width 32 length 32
sorry if this was confusing
Given: The length of a rectangle is eight more than twice its width. The perimeter is 88 feet.
To find: The dimensions of the rectangle.
It is given that the length of a rectangle is eight more than twice its width.
Let the width of the rectangle be w.
So, the length of the rectangle
We know that the perimeter of a rectangle is
Here, perimeter of rectangle
So, we have
Therefore, width of the rectangle is 12 feet
Hence, length of the rectangle is 32 feet and width of the rectangle is 12 feet.
48 = x + x + x + 8 + x + 8
48 = 4x + 16
32 = 4x
x = 8
length = 16
width = 8
24 and 8
The formula for the perimeter of the rect. is P = 2W + 2L.
Here, L = 2W + 8, so the area formula becomes:
P = 2W + 2(2W + 8) = 64
Dividing all terms by 2 reduces this to:
W + 2W + 8 = 32, or 3W = 24, or W = 8.
Then: L = 2W + 8 = 2(8) + 8 = 24
Check: does 2W + 2L = 64 when these values for W and L are substituted?
Is 2(8) + 2(24) = 64 true? Yes
The length and width are 24 and 8 respectively.