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The slant height of a frustum of a cone is 4 cm and the perimeters of its circular ends are $18\ cm$ and $6\ cm$. Find the curved surface area of the frustum.
Given: The slant height of a frustum of a cone is 4 cm and the perimeters of its circular ends are $18\ cm$ and $6\ cm$.
To do: To find the curved surface area of the frustum.
Solution:
$l=4\ cm$
Circumference of a circular end $=18\ cm$
$\Rightarrow 2\pi r_{1}=18$
$\Rightarrow \pi \times r_{1}=\frac{18}{2}=9$ ............. $( 1)$
Circumference of other circular end $=6\ cm$
$\Rightarrow 2\pi r_{2}=6$
$\Rightarrow \pi r_{2}=\frac{6}{2}=3$ ............. $( 2)$
Adding $( 1)$ and $( 2)$
$\pi r_{1}+\pi r_{2}=9+3=12$
Curved surface area$=\pi(r_{1}+r_{2})l$
$=(9+3)\times 4$
$=48\ cm^2$
Thus, the curved surface area of the frustum is $48\ cm^{2}$.
Updated on: 10-Oct-2022
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