The external fencing of a circular path around a circular plot of land is 33 m more than its inerior fencing. The width of the path around the plot is

Options

A. 5.52 m
B. 5.25 m
C. 2.55 m
D. 2.25 m

Answer

To solve the problem step by step, we need to find the width of the circular path around the circular plot of land.

### Step 1: Understand the Problem
We have two circles:
- The inner circle with radius \( r \) (the plot of land).
- The outer circle with radius \( R \) (the outer edge of the circular path).

The problem states that the external fencing (circumference of the outer circle) is 33 meters more than the internal fencing (circumference of the inner circle).

### Step 2: Write the Formulas for Circumference
The circumference of a circle is given by the formula:
\[ C = 2\pi r \]
For the inner circle (plot):
\[ C_{inner} = 2\pi r \]
For the outer circle (path):
\[ C_{outer} = 2\pi R \]

### Step 3: Set Up the Equation
According to the problem, the difference in the circumferences is 33 meters:
\[ C_{outer} - C_{inner} = 33 \]
Substituting the formulas for circumference:
\[ 2\pi R - 2\pi r = 33 \]

### Step 4: Factor Out Common Terms
We can factor out \( 2\pi \) from the left side:
\[ 2\pi (R - r) = 33 \]

### Step 5: Solve for the Width of the Path
The width of the path is defined as \( R - r \). We can isolate this term:
\[ R - r = \frac{33}{2\pi} \]

### Step 6: Substitute the Value of \( \pi \)
Using \( \pi \approx \frac{22}{7} \):
\[ R - r = \frac{33}{2 \times \frac{22}{7}} = \frac{33 \times 7}{44} = \frac{231}{44} \]

### Step 7: Simplify the Fraction
Now, we can simplify \( \frac{231}{44} \):
\[ \frac{231}{44} \approx 5.25 \]

### Conclusion
The width of the path around the plot is approximately 5.25 meters.

### Final Answer
The width of the path is **5.25 meters**.