1. **State the problem:** We have a right-angled triangle with adjacent sides of lengths $x$ cm and $(x-1)$ cm, and the hypotenuse length is $\sqrt{13}$ cm. We need to find the value of $x$. 2. **Formula used:** In a right-angled triangle, by Pythagoras' theorem: $$\text{(side)}^2 + \text{(side)}^2 = \text{(hypotenuse)}^2$$ 3. **Apply the formula:** $$x^2 + (x-1)^2 = (\sqrt{13})^2$$ 4. **Expand and simplify:** $$x^2 + (x^2 - 2x + 1) = 13$$ $$x^2 + x^2 - 2x + 1 = 13$$ $$2x^2 - 2x + 1 = 13$$ 5. **Bring all terms to one side:** $$2x^2 - 2x + 1 - 13 = 0$$ $$2x^2 - 2x - 12 = 0$$ 6. **Simplify by dividing entire equation by 2:** $$\cancel{2}x^2 - \cancel{2}x - \cancel{12} = 0$$ $$x^2 - x - 6 = 0$$ 7. **Factorize the quadratic:** $$x^2 - x - 6 = (x - 3)(x + 2) = 0$$ 8. **Solve for $x$:** $$x - 3 = 0 \Rightarrow x = 3$$ $$x + 2 = 0 \Rightarrow x = -2$$ 9. **Interpret the solution:** Since length cannot be negative, $x = 3$ cm. **Final answer:** $\boxed{3}$