The Fencing Problem. My aim is to determine which shape will give me the largest area with the same perimeter. I will be using squares and other shapes, such as triangles, pentagons and etc.

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Nadeem Hussein 110                                7070

The Fencing Problem

Aim

 My aim is to determine which shape will give me the largest area with the same perimeter. I will be using squares and other shapes, such as triangles, pentagons and etc. After I have collected my data from a certain shape, (e.g. rectangles), I will draw a graph to show what my results show and make observations on the results. I will also take more than one type of data for every shape, so I will use the trial and improvement method to see what type of a certain shape gives me the largest area. An example would be me taking 10 different types of rectangles to see which one will give the largest Area. I will also try to find out a rule from my results and if there is I will find an algebraic formula to show this. My own hypothesis is that the shape that is regular will have the largest area. The perimeter that I will be able to use is 1000m. I will present my data in a table, graph and a diagram where you will be able to see the calculation and diagram of the certain object. I will increase and decrease the lengths and widths of the shape with a fixed distance so that my graph shows the data easier to interpret.

Rectangles + Square

Analysis

From this graph I can tell that there is an increase in Area as I decrease the gap between the length and width of the square. Furthermore, the results show that the rectangle with the highest area is the regular rectangle, square. This goes with my theory as I said that the highest area will be found from the regular shapes. It proves that as I decrease the distance between the length and width the area will increase. I can also prove this with shape 1 + 2. In shape 1, which had a length of 10 and a width of 490, the area was 4900, while with shape 2, which had a length of 20 and width of 480, the area was 9600. I can also prove that it is not that the length which affects the area. Some may think this as I increase the length the area increases, however this is false. I can say this because I would have to times the length by the width, which would decrease. An example would be Shape 24, which has a length of 240 and a width of 260, and a shape 26, which would have a length of 260 and width of 240, they would both produce the same area, which is in this case 62400.

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Triangles

Isosceles Triangles

Scalene Triangles

        

Equilateral Triangle

        

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