Question

The height of a right circular cone is $20cm$. A small cone is cut off at the top by a plane parallel to the base. It its volume be $\frac{1}{8}$ of the volume of the given cone, at what height above the base is the section made?

Answer 1

VAB be a cone of height $h_1=V{O}_1=20cm$

Front triangles $V{O}_{1}A$ and $VO{A}_1$

$\frac{V{O}_1}{VO}=\frac{O_{1}A}{O{A}_1}$

$\frac{20}{VO}=\frac{O_{1}A}{O{A}_1}$

Volume of cone $V{A}_{1}O=\frac{1}{8}\times$Volume of cone $VAB$

We have

$\frac{1}{3}\pi \times O{A}_1^2\times VO=\frac{1}{8}\times 13\pi \times O_1{A}_1^2\times 20$

$\Rightarrow {\left(\frac{O{A}_1}{O_{1}A}\right)}^2\times VO=\frac{20}{8}$

$\Rightarrow {\left(\frac{VO}{20}\right)}^2\times VO=\frac{20}{8}$

$\Rightarrow (VO{)}^3=\frac{20\times 400}{8}$

$\Rightarrow (VO{)}^3=1000$

$\Rightarrow VO=10$

Height at which section is made $=20-10=10cm$