question_answer

A cone is cut at mid-point of its height by frustum parallel to its base. The ratio between the two parts of cone would be

A) 1:1                               

B) 1: 8

C) 1: 4                              

D) 1: 7

Correct Answer: D

Solution :

Since, \[\Delta ADE\approx \Delta ABC\]

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

\[\therefore \]\[AD=\frac{AB}{2}\]and \[DE=\frac{BC}{2}\]

Required Ratio

\[=\frac{\text{Volume}\,\,\text{of}\,\,\text{cone}}{\text{Volume}\,\,\text{of}\,\,\text{frustum}}\]

\[=\frac{\frac{1}{3}\times \pi {{(DE)}^{2}}\times AD}{\frac{1}{3}\pi B{{C}^{2}}\times AB-\frac{1}{3}\pi {{(DE)}^{2}}\times AD}\]

\[=\frac{D{{E}^{2}}\times AD}{B{{C}^{2}}\times AB-D{{E}^{2}}\times AD}\]

\[=\frac{\frac{B{{C}^{2}}}{4}\times \frac{AB}{2}}{B{{C}^{2}}\times AB-\frac{B{{C}^{2}}}{4}\times \frac{AB}{2}}=\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac{1}{7}=1:7\]