Given:
Carrie has 180 ft of the fencing and wishes the fence to be 4times as long as it is wide.
To Find:
The area in square feet that the fencing encloses.
Answer:
The fence encloses 3600 sq ft.
Step-by-step explanation:
Let x denote the length of the dog run and y denote the width of the dog run.
Given that the garage wall serves as one side of the dog run, we are left with 3 other sides to instal the fence.
Carrie has 180ft of fencing with her, so the sum of the lengths of the 3 sides has to be equal to 180. We can represent this in the form of an equation as
![x+y+y=180\\\\x+2y=180\\\\x=180-2y](https://tex.z-dn.net/?f=x%2By%2By%3D180%5C%5C%5C%5Cx%2B2y%3D180%5C%5C%5C%5Cx%3D180-2y)
We are also given that Carrie wishes the fence to be 4 times as long as it is wide.
So,
![x=4y](https://tex.z-dn.net/?f=x%3D4y)
Replacing this value into the first equation, we have
![4y=180-2y\\\\ 6y=180\\\\ y=30](https://tex.z-dn.net/?f=4y%3D180-2y%5C%5C%5C%5C%0A6y%3D180%5C%5C%5C%5C%0Ay%3D30)
Therefore,
![x=(4)(30)=120](https://tex.z-dn.net/?f=x%3D%284%29%2830%29%3D120)
Thus, the length of the dog run is 120ft and the width is 30ft.
The area enclosed will be equal to the length multiplied with the width. So,
![(120)(30)=3600](https://tex.z-dn.net/?f=%28120%29%2830%29%3D3600)
The fence encloses 3600 sq ft.