Question : The sides of a right triangle $\triangle ABC$ are a, b, and c where c is the hypotenuse. What will be the radius of the circle of this triangle?
Option 1: $\frac{\left (a+b+c \right)}{2}$
Option 2: $\frac{\left (a+b-c \right)}{2}$
Option 3: $\frac{\left (b+c-a \right)}{2}$
Option 4: $\frac{\left (a+c-b \right)}{2}$
Correct Answer: $\frac{\left (a+b-c \right)}{2}$
Solution : Let the circle touch the sides BC, CA, and AB of the right triangle ABC at D, E, and F, respectively, where BC = a, CA = b and AB = c Since lengths of tangents drawn from an external point are equal. ⇒ AE = AF and BD = BF OE is perpendicular to AC, OD is perpendicular to CD [since the radius is perpendicular to tangents] $\therefore$ OECD is a square. So, CE = CD = r And b – r = AF, a – r = BF Therefore, AB = AF + BF Here, c = (b – r) + (a – r) $\therefore$ r = $\frac{a+b–c}{2}$ Hence, the correct answer is $\frac{a+b–c}{2}$.
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