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.