Solution:

Let the number of sides be $n$.

Let $a_1 = 48~\mathrm{cm}$ and $a_2 = 55~\mathrm{cm}$ be the side lengths of the two polygons.

The area $A$ of a regular polygon with $n$ sides and side length $a$ is:
$$A = \frac{n a^2}{4} \cot \frac{\pi}{n}$$

Let $a_3$ be the side length of the required polygon whose area is equal to the sum of the areas of the given polygons.

So,
$$\frac{n a_3^2}{4} \cot \frac{\pi}{n} = \frac{n a_1^2}{4} \cot \frac{\pi}{n} + \frac{n a_2^2}{4} \cot \frac{\pi}{n}$$

Since $n$ and $\cot \frac{\pi}{n}$ are the same for all polygons, we can divide both sides by $\frac{n}{4} \cot \frac{\pi}{n}$:
$$a_3^2 = a_1^2 + a_2^2$$

Therefore,
$$a_3 = \sqrt{a_1^2 + a_2^2} = \sqrt{48^2 + 55^2} = \sqrt{2304 + 3025} = \sqrt{5329} = 73$$

So, the required side length is $73~\mathrm{cm}$.