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


Given: Two similar cones have volumes $12\pi$ cu. units and $96 \pi$ cu. units. If the curved surface area of the smaller cone is $15 \pi$ sq. units.

To do: To find the curved surface area of the larger one.

Solution:

Volume of a cone$=( \frac{1}{3})\pi r^2h$

Two similar cones $\Rightarrow  \frac{r_2}{r_1}=\frac{h_2}{h_1}=k$

Volume of one cone $=( \frac{1}{3})\pi r_2^2 h_2=96$

Volume of Another cone $=( \frac{1}{3})\pi r_1^2 h_1=12$

$\Rightarrow  \frac{( \frac{1}{3})\pi r_2^2 h_2}{( \frac{1}{3})\pi r_1^2 h_1}=\frac{96}{12}$

$\Rightarrow  (\frac{r_2}{r_1})^2\times( \frac{h_2}{h_1})=8$

$\Rightarrow  k^3\times k=2^3$

$\Rightarrow  k^3=2^3$

$\Rightarrow  k=2$

$\Rightarrow  \frac{r_2}{r_1}\Rightarrow r_2=2r_1$

Curved surface Area of smaller cone $=\pi r_1( \sqrt{r_1^2  + h_1^2})=15$

Curved surface Area of Larger cone $=\pi r_2( \sqrt{r_2^2  + h_2^2})$

$=\pi ( 2r_1)( ( \sqrt{2r_1})^2 +( \sqrt{2h_1})^2)$

$=2\pi r_1\times2(\sqrt{r_1^2  + h_1^2})$

$=4\pi r_1(\sqrt{r_1^2  + h_1^2})$

$=4\pi\times15$

$=60\pi$ sq. units

Thus, Curved surface Area of Larger cone$=60\pi$ sq. units.

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Updated on: 10-Oct-2022

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