Question

Pory is working a buliting project. The
that prepact requires sing one
23-9-22 What are the two zeroes of
may not all be used
13
5. The area of a rubber coating for a fhat roof is 961. The rectangular frame the carpenter built for the flat roof has dimensions such that the length is fet langer than the width
Part A: Wime an equation that could be used to represent the area of the frame
Part B. Determine the dimensions of the frame.
Langth
Width:

Answer 1

To solve your problem, let's break it down into two parts as specified.

### Part A: Write an equation for the area of the frame.

Let:
- \( w \) = width of the frame (in feet)
- Then, the length \( l \) = \( w + 10 \) (since the length is 10 feet longer than the width)

The area \( A \) of the rectangular frame can be represented as:
\[
A = l \times w
\]
Substituting \( l \) in the area equation gives:
\[
A = (w + 10) \times w
\]
We know the area is 961 square feet, so we can set up the equation:
\[
w(w + 10) = 961
\]

### Part B: Determine the dimensions of the frame.

Now, expand and rearrange the equation from Part A:
\[
w^2 + 10w - 961 = 0
\]

We can solve this quadratic equation using the quadratic formula:
\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 10 \), and \( c = -961 \).

Calculating the discriminant:
\[
b^2 - 4ac = 10^2 - 4 \times 1 \times -961 = 100 + 3844 = 3944
\]

Now applying the quadratic formula:
\[
w = \frac{-10 \pm \sqrt{3944}}{2 \times 1} \\
w = \frac{-10 \pm 62.78}{2}
\]

Calculating the two possible solutions for \( w \):
1. \( w = \frac{-10 + 62.78}{2} \approx \frac{52.78}{2} \approx 26.39 \)
2. \( w = \frac{-10 - 62.78}{2} \approx \frac{-72.78}{2} \approx -36.39 \) (not a valid solution for width)

Therefore, we take \( w \approx 26.39 \) feet.

To find the length \( l \):
\[
l = w + 10 \approx 26.39 + 10 = 36.39 \text{ feet}
\]

### Conclusion

The dimensions of the frame are approximately:
- Width: **26.39 feet**
- Length: **36.39 feet**