The base of a straight pyramid is an equilateral triangle. The height of the pyramid is $11 \mathrm{~cm}$, and the area of one of the side faces is $210 \mathrm{~cm}^{2}$. Calculate the length of one of the base edges of the pyramid and the volume.

Let the height of the pyramid be denoted by $AB$, and the midpoint of one of the base edges by $C$. Then, the height of one of the lateral faces is

$$
AC = \frac{2 \cdot 210}{a} \quad \text{and} \quad BC = \frac{a}{6} \sqrt{3}
$$

Thus, in the right triangle $ABC$,

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

or

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

From this, the positive value of $a^{2}$ is 900, so $a = 30 \, \text{cm}$.

Therefore, the volume of the pyramid is

$$
K = \frac{a^{2}}{4} \sqrt{3} \cdot \frac{11}{3} \, \text{cm}^{3} = 825 \sqrt{3} \, \text{cm}^{3}
$$