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What is the surface area of a frustum of cone, whose larger and smaller radius is $6\ cm$ and $2\ cm$ The height of the cone is $3\ cm.\ ( Use\ \pi=3.14)$
Given: Larger and smaller radius is $6\ cm$ and $2\ cm$ The height of the cone is $3\ cm$.
To do: To find the surface area of a frustum of cone.
Solution:
Surface area of frustum cone $= \pi (r+R)\sqrt{( R-r)^2+h^2}+\pi r^2+\pi R^2$
$=\pi (6+2)\sqrt{(6-2)^2+3^2}+\pi \times6^2+\pi \times2^2$
$=3.14\times 8\sqrt{16+9}+3.14\times 36+3.14\times 4$
$=25.12\times 5+113.04+12.56$
$=251.2\ cm^2$
Updated on: 10-Oct-2022
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