Question

Find the total surface area of frustum in the given figure.

Answer 1

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 $\Delta ABG$ and $\Delta ADF,DF||BG$
$\therefore \Delta ABG\sim \Delta ADF\frac{DF}{BG}=\frac{AF}{AG}=\frac{AD}{AB}\frac{r_2}{r_1}=\frac{h_1-h}{h_1}=\frac{l_1-l}{l_1}\frac{r_2}{r_1}=1-\frac{r_2}{r_1}=\frac{r_1-r_2}{r_1}l=\frac{l}{l_1}=\frac{r_2}{r_1}\frac{l}{l_1}=1-\frac{r_2}{r_1}=\frac{r_1-r_2}{r_1}\frac{l_1}{l}=\frac{r_1}{r_1-r_2}{l}_1=\frac{r_{1}l}{r_1-r_2}$
Regards
CSA of frustum DECB =CSA of cone ABC-CSA cone ADE
$=\pi r_1{l}_1-\pi t_2(l_1-l)=\pi r_1\left(\frac{l{r}_2}{r_1-r_2}\right)-\pi r_2[\frac{r_{1}l}{r_1-r_2}-l]=\frac{\pi r_1^{2}l}{r_1-r_2}-\pi r_2\left(\frac{r_{1}l-r_{1}l+r_{2}l}{r_1-r_2}\right)=\frac{\pi r_1^{2}l}{r_1-r_2}-\frac{\pi r_2^{2}l}{r_1-r_2}=\pi l\left[\frac{r_1^2-r_2^2}{r_1-r_2}\right]$
CSA of frustum = $\pi (r_1+r_2)l$
Total surface area of frustum = CSA of frustum + ARea of upper circular end +Area of lower circular end
$=\pi \left(r_1{r}_2\right)l+\pi r_2^2+\pi r_1^2=\pi \left[\right(r_1+r_2)l+r_1^2+r_2^2]$