Question

A cone is cut into three parts by planes parallel to the base such that the height is
trisected. Find ratio of the volumes of the three parts so obtained.​

Answer 1

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

Answers

The height of a Cone, 3h is trisected by2 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 ABG ~ tri ACF ~ tri ADE ( by AAA similarity criterion. So corresponding sides are to be proportional.

So, AB/AC = h/2h = 1/2 = BG/CF = r/2r

AB/AD = h/3h = 1/3 = BG/ DE = r/ 3r

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

Volume of Cone ABG = 1/3 pi r² h ………….(1)

Volume of middle frustum =

1/3 pi ( r² + 4r² + 2r² ) h

= 1/3 pi 7r² h ……………….…….(2)

Volume of next frustum =

1/3 pi ( 4r² + 9r² + 6r²) h

= 1/3 pi 19r² h …………………….(3)

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

we get, (1/3pi r² h) : (1/3 pi 7r² h) : (1/3 pi 19r² h)

= 1:7:19