Let the sides of the smaller square be $$x$$ units and the sides of the larger square be $$y$$ units. Construct diagonal HF on the larger square. Let the point of intersection of AD and diagonal be I and point of intersection of BC and diagonal be J.  
We know that since EFGH is a square, the diagonal will bisect the angle. Therefore $$angle EHF=45^{o}$$ Observe $$ riangle AEH$$ and $$ riangle DHG$$ :  AE=AH=DH=DG = $$x$$ units and EH=HG=$$y$$ units.  Thus by SSS property, $$ riangle AEH cong riangle DHG$$ Thus, we know that $$angle$$AHE=$$angle$$DHG Also, $$angle$$AHE+$$angle$$DHG+$$angle$$AHD= $$90^{o}$$ (angle of a square) $$ Rightarrow (2 imes angle AHE)+ 60^{o} = 90^{o}$$ (angle of an equilateral triangle) $$ Rightarrow (2 imes angle AHE) = 30^{o}$$ $$ Rightarrow angle AHE = 15^{o}$$ Since $$angle AHE = 15^{o}, angle AHI = 45^{o}-15^{o} = 30^{o} $$ Since $$30^{o}= dfrac{60^{o}}{2}$$ we can say that HF is the angle bisector of $$angle$$AHD  Since $$ riangle$$AHD is an equilateral triangle, we know that the angle bisector, median and altitude will all be the same line ie HI in this case.  By symmetry, we know that JF will also be the angle bisector, median and altitude.  Also, length of altitude of equilateral triangle ie HI and JF = $$dfrac{sqrt{3} imes x}{2} $$ Length of diagonal HF = $$sqrt{2} imes y$$ From the figure, we can express the length of diagonal HF as given :  $$ sqrt{2} imes y = dfrac{sqrt{3} imes x}{2} + x + dfrac{sqrt{3} imes x}{2} $$ $$ Rightarrow sqrt{2} imes y = (sqrt{3} imes x) + x $$ $$ Rightarrow sqrt{2} imes y = x imes (sqrt{3} + 1) $$ $$ Rightarrow y = x imes (dfrac{sqrt{3} + 1}{sqrt{2}}) $$
Ratio of areas asked = $$(dfrac{y}{x})^{2} = (dfrac{sqrt{3} + 1}{sqrt{2}})^{2} = dfrac{4+2sqrt{3}}{2} = 2+sqrt{3}$$