In the given figure, a hollow cone is cut by a plane parallel to the base. If the curved surface area of the frustum obtained is 8 times of the curved surface area of the smaller cone obtained, then which of the following is true?

r_{2} = 9 r_{1}

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r_{2}^{2} = 3 r_{1}^{2}

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r_{2} = 3 r_{1}

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3 r_{2}^{2} = r_{1}^{2}

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Solution

The correct option is C $r_{2} = 3 r_{1}$

Let $r_{1} a n d r_{2}$ be the radius of upper end and lower end of the frustum respectively, and $l_{1} a n d l_{2}$ be the slant height of the smaller cone and the frustum respectively.

CSA of frustum = 8 (CSA of smaller cone)

$\Rightarrow π (r_{1} + r_{2}) l_{2} = 8 (π r_{1} l_{1})$

$\Rightarrow \frac{r_{1} + r_{2}}{r_{1}} = 8 \frac{l_{1}}{l_{2}}$ ...(i)

Also, the smaller and the bigger cones are similar,

$\Rightarrow \frac{l_{1}}{l_{1} + l_{2}} = \frac{r_{1}}{r_{2}}$ (Ratio of the corresponding sides of similar triangles are equal)

$\Rightarrow \frac{1}{1 + \frac{l_{2}}{l_{1}}} = \frac{r_{1}}{r_{2}}$

$\Rightarrow \frac{r_{2}}{r_{1}} = 1 + \frac{l_{2}}{l_{1}}$

$\Rightarrow \frac{l_{2}}{l_{1}} = \frac{r_{2}}{r_{1}} - 1 = \frac{r_{2} - r_{1}}{r_{1}}$

$\Rightarrow \frac{l_{1}}{l_{2}} = \frac{r_{1}}{r_{2} - r_{1}}$ ...(ii)

Substitute (ii) in (i), we get

$\frac{r_{1} + r_{2}}{r_{1}} = 8 [\frac{r_{1}}{r_{2} - r_{1}}]$

$\Rightarrow r_{2}^{2} - r_{1}^{2} = 8 r_{1}^{2}$

$\Rightarrow r_{2}^{2} = 9 r_{1}^{2}$

$\Rightarrow r_{2} = 3 r_{1}$

Hence, the correct answer is option (c).

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