To solve this problem, we need to:

Step 1: Define the variables

Let:

• $w$ = the width of the rectangle (in yards),
• $l$ = the length of the rectangle.

We are told that: $l = 2w - 5$ and the area of the rectangle is given as: $\text{Area} = l \times w = 273 \, \text{yd}^2.$

Step 2: Substitute $l = 2w - 5$ into the area equation

$(2w - 5) \cdot w = 273.$

Step 3: Expand and simplify

$2w^2 - 5w = 273.$

Rearrange into standard quadratic form: $2w^2 - 5w - 273 = 0.$

Step 4: Solve the quadratic equation

We use the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 2$, $b = -5$, and $c = -273$.

Substitute into the formula: $w = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-273)}}{2(2)}.$

Simplify: $w = \frac{5 \pm \sqrt{25 + 2184}}{4}.$ $w = \frac{5 \pm \sqrt{2209}}{4}.$ $w = \frac{5 \pm 47}{4}.$

Solve for the two possible values of $w$:

1. $w = \frac{5 + 47}{4} = \frac{52}{4} = 13$,
2. $w = \frac{5 - 47}{4} = \frac{-42}{4} = -10.5$ (not possible, as width cannot be negative).

So, $w = 13 \, \text{yd}$.

Step 5: Find the length

Substitute $w = 13$ into $l = 2w - 5$: $l = 2(13) - 5 = 26 - 5 = 21 \, \text{yd}.$

Step 6: Calculate the perimeter

The perimeter of a rectangle is: $P = 2(l + w).$ Substitute $l = 21$ and $w = 13$: $P = 2(21 + 13) = 2(34) = 68 \, \text{yd}.$

Final Answer:

The perimeter is 68 yards.

Would you like a detailed explanation of any step or have further questions?

Here are some related questions for further exploration:

1. How do you solve quadratic equations by factoring?
2. What are other methods to calculate the perimeter of shapes?
3. Can the quadratic formula be used for all quadratic equations?
4. How do changes in area affect the perimeter of a rectangle?
5. What is the relationship between length and width in similar problems?

Tip: Always verify your solutions by substituting back into the original equations!