Question

A right circular cone is divided into three parts by trisecting its height by two planes drawn parallel to the base. Show that the volumes of the three portions starting from the top are in the ratio 1 : 7 : 19.

Answer 1

Let the radii of three cones from top be $r_1,r_2$ and $r_3$ respectively.

Let the height of given cone be 3 h.

so, the height of cone ADE = 2 h and height of cone ABC = h

$△ABC\sim △ADE,\frac{r_1}{r_2}=\frac{h}{2h}\Rightarrow 2r_1=r_2△ABC\sim △AFG,\frac{r_1}{r_3}=\frac{h}{3h}\Rightarrow 3r_1=r_3$

Volume of cone ABC = $\frac{1}{3}\pi r_1^{2}h$

Volume of cone ADE = $\frac{1}{3}\pi r_2^2.2h=\frac{1}{3}\pi (2r_1{)}^2.2h$

Volume of frustum BCED =$\frac{1}{3}\pi 4r_1^2.2h-\frac{1}{3}\pi r_1^{2}h=\frac{7}{3}\pi r_1^{2}h$

Volume of frustum DEGF = $\frac{1}{3}\pi r_3^2.3h-\frac{1}{3}\pi r_2^2.2h=\frac{1}{3}\pi (3r_1{)}^2.3h-\frac{1}{3}\pi (2r_1{)}^2.2h=\frac{1}{3}\pi r_1^{2}h(27-8)=\frac{19}{3}\pi r_1^{2}h$

Ratio = $\frac{1}{3}\pi r_1^{2}h:\frac{7}{3}\pi r_1^{2}h:\frac{19}{3}\pi r_1^{2}h=1:7:19$

Hence, required ratio = 1 : 7 : 19.