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In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio
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The correct answer is:
$1: 2: 3$
We have, $\Delta=\frac{\sqrt{3}}{4} a^{2}, s=\frac{3 a}{2}$
Inradius $r=\frac{\Delta}{s}=\frac{a}{2 \sqrt{3}}$
Circumradius $\mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}=\frac{\mathrm{a}^{3}}{\sqrt{3} \mathrm{a}^{2}}=\frac{\mathrm{a}}{\sqrt{3}}$
and exradii $r_{1}=\frac{\Delta}{s-a}=\frac{\sqrt{3} / 4 a^{2}}{a / 2}$
$$
\begin{array}{l}
=\frac{\sqrt{3}}{2} \mathrm{a} \\
\therefore \text { Required ratio }=r: R: r_{1} \\
=\frac{a}{2 \sqrt{3}}: \frac{a}{\sqrt{3}}: \frac{\sqrt{3}}{2} a=1: 2: 3
\end{array}
$$
Inradius $r=\frac{\Delta}{s}=\frac{a}{2 \sqrt{3}}$
Circumradius $\mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}=\frac{\mathrm{a}^{3}}{\sqrt{3} \mathrm{a}^{2}}=\frac{\mathrm{a}}{\sqrt{3}}$
and exradii $r_{1}=\frac{\Delta}{s-a}=\frac{\sqrt{3} / 4 a^{2}}{a / 2}$
$$
\begin{array}{l}
=\frac{\sqrt{3}}{2} \mathrm{a} \\
\therefore \text { Required ratio }=r: R: r_{1} \\
=\frac{a}{2 \sqrt{3}}: \frac{a}{\sqrt{3}}: \frac{\sqrt{3}}{2} a=1: 2: 3
\end{array}
$$
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