The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is

Let the base of the regular pyramid be a square $ABCD$ of side length $a=20\text{cm}$ and let $V$ be the vertex of the pyramid.

The center of the square base, $O$, lies directly below the vertex $V$, so the vertical height of the pyramid is the length $VO$.

First, calculate the distance from the center $O$ of the square base to one of its corners, say $A$. The diagonal of the square base is:
$AC=a\sqrt{2}=20\sqrt{2}\text{cm}$
Thus, the distance from the center $O$ to corner $A$ is:
$OA=\frac{AC}{2}=10\sqrt{2}\text{cm}$

Since the lateral faces are equilateral triangles of side 20 cm, the slant edge length is $VA=20\text{cm}$.

Using the Pythagorean theorem in the right-angled triangle $VOA$:
$V{O}^2+O{A}^2=V{A}^2$
$V{O}^2+{\left(10\sqrt{2}\right)}^2={20}^2$
$V{O}^2+200=400$
$V{O}^2=200$
$VO=\sqrt{200}=10\sqrt{2}\text{cm}$