The height of a cone is 40 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be $\dfrac{1}{64}$ of the volume of the given cone, at what height above the base is the section cut?

Let OAB be the given cone of height 40 cm and base radius R cm. Let this cone be cut by the plane CND (parallel to the base plane AMB) to obtain cone OCD with height h cm and base radius r cm as shown in the figure below :

From figure,

∠NOD = ∠MOB (Common angle)

∠OND = ∠OMB = 90° (Heights are perpendicular to radii)

∠ODN = ∠OBM (Corresponding angles since ND || MB)

∴ △OND ~ △OMB (By AA similarity)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

∴ $\dfrac{\text{r}}{\text{R}} = \dfrac{\text{h}}{40}$ …(1)

According to given,

Volume of cone OCD = $\dfrac{1}{64}$ Volume of cone OAB

∴ $\dfrac{1}{3}$ πr²h = $\dfrac{1}{64} \times \dfrac{1}{3}$ πR² × 40

Dividing both sides by π and multiplying by 3 we get,

$\Rightarrow \text{r}^2 \text{h} = \dfrac{40}{64} \times \text{R}^2 \\[1em] \Rightarrow \dfrac{\text{r}^2}{\text{R}^2} = \dfrac{5}{8 \text{h}} \\[1em] \Rightarrow \Big(\dfrac{\text{r}}{\text{R}}\Big)^2 = \dfrac{5}{8 \text{h}} \\[1em]$

Using eq.(1),

$\Rightarrow \Big(\dfrac{\text{h}}{40}\Big)^2 = \dfrac{5}{8 \text{h}} \\[1em] \Rightarrow \dfrac{\text{h}^2}{1600} = \dfrac{5}{8 \text{h}} \\[1em] \Rightarrow \text{h}^3 = \dfrac{5 \times 1600}{8} \\[1em] \Rightarrow \text{h}^3 = 5 \times 200 \\[1em] \Rightarrow \text{h}^3 = 1000 \\[1em] \Rightarrow \text{h} = \sqrt[3]{1000} \\[1em] \Rightarrow \text{h} = 10 \text{ cm.}$

The height of the cone OCD = 10 cm

∴ The section is cut at the height of 40 - 10 = 30 cm.

Hence, the section is cut above 30 cm from the base.