cone of base radius 4 cm is divided Into two parts by drawing a plane through the mid point of its height and parallel to its base. Compare the volume of the two parts.
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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 = πr2H/3
Again, volume of the frustum = Volume of the cone ABC - Volume of the cone AED
= πr2H/3 - {π(R/2)2 (H/2)}/3
= πr2H/3 - {πR2 H}/(8*3)
= (πr2H/3)*(1 - 1/8)
= (πr2H)/{3 *(7/8)}
= (7πr2H)/{3 *8}
Now, volume of the part taken out/volume of the remaining part of the cone = {(1/8) * πr2H/3}/{(7πr2 H)/(3 *8)}
= 1/7
So, the required ratio is 1 : 7