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Question

The height of a solid cone is 12 cm and the area of the circular base is 64πcm2. A plane parallel to the base of the cone cuts through the cone 9 cm above the vertex of the cone, the areas of the base of the new cone so formed is

(a) 9π cm2

(b) 16π cm2

(c) 25π cm2

(d) 36π cm2

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Solution

If a cone is cut into two parts by a plane parallel to the base, the portion that contains the base is called the frustum of a cone

Let ‘r’ be the top radius

R’ be the radius of the base

h’ be the height of the frustum

l’ be the slant height of the frustum.

H’ be the height of the complete cone from which the frustum is cut

Then from similar triangles we can write the following relationship

Here it is given that the area of the base is 64π cm2.

The area of the base with a base radius of ‘r’ is given by the formula

Area of base = πr2

Substituting the known values in this equation we get

64 π = πr2

r2 = 64

r = 8

Hence the radius of the base of the original cone is 8 cm.

So, now let the plane cut the cone parallel to the base at 9 cm from the vertex.

Based on this we get the values as

R = 8

H = 12

H – h = 9

Substituting these values in the relationship mentioned earlier

Hence the radius of the new conical part that has been formed is 6 cm.

And the area of this base of this conical part would be

Area of the base = πr2

= 36π

Hence the correct choice is option (d).


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