The ratio of semiperimeter of a parallelogram to its longer side is same as the ratio of its longer side to its shorter side. If the shorter side measures 2 cm, find the maximum possible area of the parallelogram.

Options

A. `2(sqrt5+2)`
B. `2(1+sqrt5)`
C. `4(1+sqrt5)`
D. data insufficient

Answer

To solve the problem, we need to find the maximum possible area of a parallelogram given the conditions stated. Let's break it down step by step.

### Step 1: Define the variables
Let:
- \( L \) = longer side of the parallelogram
- \( B \) = shorter side of the parallelogram (given as \( B = 2 \) cm)

### Step 2: Calculate the semiperimeter
The semiperimeter \( S \) of a parallelogram is given by the formula:
\[ S = \frac{L + B + L + B}{2} = L + B \]
Substituting \( B = 2 \):
\[ S = L + 2 \]

### Step 3: Set up the ratio condition
According to the problem, the ratio of the semiperimeter to the longer side is equal to the ratio of the longer side to the shorter side:
\[ \frac{S}{L} = \frac{L}{B} \]
Substituting the expressions we have:
\[ \frac{L + 2}{L} = \frac{L}{2} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ (L + 2) \cdot 2 = L \cdot L \]
This simplifies to:
\[ 2L + 4 = L^2 \]

### Step 5: Rearrange the equation
Rearranging the equation gives us:
\[ L^2 - 2L - 4 = 0 \]

### Step 6: Solve the quadratic equation
We can solve the quadratic equation using the quadratic formula:
\[ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \), and \( c = -4 \):
\[ L = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ L = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5} \]

### Step 7: Choose the positive value for length
Since length cannot be negative, we take:
\[ L = 1 + \sqrt{5} \]

### Step 8: Calculate the area of the parallelogram
The area \( A \) of the parallelogram can be calculated using the formula:
\[ A = L \times B \]
Substituting the values we found:
\[ A = (1 + \sqrt{5}) \times 2 = 2(1 + \sqrt{5}) = 2 + 2\sqrt{5} \]

### Conclusion
Thus, the maximum possible area of the parallelogram is:
\[ \boxed{2 + 2\sqrt{5}} \text{ cm}^2 \]