The perimeter of a rectangular lot is 190 ft. The width of the lot is 15 ft more than the length. What are the dimensions of the lot?
P = 2L + 2W
190 = 2L + 2(L + 15)
190 = 4L + 30
160 = 4L
40 = L
To find the dimensions of the rectangular lot, we can set up equations based on the given information.
Let's assume the length of the lot is L feet.
According to the problem, the width is 15 feet more than the length, so the width would be L + 15 feet.
The formula to find the perimeter of a rectangle is P = 2(L + W), where P is the perimeter, L is the length, and W is the width.
Given that the perimeter of the lot is 190 feet, we can now create an equation:
190 = 2(L + (L + 15))
Simplifying the equation, we have:
190 = 2(2L + 15)
190 = 4L + 30
160 = 4L
Dividing both sides of the equation by 4, we find:
L = 40
Now we can substitute 40 back into the expression to find the width:
W = L + 15
W = 40 + 15
W = 55
Therefore, the dimensions of the lot are length = 40 feet and width = 55 feet.
Let's denote the length of the lot as x feet.
According to the given information, the width of the lot is 15 feet more than the length, so the width can be expressed as (x + 15) feet.
The perimeter of a rectangle is calculated by adding the lengths of all sides, so the equation to find the perimeter can be written as:
2(length) + 2(width) = perimeter
Substituting the given values, we get:
2(x) + 2(x + 15) = 190
Simplifying the equation:
2x + 2x + 30 = 190
Combine like terms:
4x + 30 = 190
Subtract 30 from both sides:
4x = 160
Divide both sides by 4:
x = 40
So, the length of the lot is 40 feet.
The width can be calculated by adding 15 to the length:
Width = 40 + 15 = 55 feet.
Therefore, the dimensions of the lot are 40 feet (length) and 55 feet (width).