Calculate the semi-perimeter s of the triangle.

The side lengths are $a = 10$ cm, $b = 13$ cm, and $c = 8$ cm.

$$s = \frac{a+b+c}{2}$$ $$s = \frac{10 + 13 + 8}{2}$$ $$s = \frac{31}{2}$$ $$s = 15.5 \text{ cm}$$

Heron's formula is given by:

$$A = \sqrt{s(s-a)(s-b)(s-c)}$$

Substitute the values of $s$, $a$, $b$, and $c$:

$$A = \sqrt{15.5(15.5-10)(15.5-13)(15.5-8)}$$ $$A = \sqrt{15.5(5.5)(2.5)(7.5)}$$

$$A = \sqrt{15.5 \times 5.5 \times 2.5 \times 7.5}$$ $$A = \sqrt{1598.4375}$$

$$A \approx 39.98046$$

Rounding to two decimal places:

$$A \approx 39.98 \text{ cm}^2$$

The area of triangle ABC is:

$$\text{39.98 cm}^2$$