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Maths GCSE Courswork
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Introduction
Introduction
A farmer has exactly 1000 metres of fencing and wants 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 1000m. So it could be
or anything with a perimeter (or circumference) of 1000m.
She wishes to fence off the plot of land which contains the maximum area.
Investigate the shape, or shapes, that could be used to fence in the maximum area using exactly 1000 metres of fencing each time.
Triangles
I will start the investigation with triangles. The triangle is one of the basic shapes as it has only three sides to it. I will calculate the maximum areas for two types of triangles:
- Equilateral Triangle
- Isosceles Triangles
An equilateral triangle is a triangle with all three of its sides the same length. An isosceles triangle is a triangle which has two of its sides the same length and the other is different. It is usually the base of an isosceles triangle which has a different length to the other two sides.
Here are some of the formulae I will need to find the maximum areas of each triangle:
- Area: base x height / 2
- Pythagoras Theorem: a2 + b2 = c2
- Heron’s Formula:
Equilateral Triangle
This shape has a perimeter of 1000m:
Perimeter = 333.33 + 333.33 + 333.33 = 1000m
Middle
The area of this isosceles triangle is 47433m2.
...read more.
Below, is a table of the areas of isosceles triangles which have bases of lengths ranging from 400m to 200m:
Length (m) | Base (m) | Height (m) | Area (m2) | Perimeter (m) |
300 | 400 | 223.6 | 44720 | 1000 |
310 | 380 | 244.95 | 46540.5 | 1000 |
320 | 360 | 264.57 | 47622.6 | 1000 |
330 | 340 | 282.84 | 48082.8 | 1000 |
333.33 | 333.33 | 288.67 | 48111.186 | 1000 |
340 | 320 | 300 | 48000 | 1000 |
350 | 300 | 316.22 | 47433 | 1000 |
360 | 280 | 331.66 | 46432.4 | 1000 |
370 | 260 | 346.41 | 45033.3 | 1000 |
380 | 240 | 360.55 | 43266 | 1000 |
390 | 220 | 374.16 | 41157.6 | 1000 |
400 | 200 | 387.29 | 38729 | 1000 |
I will now create a line graph to represent these results (see next page).
From the results in the table and the graph on the next page I have worked out that the equilateral triangle which has all three sides of length 333.33m is the triangle which gives the maximum area, 48111.186m2, out of all the triangles investigated in this section.
Rectangles
The next set of shapes that I am going to be investigating is rectangles. They are shapes with four sides and two sets of parallel lines. In this section, I will be calculating the area of two types of rectangles:
- Square
- Oblongs
A square is a type of rectangle where all four of its sides are equal and are the same length. An oblong is a rectangle which has two pairs of equal sides.
The only formula I will need to work out the areas of the rectangles is:
Area = length x width
Square
The square of 1000m that I am going to be investigating has four sides, each measuring up to 250m. This is the only square possible which matches the principle of having the perimeter of 1000m:
Perimeter = 250 + 250 + 250 + 250 =
Conclusion
1000000
79577.471
1000
So far, the shape which has given me the maximum area is the shape with 1000000 sides. However, there is still one more shape we need to find the area of and that is the circle.
Circle
The last shape that I will calculate the area of is the circle.
The circle above must have a circumference of 1000m. To find the area, I will need to use the formula:
Area = πr2
But before I can work out the area of the circle, I need to calculate the radius of the circle which is until now, unknown.
Diameter of circle = circumference / pi
= 1000 / 3.141592654
= 318.3098862
Radius of circle = diameter / 2
= 318.3098862 / 2
= 159.1549431m
Now I can calculate the area of the circle:
Area = πr2
= π x (159.15494312)
= π x 25330.29591
= 79577.471m2
A graph of all the shapes that I have investigated in this coursework:
Evaluation
There is a trend in the areas of all the shapes that I have investigated. As the number of sides of the shapes increases, the area increases. However, there is a point of the graph where the area cannot get any bigger. This assumption concerns the last three shapes that I investigated; the 100000 sided shape, 1000000 sided shape and circle. This tells me that the area that these three shapes obtained is the maximum area that can be achieved using a shape which has a perimeter of 1000m.
Therefore, I can come to a conclusion that the farmer can fence off a plot of level land which has the shape of a circle and a circumference of 1000m.
This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.
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