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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.
Updated on: 10-Oct-2022
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