Inradius of a Right Triangle — r = (a+b−c)/2
The inscribed circle of a right triangle — its radius? Just (a+b−c)/2 from the three sides. One tangent fact does it.
Problem
Express the inradius r of a right triangle with legs a, b and hypotenuse c in terms of its sides.
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The inscribed circle of a right triangle (legs a, b, hypotenuse c). Radius r?
Why this approach works
Use the fact that two tangents from a point are equal. The incircle touches both legs, so from the right-angle vertex both tangent lengths are r (an r×r square sits in the corner). The remaining tangents along the legs are b−r and a−r, and these equal the two tangents on the hypotenuse. So c = (b−r)+(a−r) = a+b−2r, giving r = (a+b−c)/2. For a 3-4-5 triangle, r = (3+4−5)/2 = 1.
Proof
Equal tangents from a point: the right-angle vertex gives r and r, the hypotenuse gives a−r and b−r. So c = a+b−2r, hence r = (a+b−c)/2.
Try it yourself
For 3-4-5, r = 1. For 5-12-13, r = (5+12−13)/2 = 2 — always an integer for integer triples.
Related mathematics
Beyond MathVoyage
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