Question

A Right circular cone is divided into three parts by trisecting its height by two planes drown parallel to the base show that the volumes of three portions starting from the top are in ratio 1:7:19

Answer 1

hope this helps you !

Answer 2

AAA Similarity is a Helpful Tool

Step-by-step explanation:

• The height of a Cone, 3h is trisected by 2 planes // to the base of the cone at equal distances.

• So, the cone is divided into a smaller cone & 2 frustums of the cone. The height of each piece is ‘h’ unit

• Since, Right Triangle ΔABG ~  ΔACF ~  ΔADE ( by AAA (or Angle-Angle-Angle) similarity criterion. So corresponding sides are to be proportional.

So,  

$\frac {AB}{AC} = \frac {h}{2h} =  \frac {1}{2} = \frac {BG}{CF} = \frac {r}{2r}$

= $\frac {AB}{AD} = \frac {h}{3h} =  \frac {1}{3} = \frac {BG}{DE} = \frac {r}{3r}$

• Now, we find the volume of each piece.. a smaller cone & 2 frustums

The volume of Cone ABG =  $\frac {1}{3} \pi r^2 h$ ………….(1)

The volume of middle frustum =

$\frac {1}{3} \pi ( r^2 + 4r^2 + 2r^2 ) h$

= $\frac {1}{3} \pi 7r^2 h$ ……………….…….(2)

The volume of next frustum =

$\frac {1}{3} \pi ( 4r^2 + 9r^2 + 6r^2) h$

= $\frac {1}{3} \pi 19r^2 h$ …………………….(3)

Now, by finding the ratio of (1),(2)&(3)

we get,

= $(\frac {1}{3} \pi r^2 h) : (\frac {1}{3} \pi 7r^2 h) : (\frac {1}{3} \pi 19r^2 h)$

= 1:7:19 Proven .!