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

(N/A) Let $ABC$ be a cone. $A$ frustum $DECB$ is obtained by cutting the cone with a plane parallel to its base. Let $r_{1}$ and $r_{2}$ be the radii of the circular ends of the frustum,$h$ be the height of the frustum,and $l$ be the slant height of the frustum.
In $\triangle ABG$ and $\triangle ADF$,since $DF \parallel BG$,by the property of similar triangles:
$\triangle ABG \sim \triangle ADF$
Therefore,$\frac{DF}{BG} = \frac{AF}{AG} = \frac{AD}{AB}$
Let $h_{1}$ be the height of the full cone $ABC$ and $l_{1}$ be its slant height.
$\frac{r_{2}}{r_{1}} = \frac{h_{1}-h}{h_{1}} = \frac{l_{1}-l}{l_{1}}$
From $\frac{r_{2}}{r_{1}} = 1 - \frac{l}{l_{1}}$,we get $\frac{l}{l_{1}} = 1 - \frac{r_{2}}{r_{1}} = \frac{r_{1}-r_{2}}{r_{1}}$
So,$l_{1} = \frac{r_{1}l}{r_{1}-r_{2}}$ and $(l_{1}-l) = \frac{r_{2}l}{r_{1}-r_{2}}$
Curved Surface Area $(CSA)$ of frustum $DECB$ = $CSA$ of cone $ABC$ - $CSA$ of cone $ADE$
$CSA = \pi r_{1}l_{1} - \pi r_{2}(l_{1}-l)$
$CSA = \pi r_{1} \left( \frac{r_{1}l}{r_{1}-r_{2}} \right) - \pi r_{2} \left( \frac{r_{2}l}{r_{1}-r_{2}} \right)$
$CSA = \frac{\pi l (r_{1}^{2} - r_{2}^{2})}{r_{1}-r_{2}} = \pi l (r_{1} + r_{2})$
Total Surface Area $(TSA)$ = $CSA$ + Area of upper circular end + Area of lower circular end
$TSA = \pi (r_{1} + r_{2})l + \pi r_{1}^{2} + \pi r_{2}^{2}$