A piece of wire is bent in the shape of an equilateral triangle each of whose sides measures8 cm. This wire is rebent to form a circular ring. What is the diameter of the ring?

Answer

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

### Step 1: Calculate the perimeter of the equilateral triangle.
The formula for the perimeter \( P \) of an equilateral triangle is given by:
\[ P = 3 \times \text{side} \]
Given that each side of the triangle measures 8 cm, we can substitute this value into the formula:
\[ P = 3 \times 8 = 24 \text{ cm} \]

### Step 2: Understand that the length of the wire remains the same.
Since the wire is bent into the shape of an equilateral triangle and then re-bent into a circular ring, the length of the wire (which is the perimeter of the triangle) will be equal to the circumference of the circle.

### Step 3: Set the perimeter of the triangle equal to the circumference of the circle.
The circumference \( C \) of a circle is given by the formula:
\[ C = 2\pi r \]
Where \( r \) is the radius of the circle. Since we found the perimeter of the triangle to be 24 cm, we can set this equal to the circumference:
\[ 24 = 2\pi r \]

### Step 4: Solve for the radius \( r \).
To find the radius, we will rearrange the equation:
\[ r = \frac{24}{2\pi} \]
Substituting \( \pi \) with \( \frac{22}{7} \):
\[ r = \frac{24}{2 \times \frac{22}{7}} = \frac{24 \times 7}{44} = \frac{168}{44} = \frac{42}{11} \approx 3.818 \text{ cm} \]

### Step 5: Calculate the diameter of the circle.
The diameter \( d \) of the circle is twice the radius:
\[ d = 2r = 2 \times \frac{42}{11} = \frac{84}{11} \approx 7.636 \text{ cm} \]

### Final Answer:
The diameter of the circular ring is approximately \( 7.64 \) cm.

---