# A cone is divided into two parts by drawing a plane through the midpoint of its axis, parallel to its base. Compare the volumes of two parts.

Solution: Let R and H are the base radius and height of large cone respectively .

So , its volume is $\\ V=1/3\pi R^2H$

Now , the small cone that is generated

after dividing the large cone will have height H/2 and radius R/2

Therefore , its volume will be $\\ v=1/3\pi (R/2)^2(H/2)=1/8[1/3\pi R^2H]=V/8$

So ,the frustum so generated will have Volume =$V-V/8=7V/8$

Therefore , the ratio of the volume of small cone and the frustum generated will be V/8:7V/8=1:7

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