Let $ABC$ be an equilateral triangle of side length 1. A square is constructed outward on each side, and $E,F,G$ are the centers of the squares on $\overline{AB},\overline{BC},\overline{CA}$, respectively. What is the ratio of the area of $\triangle EFG$ to the area of $\triangle ABC$?

1. By symmetry, $EFG$ is equilateral. At $B$, the square centers satisfy $BE=BF=\dfrac1{\sqrt2}$ and $\angle EBF=150^\circ$.
2. By the Law of Cosines, $EF^2=2\left(\dfrac12\right)(1-\cos150^\circ)=1+\dfrac{\sqrt3}2$.
3. Since both triangles are equilateral, $\dfrac{[EFG]}{[ABC]}=\dfrac{EF^2}{AB^2}=\boxed{\dfrac{2+\sqrt3}2}$.