With usual notations in a triangle ABC, if $r_{1} = 2 r_{2} = 2 r_{3}$ then -

4 a = 3 b

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3 a = 2 b

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4 b = 3 a

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2 a = 3 b

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Solution

The correct option is B $4 b = 3 a$
$r_{1} = 2 r_{2} = 2 r_{3} \frac{Δ}{s - a} = \frac{2 Δ}{s - b} = \frac{2 Δ}{s - c} \frac{Δ}{s - a} = \frac{2 Δ}{s - b} s - b = 2 s - 2 a \Rightarrow 3 a - 3 b = c . . . . . . (i) \frac{2 Δ}{s - b} = \frac{2 Δ}{s - c} \Rightarrow b = c$

Substituting in $(i)$

$3 a - 3 b = b 3 a = 4 b$

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