SOLUTION: Ajax High School is putting a fence around the football field. The distance between the field and the fence is going to be 60 feet on end zone side and 45 feet on the sideline. Ho

Question 6918: Ajax High School is putting a fence around the football field. The distance between the field and the fence is going to be 60 feet on end zone side and 45 feet on the sideline. How many feet of fence will be needed? The football field measures 360 ft by 150 ft. Do I use P=2l+2w (720+300)? and what does the distance between field and fence have to do with the problem? Thanks, Sandy
Found 2 solutions by bonster, youngd:
Answer by bonster(299) (Show Source): You can put this solution on YOUR website!
The reason why the info about the distance between the field and the fence is important is because the fence is not being built along the edge of the football field. The problem doesn't ask how many feet of fence will be needed if the fence were to be put up right along the border; It asks you to take into consideration the spacing distance that will be used.
So let's solve this problem!
1. we know that the field measures 360 ft by 150 ft.
2. The perimeter of the football field is 360 + 150 + 360 + 150
3. We need to add the distance between the sides and the fence: so (360+45) + (150+60) + (360+45) + (150+60) or: 2(360+45) + 2(150+60)
4. 405ft + 210ft + 405ft + 210ft= 1230ft or: 2(405)+2(210)= 810+420=1230ft
1230 feet of fence will be needed.
Answer by youngd(1) (Show Source): You can put this solution on YOUR website!
bonster(299)'s answer is incorrect.

The field is 360 feet long by 150 feet wide, and the fence is being placed 60 feet from EACH end zone and 45 from EACH sideline. Therefore, the solution is;
P=l+w+l+w or 2l +2w
P=(60+360+60)+(45+150+45)+(60+360+60)+(45+150+45)
P = 2*(480)+ 2*(240)
P = 960 + 480
P = 1440
1440 feet of fence will be needed.
The key is adding the additional fence on both ends.
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