Derive the formula for the curved surface area and total surface area of the frustum of a cone.

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Solution

Let $A B C$ be a cone. A frustum $D E C B$ is cut by a plane parallel to its base.
Let $r_{1}$ and $r_{2}$ be the radii of the ends of the frustum of the cone and $h$ be the height of the frustum of the cone.

In $∠ A B G$ and $∠ A D F$ , $D F ∥ B G$

Therefore, $∠ A B G \sim ∠ A D F$
$\frac{D F}{B G} = \frac{A F}{A G} = \frac{A D}{A B}$

$\frac{r_{2}}{r_{1}} = \frac{h_{1} - h}{h_{1}} = \frac{l_{1} - l}{1_{1}}$

$\frac{r_{2}}{r_{1}} = 1 - \frac{h}{h_{1}} = 1 - \frac{l}{l_{1}}$

$1 - \frac{l}{l_{1}} = \frac{r_{2}}{r_{1}}$

$1 - \frac{l}{l_{1}} = 1 - \frac{r_{2}}{r_{1}} = \frac{r_{1} - r_{2}}{r_{1}}$

$\frac{l_{1}}{l} = \frac{r_{1}}{r_{1} - r_{2}}$

$l_{1} = \frac{r_{1} l}{r_{1} - r_{2}}$

CSA of frustum $D E C B =$ CSA of cone $A B C -$ CSA cone $A D E$
$= π r_{1} l_{1} - π r_{2} (l_{1} - l)$

$= π r_{1} (\frac{r_{1}}{r_{1} - r_{2}}) - π r_{2} (\frac{r_{1}}{r_{1} - r_{2}} - l)$

$\frac{π r_{1}^{2} l}{r_{1} - r_{2}} - π r_{2} (\frac{r_{1} l - r_{1} l + r_{2} l}{r_{1} - r_{2}})$

$= \frac{π r_{1}^{2} l}{r_{1} - r_{2}} - \frac{π r_{2}^{2} l}{r_{1} - r_{2}}$

$= π l [\frac{r_{1}^{2} - r_{2}^{2}}{r_{1} - r_{2}}]$

CSA of frustum = $π (r_{1} + r_{2}) l$ .

Total surface area of frustum = CSA of frustum + Area of upper circular end + Area of lower circular end.
$= π (r_{1} + r_{2}) l + π r_{2}^{2} + π r_{1}^{2}$
$= π [(r_{1} + r_{2}) l + r_{1}^{2} + r_{2}^{2}]$

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