Question

A cone has a radius of 4 and a height of 8. The smaller cone is 1/8 of the bigger cones volume. What are the dimensions of the smaller cone.

Answer 1

A cone has a radius of 4 and a height of 8. The smaller cone is 1/8 of the bigger cones volume. What are the dimensions of the smaller cone?

**Solution:**

We have

$$V_b = \frac{1}{3} \pi R_b^2 h_b$$

where $V_b, R_b, h_b$ is the volume, radius and height of the bigger cone. Because $R_b = 4$ and $h_b = 8$ then

$$\frac{h_b}{R_b} = \frac{8}{4} = 2 \Rightarrow h_b = 2 R_b.$$

Thus

$$V_b = \frac{1}{3} \pi R_b^2 \cdot 2 R_b,$$ $$V_b = \frac{2}{3} \pi R_b^3.$$

Denote $V_s, R_s, h_s$ is the volume, radius and height of the smaller cone. So

$$h_s = 2 R_s, V_s = \frac{2}{3} \pi R_s^3.$$

Thus we have

$$\frac{V_s}{V_b} = \frac{\frac{2}{3} \pi R_s^3}{\frac{2}{3} \pi R_b^3} = \frac{1}{8},$$ $$\frac{R_s^3}{R_b^3} = \frac{1}{8},$$ $$\left(\frac{R_s}{R_b}\right)^3 = \left(\frac{1}{2}\right)^3,$$ $$\frac{R_s}{R_b} = \frac{1}{2},$$ $$R_s = \frac{1}{2} R_b = \frac{1}{2} \cdot 4 = 2.$$

Also

$$h_s = 2 R_s = 4.$$

Answer: $R_s = 2, h_s = 4$.