GCSE Maths Coursework:-
By Sarah Wolfe, 10Ba2
A farmer has exactly 1000 metres of fencing and wants t fence off plot of level land. I am going to come up with a range of shapes which has a perimeter or circumference of 1000 metres. For each shape I draw I will draw different size ones and find their areas. The reason for finding their areas is because the farmer wants to find a shape with the maximum area. To get started on this exercise, I am first of all going to come with a shape which I think is easiest to use; this will be a rectangle.
) Perimeter = 450 + 50 + 450 + 50
= 1000m
Area = 450 x 50
= 22,500m
2) Perimeter = 400 + 100 + 400 + 100
= 1000m
Area = 400 x 100
= 40,000m
3) Perimeter = 385 + 115 + 385 +115
= 1000m
Area = 385 x 115
= 44,275m
4) Perimeter = 375 + 125 + 375+ 125
= 1000m
Area = 375 x 125
= 46,875m
5) Perimeter = 350 + 150 +350 +150
= 1000m
Area = 350 x 150
= 52,500m
6) Perimeter = 325 +175+325 +175
= 1000m
Area = 325 x 175
=56,875m
Whilst finding the areas of these shapes, I'm finding that when I decrease the size of the length and increase the width, the area seems to get bigger.
7) Perimeter = 275 + 225 + 275 +225
= 1000m
Area = 275 x 225
= 61, 875m
8) Perimeter = 265 +235 + 265+ 235
= 1000m
Area = 265 x 235
= 62,275m
9) Perimeter = 250 +250 +250+250
= 1000m
Area = 250 x 250
= 62,500m
To make the process easier and quicker, I am now going to put the areas I have already found into a table.
Table to show areas of rectangles
Length (m)
Width (m)
Area (m )
450
50
22,500
400
00
40,000
385
15
44,275
375
25
46,875
350
50
52,500
325
75
56,875
275
225
61,875
265
235
62,275
250
250
62,500
With these figures, I am now going to produce a graph to prove that 250m is the length that gives the maximum area. (see next page)
Even if you change the lengths around, so that the lengths are the widths and the widths are the lengths, you will still get the same area. This shows on the graph as the graph is symmetrical.
After drawing all the possible rectangles and producing a graph, it shows that the square (which is part of the rectangle family) gives the biggest area.
250m
250m
I ...
This is a preview of the whole essay
With these figures, I am now going to produce a graph to prove that 250m is the length that gives the maximum area. (see next page)
Even if you change the lengths around, so that the lengths are the widths and the widths are the lengths, you will still get the same area. This shows on the graph as the graph is symmetrical.
After drawing all the possible rectangles and producing a graph, it shows that the square (which is part of the rectangle family) gives the biggest area.
250m
250m
I am now going to draw other quadrilaterals to see if they give an even bigger area than the square.
Now I have found the areas of other quadrilaterals, it has shown me that they do not give the biggest area.
My next shape I am going to use is triangles. The reason for this is because it is still one of the basic shapes to use and I also want to find out that if I decrease the number of sides, will it give me a bigger area?
I am going to stick to using 'iscoseleles' triangles as they are the easiest to use. I noticed if I was to find the areas of some 'scalene' triangles, that I would have to use a length of 500 or less, otherwise it would not work.
Scale: 1cm = 100m
)
450 = 50 + h
202,500 = 2,500 + h
h = 202,500 - 2,500
h = 200,000
h = 200,000
h = 447m
Area = b x h
2
= 100 x 447
2
Area = 22,350m
2) 400 = 100 + h
160,000 = 10,000 + h
h = 160,000-10,000
h = 150,000
h = 150,000
h = 387m
Area = b x h
2
= 200 x 387
2
Area = 38,700m
3) 350 = 150 + h
122,500 = 22,500 + h
h = 122,500 - 22,500
h = 100,000
h = 100,000
h = 316m
Area = b x h
2
= 300 x 316
2
Area = 47,400m
325 = 175 + h
4) 105,625 = 30,625 + h
h = 105,625 - 30,625
h = 75,000
h = 75,000
h = 273m
5) 300 = 200 + h
90,000 = 40,000 + h
h = 90,000 - 40,000
h = 50,000
h = 50,000
h = 224m
Area = b x h
2
= 400 x 224
2
Area = 44, 800m
You would think that this triangle would give a bigger area than the one above, so obviously the lengths which are nearer to 300 but less then 400 will give a bigger area.
I have also noticed that as all of the sides which have a near or enough value, the area of that triangle is bigger.
I predict that, like the square, the equilateral triangle will give the biggest area.
333.3 = 166.6 + h
111,111.1 = 277,7.7+ h
h = 111,111.1- 277,7.7
h = 108,333.4
h = 108,333.4
h = 329.1m
Area = b x h
2
= 333.3 x 329.1
2
Area = 54,850 m (to 1d.p)
I am going to do another table, so I can see clearly the areas I have already found.
Base (m)
Height (m)
Area (m )
00
450
22,350
200
400
38,700
300
350
47,400
350
325
47,950
33.3
33.3
54,850
400
300
44,800
My predication was right and the reason I know this is because when you increase the length higher 33.3m, then the area starts to decrease, as you can see in my table. I am not going to do a graph, as I know it will give me the same sort of graph when I did one for the square; the graph will be symmetrical. I think that this will be the pattern for all of the shapes I do; the one which is equilateral will give the biggest area of that family.
It seems to me that the shapes with the smaller sides, give the smallest area, so I am now going to move on to using shapes with more sides; the first one I will start with will be a pentagon.
I am going to stick to finding the area of a regular pentagon because, before when finding the areas of the rectangles and triangles, my conclusion has been that the shape which is equilateral, gives the biggest area. The pattern is probably going to be the same throughout.
To work out the length of each side:
000 5 = 200m
Now I have found each side, I need 2 split the pentagon into sections, otherwise it will be difficult to find the area. I am going to split it up in triangles. When I have done that I then need to work out the angle of each triangle: 360 5 = 72
tan 36 = 100
h
h = 100
tan 36
h = 137.6m (to 1d.p)
Area = b x h Area of pentagon
2 = 5 x 137,60
= 200 x 137.6 = 68,800m
2
= 27,520
2 Area (of one triangle) = 13,760m
As you can see the area is bigger, so I am now going to find the area of a regular hexagon.
Tan30 = 83.3
h
h = 83.3
tan 30
h = 144.3m(to 1dp)
Area of one triangle = b x h Area of hexagon
2 = 6 x 120.25
A= 166.6 x 144.3 = 72.150m
2
A = 120,25m
Again, the area seems to get bigger, the more sides the shape has, so I am now going on to a regular octagon.
Tan 22.5= 62.5
h
h = 62.5
tan 22.5
h = 150.9m (to 1d.p)
Area of one triangle = b x h
2 Area of octagon
A= 125 x 150.9 = 8 x 9431.25
2 = 75.450m
A = 9431.25m
I am now going to increase the number of sides a lot higher. I am not going to do the ones in between as I know already that the more sides I have, the area will be bigger. The next one I am going to do is a 10-sided shape.
I am not going to draw the diagram as it will be very difficult to draw, so I am going the find the area of one of the triangles in that shape and then times it by the number of sides in that shape.
Length of one side = 1000 10
= 100m
Central angle = 360 10
= 36
100m
Tan 18 = 50 Area of one triangle = 100x 153.9
h = 153.90 m
h h = 50
tan 18
50m h = 153.9m (to 1d.p) Area of 10-sided shape
= 10 x 153.90
= 153.90m
My conclusion is that more sides the shape has, the bigger the area is. I am not going to go any further to work out shapes which have more sides than the ones I have already done, as I am know what to expect. However, my thought is that as the shape expands with the more sides it has, it will eventually become a circle. My observation is that the circle will be the answer, which gives the biggest area out of all the shapes I have investigated. To make sure that my prediction is right, I am going to work out the area of the circle which has a circumference of 1000m.
(Not to scale) Circumference = 1000m
In order to find the area, I need to find the diameter
the circle, which has a circumference of 1000m.
Diameter = 318m
Diameter = C
D = 1000
3.14
D = 318.3 m ( to 1d.p)
(Radius = 159.15m)
Area = x (radius)
= x 159.15
= 795.72m
To work out the areas of the hexagon, pentagon and octagon I had to do each shape in sections. Here are a set of instructions of how I did it.
) From the centre of 2) Then, find the central
the shape, split it up angle. (360 the
into equal triangles. number of sides)
Then label the length of
each side. (1000m
the number of sides)
4) Once you have found 3) Now find the area
the area of that triangle, of one of the triangles;
multiply the area by the to do this, you need to
number of sides. This find the height first,
now gives the final area using trigonometry.
of that shape.
I could also use a formula; e.g if I were to find the area of a 20 sided shape.
I could use 'n' as the number of sides
) Split it up into 20 triangles
2) Find central angle - 360 = ?
20
3) Find area of each triangle- need height (trigonometry)
4) 'n' (20) x area of one triangle = total area of shape
Now I have got worked out the area of the all possible shapes, I am now going to put the shapes with their biggest area, into a table.
Shape
Area m (best area of shape)
Rectangle (square)
62,500m
Triangle
54,850m
Pentagon
68,800m
Hexagon
72,150m
Octagon
75,450m
0 sided shape
20 sided shape
Circle
79,572m
I am now going to increase the number of sides a lot higher. I am not going to do the ones in between as I know already that the more sides I have, the area will be bigger. I am going to do a few more examples
