A common cone's lateral surface area is in the ratio of $25: 32$ to its total surface area. What is the surface area and volume of the cone if the height $m=96 \mathrm{~cm}$?
Let the slant height of the cone be $l$, and the radius of its base circle be $r$. According to the problem,

$$
\operatorname{lr} \pi: \operatorname{lr} \pi+r^{2} \pi=25: 32
$$

from which

$$
r=\frac{7}{25} l
$$

Furthermore,

$$
l^{2}-r^{2}=l^{2}-\frac{49}{625} l^{2}=96^{2}
$$

thus

$$
l=100 \mathrm{~cm}, \mathrm{~s} \text { so } r=28 \mathrm{~cm}
$$

Therefore, the surface area of the cone is

$$
F=l r \pi+r^{2} \pi=3584 \pi \mathrm{cm}^{2}
$$

and its volume is

$$
K=\frac{r^{2} \pi}{3} m=25088 \pi \mathrm{cm}^{3}
$$

(Bauer Jakab, Zenta.)