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 |
\[=\frac{\text{Volume}\,\,\text{of}\,\,\text{cone}}{\text{Volume}\,\,\text{of}\,\,\text{frustum}}\] |
\[=\frac{\frac{1}{3}\times \pi {{(DE)}^{2}}\times AD}{\frac{1}{3}\pi B{{C}^{2}}\times AB-\frac{1}{3}\pi {{(DE)}^{2}}\times AD}\] |
\[=\frac{D{{E}^{2}}\times AD}{B{{C}^{2}}\times AB-D{{E}^{2}}\times AD}\] |
\[=\frac{\frac{B{{C}^{2}}}{4}\times \frac{AB}{2}}{B{{C}^{2}}\times AB-\frac{B{{C}^{2}}}{4}\times \frac{AB}{2}}=\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac{1}{7}=1:7\] |
You need to login to perform this action.
You will be redirected in
3 sec