A farmer has exactly 1000 metres of fencing, with it she wishes to fence off a plot of level land. She is not concerned about the shape of the plot, but it must have a perimeter of 1000 metres. I am going to consider using rectangles first as this shape is very simple and it would be easy for me to work out the maximum area and perimeter.
Rectangle 5 is a square. After this the order of the rectangles are the same as before but in a reverse order, this is shown in rectangle 6. To work out the area of each rectangle I am going to multiply the length of the rectangle by the width of the rectangle. Here is a table to show you my results.
Length (m)
Width (m)
Area (m )
450
50
22500
400
00
40000
350
50
52500
300
200
60000
250
250
62500
200
300
60000
50
350
52000
00
400
40000
50
450
22500
The square gives me the maximum area I have highlighted this in my results table in red. I have produced a graph to show you my results.
I am now going to consider using triangles. I am going to find out the maximum area and the most suitable triangle to do this. To find out the most suitable triangle I am going to use a string. To do this I am going to tie the ends together, from this I am going to keep the base the same and ask my partner to move it into a triangle and ask her to change it until I see a triangle, which is suitable.
Picture 1
In picture one I have shown you how I have found the isosceles triangles as I think this kind of triangle will give me the maximum area whereas a scalene triangle would not.
475m 475m 450m 450m 425m 425m
50m
100m 150m
400m 400m
375m 375m
350m 350m
200m
250m
300m
333.3m 333.3m
325m 325m
333.3m
350m
In triangle seven it shows us an equilateral triangle. This is where I should stop as I am just investigating isosceles triangles. To find out the area of each triangle I am going to half the base and multiply it with the height. As I am not given the height I am going to have to fine the height by using Pythagoras theorem. Here are three examples.
Rectangle 5 is a square. After this the order of the rectangles are the same as before but in a reverse order, this is shown in rectangle 6. To work out the area of each rectangle I am going to multiply the length of the rectangle by the width of the rectangle. Here is a table to show you my results.
Length (m)
Width (m)
Area (m )
450
50
22500
400
00
40000
350
50
52500
300
200
60000
250
250
62500
200
300
60000
50
350
52000
00
400
40000
50
450
22500
The square gives me the maximum area I have highlighted this in my results table in red. I have produced a graph to show you my results.
I am now going to consider using triangles. I am going to find out the maximum area and the most suitable triangle to do this. To find out the most suitable triangle I am going to use a string. To do this I am going to tie the ends together, from this I am going to keep the base the same and ask my partner to move it into a triangle and ask her to change it until I see a triangle, which is suitable.
Picture 1
In picture one I have shown you how I have found the isosceles triangles as I think this kind of triangle will give me the maximum area whereas a scalene triangle would not.
475m 475m 450m 450m 425m 425m
50m
100m 150m
400m 400m
375m 375m
350m 350m
200m
250m
300m
333.3m 333.3m
325m 325m
333.3m
350m
In triangle seven it shows us an equilateral triangle. This is where I should stop as I am just investigating isosceles triangles. To find out the area of each triangle I am going to half the base and multiply it with the height. As I am not given the height I am going to have to fine the height by using Pythagoras theorem. Here are three examples.