Official solution
\[\tan 2\theta = 2 \Rightarrow {{2\tan \theta } \over {1 - {{\tan }^2}\theta }} = 2\]
\[\tan \theta = {{\sqrt 5 - 1} \over 2}\] (as \[\theta\] is acute)
Area \[ = {1 \over 2}{L^2}\sin 2\theta = {1 \over 2}.\,{5 \over {{{\tan }^2}\theta }}.\,2\sin \theta \cos \theta \]
\[ = {{5\sin \theta \cos \theta } \over {{{\sin }^2}\theta }}.\,{\cos ^2}\theta \]
\[ = 5\cot \theta .\,{\cos ^2}\theta \]
\[ = 5.\,{2 \over {\sqrt 5 - 1}}.\,{1 \over {1 + {{\left( {{{\sqrt 5 - 1} \over 2}} \right)}^2}}}\]
\[ = {{10} \over {\sqrt 5 - 1}}.\,{4 \over {4 + 6 - 2\sqrt 5 }}\]
\[ = {{40} \over {2\sqrt 5 {{(\sqrt 5 - 1)}^2}}} = {{4\sqrt 5 } \over {6 - 2\sqrt 5 }}\]
\[ = {{4\sqrt 5 (6 + 2\sqrt 5 )} \over {16}}\]
\[ = {{\sqrt 5 (3 + \sqrt 5 )} \over 2}\]