Solution: For a right triangle with legs $ a $, $ b $, hypotenuse $ c $, the inradius is $ r = \frac{a + b - c}{2} $. Using the Pythagorean theorem $ c = \sqrt{a^2 + b^2} $, substitute $ b = \frac{2r + c - a}{1} $ from rearranging the inradius formula. Squaring and solving yields $ c = a + \frac{2r^2 + 2ar}{2r + 2a} $. Simplifying, $ c = \frac{a(2r + a)}{r} $. Thus, the hypotenuse is $ \boxed{\dfrac{a(a + 2r)}{r}} $.

Sat, 02 May 2026 20:09:07 +0000

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