If a, b, c are the sides of a right triangle where c is the hypotenuse, then prove that radius r of the circle touches the sides of the triangle is given by r = a+b+c/2
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let the circle touches the sides AB,BC and CA of triangle ABC at D, E and F
as lengths of tangents drawn from an external point are equal ...we have
AD=AF, BD=BE and CE=CF
same as.... EB=BD=r
then... c = AF+FC
∴ c = AD+CE
⇒ c = (AB-DB)(CB-EB)
⇒ c = a-r +b-r
⇒ 2r =a+b-c
⇒ r =( a+b-c)/2
Anonymous:
good answer
Sol:
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
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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