Vertical height of regular pyramid

The base is a square of side $a = 20\text{ cm}$. Distance from the center of the base to any vertex $= \frac{\text{diagonal}}{2} = \frac{20\sqrt{2}}{2} = 10\sqrt{2}\text{ cm}$.

The slant edge $e = 20\text{ cm}$ (since each lateral face is an equilateral triangle of side $20\text{ cm}$).

In the right triangle formed by vertical height $h$, distance from center to vertex, and slant edge: $h^2 + (10\sqrt{2})^2 = 20^2$ $h^2 + 200 = 400 \implies h^2 = 200 \implies h = 10\sqrt{2}\text{ cm}$