The problem describes a swimming pool drawing with dimensions of 5 inches and 2 inches, with a scale of 1 inch = 15 feet in real life. We are asked to find the perimeter of the actual swimming pool.

Step-by-Step Solution:

1. Scale Conversion:

   • The dimensions in the drawing are 5 inches and 2 inches.
   • Using the scale 1 inch = 15 feet, the actual dimensions of the swimming pool will be:
     • Length: $5 \, \text{inches} \times 15 = 75 \, \text{feet}$
     • Width: $2 \, \text{inches} \times 15 = 30 \, \text{feet}$

2. Perimeter of a Rectangle Formula: The perimeter $P$ of a rectangle is given by: $P = 2 \times (\text{Length} + \text{Width})$

3. Calculate the Perimeter: Using the actual dimensions: $P = 2 \times (75 \, \text{feet} + 30 \, \text{feet}) = 2 \times 105 = 210 \, \text{feet}$

Thus, the perimeter of the actual rectangular swimming pool is 210 feet.

Would you like further details or have any questions? Here are some related problems to expand on this:

1. What would the area of the swimming pool be in square feet?
2. How would the perimeter change if the scale was 1 inch = 10 feet?
3. What would be the dimensions in real life if the drawing was enlarged to 10 inches by 4 inches?
4. How do we calculate the diagonal length of the swimming pool?
5. Can we determine the swimming pool's perimeter if the shape were irregular?

Tip: Always double-check that the correct units (feet, inches, etc.) are used when performing calculations with scale models.