The height of a right circular cone is trisected by two planes drawn parallel to the base. Show that the volumes of the three portions starting from the top are in the ratio 1:7:19.

Explanation

Since height is trisected , therefore by basic proportionality there m, base radii of three cones VCD,VA'B' and VAB are also in the ratio 1:2:3.
Let volume of cone VCD=V =`(1)/(3)pir_(2)^(2)h`
Volume of cone VAB=`(1)/(3)pir_(1)^(2)(2h)=8xx(1)/(3)pir_(2)^(2)h=8V` ltbRgt and volume of cone VAB=`(1)/(3)pir^(2)(3h)=(1)/(3)pi(3r_(2))^(2)(3h)=27xx(1)/(3)xx(1)/(3)pir_(2)^(2)h=27V`
Ratio of volume of 3 portions
=volume (VCD): volume (CDB'A'): volume (A'B'BA)`
=V:8V:27V-8V=:19