Question

Qid : 61 - if a cone is divided into two parts by drawing a plane through the midpoints of its axis, then the ratio of the volume of the 2 parts of the cone is

Answer 1

Let OAB be the cone of the base radius r cm and height h cm. Let P be the midpoint ot it's axis OQ, Then,

$\sf{OP=\frac{1}{2}OQ=\frac{h}{2}cm}$

$\sf{∆OQB~∆OPD}$

$\sf{\frac{OQ}{QB}=\frac{OP}{PD}}$

$\sf{\frac{h}{r}=\frac{h/2}{PD}}$

$\sf{PD=\frac{r}{2}cm}$

The plane CD divides the cone OAB into a smaller cone OCD and a frustum of cone ABDC

$\longrightarrow$$\sf{\frac{Volume\:of\:the\:small\:cone}{Volume\:of\:the\:frustum\:of\:cone}}$

$\longrightarrow$$\sf{\frac{ \frac{1}{3}\pi { (\frac{r}{2})( \frac{h}{2} ) }^{2} }{ \frac{1}{3}\pi (\frac{h}{2} )( {r}^{2} + {( \frac{r}{2}) + }^{2} + r \times \frac{r}{2} )} }$

$\longrightarrow$$\sf{\frac{ \frac{ {r}^{2} }{4} }{ {r}^{2}(1 + \frac{1}{4} + \frac{1}{2}) } }$

$\longrightarrow$$\sf{ \frac{1}{4} \times \frac{4}{7} = \frac{1}{7} }$

$\longrightarrow$$\sf{1:7}$