R = rac{s\sqrt{3}}{3}

February 22, 2026

$R = rac{s\sqrt{3}}{3}$

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R = rac{s}{\sqrt{3}} \cdot rac{2}{1} \quad ext{Incorrect.}
Correct derivation: The height $ h $ of an equilateral triangle is $ rac{\sqrt{3}}{2}s $. The centroid (which coincides with the circumcenter) is at $ rac{2}{3} $ of the height from a vertex, so:
R = rac{2}{3} \cdot rac{\sqrt{3}}{2}s = rac{s}{\sqrt{3}} \cdot rac{\sqrt{3}}{\sqrt{3}} = rac{s\sqrt{3}}{3}
The area $ A $ of the circular dome is:
A = \pi R^2 = \pi \left( rac{s\sqrt{3}}{3}
ight)^2 = \pi \cdot rac{3s^2}{9} = rac{\pi s^2}{3}

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