Question
. The area of a rubber coating for a fhat roof is 96ft2. 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:
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:
Answers
GPT-4o mini
To solve the problem, let's define the width as \( w \) feet. According to the problem, the length of the frame is 6 feet longer than the width, so we can express the length as:
\[
l = w + 6
\]
### Part A: Write an equation for the area of the frame
The area \( A \) of a rectangle is given by the formula:
\[
A = \text{length} \times \text{width}
\]
Substituting our expressions for length and width into the area formula, we have:
\[
96 = (w + 6) \times w
\]
This expands to:
\[
96 = w^2 + 6w
\]
Rearranging gives us the quadratic equation:
\[
w^2 + 6w - 96 = 0
\]
### Part B: Determine the dimensions of the frame
To find the width \( w \), we can use the quadratic formula, which is:
\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our equation \( w^2 + 6w - 96 = 0 \), we have:
- \( a = 1 \)
- \( b = 6 \)
- \( c = -96 \)
Let's calculate the discriminant:
\[
b^2 - 4ac = 6^2 - 4(1)(-96) = 36 + 384 = 420
\]
Now applying the quadratic formula:
\[
w = \frac{-6 \pm \sqrt{420}}{2(1)} = \frac{-6 \pm \sqrt{420}}{2}
\]
Calculating \( \sqrt{420} \):
\[
\sqrt{420} \approx 20.49
\]
Now substituting back into the formula:
\[
w = \frac{-6 \pm 20.49}{2}
\]
Calculating both possible values of \( w \):
1.
\[
w = \frac{14.49}{2} \approx 7.245 \quad (\text{this is a valid width})
\]
2.
\[
w = \frac{-26.49}{2} \approx -13.245 \quad (\text{not valid since width cannot be negative})
\]
Thus, we accept \( w \approx 7.245 \).
Now, we can find the length:
\[
l = w + 6 \approx 7.245 + 6 \approx 13.245
\]
### Final Dimensions
- Width \( \approx 7.25 \) feet (approximately)
- Length \( \approx 13.25 \) feet (approximately)
So, the dimensions of the frame are approximately:
- **Length:** \( 13.25 \) feet
- **Width:** \( 7.25 \) feet
\[
l = w + 6
\]
### Part A: Write an equation for the area of the frame
The area \( A \) of a rectangle is given by the formula:
\[
A = \text{length} \times \text{width}
\]
Substituting our expressions for length and width into the area formula, we have:
\[
96 = (w + 6) \times w
\]
This expands to:
\[
96 = w^2 + 6w
\]
Rearranging gives us the quadratic equation:
\[
w^2 + 6w - 96 = 0
\]
### Part B: Determine the dimensions of the frame
To find the width \( w \), we can use the quadratic formula, which is:
\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our equation \( w^2 + 6w - 96 = 0 \), we have:
- \( a = 1 \)
- \( b = 6 \)
- \( c = -96 \)
Let's calculate the discriminant:
\[
b^2 - 4ac = 6^2 - 4(1)(-96) = 36 + 384 = 420
\]
Now applying the quadratic formula:
\[
w = \frac{-6 \pm \sqrt{420}}{2(1)} = \frac{-6 \pm \sqrt{420}}{2}
\]
Calculating \( \sqrt{420} \):
\[
\sqrt{420} \approx 20.49
\]
Now substituting back into the formula:
\[
w = \frac{-6 \pm 20.49}{2}
\]
Calculating both possible values of \( w \):
1.
\[
w = \frac{14.49}{2} \approx 7.245 \quad (\text{this is a valid width})
\]
2.
\[
w = \frac{-26.49}{2} \approx -13.245 \quad (\text{not valid since width cannot be negative})
\]
Thus, we accept \( w \approx 7.245 \).
Now, we can find the length:
\[
l = w + 6 \approx 7.245 + 6 \approx 13.245
\]
### Final Dimensions
- Width \( \approx 7.25 \) feet (approximately)
- Length \( \approx 13.25 \) feet (approximately)
So, the dimensions of the frame are approximately:
- **Length:** \( 13.25 \) feet
- **Width:** \( 7.25 \) feet