Question

In an equilateral triangle, the inradius $r$, circumradius $R$ and the ex-radius $r_1$ are in?

• A. $A.P$
• B. $G.P$
• C. $H.P$
• D. None of the above

Answer 1

The correct option is A $A.P$

We have inradius $r=\frac{\Delta }{s}$
Circumradius $R=\frac{abc}{4\Delta }$
Ex-radius $r_1=\frac{\Delta }{s-a}$
For equilateral triangle , $\Delta =\frac{\sqrt{3}}{4}{a}^2$, $2s=3a$
So, $r=\frac{\frac{\sqrt{3}}{4}{a}^2}{\frac{3a}{2}}=\frac{a}{2\sqrt{3}}$
$R=\frac{a^3}{4.\frac{\sqrt{3}{a}^2}{4}}$
$\Rightarrow R=\frac{a}{\sqrt{3}}=\frac{2a}{2\sqrt{3}}$
$r_1=\frac{\frac{\sqrt{3}{a}^2}{4}}{\frac{a}{2}}$
$\Rightarrow r_1=\frac{\sqrt{3}}{2}a=\frac{3a}{2\sqrt{3}}$
Now, $r+r_1=\frac{a}{2\sqrt{3}}+\frac{3a}{2\sqrt{3}}$
$\Rightarrow r+r_1=\frac{4a}{2\sqrt{3}}$
$\Rightarrow r+r_1=2R$
$\Rightarrow r,R,r_1$ are in AP.