The length of a rectangle is 8 feet more than its width. If the width is increased by 4 feet and the length is decreased by 5 feet, the area will remain the same. Find the dimensions of the original rectangle.
To find the dimensions of the original rectangle, let's assign variables to the dimensions.
Let "x" represent the width of the rectangle.
Since the length is 8 feet more than the width, the length is x + 8.
We're given information:
If the width is increased by 4 feet, the new width is x + 4.
If the length is decreased by 5 feet, the new length is x + 8 - 5 = x + 3.
The area of a rectangle is calculated by multiplying its length by its width.
So, the area of the original rectangle is x * (x + 8).
According to the problem, if we increase the width by 4 feet and decrease the length by 5 feet, the area will remain the same.
So, the area of the modified rectangle is (x + 4) * (x + 3).
Setting up the equation:
x * (x + 8) = (x + 4) * (x + 3)
Expanding the equation:
x^2 + 8x = x^2 + 7x + 4x + 12
Simplifying the equation:
x^2 + 8x = x^2 + 11x + 12
Rearranging terms:
x^2 + 8x - x^2 - 11x - 12 = 0
Combining like terms:
-3x - 12 = 0
Solving for x:
-3x = 12
x = -12 / -3
x = 4
Therefore, the width of the original rectangle is 4 feet.
To find the length, we substitute the value of x into the expression x + 8:
Length = 4 + 8 = 12 feet
So, the dimensions of the original rectangle are:
Width = 4 feet
Length = 12 feet
I NEED HALP
Length: 20
Width: 12
Width = W feet.
Length = W+8 Ft.
Width = W+4.
Length = W+3.
A1 = A2, (W+8)*W = (W+3)*(W+4),
W^2 + 8W = W^2 + 4W + 3W + 12,
W = 12 Ft., W+8 = 20 Ft.