Let the width of the rectangle be denoted by $w$, and the length of the rectangle be $l$.

According to the problem:

• The length is 22 inches more than twice the width:
  $l = 2w + 22$

We also know that the diagonal of the rectangle is given as 1313 inches. Using the Pythagorean theorem, we can relate the width, length, and diagonal as follows: $l^2 + w^2 = 1313^2$

Now, substitute $l = 2w + 22$ into the equation: $(2w + 22)^2 + w^2 = 1313^2$

We will now solve this equation to find the value of $w$, and subsequently calculate $l$.

First, expand $(2w + 22)^2$: $(2w + 22)^2 = 4w^2 + 88w + 484$

Now substitute back into the equation: $4w^2 + 88w + 484 + w^2 = 1313^2$

Simplify the equation: $5w^2 + 88w + 484 = 1723169$

Now, subtract 484 from both sides: $5w^2 + 88w = 1722685$

At this point, we can solve this quadratic equation. Let me calculate the value of $w$.The two solutions for the width $w$ are:

$w = \frac{-44 + \sqrt{8615361}}{5} \quad \text{or} \quad w = \frac{-44 - \sqrt{8615361}}{5}$

Since the width must be a positive value, we take the positive solution: $w = \frac{-44 + \sqrt{8615361}}{5}$

Let me calculate this value for $w$.The width $w$ of the rectangle is approximately $578.24$ inches.

Now, using the relation $l = 2w + 22$, let's calculate the length $l$.The length $l$ of the rectangle is approximately $1178.48$ inches.

Thus, the dimensions of the rectangle should be:

• Width: $578.24$ inches
• Length: $1178.48$ inches

Would you like further details or clarification?

Here are 5 related questions you might find interesting:

1. How does the Pythagorean theorem apply to non-right triangles?
2. Can you derive the quadratic formula from the standard form of a quadratic equation?
3. How would the problem change if the diagonal was a different length?
4. What if the length was described as 10 inches more than three times the width?
5. How would you calculate the perimeter of this rectangle?

Tip: Always ensure that the values you obtain in a word problem are reasonable based on the context before finalizing your answer.