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  4. In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio

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Q. In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio

VITEEEVITEEE 2011

Solution:

We have, $\Delta=\frac{\sqrt{3}}{4}a^{2}, s=\frac{3a}{2}$
Inradius $r=\frac{\Delta}{s}=\frac{a}{2\sqrt{3}}$
Circumradius $R=\frac{abc}{4\Delta}=\frac{a^{3}}{\sqrt{3}a^{2}}=\frac{a}{\sqrt{3}}$
and exradii $r_{1}=\frac{\Delta}{s-a}=\frac{\sqrt{3}/ 4a^{2}}{a / 2}$
$= \frac{\sqrt{3}}{2}a$
$\therefore $ Required ratio $=r: R: r_{1}$
$= \frac{a}{2\sqrt{3}} : \frac{a}{\sqrt{3}} : \frac{\sqrt{3}}{2}$
$ a=1 : 2 : 3.$