To find the dimensions of the rectangular window, we can set up an equation based on the given information.

Let's denote:

• The width of the window as $w$ feet.
• The length of the window as $l = w + 2$ feet (since the length is two feet more than the width).

The area $A$ of a rectangle is given by the formula: $A = \text{length} \times \text{width}$ Given that the area is 143 square feet, we have: $w \times (w + 2) = 143$

This is a quadratic equation. Expanding and rearranging it gives: $w^2 + 2w - 143 = 0$

We can solve this quadratic equation using the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 2$, and $c = -143$.

Substituting the values: $w = \frac{-2 \pm \sqrt{2^2 + 4(1)(143)}}{2(1)}$ $w = \frac{-2 \pm \sqrt{4 + 572}}{2}$ $w = \frac{-2 \pm \sqrt{576}}{2}$ $w = \frac{-2 \pm 24}{2}$

This gives us two possible solutions: $w = \frac{-2 + 24}{2} = \frac{22}{2} = 11 \quad \text{(valid dimension)}$ $w = \frac{-2 - 24}{2} = \frac{-26}{2} = -13 \quad \text{(not a valid dimension, since width can't be negative)}$

So, the width $w$ is 11 feet. The length $l$ is: $l = w + 2 = 11 + 2 = 13 \text{ feet}$

Dimensions of the window:

• Width: 11 feet
• Length: 13 feet

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the perimeter of the window?
2. If the length was three feet more than the width, what would the dimensions be?
3. How would the area change if both the length and width were doubled?
4. What would the dimensions be if the area was 200 square feet?
5. How can you use the quadratic formula to solve real-life problems?

Tip: When solving quadratic equations, always check both possible solutions to ensure they make sense in the context of the problem.