Maths investigation - The Fencing Problem

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Maths investigation – The Fencing Problem                                

Aim – to investigate which geometrical enclosed shape would give the largest area when given a set perimeter.

In the following shapes I will use a perimeter of 1000m. I will start with the simplest polygon, a triangle.

Since in a triangle there are 3 variables i.e. three sides which can be different. There is no way in linking all three together, by this I mean if one side is 200m then the other sides can be a range of things. I am going to fix a base and then draw numerous triangles off this base. I can tell that all the triangles will have the same perimeter because using a setsquare and two points can draw the same shape. If the setsquare had to touch these two points and a point was drawn at the 90 angle then a circle would be its locus. Since the size of  the set square never changes the perimeter must remain the same.

The area of a triangle depends on two things: the height and the base. The base is fixed in this example so the triangle that has the biggest height, i.e. the middle triangle, will have the biggest area. The middle triangle turns out to be an icosoles triangle.  

I am going to focus only on icosoles triangles. I have constructed a formula linking all three sides in and icosoles triangle.

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                                X=any number which is greater than 250 and less than 500

Using Pythagoras theorem I can find and equation linking a side to the area.

½(1000 – 2X)² + H² = X²

H² = X² + (X –500)²                                        H = height

As you can see from the table the maximum area is when X = 333 1/3. When this number is plugged into the formula we see that this is actually an equilateral triangle.

The next simplest shape is the 4-sided shape, namely a rectangle. I ...

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