In triangle ABC, ∠C = 90°, AC = 6 cm and BC = 8 cm. What is the distance between its incentre and its circumcentre?

Put C at the origin with CB along the $x$-axis (8) and CA along the $y$-axis (6). The inradius is $\frac{6+8-10}{2}=2$, so the incentre is (2, 2). The circumcentre is the midpoint of the hypotenuse, (4, 3). Distance $=\sqrt{2^2+1^2}=\sqrt{5}$ cm.