A metal wire, when bent in the form of an equilateral triangle of largest area, encloses an area of $484{\sqrt3} \text{ cm}^2$. If the same wire is bent into the form of a circle of largest area, find the area of this circle.

Given:

Area of the equivalent triangle = $484{\sqrt3} \text{ cm}^2$

Let a be the length of the equivalent triangle.

Area of equivalent triangle = $\dfrac{\sqrt{3}}{4} \times a^2$

$⇒ \dfrac{\sqrt{3}}{4} \times a^2 = 484{\sqrt3} \text{ cm}^2\\[1em] ⇒ \dfrac{\cancel{\sqrt{3}}}{4} \times a^2 = 484 \cancel{\sqrt{3}} \text{ cm}^2\\[1em] ⇒ \dfrac{1}{4} \times a^2 = 484 \text{ cm}^2\\[1em] ⇒ a^2 = 484 \times 4 \text{ cm}^2\\[1em] ⇒ a^2 = 1,936 \text{ cm}^2\\[1em] ⇒ a = \sqrt{1,936} \text{ cm}^2\\[1em] ⇒ a = 44 \text{ cm}$

Perimeter of the triangle = 3 x side

= 3 x 44 cm = 132 cm

Perimeter of triangle = Circumference of circle

$⇒ 2πr = 132\\[1em] ⇒ 2 \times \dfrac{22}{7} \times r = 132\\[1em] ⇒ \dfrac{44}{7} \times r = 132\\[1em] ⇒ r = \dfrac{7 \times 132}{44}\\[1em] ⇒ r = \dfrac{7 \times 3}{1}\\[1em] ⇒ r = 21$

Area of circle = πr²

$= \dfrac{22}{7} \times 21^2\\[1em] = \dfrac{22}{7} \times 441\\[1em] = 22 \times 63\\[1em] = 1,386 \text{ cm}^2$

Hence, the area of the circle is 1,386 cm².