question_answer

The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height Find the ratio of the volumes of the two parts.

Answer:

Let \[BC=r\text{ }cm\] & \[DE=R\text{ }cm\]

Since B is mid-point of AD & \[BC\parallel DE\]

\[\therefore \] C is mid-point of AE or \[AC=CE\]

Also \[\Delta \,ABC\sim \Delta \,ADE\]

\[\therefore \]                  \[\frac{AB}{AD}=\frac{BC}{DE}=\frac{AC}{AE}=\frac{1}{2}\]

\[BC=\frac{1}{2}DE\]

\[r=\frac{1}{2}R\] or \[R=2r\]

Now,     \[\frac{Volume\,\,of\,\,cone}{Volume\,\,of\,\,frustum}=\frac{\frac{1}{3}\pi {{r}^{2}}\left( \frac{h}{2} \right)}{\frac{1}{3}\pi \left( \frac{h}{2} \right)\left( {{R}^{2}}+{{r}^{2}}+Rr \right)}\]

\[=\frac{{{r}^{2}}}{\left( {{R}^{2}}+{{r}^{2}}+Rr \right)}=\frac{{{r}^{2}}}{4{{r}^{2}}+{{r}^{2}}+2r\cdot r}\]

\[=\frac{{{r}^{2}}}{7{{r}^{2}}}=\frac{1}{7}\] or \[1:7\]