A frustum of a right circular cone with a height of $16 \mathrm{cm}$, with circular end radii of $8 \mathrm{cm}$ and $20 \mathrm{cm}$. The slant height of  is equal to:

To do so, we will first determine the slant height of the cone and then apply the surface area and volume formulas for frustum.

Given:

$\mathbf{\text{H}}\mathbf{=}\mathbf{16}\mathbf{ }\mathbf{\text{cm}}$

$\text{R}=20 \text{cm}$

$\text{r}=8 \text{cm}$

Let the slant height of the frustum of cone be 1 ,

$1=\sqrt{h^2+{(R-r)}^2}$

$=\sqrt{{\left(16\right)}^2+{(20-8)}^2}$

$=\sqrt{256+144}$

$=\sqrt{400}$

$=20 \text{cm}.$