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In a triangle $ABC, r_{1} r_{2} + r_{3} =$

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$\frac{a b}{c}$

$a b c$

$ab$

$\frac{(a + b)}{c}$

answer is C.

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Detailed Solution

$\begin{array}{l} r_{1} r_{2} + r_{3} = \frac{Δ^{2}}{(s - a) (s - b)} + \frac{Δ^{2}}{s (s - c)} \\ = \frac{Δ^{2} [s (s - c) + (s - a) (s - b)]}{s (s - a) (s - b) (s - c)} \\ = 2 s^{2} - (a + b + c) s + a b = a b \end{array}$

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