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

Let AB = BC = CA = x cm ⇒ BD = CD = Suppose the radius of the inscribed circle be r cm.

Area of the inscribed circle = 154 cm2  (Given) In ΔBOD, • 76

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.

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

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

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