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?](https://www.tinkutara.com/question/Q134002.png)
$${What}\:{will}\:{be}\:{the}\:{minimum}\:{area}\:{of}\:{a}\:{heptagon}\:{inscribed}\:{in} \\ $$$${an}\:{unit}\:{square}? \\ $$
Commented by Dwaipayan Shikari last updated on 26/Feb/21
![](https://www.tinkutara.com/question/19116.png)
Answered by mr W last updated on 26/Feb/21
![](https://www.tinkutara.com/question/19118.png)
Commented by mr W last updated on 26/Feb/21
![side length of heptagon = a θ=(((7−2)×180°)/7)=((900°)/7) α=90°−(θ/2)=((180°)/7) β=θ−2α=((540°)/7) γ=180°−45°−β=((405°)/7) DB=b=2a cos α=2a cos ((180°)/7) AB=(a/( (√2))) BC=b sin γ=2a cos ((180°)/7) sin ((405°)/7) AC=((1/( (√2)))+2 cos ((180°)/7) sin ((405°)/7))a=1 ⇒a=(1/((1/( (√2)))+2 cos ((180°)/7) sin ((405°)/7))) area of heptagon A=((7a^2 )/(4 tan ((180°)/7))) =(7/(4 tan ((180°)/7)((1/( (√2)))+2 cos ((180°)/7) sin ((405°)/7))^2 )) ≈0.728 878 175 check: ED=(b/( (√2))) DC=b cos γ EC=b((1/( (√2)))+cos γ)=2a cos ((180°)/7)((1/( (√2)))+cos ((405°)/7)) =2×(1/((1/( (√2)))+2 cos ((180°)/7) sin ((405°)/7)))×cos ((180°)/7)((1/( (√2)))+cos ((405°)/7)) =1](https://www.tinkutara.com/question/Q134013.png)
$${side}\:{length}\:{of}\:{heptagon}\:=\:{a} \\ $$$$\theta=\frac{\left(\mathrm{7}−\mathrm{2}\right)×\mathrm{180}°}{\mathrm{7}}=\frac{\mathrm{900}°}{\mathrm{7}} \\ $$$$\alpha=\mathrm{90}°−\frac{\theta}{\mathrm{2}}=\frac{\mathrm{180}°}{\mathrm{7}} \\ $$$$\beta=\theta−\mathrm{2}\alpha=\frac{\mathrm{540}°}{\mathrm{7}} \\ $$$$\gamma=\mathrm{180}°−\mathrm{45}°−\beta=\frac{\mathrm{405}°}{\mathrm{7}} \\ $$$${DB}={b}=\mathrm{2}{a}\:\mathrm{cos}\:\alpha=\mathrm{2}{a}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}} \\ $$$${AB}=\frac{{a}}{\:\sqrt{\mathrm{2}}} \\ $$$${BC}={b}\:\mathrm{sin}\:\gamma=\mathrm{2}{a}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}} \\ $$$${AC}=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}}\right){a}=\mathrm{1} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}}} \\ $$$${area}\:{of}\:{heptagon} \\ $$$${A}=\frac{\mathrm{7}{a}^{\mathrm{2}} }{\mathrm{4}\:\mathrm{tan}\:\frac{\mathrm{180}°}{\mathrm{7}}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{4}\:\mathrm{tan}\:\frac{\mathrm{180}°}{\mathrm{7}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}}\right)^{\mathrm{2}} } \\ $$$$\approx\mathrm{0}.\mathrm{728}\:\mathrm{878}\:\mathrm{175} \\ $$$$ \\ $$$${check}: \\ $$$${ED}=\frac{{b}}{\:\sqrt{\mathrm{2}}} \\ $$$${DC}={b}\:\mathrm{cos}\:\gamma \\ $$$${EC}={b}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{cos}\:\gamma\right)=\mathrm{2}{a}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{cos}\:\frac{\mathrm{405}°}{\mathrm{7}}\right) \\ $$$$=\mathrm{2}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{405}°}{\mathrm{7}}}×\mathrm{cos}\:\frac{\mathrm{180}°}{\mathrm{7}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{cos}\:\frac{\mathrm{405}°}{\mathrm{7}}\right) \\ $$$$=\mathrm{1} \\ $$
Commented by Dwaipayan Shikari last updated on 26/Feb/21
![Great sir!](https://www.tinkutara.com/question/Q134025.png)
$${Great}\:{sir}! \\ $$