If a, b and c are the sides of a right angled triangle where c is the hypotenuse, then prove that the radius r of the circle which touches the sides of triangle is given by $r = \frac{a + b - c}{2}$

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Solution

Let the circle touches the sides BC, CA, AB of the right triangle ABC at D, E and F respectively, where BC = a, CA = b and AB = c. Then AE = AF and BD = BF. Also CE = CD = r.
i.e., b - r = AF, a - r = BF
or AB = c = AF + BF = b - r + a - r
This gives $r = \frac{(a + b - c)}{2}$

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If a, b and c are the sides of a right angled triangle where c is the hypotenuse, then prove that the radius r of the circle which touches the sides of triangle is given by r=a+b−c2

Let the circle touches the sides BC, CA, AB of the right triangle ABC at D, E and F respectively, where BC = a, CA = b and AB = c. Then AE = AF and BD = BF. Also CE = CD = r. i.e., b - r = AF, a - r = BF or AB = c = AF + BF = b - r + a - r This gives r=(a+b−c)2

If a, b and c are the sides of a right angled triangle where c is the hypotenuse, then prove that the radius r of the circle which touches the sides of triangle is given by $r = \frac{a + b - c}{2}$

Let the circle touches the sides BC, CA, AB of the right triangle ABC at D, E and F respectively, where BC = a, CA = b and AB = c. Then AE = AF and BD = BF. Also CE = CD = r.
i.e., b - r = AF, a - r = BF
or AB = c = AF + BF = b - r + a - r
This gives $r = \frac{(a + b - c)}{2}$