the perimeter of a dog park is 156 ft. If the length of the park is 2 ft. less than 3 times the width, what are the dimensions of the dog park?
width ---- w
length ---- 3w - 2
2w + 2(3w-2) = 156
2w + 6w - 4 = 156
8w = 160
w = 20
park is 20 by 58
The dimensions of the dog part are 20ft by 56 ft.
Set up the problem
w + w + ( 3w - 2 ) + ( 3w - 2 ) = 156
P = 2L + 2W
156 = 2(3W - 2) + 2W
156 = 6W - 4 + 2W
160 = 8W
20 = W
~~~~~~~~~~~~~~~~`
L = 58
sorry I post 56 instead of 58, the dimensions are 20ft by 58ft
How did you get the length to be 58. Where did that number come from?
ohhhhh nvrm i got
To find the dimensions of the dog park, we need to set up a system of equations. Let's denote the width of the park as "w" and the length as "l".
From the given information, we can create two equations:
1. The perimeter of a rectangle is calculated by adding the lengths of all sides, which in this case is given as 156 ft. The perimeter equation for a rectangle is:
Perimeter = 2 * (Length + Width)
So, we can write the first equation as:
156 = 2 * (l + w)
2. The length of the park is 2 ft less than 3 times the width of the park. Mathematically, this can be expressed as:
Length = 3 * Width - 2
Now, we have the two equations:
156 = 2 * (l + w) -- (equation 1)
l = 3w - 2 -- (equation 2)
Now, we can use substitution or elimination method to solve the system of equations.
Let's solve for "l" in equation 1:
l = (156 / 2) - w
l = 78 - w
Next, substitute this value of "l" in equation 2:
78 - w = 3w - 2
Simplify the equation:
78 + 2 = 3w + w
80 = 4w
Divide both sides by 4 to solve for "w":
w = 20
Now, substitute this value of "w" back into equation 2 to solve for "l":
l = 3(20) - 2
l = 60 - 2
l = 58
Therefore, the width of the dog park is 20 ft and the length is 58 ft.