Question
A rectangle field is 125 yards long and 75 yards wide. Give the length and width of another rectangle that has the same perimeter bit a larger area
Answers
GPT-4o mini
To find a rectangle that has the same perimeter as the original rectangle but a larger area, we first need to calculate the perimeter of the original rectangle.
The formula for the perimeter \( P \) of a rectangle is given by:
\[
P = 2(\text{length} + \text{width})
\]
For the given rectangle:
- Length = 125 yards
- Width = 75 yards
Calculating the perimeter:
\[
P = 2(125 + 75) = 2(200) = 400 \text{ yards}
\]
Now, we need to find another rectangle with a perimeter of 400 yards, but with a larger area than the original rectangle.
First, we calculate the area of the original rectangle using the formula:
\[
\text{Area} = \text{length} \times \text{width}
\]
Calculating the area for the original rectangle:
\[
\text{Area} = 125 \times 75 = 9375 \text{ square yards}
\]
We need to find dimensions (length and width) for a new rectangle where:
1. The perimeter \( P \) is 400 yards.
2. The area is greater than 9375 square yards.
Let the new rectangle's dimensions be \( L \) (length) and \( W \) (width). We have:
\[
L + W = \frac{P}{2} = 200
\]
Now, we express the area \( A \):
\[
A = L \times W
\]
Substituting \( W \) with \( 200 - L \):
\[
A = L \times (200 - L) = 200L - L^2
\]
This quadratic equation represents a parabola opening downwards. Its maximum area occurs at the vertex, which can be found using the formula \( L = -\frac{b}{2a} \) for a quadratic \( ax^2 + bx + c \).
Here, \( a = -1 \) and \( b = 200 \):
\[
L = -\frac{200}{2 \times -1} = 100
\]
So at \( L = 100 \), we find \( W \):
\[
W = 200 - L = 200 - 100 = 100
\]
Calculating the area of this square:
\[
A = 100 \times 100 = 10000 \text{ square yards}
\]
This area is indeed greater than 9375 square yards.
Thus, the new rectangle with dimensions \( 100 \) yards by \( 100 \) yards has the same perimeter (400 yards) and a larger area (10,000 square yards).
Therefore, the length and width of the new rectangle are:
**Length: 100 yards, Width: 100 yards.**
The formula for the perimeter \( P \) of a rectangle is given by:
\[
P = 2(\text{length} + \text{width})
\]
For the given rectangle:
- Length = 125 yards
- Width = 75 yards
Calculating the perimeter:
\[
P = 2(125 + 75) = 2(200) = 400 \text{ yards}
\]
Now, we need to find another rectangle with a perimeter of 400 yards, but with a larger area than the original rectangle.
First, we calculate the area of the original rectangle using the formula:
\[
\text{Area} = \text{length} \times \text{width}
\]
Calculating the area for the original rectangle:
\[
\text{Area} = 125 \times 75 = 9375 \text{ square yards}
\]
We need to find dimensions (length and width) for a new rectangle where:
1. The perimeter \( P \) is 400 yards.
2. The area is greater than 9375 square yards.
Let the new rectangle's dimensions be \( L \) (length) and \( W \) (width). We have:
\[
L + W = \frac{P}{2} = 200
\]
Now, we express the area \( A \):
\[
A = L \times W
\]
Substituting \( W \) with \( 200 - L \):
\[
A = L \times (200 - L) = 200L - L^2
\]
This quadratic equation represents a parabola opening downwards. Its maximum area occurs at the vertex, which can be found using the formula \( L = -\frac{b}{2a} \) for a quadratic \( ax^2 + bx + c \).
Here, \( a = -1 \) and \( b = 200 \):
\[
L = -\frac{200}{2 \times -1} = 100
\]
So at \( L = 100 \), we find \( W \):
\[
W = 200 - L = 200 - 100 = 100
\]
Calculating the area of this square:
\[
A = 100 \times 100 = 10000 \text{ square yards}
\]
This area is indeed greater than 9375 square yards.
Thus, the new rectangle with dimensions \( 100 \) yards by \( 100 \) yards has the same perimeter (400 yards) and a larger area (10,000 square yards).
Therefore, the length and width of the new rectangle are:
**Length: 100 yards, Width: 100 yards.**