Let [tex]x[/tex] be the side length of the small square. Then, the side length of the big square is [tex]2x[/tex].
The respective areas are [tex]x^2[/tex] and [tex](2x)^2=4x^2[/tex]
So, the sum of the areas are [tex]x^2+4x^2=5x^2[/tex]
We have
[tex]5x^2=45 \iff x^2=9 \iff x=\pm 3[/tex]
We can't have negative lengts, so we choose [tex]x=3[/tex]
So, the side length of the larger square is 6.
