In triangle ABC if r1 = 2r2 = 3r3, then a:b is equal to

r1=2r2=3r3⇒Δs−a=2Δs−b=3Δs−c⇒1s−a=2s−b=3s−c=k (say) ⇒s−a=1k,s−b=2k and s−c=3k Adding, we get 3s−(a+b+c)=6k ⇒ s=6k ⇒ a=5k and b=4k ⇒ ab=54

In triangle ABC if r1 = 2r2 = 3r3, then a:b is equal to

r1=2r2=3r3⇒Δs−a=2Δs−b=3Δs−c⇒1s−a=2s−b=3s−c=k (say) ⇒s−a=1k,s−b=2k and s−c=3k Adding, we get 3s−(a+b+c)=6k ⇒ s=6k ⇒ a=5k and b=4k ⇒ ab=54

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In triangle ABC if r₁ = 2r₂ = 3r₃, then a:b is equal to

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Detailed Solution

$\begin{array}{l} r_{1} = 2 r_{2} = 3 r_{3} \\ \Rightarrow \frac{Δ}{s - a} = 2 \frac{Δ}{s - b} = 3 \frac{Δ}{s - c} \\ \Rightarrow \frac{1}{s - a} = \frac{2}{s - b} = \frac{3}{s - c} = k (say) \\ \Rightarrow s - a = \frac{1}{k}, s - b = \frac{2}{k} and s - c = \frac{3}{k} \end{array}$
$Adding, we get 3 s - (a + b + c) = \frac{6}{k} \Rightarrow s = \frac{6}{k} \Rightarrow a = \frac{5}{k} and b = \frac{4}{k} \Rightarrow \frac{a}{b} = \frac{5}{4}$