Question

A cone of radius 4 centimetre is divided into two parts by drawing a plane through midpoint of its axis and parallel to its base compare the volume of two parts

Answer 1

Let the height of the cone is H and radius of the cone is R.

Given, the cone is divided into two parts through the mid point of its axis.

So, AQ = AP/2

Since QD || PC

So, the triangle AQD is similar to the triangle APC.

Now, by the condition of similarity,

QD/PC = AQ/AP = AQ/2AQ

=> QD/PC = AQ/AP = 1/2

=> QD/R = 1/2

and QD = 2R

Now, volume of the cone = πr2 H/3

Again, volume of the frustum = Volume of the cone ABC - Volume of the cone AED

= πr2 H/3 - {π(R/2)2 (H/2)}/3

= πr2 H/3 - {πR2 H}/(8*3)

= (πr2 H/3)*(1 - 1/8)

= (πr2 H)/{3 *(7/8)}

= (7πr2 H)/{3 *8}

Now, volume of the part taken out/volume of the remaining part of the cone = {(1/8) * πr2 H/3}/{(7πr2 H)/(3 *8)}

= 1/7