Question
Fairfield has one indoor skating ring and one outdoor skating rink that cover the same amount of total area. Skaters do laps around the perimeter to warm up. on which rink is the lap longer. The outdoor rink is a square and the indoor rink is a rectangle. The outdoor or indoor rink is closer to a square so its perimeter is the same longer, shorter. One lap is longer at the indoor, outdoor rink.
Answers
GPT-4o mini
To determine which rink has a longer lap, we need to compare the perimeters of the two shapes: a square (the outdoor rink) and a rectangle (the indoor rink), given that both rinks cover the same total area.
1. **Outdoor Rink (Square)**:
- Let the side length of the square rink be \( s \).
- The area of the square is \( s^2 \).
- The perimeter of the square is \( 4s \).
2. **Indoor Rink (Rectangle)**:
- Let the dimensions of the rectangle be \( l \) (length) and \( w \) (width).
- The area of the rectangle is \( l \times w \).
- Given that both areas are equal, we have \( s^2 = l \times w \).
- The perimeter of the rectangle is \( 2l + 2w \).
Since we know \( s^2 = lw \), we can express \( w \) in terms of \( s \) and \( l \):
\[
w = \frac{s^2}{l}
\]
Now we can express the perimeter of the rectangle:
\[
\text{Perimeter of the rectangle} = 2l + 2w = 2l + 2\left(\frac{s^2}{l}\right) = 2l + \frac{2s^2}{l}
\]
To compare the two perimeters, the perimeter of the square is \( 4s \). The challenge now is to show if the perimeter of the rectangle is longer, shorter, or the same.
To find when the rectangle has the same perimeter as the square, we can set equal and solve:
\[
4s = 2l + \frac{2s^2}{l}
\]
This equality doesn't hold for all rectangles, as it depends on the values of \( l \) and \( w \). However, generally, a square has the smallest perimeter for a given area compared to other rectangles (this is due to the isoperimetric inequality).
Since the square is the most efficient shape in terms of perimeter for a given area, and rectangles that are not squares will have a longer perimeter when the area is constant, the outdoor rink (square) has a shorter lap compared to the indoor rink (rectangle).
**Thus, the lap is longer at the indoor rink.**
1. **Outdoor Rink (Square)**:
- Let the side length of the square rink be \( s \).
- The area of the square is \( s^2 \).
- The perimeter of the square is \( 4s \).
2. **Indoor Rink (Rectangle)**:
- Let the dimensions of the rectangle be \( l \) (length) and \( w \) (width).
- The area of the rectangle is \( l \times w \).
- Given that both areas are equal, we have \( s^2 = l \times w \).
- The perimeter of the rectangle is \( 2l + 2w \).
Since we know \( s^2 = lw \), we can express \( w \) in terms of \( s \) and \( l \):
\[
w = \frac{s^2}{l}
\]
Now we can express the perimeter of the rectangle:
\[
\text{Perimeter of the rectangle} = 2l + 2w = 2l + 2\left(\frac{s^2}{l}\right) = 2l + \frac{2s^2}{l}
\]
To compare the two perimeters, the perimeter of the square is \( 4s \). The challenge now is to show if the perimeter of the rectangle is longer, shorter, or the same.
To find when the rectangle has the same perimeter as the square, we can set equal and solve:
\[
4s = 2l + \frac{2s^2}{l}
\]
This equality doesn't hold for all rectangles, as it depends on the values of \( l \) and \( w \). However, generally, a square has the smallest perimeter for a given area compared to other rectangles (this is due to the isoperimetric inequality).
Since the square is the most efficient shape in terms of perimeter for a given area, and rectangles that are not squares will have a longer perimeter when the area is constant, the outdoor rink (square) has a shorter lap compared to the indoor rink (rectangle).
**Thus, the lap is longer at the indoor rink.**