A copper wire when bent in the form of an equilateral triangle has area $121\sqrt{3}$ cm². If the same wire is bent in the form of a circle, find the area enclosed by the wire.

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4}$ (side)²

Given,

Area of equilateral triangle = $121\sqrt{3}$ cm²

$\therefore \dfrac{\sqrt{3}}{4}\text{ (side)}^2 = 121\sqrt{3} \\[1em] \Rightarrow \text{(side)}^2 = \dfrac{121\sqrt{3} \times 4}{\sqrt{3}} \\[1em] \Rightarrow \text{(side)}^2 = 484 \\[1em] \Rightarrow \text{(side)}^2 = 22^2 \\[1em] \Rightarrow \text{ side } = 22 \text{ cm}.$

Perimeter of equilateral triangle = 3 x side
= 3 x 22 = 66 cm.

Circumference of the circle of same wire = Perimeter of triangle of same wire

Let radius of circle formed = r cm.

∴ 2πr = 66

$\Rightarrow 2 \times \dfrac{22}{7} \times r = 66 \\[1em] \Rightarrow r = \dfrac{66 \times 7}{2 \times 22} \\[1em] \Rightarrow r = \dfrac{21}{2} \text{ cm}^2$

Area of circle = πr²

= $\dfrac{22}{7} \times \dfrac{21}{2} \times \dfrac{21}{2}$

= 346.5 cm².

Hence, area of circle = 346.5 cm².