Question 4
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
Let ORN be the cone then given, radius of the base of the cone =r1=8cm
And height of the cone, (h) OM = 12 cm
Let P be the mid-point of OM, then
OP=PM=122=6cmNow,ΔOPD∼ΔOMN∴OPOM=PDMN⇒612=PD8⇒12=PD8⇒PD=4cm
The plane along CD divides the cone into two parts, namely
(i) a smaller cone of radius 4 cm and height 6cm and
(ii) frustum of a cone for which.
Radius of the top of the frustum r1=4cm
Radius of the bottom r2=8cm
Height of the frustum h = 6cm
Volume of smaller cone
=(13π×4×4×6)
=32πcm3
[∵Volume of the cone=13π×r2h]
Volume of the frustum of cone
=13π×6[(8)2+(4)2+8×4]=224π
[Volume of the frustum of cone=13π×h(r21+r22+r1r2)]
∴ Required ratio
= Volume of frustum : Volume of cone
=224π:32π=7:1