Question

Let ABC be a hollow cone of radius R, height H and slant height L. Let this cone be cut by a plane A'B' parallel to AB. O' is the centre of base of the cut out cone A'B'C'. Find the ratio of the height of cone to the height of frustum.

Answer 1

Let ABC be a hollow cone of radius R, height H and slant height L. Let this cone be cut by a plane A'B' parallel to AB. O' is the centre of base of the cut out cone A'B'C'
Let h be the height, r be the radius and l be the slant height of cone A'B'C
Clearly, $\Delta A^{{}^{\prime }}{B}^{{}^{\prime }}C-\Delta ABC$
$\Rightarrow \frac{h}{H}=\frac{r}{R}=\frac{l}{L}\dots \dots \left(1\right)$
We are given that
CSA of frustum $\Delta A^{{}^{\prime }}{B}^{{}^{\prime }}B$
$=\frac{8}{9}\times CSAofcone\Rightarrow \pi (R+r)(L-l)=\frac{8}{9}\pi RL\Rightarrow (R+r)(L-l)=\frac{8}{9}RL\Rightarrow \left(\frac{R+r}{R}\right)\left(\frac{L-l}{L}\right)=\frac{8}{9}\Rightarrow (1+\frac{r}{R})(1-\frac{l}{L})=\frac{8}{9}\Rightarrow (1+\frac{h}{H})(1-\frac{h}{H})=\frac{8}{9}\left(using\right(1\left)\right)\Rightarrow 1-\frac{h^2}{H^2}=\frac{8}{9}\left[\right(a+b\left)\right(a-b)=a^2-b^2]\Rightarrow \frac{h^2}{H^2}=1-\frac{8}{9}\Rightarrow \frac{h^2}{H^2}=\frac{1}{9}\Rightarrow \frac{h}{H}=\frac{1}{3}\Rightarrow h=\frac{H}{3}So,H-h=H-\frac{H}{3}=\frac{2}{3}H$
Required ratio $=\frac{h}{H-h}=\frac{\frac{H}{3}}{\frac{2}{3}H}=\frac{1}{2}$