Rectangle ABCD has area 200. An ellipse with area `200pi` passes through A and C and has foci at B and D. If the perimeter of the rectangle is P. Then the value of `(P)/(20)` is

Options

A. 2
B. 4
C. 8
D. None of these

Explanation

To solve the problem, we need to find the perimeter \( P \) of rectangle \( ABCD \) given the area and the properties of the ellipse.

### Step-by-Step Solution:

1. **Understanding the Rectangle and Its Area**:
- Let the length of rectangle \( ABCD \) be \( L \) and the width be \( W \).
- The area of the rectangle is given by:
\[ L \times W = 200 \]

2. **Ellipse Properties**:
- The area of the ellipse is given by:
\[ \text{Area of ellipse} = \pi \times a \times b = 200\pi \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.
- From this, we can deduce:
\[ a \times b = 200 \]

3. **Foci of the Ellipse**:
- The foci of the ellipse are located at points \( B \) and \( D \).
- The distance between the foci \( 2c \) is given by:
\[ c = \sqrt{a^2 - b^2} \]

4. **Relating Rectangle Dimensions to Ellipse**:
- The distance \( BD \) (the distance between the foci) can be expressed in terms of the rectangle's dimensions:
\[ BD = \sqrt{L^2 + W^2} \]
- Since \( B \) and \( D \) are the foci, we have:
\[ 2c = \sqrt{L^2 + W^2} \]

5. **Using the Area of the Rectangle**:
- From the area equation \( L \times W = 200 \), we can express \( W \) in terms of \( L \):
\[ W = \frac{200}{L} \]

6. **Substituting into the Distance Equation**:
- Substitute \( W \) into the equation for \( BD \):
\[ 2c = \sqrt{L^2 + \left(\frac{200}{L}\right)^2} \]
- This simplifies to:
\[ 2c = \sqrt{L^2 + \frac{40000}{L^2}} \]

7. **Finding the Value of \( c \)**:
- We know from the ellipse properties that:
\[ c = \sqrt{a^2 - b^2} \]
- Using the area relation \( a \times b = 200 \) and substituting \( a = \frac{200}{b} \):
\[ c = \sqrt{\left(\frac{200}{b}\right)^2 - b^2} \]

8. **Perimeter of the Rectangle**:
- The perimeter \( P \) of the rectangle is given by:
\[ P = 2(L + W) = 2\left(L + \frac{200}{L}\right) \]

9. **Finding \( P/20 \)**:
- We need to find \( \frac{P}{20} \):
\[ \frac{P}{20} = \frac{2\left(L + \frac{200}{L}\right)}{20} = \frac{L + \frac{200}{L}}{10} \]

10. **Final Calculation**:
- To find the specific value of \( \frac{P}{20} \), we can assume \( L = 20 \) and \( W = 10 \) (since \( 20 \times 10 = 200 \)):
\[ P = 2(20 + 10) = 60 \]
\[ \frac{P}{20} = \frac{60}{20} = 3 \]

### Final Answer:
\[ \frac{P}{20} = 3 \]