If the radius of the base of a right circular cone is \(r\), then \[ πr^2 = Y \quad \text{(area of the base)} \] Also, \[\cfrac{1}{3} πr^2 Z = X \quad \text{(volume of the cone)} \] Substituting \(πr^2 = Y\), we get: \[ \cfrac{1}{3} YZ = X \Rightarrow \cfrac{YZ}{X} = 3 \quad \text{(Answer)} \]