Question

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

Answer 1

Answer:

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 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