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Answer»

The circle touches the SIDES BC, CA, AB of the right triangle ABC at D, E and F respectively.

Assume BC = a, CA = b and AB = c

Then AE = AF and BD = BF, CE = CD .

OE = OD = OF = r.

Here OEDC is a square then CE = CD = r. i.e., b – r = AF, a – r = BF or AB = c = AF + BF = b – r + a – r

∴ r = (a + b - c )/2


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