• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
• Level: GCSE
• Subject: Maths
• Word count: 1242

# The Fencing Problem.

Extracts from this document...

Introduction

MATHS GSCE COURSEWORK – EDEXCEL 2003 DISPATCH 1

THE FENCING PROBLEM

YEAR 4-LB

The Fencing Problem

The fencing problem requires me to investigate the shape with the maximum area using 1000 metres of fencing. I have decided to start my search on triangles. A triangle being a shape with the minimum sides logically concludes that I begin there, as they are the most simple. From then on I will continue shape by shape as the number of sides increase; quadrilaterals, pentagons etc…

The area of a triangle is measured by calculating: ½ (base * height)

Therefore if I was to choose the length of the base then the area would be automatically decided by the height of the triangle. The type of triangle which I am going to investigate is the ‘isosceles’ triangle because as I vary the base, the other two sides would remain equal and this would give the maximum height for the triangle, thus giving the maximum area.

Formula to find out the length of the other two sides:

((1000 – (variable base))/2)

So as I have chosen to do isosceles triangles, the main variable is the base.

Middle

Now that parallelograms have been eliminated, I can just concentrate on finding the maximum area for a rectangular shape using different base sizes. The formula is easier to work out for quadrilaterals than that of triangles:  AREA = (base * height).

AREA = (variable base * ((1000- (2*variable base))/2)

From the last two investigations on triangles and quadrilaterals, the maximum area has been found by using the regular shape, i.e. an equilateral, a square; where all the sides are equal. Therefore for the remainder of my investigation I realise that it would be sensible to just concentrate on the regular shape. Pentagon is the name given to a five-sided shape and the area of this is found by splitting the shape into 5 regular triangles.  The area of one of these triangles is worked out by splitting it further in half, once this is worked out it can be multiplied by five and the area of the total shape can be calculated.

I have found the maximum area for a five-sided shape, so now I am going to try another shape.

Conclusion

I can now go further and investigate what shape would have the maximum area using 1000 metres of fencing.

My analysis shows that as the number of sides increase so does the maximum area. However, with increasing number of sides, the rate of rise in area gets smaller and smaller. As the number of sides increase the shape becomes increasingly circular. Therefore, I am going to see what the area of a circle values at.

Circumference of circle = 2πr

1000 = 2πr

500          = πr

Area of Circle = πr²

= π(159.1549431²)

Area of Circle = 79577.47155m²

It seems to be that there is an ASYMPTOTE in this problem. Looking at my graph there seems to be a number which the graph is trying to reach but will never actually get there. This asymptote is the area of a circle; this can be proven mathematically using explanation involving radians. Firstly the area of a circle can be linked to the formula I worked out to find the maximum area of any sided shape; this is because as the number of sides in a shape increase the actual structure becomes more and more like a circle.

AREA OF CIRCLE       ≈         AREA OF ANY SIDED SHAPE!

Π * 500/π²                            ≈                           500²/ (n (tan (180/n))

500²/π                                   ≈                           500²/ (n (tan (180/n))

CONTINUING:

500²/π                                   ≈                           500²/ (n (tan (π/n))

THERFORE:

500²/π                                   ≈                           500²/ (n (π/n)

500²/π                                   ≈                           500²/π

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

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## Maths Coursework - The Fencing Problem

Conclusion I found the total area to be 57320 m�. This is less than 68800 m�, which is the total area of a regular pentagon. This proves that regular polygons have a larger area. General Formula for regular polygons I looked at formula for internal and external angles of polygons,

2. ## Fencing problem.

shown above the formula that has been represented below shall be used: Area of triangle = 1/2 � Base � Height We now are familiar with the base but not the height. As the height is unknown, we shall discover the height before the investigation is continued.

1. ## The Fencing Problem

I will construct another table of data with the side lengths having a difference of 2 between them. 1) 400 500 100 2) 410 490 100 3) 420 480 100 4) 430 470 100 5) 440 460 100 6) 450 450 100 7)

2. ## Fencing Problem

= 11858.54123 m� * 1000 / 2 = 500 m * 500 x (500 - 450) x (500 - 450) x (500 - 100) = 500000000 * 500000000 = 22360.67978 m� * PERIMETER = 1000 m AREA = 22360.67978 m� * 1000 / 2 = 500 m * 500 x (500 - 425)

1. ## Maths:Fencing Problem

So when the shape has equal sides it has maximum area. I will look at some other quadrilaterals such as rhombuses, parallelograms and trapeziums. This is a parallelogram: This becomes a rectangle but my previous work has shown that a rectangle doesn't give me the maximum area.

2. ## The Fencing Problem

48016.064 321.5 339.25 298.747 48023.642 322.0 339.00 298.329 48030.917 322.5 338.75 297.909 48037.888 323.0 338.50 297.489 48044.554 323.5 338.25 297.069 48050.913 324.0 338.00 296.648 48056.966 324.5 337.75 296.226 48062.711 325.0 337.50 295.804 48068.148 325.5 337.25 295.381 48073.276 326.0 337.00 294.958 48078.093 326.5 336.75 294.534 48082.599 327.0 336.50 294.109 48086.793 327.5

1. ## fencing problem

And the graph will back predicti Calculating the maximum area of Triangles with the perimeter of 1000m� 333.3� 333.3� 400� 400� H 333.3� = 166.7� 200� = 100� 2 2 333.3�-166.7� 400� -100� H� = 111088.89-27788.89 H� = 160000-10000 83300 =288.6 150000 = 387.2 Area= 333.3�2886 = 48095.19m� Area= 200�3872=

2. ## The Fencing Problem

A simple diagram can explain this. In the diagram, the 10 x 6 rectangle is being multiplied step by step as well as the 8 x 8 square. 10 8 20 16 30 24 40 32 50 40 60 48 - 56 - 64 As you can clearly see, when

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to