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.
Do not write an equation yet. Shake the figure first.
Play the reversal slowly
Now move the same pieces step by step and prove that the pattern you touched was not an accident.
The inscribed circle of a right triangle (legs a, b, hypotenuse c). Radius r?
Return to the exact question
Express the inradius r of a right triangle with legs a, b and hypotenuse c in terms of its sides.
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.
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