A man running round a race course note that the sum of the distances of two flag posts from him is 8 meters. The area of the path he encloses in square meters if the d is t a n c e between the flag posts is 4 is

Options

A. `15pi`
B. `12pi`
C. `18pi`
D. `8sqrt(3)pi`

Answer

To solve the problem step by step, we will analyze the information given and apply the properties of an ellipse.

### Step 1: Understand the Problem
We have two flag posts, and the sum of the distances from a man to these two flag posts is 8 meters. The distance between the flag posts is given as 4 meters.

### Step 2: Set Up the Ellipse
In the context of an ellipse, the sum of the distances from any point on the ellipse to the two foci (in this case, the flag posts) is constant. Here, the constant sum is 8 meters.

Let:
- The two flag posts be points A and B.
- The distance between the flag posts (AB) = 4 meters.
- The sum of the distances from the man to the flag posts (AC + BC) = 8 meters.

### Step 3: Identify the Semi-Major Axis (A)
In an ellipse, the sum of the distances from any point on the ellipse to the two foci is equal to 2A, where A is the semi-major axis. Therefore, we have:
\[ 2A = 8 \]
From this, we can find A:
\[ A = \frac{8}{2} = 4 \]

### Step 4: Identify the Distance Between the Foci (2C)
The distance between the two flag posts (foci) is given as 4 meters. Therefore:
\[ AB = 2C = 4 \]
From this, we can find C:
\[ C = \frac{4}{2} = 2 \]

### Step 5: Calculate the Semi-Minor Axis (B)
Using the relationship between A, B, and C in an ellipse:
\[ C^2 = A^2 - B^2 \]
Substituting the values we have:
\[ 2^2 = 4^2 - B^2 \]
\[ 4 = 16 - B^2 \]
Rearranging gives:
\[ B^2 = 16 - 4 = 12 \]
Taking the square root:
\[ B = \sqrt{12} = 2\sqrt{3} \]

### Step 6: Calculate the Area of the Ellipse
The area (A) of an ellipse is given by the formula:
\[ \text{Area} = \pi \times A \times B \]
Substituting the values of A and B:
\[ \text{Area} = \pi \times 4 \times 2\sqrt{3} \]
\[ \text{Area} = 8\sqrt{3}\pi \]

### Final Answer
The area of the path enclosed by the man is:
\[ \text{Area} = 8\sqrt{3}\pi \, \text{square meters} \]

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