Question

A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is $\frac{8}{9}$ of the curved surface of the whole cone, find the ratio of the line segments into which the cone's altitude is divided by the plane.

Answer 1

Assume that the ratio of the altitude of the bigger and the smaller cone be k:1.

Let R and r be the radii of the bigger and the smaller cone respectively.

Let H and h be the height of the bigger and the smaller cone respectively.

Consider the similar triangles $△$ AGC & $△$ AFE ,

By the property of similarity, we have,

$\frac{AG}{AF}=\frac{GC}{FE}$

$\frac{h}{H}=\frac{r}{R}=\frac{1}{k},$ where k is some constant.

Curved surface area of bigger cone = $\pi RL$, where L is the slant height of the bigger cone.

Curved surface area of smaller cone = $\pi rl$, where l is the slant height of the smaller cone.

Again by the property of similarity, we have,

$\frac{l}{L}=\frac{r}{R}=\frac{1}{k}$

Given that the ratio of the curved surface area of the frustum of the cone to the whole cone is $\frac{8}{9}.$

The ratio of the curved surface area of the smaller cone to the bigger cone is $\frac{1}{9}$.

$\frac{\pi rl}{\pi RL}=\frac{1}{k^2}=\frac{1}{9}$

$k=3$

$\frac{h}{H}=\frac{1}{3}$

Therefore, $\frac{h}{H-h}=\frac{1}{3-1}=\frac{1}{2}$

Hence, the required ratio is $1:2$.