The slant height of the frustum of a cone is 4 cm and the perimeters (i.e. circumferences) of its circular ends are 18 cm and 6 cm.

Let R and r be the radii of the top and base of the frustum of the cone, respectively, and its slant height be l.

Then,

$2 \pi R=18 \Rightarrow \pi R=9 \quad \ldots(i)$

$2 \pi \mathrm{r}=6 \Rightarrow \pi \mathrm{r}=3 \quad \ldots(i i)$

Curved surface area of the frustum $=\pi l(R+r)$

$=l \times(\pi R+\pi r)$

$=4 \times(9+3) \quad[$ Since $l=4 \mathrm{~cm}] \quad($ from $(i)$ and $(i i))$

$=48 \mathrm{~cm}^{2}$