Question 6
Derive the formula for the curved surface area and total surface area of the frustum of cone.

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Solution

Let ABC be a cone. A frustum DECB 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 $Δ A B G a n d Δ A D F, D F | | B G$

$∴ Δ A B G a n d Δ A D F a r e s i m i l a r$

$\frac{D F}{B G} = \frac{A F}{A G} = \frac{A D}{A B} \frac{r_{2}}{r_{1}} = \frac{h_{1} - h}{h_{1}} = \frac{l_{l} - l}{l_{1}} \frac{r_{2}}{r_{1}} = 1 - \frac{h}{h_{1}} = 1 - \frac{l}{l_{1}} 1 - \frac{l}{l_{1}} = \frac{r_{2}}{r_{1}} \frac{l}{l_{1}} = 1 - \frac{r_{2}}{r_{1}} = \frac{r_{1} - r_{2}}{r_{1}} \frac{l_{1}}{l} = \frac{r_{1}}{r_{1} - r_{2}} l_{1} = \frac{r_{1} l}{r_{1} - r_{2}}$ ......(i)

CSA of frustum DECB = CSA of cone ABC - CSA cone ADE

$= π r_{1} l_{1} - π r_{2} (l_{1} - l) = π r_{1} (\frac{l r_{1}}{r_{1} - r_{2}}) - π r_{2} [\frac{r_{1} l}{r_{1} - r_{2}} - l] (B y u s i n g e q . (i)) = \frac{π r_{1}^{2} l}{r_{1} - r_{2}} - π r_{2} (\frac{r_{1} l - r_{1} l + r_{2} l}{r_{1} - r_{2}}) = \frac{π r_{1}^{2} l}{r_{1} - r_{2}} - \frac{π r_{2}^{2} l}{r_{1} - r_{2}} = π l [\frac{r_{1}^{2} - r_{2}^{2}}{r_{1} - r_{2}}]$

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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