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A right circular cone of height h and radius R is cut by a plane parallel to the base at a distance h/4 from the base. The ratio of the curved surface areas of the resulting cone to the frustum is:

A7 : 16

B16 : 9

C16 : 7

D9 : 7

Answer:

D. 9 : 7

Read Explanation:

A plane parallel to the base cuts the cone at a height ( \frac{h}{4} ) from the base.

So:

  • Height of small top cone (=h−h4=3h4)(= h - \frac{h}{4} = \frac{3h}{4})

  • It is similar to the original cone, so the linear scale factor is (34)( \frac{3}{4} )

    Curved surface area (CSA)

For similar cones, CSA scales as the square of the linear scale factor.

So,
CSAsmall cone∝(34)2=916\text{CSA}_{\text{small cone}} \propto \left(\frac{3}{4}\right)^2 = \frac{9}{16}

Thus,

  • CSA of small cone = (916)( \frac{9}{16} ) of CSA of original cone

  • CSA of frustum = remaining part = (1−916=716)( 1 - \frac{9}{16} = \frac{7}{16} )

Required ratio:

  • CSA of cone : CSA of frustum=916:716=9:7\text{CSA of cone : CSA of frustum} = \frac{9}{16} : \frac{7}{16} = 9 : 7


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