The length of a rectangular poster is 1 ft more than the width, and a diagonal of the poster is 5 ft. Find the length and the width.

We can solve this problem using the Pythagorean theorem, which relates the sides of a right triangle. In this case, the length, width, and diagonal of the rectangular poster form a right triangle.

Let:
- \( w \) represent the width of the poster (in feet).
- \( l \) represent the length of the poster (in feet).

According to the problem:
1. The length is 1 foot more than the width, so:
   \[
   l = w + 1
   \]
2. The diagonal is given as 5 feet. By the Pythagorean theorem:
   \[
   l^2 + w^2 = 5^2
   \]
   Substituting \( l = w + 1 \) into the equation:
   \[
   (w + 1)^2 + w^2 = 25
   \]

Now, let's solve this equation step by step:

\[
(w + 1)^2 = w^2 + 2w + 1
\]
Substituting this back:
\[
w^2 + 2w + 1 + w^2 = 25
\]
\[
2w^2 + 2w + 1 = 25
\]
Subtract 25 from both sides:
\[
2w^2 + 2w - 24 = 0
\]
Divide through by 2 to simplify:
\[
w^2 + w - 12 = 0
\]

Now, we solve this quadratic equation using the quadratic formula:
\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For the equation \( w^2 + w - 12 = 0 \), we have \( a = 1 \), \( b = 1 \), and \( c = -12 \). Substituting these into the quadratic formula:
\[
w = \frac{-1 \pm \sqrt{1^2 - 4(1)(-12)}}{2(1)}
\]
\[
w = \frac{-1 \pm \sqrt{1 + 48}}{2}
\]
\[
w = \frac{-1 \pm \sqrt{49}}{2}
\]
\[
w = \frac{-1 \pm 7}{2}
\]

So, we have two possible solutions:
\[
w = \frac{-1 + 7}{2} = 3 \quad \text{or} \quad w = \frac{-1 - 7}{2} = -4
\]

Since the width can't be negative, we take \( w = 3 \) feet.

Now, using the relationship \( l = w + 1 \):
\[
l = 3 + 1 = 4 \, \text{feet}
\]

Thus, the dimensions of the poster are:
- Width: \( 3 \) feet
- Length: \( 4 \) feet

Would you like any further explanation or details on how these steps were derived?

### Related Questions:
1. How would the dimensions change if the diagonal were 6 feet instead of 5?
2. What is the area of the poster based on the dimensions we found?
3. How would you set up a similar problem if the diagonal was not provided, but the perimeter was?
4. Could you find the dimensions if the width was 2 feet more than the length instead?
5. How does the Pythagorean theorem help in determining unknown sides in general?

### Tip:
When solving quadratic equations, always check for any physical constraints (like dimensions, which cannot be negative) before selecting the appropriate solution.

Learn how to find the length and width of a rectangular poster where the length is 1 foot more than the width, and the diagonal is 5 feet. This detailed step-by-step solution uses the Pythagorean theorem and quadratic equations. Suitable for students in Grades 8-10.

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