A copper wire is bent in the form of an equilateral triangle and has area `121sqrt(3) cm^(2)`. IF the same wire is bent into the form of a circle, the area (in `cm^(2)`) enclosed by the wire is (Take `pi = (22)/(7)`)

Options

A. `364.5`
B. `693.5`
C. `346.5`
D. `639.5`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Find the side length of the equilateral triangle.

The area \( A \) of an equilateral triangle is given by the formula:
\[ A = \frac{\sqrt{3}}{4} a^2 \]
where \( a \) is the length of a side of the triangle.

Given that the area of the triangle is \( 121\sqrt{3} \, \text{cm}^2 \), we can set up the equation:
\[ \frac{\sqrt{3}}{4} a^2 = 121\sqrt{3} \]

### Step 2: Simplify the equation.

To eliminate \( \sqrt{3} \) from both sides, we divide both sides by \( \sqrt{3} \):
\[ \frac{1}{4} a^2 = 121 \]

Next, multiply both sides by 4:
\[ a^2 = 484 \]

### Step 3: Solve for \( a \).

Taking the square root of both sides gives:
\[ a = \sqrt{484} = 22 \, \text{cm} \]

### Step 4: Calculate the perimeter of the triangle.

The perimeter \( P \) of an equilateral triangle is given by:
\[ P = 3a \]
Substituting the value of \( a \):
\[ P = 3 \times 22 = 66 \, \text{cm} \]

### Step 5: Find the radius of the circle.

When the wire is bent into the form of a circle, the perimeter of the triangle becomes the circumference of the circle. The circumference \( C \) of a circle is given by:
\[ C = 2\pi r \]
Setting the circumference equal to the perimeter:
\[ 2\pi r = 66 \]

### Step 6: Solve for \( r \).

Substituting \( \pi = \frac{22}{7} \):
\[ 2 \times \frac{22}{7} \times r = 66 \]
Simplifying this gives:
\[ \frac{44}{7} r = 66 \]
Multiplying both sides by \( \frac{7}{44} \):
\[ r = 66 \times \frac{7}{44} = \frac{462}{44} = \frac{231}{22} = 10.5 \, \text{cm} \]

### Step 7: Calculate the area of the circle.

The area \( A \) of a circle is given by:
\[ A = \pi r^2 \]
Substituting the values:
\[ A = \frac{22}{7} \times (10.5)^2 \]
Calculating \( (10.5)^2 \):
\[ (10.5)^2 = 110.25 \]
Thus, substituting back:
\[ A = \frac{22}{7} \times 110.25 \]
Calculating further:
\[ A = \frac{22 \times 110.25}{7} = \frac{2425.5}{7} = 346.5 \, \text{cm}^2 \]

### Final Answer:
The area enclosed by the wire when bent into the form of a circle is \( 346.5 \, \text{cm}^2 \).

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