A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$, one hexagon on each side of the square, so that the edges of adjacent hexagons coincide. The top eight vertices of the hexagons lie in a plane parallel to the square. How far above the plane of the square is this plane?

1. Put the square at $(\pm\frac12,\pm\frac12,0)$. At a corner of the square three faces meet with angles $90^\circ,120^\circ,120^\circ$. The hexagon on one side makes an angle $\theta$ with the horizontal, where $\cos\theta=\frac1{\sqrt3}$, as shown in the problem (this is forced by the condition that adjacent hexagons share an edge).
2. A regular hexagon of side $1$ has height $\sqrt3$ measured from one side to the opposite side. The far side is therefore $\sqrt3\sin\theta$ above the plane of the square.
3. $\sin\theta=\sqrt{1-\frac13}=\sqrt{\frac23}$, so the height is $\sqrt3\cdot\sqrt{\frac23}=\boxed{\sqrt2}$.