Solution: For a right triangle with legs $ a $, $ b $, hypotenuse $ c $, the inradius is $ r = \fraca + b - c2 $. Using the Pythagorean theorem $ c = \sqrta^2 + b^2 $, substitute $ b = \frac2r + c - a1 $ from rearranging the inradius formula. Squaring and solving yields $ c = a + \frac2r^2 + 2ar2r + 2a $. Simplifying, $ c = \fraca(2r + a)r $. Thus, the hypotenuse is $ \boxed\dfraca(a + 2r)r $. - Crosslake