}{\mathrm{7}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}}\right)^{\mathrm{2}} } \\$$$$\approx\mathrm{0}.\mathrm{728}\:\mathrm{878}\:\mathrm{175} \\$$$$\\$$$${check}: \\$$$${ED}=\frac{{b}}{\:\sqrt{\mathrm{2}}} \\$$$${DC}={b}\:\mathrm{cos}\:\gamma \\$$$${EC}={b}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{cos}\:\gamma\right)=\mathrm{2}{a}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{cos}\:\frac{\mathrm{405}°}{\mathrm{7}}\right) \\$$$$=\mathrm{2}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}}}×\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{cos}\:\frac{\mathrm{405}°}{\mathrm{7}}\right) \\$$$$=\mathrm{1} \\$$
Commented by Dwaipayan Shikari last updated on 26/Feb/21
$${Great}\:{sir}! \\$$