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In a $\triangle A B C$, if $r_1=12, r_2=18$ and $r_3=36$, then $b=$
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The correct answer is:
$24$
Given, in $\triangle A B C, r_1=12, r_2=18, r_3=36$
$\because \quad \frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}$
$=\frac{1}{12}+\frac{1}{18}+\frac{1}{36}=\frac{3+2+1}{36}$
$\begin{aligned} & \Rightarrow \quad r=6 \\ & \because \quad \Delta^2=r_1 r_2 r_3=6 \times 12 \times 18 \times 36 \\ & \end{aligned}$
$\Rightarrow \quad \Delta=\sqrt{6 \times 6 \times 2 \times 6 \times 3 \times 6^2}=216$
and $r=\frac{\Delta}{s} \Rightarrow s=\frac{\Delta}{r}=\frac{216}{6}=36$.
Then, $r_2=\frac{\Delta}{s-b}=\frac{216}{36-b}$
$\Rightarrow \quad 18=\frac{216}{36-b} \Rightarrow 36-b=12$
$\Rightarrow \quad b=24$
$\because \quad \frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}$
$=\frac{1}{12}+\frac{1}{18}+\frac{1}{36}=\frac{3+2+1}{36}$
$\begin{aligned} & \Rightarrow \quad r=6 \\ & \because \quad \Delta^2=r_1 r_2 r_3=6 \times 12 \times 18 \times 36 \\ & \end{aligned}$
$\Rightarrow \quad \Delta=\sqrt{6 \times 6 \times 2 \times 6 \times 3 \times 6^2}=216$
and $r=\frac{\Delta}{s} \Rightarrow s=\frac{\Delta}{r}=\frac{216}{6}=36$.
Then, $r_2=\frac{\Delta}{s-b}=\frac{216}{36-b}$
$\Rightarrow \quad 18=\frac{216}{36-b} \Rightarrow 36-b=12$
$\Rightarrow \quad b=24$
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