Let `A_0,A_1,.......,A_(n-1)` be a n-sided polygon with vertices as `1,omega,omega^2,.......,omega^(n-1)` Let `B_1, B_1,..... B_(n -1)` be another polygon with vertices `1 , 1 + omega, 1 +omega^2,........,1+omega^(n-1)[omega=cos((2pi )/n)+i sin n((2pi)/n)] ` for `n=4,(A rdot(A_0, A_1, A_2, A_3))/(A r*(B_0, B_1, , B_3))` is `lambda` then

Options

A. `[lamda]gt3`
B. `(3lamda)/2 epsilon I^(+)`
C. `([lamda])/3 epsilon I^(+)`
D. `lamda` is irrational

Explanation

To solve the problem, we need to find the ratio of the areas of two polygons defined by their vertices. Let's break down the steps to find the solution.

### Step 1: Define the vertices of the polygons

1. **For the first polygon \(A_0, A_1, A_2, A_3\)**:
- The vertices are given by \(1, \omega, \omega^2, \omega^3\), where \(\omega = \cos\left(\frac{2\pi}{n}\right) + i \sin\left(\frac{2\pi}{n}\right)\).
- For \(n = 4\):
\[ \omega = \cos\left(\frac{2\pi}{4}\right) + i \sin\left(\frac{2\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right) = 0 + i(1) = i \]
- Thus, the vertices are:
- \(A_0 = 1\) (which is \((1, 0)\))
- \(A_1 = \omega = i\) (which is \((0, 1)\))
- \(A_2 = \omega^2 = -1\) (which is \((-1, 0)\))
- \(A_3 = \omega^3 = -i\) (which is \((0, -1)\))

2. **For the second polygon \(B_0, B_1, B_2, B_3\)**:
- The vertices are given by \(1, 1 + \omega, 1 + \omega^2, 1 + \omega^3\).
- Thus, the vertices are:
- \(B_0 = 1\) (which is \((1, 0)\))
- \(B_1 = 1 + \omega = 1 + i\) (which is \((1, 1)\))
- \(B_2 = 1 + \omega^2 = 1 - 1 = 0\) (which is \((0, 0)\))
- \(B_3 = 1 + \omega^3 = 1 - i\) (which is \((1, -1)\))

### Step 2: Calculate the area of polygon \(A\)

The area of a polygon can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| \sum_{i=0}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) \right| \]
where \((x_n, y_n) = (x_0, y_0)\).

For polygon \(A\):
- Vertices: \((1, 0), (0, 1), (-1, 0), (0, -1)\)

Calculating the area:
\[ \text{Area}_A = \frac{1}{2} \left| 1 \cdot 1 + 0 \cdot 0 + (-1) \cdot (-1) + 0 \cdot 0 - (0 \cdot 0 + 1 \cdot (-1) + 0 \cdot 0 + (-1) \cdot 1) \right| \]
\[ = \frac{1}{2} \left| 1 + 0 + 1 + 0 - (0 - 1 + 0 - 1) \right| \]
\[ = \frac{1}{2} \left| 2 + 2 \right| = \frac{1}{2} \cdot 4 = 2 \]

### Step 3: Calculate the area of polygon \(B\)

For polygon \(B\):
- Vertices: \((1, 0), (1, 1), (0, 0), (1, -1)\)

Calculating the area:
\[ \text{Area}_B = \frac{1}{2} \left| 1 \cdot 1 + 1 \cdot 0 + 0 \cdot (-1) + (-1) \cdot 0 - (0 \cdot 1 + 1 \cdot 0 + 0 \cdot 1 + (-1) \cdot 1) \right| \]
\[ = \frac{1}{2} \left| 1 + 0 + 0 + 0 - (0 + 0 + 0 - 1) \right| \]
\[ = \frac{1}{2} \left| 1 + 1 \right| = \frac{1}{2} \cdot 2 = 1 \]

### Step 4: Find the ratio of the areas

Now, we find the ratio of the areas:
\[ \lambda = \frac{\text{Area}_A}{\text{Area}_B} = \frac{2}{1} = 2 \]

### Conclusion

The value of \(\lambda\) is \(2\).