Question
Andres has 48 m of fencing to build a four-sided fence around a rectangular plot of land. The area of the land is 143 square meters. Solve for the dimensions (length and width) of the field.
Answer
Attempt 1 out of 2
The dimensions are
meters by
meters.
Answer
Attempt 1 out of 2
The dimensions are
meters by
meters.
Answers
GPT-4o mini
To find the dimensions of the rectangular plot of land, we can use the information we have regarding the perimeter and the area.
We know:
1. The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2L + 2W
\]
where \( L \) is the length and \( W \) is the width.
2. The area \( A \) of a rectangle is given by the formula:
\[
A = L \times W
\]
From the problem, we have:
- The perimeter \( P = 48 \) meters.
- The area \( A = 143 \) square meters.
### Step 1: Use the perimeter to express one variable in terms of the other.
From the perimeter equation, we can rearrange to find:
\[
2L + 2W = 48
\]
Dividing the whole equation by 2 gives:
\[
L + W = 24
\]
We can express \( W \) in terms of \( L \):
\[
W = 24 - L
\]
### Step 2: Substitute the expression for \( W \) into the area formula.
We substitute \( W \) into the area equation:
\[
A = L \times W = L \times (24 - L)
\]
We know the area is 143 square meters:
\[
L(24 - L) = 143
\]
### Step 3: Simplify and rearrange the equation.
Expanding this gives:
\[
24L - L^2 = 143
\]
Rearranging this into standard quadratic form:
\[
L^2 - 24L + 143 = 0
\]
### Step 4: Solve the quadratic equation.
We can use the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -24, c = 143 \):
\[
b^2 - 4ac = (-24)^2 - 4(1)(143) = 576 - 572 = 4
\]
Now, applying the quadratic formula:
\[
L = \frac{24 \pm \sqrt{4}}{2(1)} = \frac{24 \pm 2}{2}
\]
This gives us two possible values for \( L \):
1. \( L = \frac{26}{2} = 13 \)
2. \( L = \frac{22}{2} = 11 \)
### Step 5: Find the corresponding widths.
Using \( W = 24 - L \):
1. If \( L = 13 \), then \( W = 24 - 13 = 11 \).
2. If \( L = 11 \), then \( W = 24 - 11 = 13 \).
### Conclusion
The dimensions of the field are:
- Length \( L = 13 \) meters and Width \( W = 11 \) meters (or vice versa, as it is a rectangle).
So the dimensions of the field are:
**13 meters by 11 meters.**
We know:
1. The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2L + 2W
\]
where \( L \) is the length and \( W \) is the width.
2. The area \( A \) of a rectangle is given by the formula:
\[
A = L \times W
\]
From the problem, we have:
- The perimeter \( P = 48 \) meters.
- The area \( A = 143 \) square meters.
### Step 1: Use the perimeter to express one variable in terms of the other.
From the perimeter equation, we can rearrange to find:
\[
2L + 2W = 48
\]
Dividing the whole equation by 2 gives:
\[
L + W = 24
\]
We can express \( W \) in terms of \( L \):
\[
W = 24 - L
\]
### Step 2: Substitute the expression for \( W \) into the area formula.
We substitute \( W \) into the area equation:
\[
A = L \times W = L \times (24 - L)
\]
We know the area is 143 square meters:
\[
L(24 - L) = 143
\]
### Step 3: Simplify and rearrange the equation.
Expanding this gives:
\[
24L - L^2 = 143
\]
Rearranging this into standard quadratic form:
\[
L^2 - 24L + 143 = 0
\]
### Step 4: Solve the quadratic equation.
We can use the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -24, c = 143 \):
\[
b^2 - 4ac = (-24)^2 - 4(1)(143) = 576 - 572 = 4
\]
Now, applying the quadratic formula:
\[
L = \frac{24 \pm \sqrt{4}}{2(1)} = \frac{24 \pm 2}{2}
\]
This gives us two possible values for \( L \):
1. \( L = \frac{26}{2} = 13 \)
2. \( L = \frac{22}{2} = 11 \)
### Step 5: Find the corresponding widths.
Using \( W = 24 - L \):
1. If \( L = 13 \), then \( W = 24 - 13 = 11 \).
2. If \( L = 11 \), then \( W = 24 - 11 = 13 \).
### Conclusion
The dimensions of the field are:
- Length \( L = 13 \) meters and Width \( W = 11 \) meters (or vice versa, as it is a rectangle).
So the dimensions of the field are:
**13 meters by 11 meters.**