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Answer: 1:3


Using lengths
  

If the areas are in the ratio $1:4,$ then the sides are in the ratio $1:2$




  

$x=s+l,$ and
$\frac12 x+s+s = x\\
\Rightarrow x + 4s = 2x\\
\Rightarrow x = 4s$

So $4s = s + l \Rightarrow 3s = l$ so the ratio is $1:3$.


Using areas
 

$(l-s)^2 : (l+s)^2 = 1:4$
$\Rightarrow(l-s)^2\times4 = (l+s)^2$
$\Rightarrow4 = \dfrac{(l+s)^2}{(l-s)^2}$
$\Rightarrow2 = \dfrac{l+s}{l-s}$
$\Rightarrow 2l-2s = l+s$
$\Rightarrow l = 3s$
So the ratio is $1:3$.


This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.