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 rectangle of length 12 cm.

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript.

### Step 1: Find the side length of the equilateral triangle.
Given the area of the equilateral triangle is \(49\sqrt{3} \, \text{cm}^2\), we can use the formula for the area of an equilateral triangle:

\[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \]

Setting this equal to the given area:

\[ \frac{\sqrt{3}}{4} a^2 = 49\sqrt{3} \]

### Step 2: Simplify the equation.
We can cancel \(\sqrt{3}\) from both sides:

\[ \frac{1}{4} a^2 = 49 \]

### Step 3: Solve for \(a^2\).
Multiply both sides by 4:

\[ a^2 = 49 \times 4 = 196 \]

### Step 4: Find the value of \(a\).
Taking the square root of both sides:

\[ a = \sqrt{196} = 14 \, \text{cm} \]

### 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: Set up the rectangle using the same wire.
The wire is now bent to form a rectangle with a length of \(12 \, \text{cm}\). Let the breadth be \(b\). The perimeter of the rectangle is given by:

\[ \text{Perimeter} = 2(\text{Length} + \text{Breadth}) = 2(12 + b) \]

### Step 7: Equate the perimeters.
Since the perimeter of the rectangle is equal to the length of the wire:

\[ 2(12 + b) = 42 \]

### Step 8: Solve for \(b\).
Dividing both sides by 2:

\[ 12 + b = 21 \]

Subtracting 12 from both sides:

\[ b = 21 - 12 = 9 \, \text{cm} \]

### Step 9: Calculate the area of the rectangle.
The area \(A\) of the rectangle is given by:

\[ A = \text{Length} \times \text{Breadth} = 12 \times 9 \]

Calculating the area:

\[ A = 108 \, \text{cm}^2 \]

### Final Answer:
The largest area enclosed by the wire when bent to form a rectangle of length \(12 \, \text{cm}\) is \(108 \, \text{cm}^2\).

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