A wire when bent in the form of a square encloses an area of 484 `m^(2)`. Find the largest area enclosed by the same wire when bent to form :
an equilateral triangle.

Answer

To find the largest area enclosed by the same wire when bent to form an equilateral triangle, we can follow these steps:

### Step 1: Find the side length of the square
Given that the area of the square is 484 m², we can find the side length of the square using the formula for the area of a square:

\[ \text{Area} = \text{side}^2 \]

Let the side length of the square be \( a \). Therefore, we have:

\[ a^2 = 484 \]

Taking the square root of both sides:

\[ a = \sqrt{484} = 22 \text{ m} \]

### Step 2: Calculate the perimeter of the square
The perimeter \( P \) of a square is given by:

\[ P = 4a \]

Substituting the value of \( a \):

\[ P = 4 \times 22 = 88 \text{ m} \]

### Step 3: Set the perimeter of the equilateral triangle equal to the perimeter of the square
When the wire is bent into the shape of an equilateral triangle, the perimeter of the triangle will be equal to the perimeter of the square. Let the side length of the equilateral triangle be \( x \). The perimeter of the triangle is given by:

\[ P = 3x \]

Setting the two perimeters equal gives us:

\[ 3x = 88 \]

### Step 4: Solve for the side length of the equilateral triangle
Now we can solve for \( x \):

\[ x = \frac{88}{3} \approx 29.33 \text{ m} \]

### Step 5: Calculate the area of the equilateral triangle
The area \( A \) of an equilateral triangle can be calculated using the formula:

\[ A = \frac{\sqrt{3}}{4} x^2 \]

Substituting the value of \( x \):

\[ A = \frac{\sqrt{3}}{4} \left(\frac{88}{3}\right)^2 \]

Calculating \( \left(\frac{88}{3}\right)^2 \):

\[ \left(\frac{88}{3}\right)^2 = \frac{7744}{9} \]

Now substituting back into the area formula:

\[ A = \frac{\sqrt{3}}{4} \times \frac{7744}{9} = \frac{7744\sqrt{3}}{36} \]

### Step 6: Simplify the area
We can simplify this further:

\[ A = \frac{1936\sqrt{3}}{9} \approx 372.57 \text{ m}^2 \]

### Final Answer
Thus, the largest area enclosed by the same wire when bent to form an equilateral triangle is approximately:

\[ \boxed{372.57 \text{ m}^2} \]