Question

A cone is divided into two parts by drawing a plane through a point which divides its height in the ratio $1:2$ starting from the vertex and the plane is parallel to the base. Compare the volume of the two parts

Answer 1

Let the plane XY divide the cone ABC in the ratio $AE:EO=12$,
where AEO is the axis of the cone.
Let $r_2$ and $r_1$ be the radii of the circular section XY and the base BC of the cone respectively and let $h_{1}h$ and $h_1$ be their heights.
Then,
$\frac{h_1}{h}=\frac{3}{2}$
$=>h_1=\frac{3}{2}h$
And,
$\frac{r_1}{r_2}=\frac{h_1}{h_1-h}$
$=\frac{\frac{3}{2}h}{\frac{1}{2}h}=3$
$\therefore r_1=3r_2$
Volume of cone AXY
$=\frac{1}{3}\pi r_2^2(h_1-h)$
$=\frac{1}{3}\pi r_2^2(\frac{3}{2}{h}_1-h)$
$=\frac{1}{6}\pi r_2^{2}h$
Volume of frustum XYBC
$=\frac{1}{3}\pi h(r_1^2+r_2^2+r_1{r}_2)$
$=\frac{1}{3}\pi h(9r_2^2+r_2^2+3r_2^2)$
$=\frac{1}{3}\pi h\left(13r_2^2\right)$
$\therefore \frac{\text{Volume of cone AXY}}{\text{Volume of frustum XYBC}}=\frac{\frac{1}{6}\pi r_2^{2}h}{\frac{13}{3}\pi r_2^2{h}^2}$
$=\frac{\text{Volume of cone AXY}}{\text{Volume of frustum XYBC}}=\frac{1}{26}$
$\therefore$ The ratio between the volume of the cone AXY and the remaining portion BCYX is $1:26$.