From the diagram, it is evident that AB serves as both the height of the equilateral triangle and the slant height of the pyramid.

Given a regular triangle with side length 20 units:
- The length of the perpendicular from vertex \( A \) to base \( BC \) is: \[ AB = \frac{\sqrt{3}}{2} \times \text{side} = \frac{\sqrt{3}}{2} \times 20 = 10\sqrt{3} \]
- The median or half-diagonal from vertex \( A \) to centroid \( O \) is: \[ AO = \frac{1}{2} \times \text{side} = \frac{1}{2} \times 20 = 10 \]
Apply Pythagoras Theorem in triangle \( \triangle AOB \):
\[ OB^2 = AB^2 - AO^2 \] \[ OB^2 = (10\sqrt{3})^2 - 10^2 = 300 - 100 = 200 \] \[ OB = \sqrt{200} = 10\sqrt{2} \]
✅ Final Answer:
The height of the pyramid is: \[ \boxed{10\sqrt{2}} \]