An architect wants to draw a rectangle with a diagonal of 1313 inches. The length of the rectangle is to be 22 inches more than twice the width. What dimensions should she make the rectangle? 




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.

An architect wants to draw a rectangle with a diagonal of 1313 inches. The length of the rectangle is to be 22 inches more than twice the width. What dimensions should she make the rectangle?

Learn how to determine the dimensions of a rectangle with a diagonal of 1313 inches. The length is 22 inches more than twice the width, and the solution involves using the Pythagorean theorem and solving a quadratic equation. Suitable for students in Grades 10-12.

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