A wire when bent in the form of an equilateral triangle encloses an area of 121√3 cm^{2}. If the same wire is bent into the form of a circle, what will be the area of the circle? [Take π = 22/7]

Let the sides of equilateral triangle be a cm

Area of equilateral triangle = 121√3 cm^{2}

Area of equilateral triangle = √3/4 × a^{2}

⇒ √3/4 a^{2} = 121√3

⇒ a^{2} = 121√3 × 4/√3 = 121 × 4 cm^{2}

⇒ a^{2} = 484 cm^{2}

⇒ a = 22 cm

Perimeter of equilateral triangle = 3a

= 3 × 22 cm = 66 cm

Perimeter of equilateral triangle = Circumference of circle

Circumference of circle = 66 cm

Let the radius of circle be r

Circumference of circle = 2πr

⇒ 2πr = 66 cm

⇒ 2 × 22/7 × r = 66 cm

⇒ r = 66 × 1/2 × 7/22 cm

⇒ r = 10.5 cm

Area of circle = πr^{2} = 22/7 × 22/7 × 10.5 × 10.5 cm2

= 346.5 cm^{2}

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