the length of a rectangle is 8 inches more than twice its width. if the perimeter of the rectangle is 64 inches, what is the length of the rectangle. show equation
2(w + 2w+8) = 64
w = 8
so the length is 24
Let's start by assigning variables to the unknown values in the problem:
Let:
Length of the rectangle = L
Width of the rectangle = W
According to the given information:
The length of the rectangle is 8 inches more than twice its width:
L = 2W + 8
Also, we know that the perimeter of a rectangle is given by the formula:
Perimeter = 2(L + W)
We are given that the perimeter of the rectangle is 64 inches:
64 = 2(L + W)
Now we have two equations:
1. L = 2W + 8
2. 64 = 2(L + W)
These equations represent the relationship between the length and width of the rectangle and the perimeter of the rectangle.
To find the length of the rectangle, we need to set up an equation using the given information.
Let's say the width of the rectangle is "w" inches.
According to the problem, the length of the rectangle is "8 inches more than twice its width." So, the length can be expressed as 2w + 8 inches.
The perimeter of a rectangle is calculated by adding the lengths of all its sides. In this case, the perimeter is given as 64 inches. Since a rectangle has two pairs of equal sides, we can set up the equation:
Perimeter = 2(length + width)
Substituting in the known values, we have:
64 = 2(2w + 8 + w)
Now, we can simplify the equation and solve for w:
64 = 2(3w + 8)
64 = 6w + 16
48 = 6w
w = 8
Therefore, the width of the rectangle is 8 inches.
To find the length, we substitute the value of w back into our equation for the length:
Length = 2w + 8
Length = 2(8) + 8
Length = 16 + 8
Length = 24
Hence, the length of the rectangle is 24 inches.