Regular Tetrahedron Surface Area

Four equilateral triangles, each with area $\ frac{\sqrt{3}}{4}s^2$, sum to the memorable $A = \sqrt{3}\,s^2 \approx 1.732s^2$. It is one of the tidiest results in solid geometry: the surface of a tetrahedron of edge $s$ is smaller than the surface of a single square of side $1.32s$.

Compare it with the cube of the same edge, which carries $6s^2$ of surface, and the tetrahedron looks almost frugal. But compare surface to volume instead and the picture inverts. The tetrahedron is the worst of the Platonic solids at enclosing volume per unit of skin, which is one reason bubbles and droplets are never tetrahedral, and why the shape survives in engineering only where stiffness, not efficiency, is the goal.