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Applications of Cross Product

In a triangle...

Question

In a triangle ABC, $r_{1} = 2, r_{2} = 3, r_{3} = 6$ , then

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Solution

$\frac{1}{r} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = 1$
$⟹ r = 1$
$⟹ △^{2} = r r_{1} r_{2} r_{3} = 1 \times 2 \times 3 \times 6 = 36$
$⟹ △ = 6$
$⟹ r_{1} + r_{2} + r_{3} - r = 4 R$
$⟹ 4 R = 2 + 3 + 6 - 1 = 10$
$⟹ 2 R = 5$ .
$⟹ r = \frac{△}{s} ⟹ s = \frac{△}{r} = \frac{6}{5 / 2} = \frac{12}{5}$

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