It is given that the base of the pyramid is square and each of the four sides are equilateral triangles.

Length of each side of the equilateral triangle = 20cm

Since the side of the triangle will be common to the square as well, the side of the square = 20cm

Let h be the vertical height of the pyramid ie OA

OB = 10 since it is half the side of the square 

AB is the height of the equilateral triangle i.e 10$$\sqrt{\ 3}$$

AOB is a right angle, so applying the Pythagorean formula, we get

$$OA^2+ OB^2= AB^2$$

$$h^2$$ + 100= 300

h=10$$\sqrt{\ 2}$$