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The radii of the circular ends of a solid frustum of a cone are $ 33 \mathrm{~cm} $ and $ 27 \mathrm{~cm} $ and its slant height is $ 10 \mathrm{~cm} $. Find its total surface area.
Given:
The radii of the circular ends of a solid frustum of a cone are \( 33 \mathrm{~cm} \) and \( 27 \mathrm{~cm} \) and its slant height is \( 10 \mathrm{~cm} \).
To do:
We have to find its total surface area.
Solution:
Upper radius of the frustum $r_1 = 33\ cm$
Lower radius of the frustum $r_2 = 27\ cm$
Slant height of the frustum $l = 10\ cm$
Therefore,
Total surface area of the frustum $=\pi(r_{1}+r_{2}) l+\pi r_{1}^{2}+\pi r_{2}^{2}$
$=\pi[(r_{1}+r_{2}) l+r_{1}^{2}+r_{2}^{2}]$
$=\pi[(33+27) \times 10+(33)^{2}+(27)^{2}]$
$=\pi[60 \times 10+(33)^{2}+(27)^{2}]$
$=\frac{22}{7}[600+1089+729]$
$=\frac{22}{7} \times 2418$
$=7599.42 \mathrm{~cm}^{2}$
The total surface area of the frustum is $7599.42 \mathrm{~cm}^{2}$.
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