The perimetre of a rectangle is 132 inches.The length exceeds the width by 14 inches. What is the length and the width?
L = W + 14
P = 2L + 2W
P = 2(W + 14) + 2W
Solve for P.
15
To solve this problem, we need to set up an equation using the given information. Let's let "L" represent the length of the rectangle and "W" represent the width.
We are given that the perimeter of the rectangle is 132 inches. The formula for the perimeter of a rectangle is P = 2L + 2W.
So, we can set up the equation as follows:
2L + 2W = 132
We are also given that the length exceeds the width by 14 inches. Mathematically, this can be represented as:
L = W + 14
Now, we can substitute the value of L from the second equation into the first equation:
2(W + 14) + 2W = 132
Simplifying the equation:
2W + 28 + 2W = 132
Combining like terms:
4W + 28 = 132
Subtracting 28 from both sides:
4W = 132 - 28
4W = 104
Now, divide both sides of the equation by 4:
W = 104/4
W = 26
Now that we have the value of W, we can substitute it back into the equation for L:
L = W + 14
L = 26 + 14
L = 40
Therefore, the length of the rectangle is 40 inches and the width is 26 inches.