Question Number 72603 by aliesam last updated on 30/Oct/19
Answered by mr W last updated on 30/Oct/19
Commented by aliesam last updated on 30/Oct/19
$${thank}\:{you}\:{sir} \\ $$
Commented by mr W last updated on 30/Oct/19
$${side}\:{length}\:{of}\:{square}\:{l} \\ $$$${l}=\mathrm{2}{a}\:\mathrm{cos}\:\theta \\ $$$${l}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}+\sqrt{\mathrm{2}}{a}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right)=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}+{a}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right) \\ $$$$\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}+{a}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)=\mathrm{2}{a}\:\mathrm{cos}\:\theta \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{cos}\:\theta−\mathrm{sin}\:\theta=\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right) \\ $$$$\frac{\sqrt{\mathrm{6}}}{\mathrm{4}}=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right) \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{4}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{6}}}{\mathrm{4}} \\ $$$${Y}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{a}−{a}\:\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{a}−{a}\left(\frac{\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}}{\mathrm{4}}\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}}{\mathrm{4}}{a} \\ $$$${X}=\mathrm{2}{a}\:\mathrm{cos}\:\theta−{Y}=\mathrm{2}{a}\left(\frac{\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}}{\mathrm{4}}\right)−\frac{\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}}{\mathrm{4}}{a} \\ $$$${X}=\frac{\mathrm{3}\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}}{\mathrm{4}}{a} \\ $$$$\frac{{X}}{{Y}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}}=\frac{\mathrm{4}\sqrt{\mathrm{15}}+\mathrm{3}}{\mathrm{11}}\approx\mathrm{1}.\mathrm{681} \\ $$