Let the radius of the inscribed circle be r. Then, the area of this circle = Π r^{2} = 154 (given)

That is 22/7 x r^{2} = 154

Implies r^{2} = 154 x 7/22 = 49

r = √49 = 7 cm.

In a triangle, the center of the inscribed circle is the point of intersection of the angular bisectors and in an equilateral triangle,

these bisectors are also the altitudes and medians whose point of intersection divides the medians in the ratio 2 : 1.

Let altitude AD = h

Therefore, angle ADB = 90^{0} and OD = 1/3 AD; i.e. r = h/3

Implies, h = 3r = 3 x 7 = 21 cm

Let each side of the triangle be a, then

In right triangle ADB, AB^{2} = AD^{2} + BD^{2}

Implies, a^{2} = h^{2} + (a/2)^{2}

Implies, 4 a^{2} = 4 h^{2} + a^{2}

Implies, 3 a^{2} = 4h^{2}

Implies, a = 2h / √3

Implies, a = 2 / √3 x 21 = 2 / √3 x 21 x √3 / √3

= 14 x √3 cm

Therefore, Perimeter of the triangle = 3a = 3 x 14 √3 = 42√3

= 42 x 1.73 = 72.7 cm