In a ΔABC, s−aΔ=18, s−bΔ=112, s−cΔ=124, then 'b'=

r1=8, r2=12, r3=241r=1r1+1r2+1r3=18+112+124⇒1r=3+2+124⇒1r=624=14r=4b=(r2−r)(r3+r1)=(12−4)(24+8)=8×32=256b=16

In a ΔABC, s−aΔ=18, s−bΔ=112, s−cΔ=124, then 'b'=

r1=8, r2=12, r3=241r=1r1+1r2+1r3=18+112+124⇒1r=3+2+124⇒1r=624=14r=4b=(r2−r)(r3+r1)=(12−4)(24+8)=8×32=256b=16

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In a $Δ A B C, \frac{s - a}{Δ} = \frac{1}{8}, \frac{s - b}{Δ} = \frac{1}{12}, \frac{s - c}{Δ} = \frac{1}{24}$ , then $' b' =$

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$r_{1} = 8, r_{2} = 12, r_{3} = 24$

$\frac{1}{r} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}$

$= \frac{1}{8} + \frac{1}{12} + \frac{1}{24}$

$\Rightarrow \frac{1}{r} = \frac{3 + 2 + 1}{24}$

$\Rightarrow \frac{1}{r} = \frac{6}{24} = \frac{1}{4}$

$r = 4$

$b = \sqrt{(r_{2} - r) (r_{3} + r_{1})}$

$= \sqrt{(12 - 4) (24 + 8)}$

$= \sqrt{8 \times 32}$

$= \sqrt{256}$

$b = 16$

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In a ΔABC, s−aΔ=18, s−bΔ=112, s−cΔ=124, then 'b'=

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In a $Δ A B C, \frac{s - a}{Δ} = \frac{1}{8}, \frac{s - b}{Δ} = \frac{1}{12}, \frac{s - c}{Δ} = \frac{1}{24}$ , then $' b' =$