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In any triangle $A B C$, if $a: b: c=2: 3: 4$, then $R: r=$
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Solution:
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Verified Answer
The correct answer is:
$16: 5$
We have,
Let
$$
a: b: c=2: 3: 4
$$
$$
\begin{aligned}
& \therefore \quad R=\frac{a b c}{4 \Delta} \text { and } r=\frac{\Delta}{s} \\
& \therefore \quad \frac{R}{r}=\frac{s \cdot a b c}{4 \Delta^2} \\
& \Rightarrow \quad \frac{R}{r}=\frac{s \cdot(2 k)(3 k)(4 k)}{4 s(s-2 k)(s-3 k)(s-4 k)} \\
& \Rightarrow \quad \frac{R}{r}=\frac{6 k^3}{\left(\frac{9 k}{2}-2 k\right)\left(\frac{9 k}{2}-3 k\right)\left(\frac{9 k}{2}-4 k\right)} \\
& {\left[\because s=\frac{a+b+c}{2}\right]} \\
&
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & \frac{R}{r}=\frac{6 \cdot 2^3 \cdot k^3}{5 k \cdot 3 k \cdot k} \Rightarrow \frac{R}{r}=\frac{16}{5} \\
\therefore & R: r=16: 5
\end{array}
$$
Let
$$
a: b: c=2: 3: 4
$$
$$
\begin{aligned}
& \therefore \quad R=\frac{a b c}{4 \Delta} \text { and } r=\frac{\Delta}{s} \\
& \therefore \quad \frac{R}{r}=\frac{s \cdot a b c}{4 \Delta^2} \\
& \Rightarrow \quad \frac{R}{r}=\frac{s \cdot(2 k)(3 k)(4 k)}{4 s(s-2 k)(s-3 k)(s-4 k)} \\
& \Rightarrow \quad \frac{R}{r}=\frac{6 k^3}{\left(\frac{9 k}{2}-2 k\right)\left(\frac{9 k}{2}-3 k\right)\left(\frac{9 k}{2}-4 k\right)} \\
& {\left[\because s=\frac{a+b+c}{2}\right]} \\
&
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & \frac{R}{r}=\frac{6 \cdot 2^3 \cdot k^3}{5 k \cdot 3 k \cdot k} \Rightarrow \frac{R}{r}=\frac{16}{5} \\
\therefore & R: r=16: 5
\end{array}
$$
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