Bangle in a triangle

There is a simplified formula for the area of any triangle, $\boxed {A = rs}$ where $r$ is inradius and $s$ is semiperimeter. Let the side of equilateral trianglebe $a$ then, area of triangle = $\frac{\sqrt{3} a^2}{4}$ From the above formula we have $\sqrt{3} \times \frac{3a}{2} = \frac{\sqrt{3} a^2}{4}$ on simplifying we get $\boxed {a = 6}$ so $\boxed {3a = 18}$

In an equilateral triangle, the inradius and centroid lie on the same point. Since the centroid divides the median in the ration $2:1$ , we get the median to be $3\sqrt { 3 }$ .

Since it is an equilateral triangle, the median will also be the altitude.

Let the side length be $x$

Using Pythagoras' theorem, we have

$\left[ \frac { x }{ 2 } \right] ^{ 2 }+\left[ 3\sqrt { 3 } \right] ^{ 2 }={ x }^{ 2 }\\ \\ \Rightarrow x=6$

Hence the Perimeter = $18$

nice & easy problem :)