Suppose the square Lena draws has the following dimensions:
[tex]Side=L \\ \\ A_{s}:Area \\ \\ \\ A_{s}=L^2[/tex]
For the rectangle we have:
[tex]Base=L_{1} \\ \\ Height=L_{2} \\ \\ \\ A_{r}=L_{1}L_{2}[/tex]
Possibility 1. In order for the area of the square to be greater than the area of the rectangle, the following inequality must be true:
[tex]\boxed{\frac{L^2}{L_{1}L_{2}}>1}[/tex]
Possibility 2. If one side of the rectangle equals the side of the square, that is:
[tex]L_{1}=L[/tex]
Then, in order for the area of the square to be greater than the area of the rectangle, the following inequality must be true:
[tex]\frac{L^2}{LL_{2}}>1 \\ \\ \boxed{\frac{L}{L2}>1}[/tex]
