In $$\triangle ABC$$, the lengths of sides $$AC$$ and $$AB$$ are 12 cm and 5 cm, respectively. If the area of $$\triangle ABC$$ is 30 cm$$^2$$ and $$R$$ and $$r$$ are respectively the radii of circumcircle and incircle of $$\triangle ABC$$, then the value of $$2R + r$$ (in cm) is equal to ________.

In $$\triangle ABC$$, we have $$AC = b = 12$$, $$AB = c = 5$$, and area $$= 30$$ cm$$^2$$.

The area of a triangle is $$\frac{1}{2} \cdot AB \cdot AC \cdot \sin A = \frac{1}{2}(5)(12)\sin A = 30$$, so $$\sin A = 1$$, meaning $$A = 90°$$.

Since $$A = 90°$$, by the Pythagorean theorem, $$BC = a = \sqrt{5^2 + 12^2} = \sqrt{25+144} = 13$$.

For a right triangle, the circumradius is $$R = \frac{a}{2} = \frac{13}{2}$$.

The semi-perimeter is $$s = \frac{a+b+c}{2} = \frac{13+12+5}{2} = 15$$.

The inradius is $$r = \frac{\text{Area}}{s} = \frac{30}{15} = 2$$.

Therefore $$2R + r = 2 \times \frac{13}{2} + 2 = 13 + 2 = 15$$.

The answer is $$15$$.