The fencing problem

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Guy McALL        Page         5/7/2007

Maths coursework

The fencing problem

Aim

My aim is to find out the largest area that can be put into a field. That has a perimeter of fencing that is 1000meters long. I am going to find out the biggest area I can get using different shapes that only have a perimeter of 1000meters.

Investigating rectangles

I have started with rectangles because they are the simplest to find the area of (H*B=area)

     

The perimeter=1000 because 250 + 250 + 250 + 250 the area is 62500 meters² I used the equation h*b.

The perimeter=1000 because 400 + 400 +100 + 100 the area is 40000 meters² I used the same equation again h*b.

The perimeter=1000 because 300 + 300 + 200 + 200 the area is 60000 meters² I used the equation h*b

Conclusion for rectangles

I found that the square had the biggest area than all the other rectangles. I found that the smaller the width the smaller the area (the width 100 and length 400 the are equals 40000.) I also found that the smaller the length the smaller the area (length 100 and width 400 the area equals 40000.) On the table when the width is large the area is smaller, when the width and length are roughly the same the area gets larger.   

Investigating Triangles

Aim

I started isosceles because they are the most obvious triangle and it will be easy to plot on a field. I have started with these lengths because of the result on the squares that if all the numbers were the same the area would be the biggest.  

Join now!

The perimeter=1000meters because 450 + 450 + 100 the area= 22360.67978meters²

Because: -

C²=A²+B²

450²=50²+B²

202500=2500+B²

200000=B²

√200000=447.2135955

Height=447.2135955

The perimeter is 1000meter and the area= 75.62368 meters² because: -

 

C²=A²+B²

275²=225²+B²

75625=50625+B²

75625-50625=25000

√25000=158.113883

Height=158.113883

The perimeter is=1000 because 475+475+50 the area= 11858.54123meters²

Because: -

 

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