A man running round a race course notes that the sum of the distance of two flag - posts from him is always 10 meters and distance between the flag - posts is 8 meters. The area of the path he encloses in square meters is (1) \(15 \pi\) (2) \(12 \pi\) (3) \(18 \pi\) (4) \(8 \pi\)

An ellipse is a fascinating geometric shape, looking like an elongated circle. When you're dealing with ellipses, there are two focal points (or foci). Each point on the ellipse maintains a constant total distance to these two foci. In the given exercise, the sum of these distances is 10 meters, which is key to understanding the shape and size of the ellipse.
For more clarity, think of the racecourse the man runs as being an ellipse. The two flag-posts are the foci. The sum of the distances from the man’s position to each flag-post is 10 meters. This characteristic property of ellipses helps us define and calculate other essential measurements such as the major axis.

Distance formulas are crucial for geometric calculations. For the ellipse, there are two important distances - the major axis and the distance between the foci.
The length of the major axis is the maximum distance across the ellipse and in this exercise, it is 10 meters since this is the constant sum of the distances from any point on the ellipse to the foci.
To find the focal distance (distance between the two foci), we use the property that this distance equals 2c. Given that this distance is 8 meters, we solve for c (4 meters).
Another important equation is the relationship between the semi-major axis (a), semi-minor axis (b), and the distance to the foci (c): \( a^2 = b^2 + c^2 \). This helps us to find b when we know a and c.

Calculating the area of an ellipse combines understanding its geometry and using specific formulas. The area (A) of an ellipse is given by the formula \( A = \pi ab \), where a is the semi-major axis and b is the semi-minor axis.
We already determined that the semi-major axis (a) is 5 meters and, using the relationship \( a^2 = b^2 + c^2 \), we found that the semi-minor axis (b) is 3 meters. Plugging these values into the area formula gives:
\[ A = \pi \times 5 \times 3 = 15\pi \text{ square meters} \]
This calculation provides the area enclosed by the man's path on the racecourse, confirming that the correct answer is \( 15\pi \) square meters.