Question

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

Answer 1

We have,

Height of the given cone, H = 20 cm

Let the radius of the given cone be h and 

the height of the smaller cone be r.

Now, in ΔAQD and ΔAPC,

∠QAD = ∠PAC     (Common angle)

∠AQD = ∠APC = 90°

So, by AA criteria 

∠AQD _˜∠APC 

${\displaystyle \Rightarrow \frac{\text{AQ}}{\text{AP}}=\frac{\text{QD}}{\text{PC}}}$

${\displaystyle \Rightarrow \frac{\text{h}}{\text{H}}=\frac{\text{r}}{\text{R}} .......\left(i\right)}$

Volume of smaller cone ${\displaystyle =\frac{1}{8}\times \text{Volume of the given cone}}$

${\displaystyle \Rightarrow \frac{\text{Volume of smaller cone}}{\text{Volume of the given one}}=\frac{1}{8}}$

${\displaystyle \Rightarrow \frac{\left(\frac{1}{3}\pi {\text{r}}^2\text{h}\right)}{\left(\frac{1}{3}\pi {\text{R}}^2\text{H}\right)}=\frac{1}{8}}$

${\displaystyle \Rightarrow {\left(\frac{\text{r}}{\text{R}}\right)}^2\times \left(\frac{\text{h}}{\text{H}}\right)=\frac{1}{8}}$

${\displaystyle \Rightarrow {\left(\frac{\text{h}}{\text{H}}\right)}^2\times \left(\frac{\text{h}}{\text{H}}\right)=\frac{1}{8}}$       [Using (i)]

${\displaystyle \Rightarrow {\left(\frac{\text{h}}{\text{H}}\right)}^3=\frac{1}{8}}$

${\displaystyle \Rightarrow \frac{\text{h}}{\text{H}}=\sqrt[3]{\frac{1}{8}}}$

${\displaystyle \Rightarrow \frac{\text{h}}{\text{20}}=\frac{1}{2}}$

${\displaystyle \Rightarrow \text{h}=\frac{20}{2}}$

${\displaystyle \Rightarrow \text{h}=10 \text{cm}}$

∴ PQ = H - h = 20 - 10 cm

So, the section is made at the height of 10 cm above the base.