1. The problem asks us to find the exact value of $x$ in a right-angled triangle with sides $3 + \sqrt{10}$, $3\sqrt{2} - \sqrt{5}$, and base $x$. The right angle is at the top vertex, so these two given sides are perpendicular. 2. By the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. The side opposite the right angle is $x$, so $$x^2 = (3 + \sqrt{10})^2 + (3\sqrt{2} - \sqrt{5})^2.$$ 3. First, expand each square: $$(3 + \sqrt{10})^2 = 3^2 + 2 \times 3 \times \sqrt{10} + (\sqrt{10})^2 = 9 + 6\sqrt{10} + 10 = 19 + 6\sqrt{10}.$$ 4. Next, $$(3\sqrt{2} - \sqrt{5})^2 = (3\sqrt{2})^2 - 2 \times 3\sqrt{2} \times \sqrt{5} + (\sqrt{5})^2 = 9 \times 2 - 6\sqrt{10} + 5 = 18 - 6\sqrt{10} + 5 = 23 - 6\sqrt{10}.$$ 5. Now add the two results: $$x^2 = (19 + 6\sqrt{10}) + (23 - 6\sqrt{10}) = 19 + 23 + 6\sqrt{10} - 6\sqrt{10} = 42.$$ 6. The $6\sqrt{10}$ terms cancel out, so $$x^2 = 42,$$ therefore $$x = \sqrt{42}.$$ 7. Hence, the exact length of $x$ is $\boxed{\sqrt{42}}$ cm.