A wire is bent in the form of an equilateral triangle of largest area. If it encloses an area of `49sqrt(3)` `cm^(2)`, find the largest area enclosed by the same wire when bent to form :
a square

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the area of the equilateral triangle
We know that the area \( A \) of an equilateral triangle can be expressed as:
\[ A = \frac{\sqrt{3}}{4} a^2 \]
where \( a \) is the length of a side of the triangle.

### Step 2: Set up the equation using the given area
We are given that the area of the equilateral triangle is \( 49\sqrt{3} \) cm². Therefore, we can set up the equation:
\[ \frac{\sqrt{3}}{4} a^2 = 49\sqrt{3} \]

### Step 3: Simplify the equation
To simplify, we can divide both sides by \( \sqrt{3} \):
\[ \frac{1}{4} a^2 = 49 \]
Now, multiply both sides by 4:
\[ a^2 = 196 \]

### Step 4: Solve for \( a \)
Taking the square root of both sides gives us:
\[ a = \sqrt{196} = 14 \text{ cm} \]

### Step 5: Calculate the perimeter of the equilateral triangle
The perimeter \( P \) of an equilateral triangle is given by:
\[ P = 3a \]
Substituting the value of \( a \):
\[ P = 3 \times 14 = 42 \text{ cm} \]

### Step 6: Use the wire to form a square
The same wire is now bent to form a square. The perimeter of the square is equal to the length of the wire:
\[ \text{Perimeter of square} = 42 \text{ cm} \]

### Step 7: Find the side length of the square
The perimeter \( P \) of a square is given by:
\[ P = 4s \]
where \( s \) is the length of a side of the square. Setting the perimeter equal to 42 cm:
\[ 4s = 42 \]
Dividing both sides by 4:
\[ s = \frac{42}{4} = 10.5 \text{ cm} \]

### Step 8: Calculate the area of the square
The area \( A \) of a square is given by:
\[ A = s^2 \]
Substituting the value of \( s \):
\[ A = (10.5)^2 = 110.25 \text{ cm}^2 \]

### Final Answer
The largest area enclosed by the same wire when bent to form a square is:
\[ \boxed{110.25 \text{ cm}^2} \]

---