A piece of wire of length `4L` is bent at random, to form a rectangle. The probability that its area is at most `L^2/4` is :

Options

A. `1//4`
B. `(sqrt3)/(2)`
C. `(2 - sqrt3)/(2)`
D. `4 - sqrt3`

Answer

To solve the problem, we need to determine the probability that the area of a rectangle formed by bending a wire of length \(4L\) is at most \(\frac{L^2}{4}\).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The wire is bent to form a rectangle with length \(x\) and breadth \(y\).
- The perimeter of the rectangle is given by \(2(x + y) = 4L\), which simplifies to \(x + y = 2L\).

2. **Expressing Breadth in Terms of Length**:
- From \(x + y = 2L\), we can express \(y\) as:
\[ y = 2L - x \]

3. **Calculating the Area of the Rectangle**:
- The area \(A\) of the rectangle is given by:
\[ A = x \cdot y = x(2L - x) = 2Lx - x^2 \]

4. **Setting Up the Inequality**:
- We want to find the values of \(x\) for which the area is at most \(\frac{L^2}{4}\):
\[ 2Lx - x^2 \leq \frac{L^2}{4} \]

5. **Rearranging the Inequality**:
- Rearranging gives:
\[ x^2 - 2Lx + \frac{L^2}{4} \geq 0 \]

6. **Factoring the Quadratic**:
- This can be rewritten as:
\[ \left(x - L\right)^2 - \frac{3L^2}{4} \geq 0 \]
- This is a difference of squares:
\[ \left(x - L - \frac{\sqrt{3}}{2}L\right)\left(x - L + \frac{\sqrt{3}}{2}L\right) \geq 0 \]

7. **Finding the Critical Points**:
- The critical points are:
\[ x = L - \frac{\sqrt{3}}{2}L \quad \text{and} \quad x = L + \frac{\sqrt{3}}{2}L \]
- Simplifying gives:
\[ x = L(1 - \frac{\sqrt{3}}{2}) \quad \text{and} \quad x = L(1 + \frac{\sqrt{3}}{2}) \]

8. **Identifying the Valid Range for \(x\)**:
- The valid range for \(x\) is:
\[ 0 \leq x \leq 2L \]
- The area condition gives two intervals:
\[ x \in \left[0, L(1 - \frac{\sqrt{3}}{2})\right] \cup \left[L(1 + \frac{\sqrt{3}}{2}), 2L\right] \]

9. **Calculating the Length of the Intervals**:
- The length of the first interval:
\[ L(1 - \frac{\sqrt{3}}{2}) - 0 = L(1 - \frac{\sqrt{3}}{2}) \]
- The length of the second interval:
\[ 2L - L(1 + \frac{\sqrt{3}}{2}) = L(2 - 1 - \frac{\sqrt{3}}{2}) = L(1 - \frac{\sqrt{3}}{2}) \]
- Total length of valid intervals:
\[ 2L(1 - \frac{\sqrt{3}}{2}) \]

10. **Finding the Total Length of Possible Values for \(x\)**:
- The total length for \(x\) is:
\[ 2L \]

11. **Calculating the Probability**:
- The probability \(P\) that the area is at most \(\frac{L^2}{4}\) is given by:
\[ P = \frac{\text{Length of valid intervals}}{\text{Total length}} = \frac{2L(1 - \frac{\sqrt{3}}{2})}{2L} = 1 - \frac{\sqrt{3}}{2} \]

### Final Answer:
Thus, the probability that the area of the rectangle is at most \(\frac{L^2}{4}\) is:
\[ \boxed{1 - \frac{\sqrt{3}}{2}} \]