To find the dimensions of the rectangle, we will use the given information about the perimeter and the relationship between the length and the width.

Let's denote the width of the rectangle as $w$.

Given:

1. The perimeter of the rectangle is 54 inches.
2. The length of the rectangle is one less than three times the width.

Let the length of the rectangle be $l$. According to the problem, we have: $l = 3w - 1$

The formula for the perimeter $P$ of a rectangle is given by: $P = 2l + 2w$

Substituting the given perimeter: $54 = 2l + 2w$

Now, substitute $l = 3w - 1$ into the perimeter equation: $54 = 2(3w - 1) + 2w$ $54 = 6w - 2 + 2w$ $54 = 8w - 2$ $54 + 2 = 8w$ $56 = 8w$ $w = 7$

Now, substitute $w = 7$ back into the equation for the length: $l = 3w - 1$ $l = 3(7) - 1$ $l = 21 - 1$ $l = 20$

So, the dimensions of the rectangle are:

• Width $w = 7$ inches
• Length $l = 20$ inches

Do you want details or have any questions?

Here are some related questions:

1. What is the area of the rectangle?
2. How would the perimeter change if the width was doubled?
3. What if the length was one more than three times the width?
4. Can you find the dimensions if the perimeter was 60 inches instead?
5. What are the dimensions if the width is decreased by 2 inches?
6. How does the relationship between length and width affect the perimeter?
7. What are the dimensions if the length is doubled while keeping the perimeter the same?
8. Can you generalize the solution for any given perimeter and similar length-width relationship?

Tip: Always verify your final answer by plugging the values back into the original equation to ensure they satisfy all given conditions.