Since both objects are assumed to have the same volume, matching the formulas we'll get:
π\(r^{2}\)h = \(\frac{1}{3}\)π\(R^{2}\)h
Having both π and h at both sides of the equation, we can ignore them, so:
\(r^{2}\) = \(\frac{1}{3}\)\(R^{2}\)
And clearing R, according to the equation rules, we'll get:
\(\sqrt[2]{3 r^{2} }\) = R
