The problem states:

"A rectangular athletic field is twice as long as it is wide. If the perimeter of the athletic field is 240 yards, what are its dimensions?"

Step 1: Define Variables

Let:

• $w$ be the width of the field in yards.
• $l$ be the length of the field in yards.

According to the problem:

• The length is twice the width: $l = 2w$.

Step 2: Perimeter Formula

The perimeter of a rectangle is given by: $P = 2l + 2w$ The perimeter is 240 yards, so: $240 = 2l + 2w$

Step 3: Substitute the Expression for $l$

Since $l = 2w$, substitute this into the perimeter equation: $240 = 2(2w) + 2w$ Simplify the equation: $240 = 4w + 2w$ $240 = 6w$

Step 4: Solve for $w$

$w = \frac{240}{6} = 40$

Step 5: Find $l$

Since $l = 2w$: $l = 2(40) = 80$

Final Answer

• The width of the field is 40 yards.
• The length of the field is 80 yards.

Would you like more details or have any questions?

Related Questions:

1. How would the dimensions change if the perimeter was 300 yards instead?
2. What is the area of the athletic field given these dimensions?
3. If the field was three times as long as it is wide, how would that affect the dimensions?
4. How would you solve this problem if only the area of the field was given instead of the perimeter?
5. Can you derive a general formula for the width and length given any perimeter?

Tip:

Always start by defining variables and writing equations based on the problem statement. This makes it easier to identify relationships and solve for unknowns.