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• Level: GCSE
• Subject: Maths
• Word count: 3964

# The Fencing Problem

Extracts from this document...

Introduction

Maths Coursework: The Fencing Problem

There is a need to make a fence that is 1000m long. The area inside the fence has to have the maximum area. I am investigating which shape would give this.

Triangles: Scalene

The diagram above is not to scale. Instead of having the perimeter to 1000m, only in this diagram, I have made the perimeters of the shape to 10, only to make this part of the investigation easier to understand. We know that the base of all the shapes is 2. The lengths for the equilateral triangle are 4 on each side. This part of the investigation is to explain why the triangle with the longest height cannot have the same base. The tallest triangle also has a perimeter of 10. One of the sides for the tallest triangle is 5, which is understandable. However the other side is 3. This is literally impossible to be because if this triangle was drawn to scale, then the side that is 3 will not end up reaching the base. The isosceles triangle has a side of 4 and it looks shorter than the side of 3. The only way the higher triangle will reach is if the base is shortened. So in the formula ‘h x b ÷ 2’, in the case of the higher triangle, the height will be longer but the base will be shorter.

Looking at this diagram, there is no need to draw out tables to find out whether or not a scalene triangle is bigger than an equilateral or an isosceles in terms of area. I have made it so that the base is the same width for all triangles. The lines that are going from top to bottom on each triangle represent the height.

Middle

My two diagrams have shown that the rectangle is the shape that will have the largest area no matter what the height or width of the other quadrilateral is.

This is a basic example of a rectangle with a perimeter of 1000m. This is the formula to calculate the area of a rectangle: height x width

height x width = area

200 x 300 = 60000m²

Therefore, the area of the triangle above is 60000m². Now I must do this for lots of rectangles using the same method as I did for triangles; fill out some tables zooming in on each table every time, and then showing the results on a chart. I’m starting by going up by 50m each time:

 BASE(m) HEIGHT(m) AREA(m²) 50 450 22500 100 400 40000 150 350 52500 200 300 60000 250 250 62500 300 200 60000

Already the area has started to decrease somewhere around the 200m-300m spot so I will zoom in to find a closer answer:

 BASE(m) HEIGHT(m) AREA(m²) 210 290 60900 220 280 61600 230 270 62100 240 260 62400 250 250 62500 260 240 62400

The area is biggest around the 240m-260m spot so I shall zoom in again:

 BASE(m) HEIGHT(m) AREA(m²) 241 259 62419 242 258 62436 243 257 62451 244 256 62464 245 255 62475 246 254 62484 247 253 62491 248 252 62496 249 251 62499 250 250 62500 251 249 62499

I have to zoom in one last time to finally figure out which rectangle has the largest area. This time I am going up in 0.1 of a metre:

 BASE(m) HEIGHT(m) AREA(m²) 249.1 250.9 62499.19 249.2 250.8 62499.36 249.3 250.7 62499.51 249.4 250.6 62499.64 249.5 250.5 62499.75 249.6 250.4 62499.84 249.7 250.3 62499.91 249.8 250.2 62499.96 249.9 250.1 62499.99 250 250 62500 250.1 249.9 62499.99

In conclusion, I have investigated that the rectangle with the largest area (that has a perimeter of 1000m), has a base length of 250m and a height length of 250m. On the next page is a line graph showing the relationship between the base and the area of the rectangles. It also shows that the rectangle with the base of 250m will have the largest area:

Conclusion:

Just like the equilateral, coincidentally, this shape happens to be a square; which is a regular polygon.

Conclusion

Comparing the formula for the area of a circle with the area of n

There is an interesting relationship between the two formulas. The formula for the area of n is:

A=     250,000

n tan (180)

n

The formula for the area of a circle that has a 1000m perimeter is:

A= πr² = π x 100,000

4π²

= 250,000

π

There is a relationship between the denominators. Let’s take 5 as an example for n. the denominator for the first equation would then be:

5 tan (180)

5

3.149591887

The value of π goes on forever in terms of decimal places but on a calculator, it is:

3.141592654

As you can see the values are very close between the two denominators. Now I shall try substituting n with a lower number than 5. I will try using 3:

3 tan (180)

3

3.144466757

As you can see the answer is getting closer to π but it isn’t going lower. Below is a table showing that as the numbers are decreasing, the value is getting closer to π but never reaching it. It is acceptable to even use numbers like 0.1 to represent n because we know that a circle doesn’t have 0.1 sides, it has an infinite number of sides:

 n n tan (180)         n 5 3.149591887 4 3.146706537 3 3.144466757 2 3.142869254 1 3.141911687 0.1 3.141595844

Even when n is 0.1, the value doesn’t reach the same as π. We can then say that π is a limit which cannot be reached but can get closer and closer to being reached.

The final part of this investigation will be a graph showing the ‘shaded’ part of the circle against the number of sides. To find the area of the shaded part, I have to subtract the area of the regular polygon from the area of the circle:

In conclusion, my investigation has shown that a farmer should have a regular circular fence as the shape of the plot.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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