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Area of a Triangle

If in a trian...

Question

If in a triangle $A B C$ , $r_{1} = 2, r_{2} = 3$ and $r_{3} = 6$ , then $a =$

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Solution

We know that
$r_{1} = \frac{Δ}{s - a} = 2 \Rightarrow a = s - \frac{Δ}{2} . . . (i)$
$r_{2} = \frac{Δ}{s - b} = 3 \Rightarrow b = s - \frac{Δ}{3} . . . (i i)$
$r_{3} = \frac{Δ}{s - c} = 6 \Rightarrow c = s - \frac{Δ}{6} . . . (i i i)$
Adding equation $(i), (i i)$ and $(i i i)$ , we get
$a + b + c = 3 s - Δ$
$\Rightarrow s = Δ$
Now, Area of triangle $A B C$
$\Rightarrow Δ = \sqrt{Δ (\frac{Δ}{2}) (\frac{Δ}{3}) (\frac{Δ}{6})}$
$\Rightarrow Δ = \frac{Δ^{2}}{6}$
Now, $Δ = 6 = s$
$r_{1} = \frac{Δ}{s - a}$
$2 = \frac{6}{6 - a}$
$∴ a = 3$

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