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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Updated on: 10-Oct-2022

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