The swimming pool at Spring Valley High School is a rectangle with a width of 65 meters and a length of 30 meters. Around the perimeter of the pool is a tiled floor that
extends w meters from the pool on all sides. Find an expression for the area of the tiled
floor.

The area of the tiled floor is given by:

area =  (190m)*w + 4*w^2

Step-by-step explanation:

Remember that the area of a rectangle of length L and width W is equal to:

A = W*L

For the case of the problem, we want to find the area of the tiled floor.

Then we need to find the difference between the area of the pool including the tiled floor and the area of the pool alone.

We know that the length of the tiled floor is w.

Then if the length of the pool is 30m, the length of the pool including the tiled floor is 30m + 2*w (because we have tiles in both sides)

And if the width of the pool is 65m, then the width of the pool including the tiled floor is 65m + 2*w

Now we want to find the difference between these areas, we have:

Area of the pool alone = 65m*30m = 1950m^2

Area of the pool including the tiles = (65m + 2*w)*(30m + 2*w)

The difference is:

D =  (65m + 2*w)*(30m + 2*w) - 1950m^2

D = 65m*30m + 65m*2w + 30m*2w + 4*w^2 - 1950m^2

D = (65m + 30m)*2w + 4*w^2 + (65m*30m -  1950m^2)

D = 95m*2w + 4*w^2

D = (190m)*w + 4*w^2

This is the area of the tiled floor.