Question

A right circular cone is cut by two planes parallel to the base and trisecting the altitude. What is the ratio of the volumes of the three parts; top, middle and bottom respectively?

Answer 1

The cone is divided into three parts by the two planes trisecting the cone. Let the height of each of the trisected portion be h units.

It is obvious the \(\bigtriangleup\)ADE, \(\bigtriangleup\)AFG, and \(\bigtriangleup\)ABC are similar. Let DE = 2x.

As the three given \(\bigtriangleup\)s are similar, \(\frac{FG}{DE}=\frac{AQ}{AP}= \frac{2h}{h}\) 

⇒ FG = 2DE = 4x

Similarly BC = 3DE = 6x 

\ PE = x, QG = 2x and RC = 3x

Now, Volume of cone ABC = \(\frac{1}{3}\) ×π × (3x)² × (3h) =  9πx²h ...(i) 

Volume of cone AFG =  \(\frac{1}{3}\) ×π × (2x)² × (2h) =  \(\frac{8}{3}\)πx²h ...(ii) 

Volume of cone ADE =  \(\frac{1}{3}\) ×π × (x)² × (h) =  \(\frac{1}{3}\)πx²h  ...(iii)

Volume of middle portion DEFG = Vol. of cone AFG – Vol. of cone ADE =   \(\frac{8}{3}\)πx²h - \(\frac{\pi x^2 h}{3}\) 

 = \(\frac{7\pi x^2h}{3}\)      ... (iv)

Volume of the lowermost portion FGBC = Vol. of cone ABC – Vol. of cone AFG

= \(9\pi x^2h - \frac{8}{3} \pi x^2 h = \frac{19\pi x^2 h}{3}\) 

Reqd. ratio = \(\frac{\pi x^2 h}{3}\) : \(\frac{7\pi x^2h}{3}\) : \(\frac{19\pi x^2 h}{3}\) = 1:7:19