A wire 34 cm long is to he bent in the form of a quadrilateral of which each angle is 90°. What is the maximum area which can be enclosed inside the quadrilateral.

Options

A. `68 cm^(2)`
B. `70 cm^(2)`
C. `71.25 cm^(2)`
D. `72.25 cm^(2)`

Answer

To find the maximum area that can be enclosed by a quadrilateral formed by bending a wire of length 34 cm, we can follow these steps:

### Step 1: Understand the Problem
We have a wire of length 34 cm that is to be bent into a quadrilateral with each angle being 90°. The quadrilateral can be a rectangle or a square since all angles are right angles.

### Step 2: Set Up the Equation for Perimeter
Let’s denote the lengths of the sides of the quadrilateral as \( x \) and \( y \). Since it is a quadrilateral with right angles, the perimeter \( P \) can be expressed as:
\[ P = 2x + 2y = 34 \]
This simplifies to:
\[ x + y = 17 \quad \text{(1)} \]

### Step 3: Express Area in Terms of One Variable
The area \( A \) of the quadrilateral (rectangle) can be expressed as:
\[ A = x \cdot y \]
From equation (1), we can express \( y \) in terms of \( x \):
\[ y = 17 - x \]
Substituting this into the area formula gives:
\[ A = x(17 - x) = 17x - x^2 \quad \text{(2)} \]

### Step 4: Find the Maximum Area
To find the maximum area, we need to differentiate the area function \( A \) with respect to \( x \) and set the derivative to zero:
\[ \frac{dA}{dx} = 17 - 2x \]
Setting the derivative equal to zero:
\[ 17 - 2x = 0 \]
Solving for \( x \):
\[ 2x = 17 \implies x = \frac{17}{2} = 8.5 \]

### Step 5: Find the Corresponding Value of \( y \)
Using equation (1) to find \( y \):
\[ y = 17 - x = 17 - 8.5 = 8.5 \]

### Step 6: Calculate the Maximum Area
Now, substituting \( x \) and \( y \) back into the area formula:
\[ A = x \cdot y = 8.5 \cdot 8.5 = 72.25 \, \text{cm}^2 \]

### Conclusion
The maximum area that can be enclosed by the quadrilateral formed by the wire is:
\[ \boxed{72.25 \, \text{cm}^2} \]