A field is six times as long as it is wide. If the perimeter is 420 FT, find the dimensions of the feild.
2(x + 6x) = 420
x = 30
Let's assume that the width of the field is represented by 'w'.
According to the problem, the length of the field is six times the width. So, the length can be represented as '6w'.
Perimeter of a rectangle is given by the formula: P = 2(l + w), where P is the perimeter, l is the length, and w is the width.
In this case, the perimeter is 420 ft, so we can write the equation as:
420 = 2(6w + w)
Simplifying the equation:
420 = 2(7w)
420 = 14w
Now, let's solve for 'w':
w = 420/14
w = 30
Therefore, the width of the field is 30 ft.
The length can be found by multiplying the width by 6:
Length = 6w = 6 * 30 = 180
So, the dimensions of the field are 30 ft (width) by 180 ft (length).
To find the dimensions of the field, we can start by defining the variables. Let's say the width of the field is represented by "w" and the length is represented by "l".
Given that the field is six times as long as it is wide, we can express this as:
l = 6w
Next, we are given that the perimeter of the field is 420 ft. The perimeter of a rectangle is given by the formula:
perimeter = 2*(length + width)
Substituting the values we have:
420 = 2*(l + w)
We can substitute the value of "l" from the first equation into the second equation and solve for "w":
420 = 2*(6w + w)
Simplifying further:
420 = 2*(7w)
Dividing both sides by 2:
210 = 7w
Finally, we can solve for "w":
w = 210 / 7
w = 30
Therefore, the width of the field is 30 ft.
To find the length, we can substitute the value of "w" back into the first equation:
l = 6w = 6 * 30 = 180
Therefore, the length of the field is 180 ft.
So, the dimensions of the field are 30 ft by 180 ft.