The length of a rectangle is two more than four times its width. Its perimeter is 54 inches. Find its dimensions length and width.
P = 2L + 2W
54 = 2(4W + 2) + 2W
54 = 10W + 4
50 = 10W
5 = W
To find the dimensions of the rectangle (length and width), we can set up two equations based on the information provided in the problem.
Let's assume that the width of the rectangle is represented by "w" inches. According to the problem, the length of the rectangle is two more than four times its width, which can be represented by the equation:
Length (L) = 4w + 2
The perimeter of a rectangle is calculated by adding all four sides together: P = 2L + 2w. In this problem, the perimeter is given as 54 inches. So we can set up the equation:
2L + 2w = 54
Substituting the value of L from the first equation into the second equation, we get:
2(4w + 2) + 2w = 54
8w + 4 + 2w = 54
10w + 4 = 54
10w = 54 - 4
10w = 50
w = 50 / 10
w = 5
Now that we have the width of the rectangle as 5 inches, we can substitute this value back into the first equation to find the length:
L = 4w + 2
L = 4(5) + 2
L = 20 + 2
L = 22
Therefore, the dimensions of the rectangle are:
Length = 22 inches
Width = 5 inches