A regular 3-sided pyramid has a height of $m=11 \mathrm{~cm}$, and the area of one side face is $210 \mathrm{~cm}^{2}$. Let's calculate the volume of the pyramid.
Given the base edge denoted by $a$, the sought volume is

$$
K=\frac{a^{2} \sqrt{3}}{4} \cdot \frac{m}{3}=\frac{a^{2} 11 \sqrt{3}}{12}
$$

Thus, only $a^{2}$ needs to be determined from the given data. Denoting the radius of the circle inscribed in the base by $\varrho$ and the height of the lateral face by $m_{0}$. According to the Pythagorean theorem, the following relationship holds for every regular pyramid:

$$
\varrho^{2}+m^{2}=m_{0}^{2}
$$

In this case, $\varrho$ is one-third of the slant height, i.e., $\varrho=\frac{1}{3} \frac{a \sqrt{3}}{2}=\frac{a \sqrt{3}}{6}$, and according to the problem, $\frac{a m_{0}}{2}=210$, from which $m_{0}=\frac{420}{a}$, and thus (1) becomes

$$
\left(\frac{a \sqrt{3}}{6}\right)^{2}+11^{2}=\left(\frac{420}{a}\right)^{2}
$$

After rearrangement,

$$
a^{4}+1452 a^{2}-2116800=0
$$

From which (ignoring the negative root),

$$
a^{2}=900
$$

and thus

$$
K=\frac{100 \cdot 11 \sqrt{3}}{12}=825 \sqrt{3} \approx 1429 \mathrm{~cm}^{3}
$$

Vida Piroska (Pannonhalma, Bencés g. IV. o. t.)