Gcse Maths Coursework Fencing
Math Coursework - The Fencing Problem
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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.
X=any number which is greater than 250 and less than 500
1000 - 2X
Using Pythagoras theorem I can find and equation linking a side to the
Â½(1000 - 2X)Â² + HÂ² = XÂ²
HÂ² = XÂ² + (X -500)Â² H = height
500 - X
XÂ² - (500-X)Â²
How to Cite this Page
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Math Coursework Fencing Triangle Base Theorem Shapes Sides
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
have constructed another formula linking the sides.
500 - X
500 - X
From the diagram the area must equal -XÂ² +500X
Unlike in the previous example, this turns out to be a quadratic
equation so I can plot it on a graph.
As you can see from the graph the maximum point is when X = 250. When
this number is plugged into the formula the rectangle is really a
What do a square and an equilateral triangle have in common? They are
both regular shapes i.e. all angles equal, all sides equal.
Why is this?
Make sides same length
Make sides same length
[IMAGE]Lets take the triangle example first. When you make one side
longer you will make the other shorter. This will decrease the height,
which means the area will be smaller. When both sides are the same
length they extend the height to its highest possible. Why does an
equilateral triangle have a larger area than an icosoles triangle? you
could think of it like this.
Rotate triangle onto side
Lets take the square example. Obviously the longer the sides the
bigger the area. This means the bigger the length and the bigger the
height, the bigger the area. In this investigation we have been given
a set perimeter. To make the length longer means you have to sacrifice
the height. To make the height bigger you have to sacrifice the
length. To get the biggest area you need the sides to be as long as
possible. When the sides are equal, it means that the sides are at the
biggest they could be simultaneously. This means The closer the sides
are in the ratio of 1:1, the bigger the area. Shapes with a ratio of
sides that is 1:1 are said to be regular. Regular shapes have numerous
properties; they can be split up into icosoles triangles. Irregular
polygons can be only split up into scalene triangles. I have already
proved why icosoles triangles have a larger area than scalene. This
means regular shapes will have a larger area than irregular shapes.
From now on I am going to find out which shape has the biggest area
with a given perimeter. I will investigate only the regular shapes
because I have proved that the regular polygon has the biggest area
out of all the irregular polygons with the same perimeter.
Text Box: 333.3
Using trigonometry. Area = Â½ x 333.3 x 333.3 x sin 60 = 48112.5
Area = 250 x 250 = 62500
Each side = 200
Using trigonometry. 200/sin 72 = Y /sin 54
Area = 5x( Â½ x 170.1 x 170.1 x sin 72)
Each side = 166 2/3
Â½ x 6 x 166 2/3 x 166 2/3 x sin 60
Area = 72168.8
Each side = 142.9
142.9/sin 51.4 = Y/sin 64.3
Y = 164.8
Area = 7x( Â½ x 164.8 x 164.8 x sin 51.4)
Area = 74288.7
Each side = 125
125/sin 45 = Y/sin 67.5
Y = 163.3
Area = 8x( Â½ x 163.3 x 163.3 x sin 45)
Area = 75425.4
Lets put all these results in a table
Number of sides
Maximum area with perimeter of 1000M
As you can see these results will keep on increasing and increasing.
This means the shape that can have the largest area must have infinite
sides. What shape has infinite sides? I will use the regular polygon
A triangle has three sides and it has 3 lines of symmetry.
A square has 4 sides and it has 4 lines of symmetry.
A pentagon has 5 sides and 5 lines of symmetry.
A hexagon has 6 sidesâ€¦.
You get the idea. The shape with infinite sides must have infinite
lines of symmetry. The only shape that has infinite lines of symmetry
is the circle. Lets find out the area of a circle with circumference
2pr = 1000
pr = 500
r = 159.2
A = p159.2Â²
A = 79622.53
Lets add this to our table of results.
Number of sides
Maximum area with perimeter of 1000M
The circle has the biggest area with a 1000M perimeter out of all the
Why is this?
When a shape is split up into triangles, the more sides it has, the
more triangles there will be yet these triangles will become smaller
as the number of sides increase. The amount at which the area of the
triangle decreases is not as great as the amount the side increases
by. When you split the shape into triangle, the more sides the shape
has the smaller the angle gets in between the two equal sides but the
perimeter of these triangles increase as the shape has more sides. The
higher the perimeter, the larger area you can make providing the
perimeter is well used i.e. the triangle is in the form of an icosoles
triangle. A circle would have infinite sides and its angles are bigger
since bigger angles can encompass more.
You didn't mention calculus so I am going to solve this for you without using calculus. I am going to assume that the field is a rectangle and has sides of length x meters and y meters.
Since the farmer has 1000 m of fencing, 2x + 2y = 1000 and thus x + y = 500.
If x and y are equal they are each 250 m and the field is a square. I want to think in terms of "How close is the field to a square?" Suppose that x = 250 + h then, since x + y = 500, y = 250 - h. Hence the area is
(250 + h) (250 - h) = 2502 - h2 square meters
In this form you can plainly see that 2502 - h2 is a maximum when h = 0, that is when the field is a square.