Face off!

I based my response on a bit of physics and biology. We know that in the real world, fluids like to form into shapes that minimize their surface area. This can be seen with water droplets and biological cells. A sphere can be thought of as a regular polyhedron with infinitely many sides, meaning that as polyhedron increase their number of sides, they will approach a spherical shape. Hence, because the cube has more sides (and is more sphere-like) than the tetrahedron, it has a smaller surface area.

Since the volume of a regular tetrahedron with side length $a$ is $\frac{a^3}{6\sqrt{2}}$ and the volume of the cube with side length $s$ is $s^3$ , this implies

$1 \text{ cm}^3 = \frac{a^3}{6\sqrt{2}} = s^3 \Rightarrow \left( \frac{a}{s} \right)^3 = \sqrt{72} .$

Since the surface area of a regular tetrahedron with side length $a$ is $4 \times \text{(area of one face)}= 4 \times \frac{\sqrt{3}}{4} a^2$ and the surface area of a cube with side length $s$ is $6 \times \text{(area of one face)} = 6 s^2,$ this implies

$\frac{ \text{Surface area of regular tetrahedron} }{\text{Surface area of cube}} = \frac{ \sqrt{3}a^2 }{ 6s^2 }= \frac{\sqrt{3}}{6} \left( \left( 72 \right)^{\frac{1}{6}} \right)^2 = \frac{\sqrt{3}}{6} \left( 72 \right)^{\frac{1}{3}} .$

Since this is greater than 1, the surface area of the regular tetrahedron is strictly larger than the surface area of the cube, so the cube has the smaller surface area.

The cube will have a smaller surface area because both objects have the same volume. Volume is like the measure of space inside of an object. The area of one face of a triangle is (1/2) * b h . The area of a face of a cube is w*l. Therefore, triangles will have smaller surface areas, in general. Since they have smaller surface areas in general and surface area is proportional to volume for both shapes, the cube must have a smaller surface area in order for the volumes to be equal.

If the cube has side length $a$ , then its volume is $V_C = a^3 = 1$ . So $a = 1$ and the area of such a cube is: $A_C = 6a^2 = 6(1) = 6$ If the regular tetrahedron has side length $b$ , then its volume is $V_T = \dfrac{\sqrt{2}b^3}{12} = 1$ so that $b = \sqrt[3]{6\sqrt{2}}$ . The area of such a tetrahedron is: $A_T = 4 \times \dfrac{1}{2} \times b \times \dfrac{b\sqrt{3}}{2} = b^2 \sqrt{3} \approx 7.2 > A_C$