Question

A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base Find the ratio of the volumes of two parts

Answer 1

Let ORN be the cone then , given radius of the base of the bases of the coner `r_(1) = 8 cm`.

and height of the cone, (h) OM = 12 cm
Let P be the mid-point of OM, then
`OP = PM = (12)/(2) = 6 cm`
Now, `Delta OPD ~ Delta OMN `
`therefore" " (OP)/(OM) = (PD)/(MN)`
`rArr" "(6)/(12) = (PD)/(8) rArr (1)/(2) = (PD)/(8)`
`rArr " "PD = 4cm`
The plane along CD divides the cone into two parts , namely
(i) a smalle cone of radius 4 cm and height 6 cm and (ii) frustum of a cone for which
Radius of the top the frustum , `r_(1) = 4 cm`
Radius of the bottom, `r_(2) = 8 cm `
and height of the frustum, h = 6 cm
`therefore` Volume of smaller cone `=((1)/(3)pi xx 4 xx 4xx6) = 32 pi cm^(3)`
and volume of the frustum of cone `= (1)/(2)xx pi xx 6[(8)^(2) + (4)^(2) + 8 xx4]`
`= 2 pi (64 + 16 +32) = 224 pi cm^(3)`
`therefore` Required ratio = volume of furstum : Volume of cone`= 224 pi : 32 pi = 1:7`