The area of a circle inscribed in an equilateral triangle is 154 cm2.

Let the radius of the inscribed circle be r cm.
Given:

Area of the circle $=154 \mathrm{~cm}^{2}$

We know

Area of the circle $=\pi r^{2}$

$\Rightarrow 154=\frac{22}{7} r^{2}$

$\Rightarrow \frac{154 \times 7}{22}=r^{2}$

$\Rightarrow r^{2}=49$

$\Rightarrow r=7$

In a triangle, the centre of the inscribed circle is the point of intersection of the medians and altitudes of the triangle. The centroid divides the median of a triangle in the ratio 2:1.
Here,
AO:OD = 2:1  

Now,
Let the altitude be h cm.
We have:

$\angle A D B=90^{\circ}$

$O D=\frac{1}{3} A D$

$O D=\frac{h}{3}$

$\Rightarrow h=3 r$

$\Rightarrow h=21$

Let each side of the triangle be a cm.

In the right - angled $\triangle \mathrm{ADB}$, we have :

$A B^{2}=A D^{2}+D B^{2}$

$a^{2}=h^{2}+\left(\frac{a}{2}\right)^{2}$

$4 a^{2}=4 h^{2}+a^{2}$

$3 a^{2}=4 h^{2}$

$a^{2}=\frac{4 h^{2}}{3}$

$a=\frac{2 h}{\sqrt{3}}$

$a=\frac{42}{\sqrt{3}}$

∴ Perimeter of the triangle = 3a

$=3 \times \frac{42}{\sqrt{3}}$

$=\sqrt{3} \times 42$

$=72.66 \mathrm{~cm}$