The height of a cone is $12 \mathrm{~cm}$. If we unfold the lateral surface, we get a sector with an angle of $100^{\circ} 48^{\prime}$. What is the surface area and volume of the cone?
Let's denote the slant height of the cone by \( l \) and the radius of its base circle by \( r \). Then,

\[
2 l \pi \frac{100.8}{360} = 2 r \pi
\]

or

\[
r = \frac{7}{25} l
\]

Furthermore,

\[
12^{2} = l^{2} - r^{2} = l^{2} - \frac{49}{625} l^{2}
\]

from which

\[
l = \frac{25}{2} \text{ cm and thus } r = \frac{7}{2} \text{ cm}
\]

The surface area of the cone is

\[
F = r^{2} \pi + r \pi l = 56 \pi \text{ cm}^2
\]

and its volume is

\[
K = \frac{r^{2} \pi}{3} \cdot 12 = 49 \pi \text{ cm}^3
\]

(Ferenc Grünhut, Budapest.)