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

 Let the radius of the inscribed circle be r. Then, the area of this circle = Π r² = 154 (given)

That is 22/7 x r² = 154

Implies r² = 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 = 90⁰ 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, AB² = AD² + BD²

Implies, a² = h² + (a/2)²

Implies, 4 a² = 4 h² + a²

Implies, 3 a² = 4h²

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