A piece of wire of length 108 cm is bent to form a semicircular arc bounded by its diameter. Find its radius and enclosed area.

Answer

To solve the problem step by step, we will find the radius of the semicircle formed by the wire and then calculate the area enclosed by it.

### Step 1: Understand the problem
We have a piece of wire of length 108 cm that is bent to form a semicircular arc, which means the length of the wire represents the perimeter of the semicircle plus the diameter.

### Step 2: Set up the equation for the perimeter
The perimeter \( P \) of a semicircle can be expressed as:
\[ P = \text{Length of arc} + \text{Diameter} \]
The length of the arc (half the circumference of a full circle) is given by:
\[ \text{Length of arc} = \pi r \]
The diameter \( D \) is:
\[ D = 2r \]
Thus, the total length of the wire can be written as:
\[ P = \pi r + 2r \]
Given that the total length of the wire is 108 cm, we have:
\[ \pi r + 2r = 108 \]

### Step 3: Substitute the value of \( \pi \)
Using \( \pi \approx \frac{22}{7} \), we substitute:
\[ \frac{22}{7} r + 2r = 108 \]

### Step 4: Combine like terms
To combine the terms, we first express \( 2r \) with a common denominator:
\[ \frac{22}{7} r + \frac{14}{7} r = 108 \]
This simplifies to:
\[ \frac{36}{7} r = 108 \]

### Step 5: Solve for \( r \)
To find \( r \), multiply both sides by \( \frac{7}{36} \):
\[ r = 108 \times \frac{7}{36} \]
Calculating this gives:
\[ r = 3 \times 7 = 21 \text{ cm} \]

### Step 6: Calculate the area of the semicircle
The area \( A \) of a semicircle is given by:
\[ A = \frac{1}{2} \pi r^2 \]
Substituting the values of \( \pi \) and \( r \):
\[ A = \frac{1}{2} \times \frac{22}{7} \times (21)^2 \]
Calculating \( (21)^2 \):
\[ (21)^2 = 441 \]
Thus, the area becomes:
\[ A = \frac{1}{2} \times \frac{22}{7} \times 441 \]
Calculating further:
\[ A = \frac{22 \times 441}{14} = \frac{9702}{14} = 693 \text{ cm}^2 \]

### Final Answer
The radius of the semicircle is **21 cm** and the enclosed area is **693 cm²**.

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