#### Questions

The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.

A cone of radius 10 cm is divided into two parts by a plane parallel to its base through the mid-point of its height. Compare the volumes of the two parts

#### Solution

WeWe have,

Radius of the solid cone, R = CP

Height of the solid cone, AP = H

Radius of the smaller cone, QD = r

Height of the smaller cone, AQ = h

Also, `AQ = "AP"/2 "i.e" h = H/2 or H = 2h` ....(1)

Now, in ∆AQD and ∆APC,

∠QAD = ∠PAC (Common angle)

∠AQD = ∠APC = 90°

So, by AA criteria

∆AQD ~ APC

`=> (AQ)/(AP) = (QD)/(PC)`

`=> h/H = r/R`

`=>h/(2h) = r/R ["Using (i)"]`

`=>1/2= r/R`

`=> R = 2r` ....(ii)

As

Volume of smaller cone = `1/3pir^2h`

And

Volume of solid cone=`1/3πR^2H`

`= 1/3 pi(2r)^2 xx (2h)` [Usinmg i and ii]

`= 8/3 pir^2h`

so,

Volume of frustum = Volume of solid cone − Volume of smaller cone

`= 8/3 pir^2h - 1/3pir^2h`

`= 7/3pir^2h`

Now, the ratio of the volumes of the two parts = `"Volume of the smaller cone"/"Volume of the frustum"`

`= (1/3pir^2h)/(7/3pir^2h)`

= `1/7`

= 1 :7

So, the ratio of the volume of the two parts of the cone is 1 : 7