If the radius of the base of the right circular cone is \(r\), then \[ \pi r^2 = A \quad \text{(base area)} \] Also, \[ \frac{1}{3} \pi r^2 H = V \quad \text{(volume of the cone)} \Rightarrow \frac{1}{3} AH = V \quad [\text{Substituting } \pi r^2 = A] \Rightarrow \frac{AH}{3V} = 1 \quad \text{(Answer)} \]