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

We have a pyramid where the base is a square with side length 20 cm, each of the 4 triangular faces is equilateral with side length 20 cm, and we need to find the vertical height.

Since all edges are 20 cm, this means every edge of the pyramid (base edges AND edges from base to apex) has the same length. This is a very special pyramid!

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Let us place the square base in a coordinate system with base vertices: A(10, 10, 0), B(-10, 10, 0), C(-10, -10, 0), D(10, -10, 0) and apex (top point): P(0, 0, h) where h is the height we want.

The apex must be directly above the center of the square base (at origin) for the pyramid to be regular.

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Since each triangular face is equilateral with side 20 cm, the distance from the apex P to any base vertex must also be 20 cm.

Let us use vertex A(10, 10, 0):

Distance from P(0, 0, h) to A(10, 10, 0) = $\sqrt{(0-10)^2 + (0-10)^2 + (h-0)^2}$

= $\sqrt{100 + 100 + h^2}$

= $\sqrt{200 + h^2}$

Since this distance equals 20 cm:

$\sqrt{200 + h^2} = 20$

$200 + h^2 = 400$

$h^2 = 200$

$h = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}$

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Let us check: If h = $10\sqrt{2}$, then the slant edge length is:

$\sqrt{200 + (10\sqrt{2})^2} = \sqrt{200 + 200} = \sqrt{400} = 20$

Therefore, the vertical height of the pyramid is $10\sqrt{2}$ cm ≈ 14.14 cm

When all edges of a pyramid are equal, you can use the distance formula between the apex and any base vertex to find the height. This problem demonstrates the beautiful relationship between 3D geometry and the Pythagorean theorem!