Question

What's the area of square?

 

Answer 1

 

\begin{aligned}
&\frac{x+\sqrt{36-x^{2}}}{10}=\frac{6}{x+\sqrt{36-x^{2}}} \\
&x=\sqrt{15}-\sqrt{3} \\
&(\sqrt{15}-\sqrt{3})^{2}=18-6 \sqrt{5}
\end{aligned}

Comments

 

 

\begin{aligned}
&\frac{4}{\sin \left(90^{\circ}-\alpha\right)}=\frac{x \sqrt{2}}{\sin \left(45^{\circ}\right)}=>x=\frac{2}{\cos (\alpha)} \\
&\frac{x}{\sin (\alpha)}=\frac{6}{\sin \left(90^{\circ}\right)}=>x=6 \sin (\alpha) \\
&6 \sin (\alpha)=\frac{2}{\cos (\alpha)}=>\sin (2 \alpha)=\frac{2}{3} ; \cos (2 \alpha)=\frac{\sqrt{5}}{3} \\
&\sin ^{2}(\alpha)=\frac{1-\cos (2 \alpha)}{2}=\frac{3-\sqrt{5}}{6} \\
&\text { Area }=x^{2}=6^{2} * \sin ^{2}(\alpha)=6(3-\sqrt{5})
\end{aligned}