In an equilateral triangle ABC, the ratio r : | vedclass

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Question

(A) In an equilateral triangle, let the side length be $a$. Then, the semi-perimeter $s = \frac{3a}{2}$ and the area $\Delta = \frac{\sqrt{3}}{4} a^2$

Vedclass · Accepted Answer

$1 : 2 : 3$

Vedclass · Answer

(A) In an equilateral triangle, let the side length be $a$. Then, the semi-perimeter $s = \frac{3a}{2}$ and the area $\Delta = \frac{\sqrt{3}}{4} a^2$. The inradius $r = \frac{\Delta}{s} = \frac{\frac{\sqrt{3}}{4} a^2}{\frac{3a}{2}} = \frac{a}{2\sqrt{3}}$. The circumradius $R = \frac{abc}{4\Delta} = \frac{a^3}{4(\frac{\sqrt{3}}{4} a^2)} = \frac{a}{\sqrt{3}}$. The exradius $r_1 = \frac{\Delta}{s-a} = \frac{\frac{\sqrt{3}}{4} a^2}{\frac{3a}{2} - a} = \frac{\frac{\sqrt{3}}{4} a^2}{\frac{a}{2}} = \frac{\sqrt{3}}{2} a$. Now, the ratio $r : R : r_1 = \frac{a}{2\sqrt{3}} : \frac{a}{\sqrt{3}} : \frac{\sqrt{3}a}{2}$. Multiplying by $\frac{2\sqrt{3}}{a}$, we get $1 : 2 : 3$.

In an equilateral triangle $ABC$, the ratio $r : R : r_1$ is

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