Question 504853
Without loss of generality, suppose the diameter of the cone is 2 (e.g. radius = 1). Then the volume of the sphere is


*[tex \LARGE V_s = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi]


and the volume of the cone is


*[tex \LARGE V_c = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h]


Equating the two volumes, we see that h = 4; i.e. the height must be four times the radius, or twice the diameter.