Solve $s=4\text{ cm}$

Step-by-step solution

1. Step 1 · Face diagonal
   $$f=\sqrt{4^2+4^2}=4\sqrt{2}$$

   Apply Pythagoras to a square face first.

2. Step 2 · Apply Pythagoras again
   $$d=\sqrt{f^2+4^2}=\sqrt{32+16}=4\sqrt{3}$$

   The face diagonal and the perpendicular third edge form a right triangle.

3. Step 3 · Check and round
   $$d\approx 6.93\text{ cm}$$

   The space diagonal is longer than the face diagonal, about 5.66 cm, and the edge, 4 cm.