If in a $Δ$ ABC, $r_{1} = 2 r_{2} = 3 r_{3}$ , then the perimeter of the triangle is equal to

3 a

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3 b

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3 c

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3 (a + b + c)

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Solution

The correct option is B $3 b$
$r_{1} = \frac{Δ}{s - a}$ ; $r_{2} = \frac{Δ}{s - b}$ ; $r_{3} = \frac{Δ}{s - c}$
$2 r_{2} = \frac{2 Δ}{s - b}$ ; $3 r_{3} = \frac{3 Δ}{s - c}$
Let,
$r_{1} = r_{2} = r_{3} = k$
$\Rightarrow \frac{Δ}{s - a} = \frac{2 Δ}{s - b} = \frac{3 Δ}{s - c} = k$

$\Rightarrow s - a = \frac{Δ}{k} . . . . . . . . . . . . . . . . . . . . . (1)$
$s - b = \frac{2 Δ}{k} . . . . . . . . . . . . . . . . . . . . (2)$
$s - c = \frac{3 Δ}{k} . . . . . . . . . . . . . . . . . . . . . (3)$
Sum of $(1), (2), (3)$ is,
$3 s - (a + b + c) = \frac{6 Δ}{k}$
$\Rightarrow 3 s - 2 s = \frac{6 Δ}{k}$
$\Rightarrow s = \frac{6 Δ}{k}$
$\Rightarrow s = 3 b$ . is the answer.

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Similar questions

△ A B C, r_{1} = {2 r}_{2} = 3 r_{3},

r_{1} = 2 r_{2} = 3 r_{3}

Δ

A B C

A, B, C

a, b, c

r_{1}, r_{2}

r_{3}

r_{1} = 2 r_{2} = 3 r_{3}

r_{1} = 2 r_{2} = 3 r_{3}

\frac{a}{b} + \frac{b}{c} + \frac{c}{a}

r_{1} = 2 r_{2} = 3 r_{3}

[100 - (4 a / 5 b)]^{2}

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