In a triangle ABC, right angled at B, the inradius is -

\frac{A B + B C - A C}{2}

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\frac{A B + A C - B C}{2}

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\frac{A B + B C + A C}{2}

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None of the above

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Solution

The correct option is A $\frac{A B + B C - A C}{2}$
Here, $b^{2} = a^{2} + c^{2} . . . . . (i)$

Thus $r = \frac{Δ}{s} = \frac{\frac{1}{2} a c}{\frac{1}{2} (a + c + b)} = \frac{a c}{a + c + b}$

$= \frac{a c}{a + c + b} . \frac{a + c - b}{a + c - b} = \frac{a c}{a^{2} + c^{2} + 2 a c - b^{2}} (a + c - b)$ , Using $(i)$

$= \frac{a + c - b}{2} = \frac{B C + A B - A C}{2}$

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