Two similar cones have volumes $12 π$ cu. units and $96 π$ cu. units. If the curved surface area of the smaller cone is $15 π$ sq. units, what is the curved surface area of the larger one?

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Solution

Small cone volume $\Rightarrow \frac{1}{3} π r_{1}^{2} h_{1} = 12 π$
Large cone volume $\Rightarrow \frac{1}{3} π r_{2}^{2} h_{2} = 96 π$
[Similar cones $\frac{r_{1}}{r_{2}} = \frac{h_{1}}{h_{2}}$ ]
$\frac{\frac{1}{3} π r_{1}^{2} h_{1}}{\frac{1}{3} π r_{2}^{2} h_{2}} = \frac{12 π}{96 π}$
$\frac{r_{1}^{2}}{r_{2}^{2}} = \frac{1}{8} \times \frac{h_{2}}{h_{1}}$ $[\frac{h_{2}}{h_{1}} = \frac{r_{2}}{r_{2}}]$
$\frac{r_{1}^{2}}{r_{2}^{2}} = \frac{1}{8} \times \frac{r_{2}}{r_{1}}$
$\Rightarrow r_{1}^{3} = 8 r_{2}^{3}$
$\Rightarrow r_{1} = 2 r_{2}$
Given surface area of smaller cone $\Rightarrow π r_{1} \sqrt{r_{1}^{2} + h^{2}} + π r_{1}^{2} = 15 π$
Large cone $= π r_{2}^{2} + π r_{2} \sqrt{r_{2}^{2} + h_{2}^{2}}$
$= π (2 r_{1})^{2} + π (2 r_{1}) \sqrt{(2 r_{1})^{2} + (2 h_{1})^{2}}$
$\Rightarrow 4 π r_{1}^{2} + π 2 r_{1} \times 2 \sqrt{r_{1}^{2} + h_{1}^{2}}$
$\Rightarrow 4 (π r_{1}^{2} + π r_{1} \sqrt{r_{1}^{2} + h_{1}^{2}}) = 4 \times 15 π$
$\Rightarrow 60 π$ .

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