Question

please solve fast
fastest gets brainlest​

Answer 1

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