Question

A cone of radius 10 cm is divided into two parts by a plane through the mid-point of its axis, parallel to the base, then the ratio of the volumes of the larger part to the smaller part is

• A. 7 : 2
• B. 7 : 1
• C. 7 : 3
• D. 7 : 4

Answer 1

The correct option is B 7 : 1

Let BC = r, DE = R = 10 cm,
AB = h, AD = 2h
In ΔABC and ΔADE,
∠ABC = ∠ADE = 90º
∠BAC = ∠DAE (Common)
$\therefore \Delta ABC\sim \Delta ADE$ (by AA Similarity)
$\frac{AB}{AD}=\frac{BC}{DE}=\frac{AC}{AE}$
$\frac{h}{2h}=\frac{r}{R}\Rightarrow R=2r$

R = 10 cm, r = 5 cm
$\frac{r}{R}=\frac{h}{2h}$
$\frac{Volumeofsmallercone}{Volumeofl\arg ercone}=\frac{\frac{\pi }{3}\times r^2\times h}{\frac{\pi }{3}\times R^2\times 2h}$
$={\left(\frac{1}{2}\right)}^2\times \frac{1}{2}=\frac{1}{8}$
Let volume of smaller cone be x
volume of bigger cone be 8x
∴ volume of frustum = 8x – x = 7x
$\frac{Volumeoffrustum}{Volumeofsmallercone}=\frac{7x}{x}=7:1$

So, option b is correct.