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

By formula,

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

Given,

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

$\therefore 121\sqrt{3} = \dfrac{\sqrt{3}}{4} \text{ (side)}^2$

⇒(side)² = $\dfrac{121 \sqrt{3} × 4}{\sqrt{3}}$

⇒ (side)² = 121 × 4

⇒ (side)² = 484

⇒ side = $\sqrt{484}$

⇒ side = 22 cm.

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

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

Let radius of circle be r cm.

∴ 2πr = 66

⇒ 2 × $\dfrac{22}{7}$ × r = 66

⇒ 44 × r = 66 × 7

⇒ r = $\dfrac{66 \times 7}{44}$

⇒ r = $\dfrac{21}{2}$ = 10.5 cm

Area of circle = πr²

= $\dfrac{22}{7}$ × (10.5)²

= $\dfrac{22}{7}$ × 10.5 × 10.5

= 22 × 1.5 × 10.5 = 346.5 cm².

Hence, area = 346.5 cm².