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