Question

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

Answer 1

$$\begin{gathered} \mathrm{In}∆\mathrm{ABC}\mathrm{and}∆\mathrm{ADE}, \\ \angle \mathrm{BAC}=\angle \mathrm{DAE}\left(\mathrm{common}\right) \\ \angle \mathrm{ABC}=\angle \mathrm{ADE}=90° \\ \therefore ∆\mathrm{ABC}~∆\mathrm{ADE} \\ \mathrm{Also}, \\ \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{BC}}{\mathrm{DE}} \\ \Rightarrow \frac{h_{\mathrm{s}}}{40}=\frac{r_s}{r_b} \end{gathered}$$
Now,
Height of the big cone (h_b) = 40 cm
Volume of the small cone = $\frac{1}{64}$Volume of the big cone
$$\begin{gathered} \Rightarrow \frac{1}{3}\mathrm{\pi }{r_s}^{\mathit{2}}{h}_s\mathit{=}\frac{1}{64}\times \frac{1}{3}\mathrm{\pi }{r_b}^{\mathit{2}}{h}_b \\ \Rightarrow {r_s}^{\mathit{2}}{h}_s\mathit{=}\frac{1}{64}\times {r_b}^{\mathit{2}}\times 40 \\ \Rightarrow \frac{{r_s}^{\mathit{2}}}{{r_b}^{\mathit{2}}}\times h_s=\frac{1}{64}\times 40 \\ \Rightarrow {\left(\frac{h_s}{40}\right)}^2\times h_s=\frac{1}{64}\times 40 \end{gathered}$$

$$\begin{gathered} \Rightarrow {h_s}^2\times h_s=\frac{40\times 40\times 40}{64} \\ \Rightarrow {h_s}^3={10}^3 \\ \Rightarrow h_s=10\mathrm{cm} \end{gathered}$$

∴ Height = 40 $-$ 10 = 30 cm