• 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
• Level: GCSE
• Subject: Maths
• Word count: 3346

# The Fencing Problem

Extracts from this document...

Introduction

The Fencing Problem We are told that a farmer has exactly 1000 metres of fencing with which she wishes to fence off a plot of land. The shape of the fence doesn't concern her, however what she does wish to do is fence off the plot of land which contains the maximum area. Therefore the point of this investigation is to investigate the shape, or shapes of the plot of land that have the maximum area. I am going to start my investigation looking at simple 4-sided shapes (the rectangles), because there are relatively easy shapes in which to find the area. I will look at special case 4-sided shapes (parallelograms and trapeziums) later if necessary. 4-Sided Shapes For a square I know that all sides are equal and therefore the length of each side must be 1000/4, if the perimeter is 1000m. To find the area of a simple four-sided shape, you do length x width. In this case the are of the square is 250 x 250 = 62,500m� Now that I have found the area of a square, I am going to investigate what happens to the area if I change the length of the sides, still keeping the perimeter to 1000m, therefore making the shape a rectangle. Area in this case = (? -50)(? +50) = 60,000m� I will continue changing the lengths of the sides, until the width is 0m, which will inevitably happen as I am going to take away 50m from the width every time. Area in this case = (?-100)(?+100) = 52,500m� Area in this case = (?-150)(?+150) = 40,000m� Area in this case = (?-200)(?+200) = 22,500m� Area in this case = (?-250)(?+250) = 0m� Here is a table of all the areas for the simple 4-sided shapes that I did. Length (m) Width (m) Area (m�) 250 250 62,500 200 300 60,000 150 350 52,000 100 400 40,000 50 450 22,500 0 500 0 Now that I have finished looking at the simple 4-sided shapes, I am going to look at simple 3-sided shapes, to see how the area varies. ...read more.

Middle

= 137.64m (to 2 d.p) Now that we know the height we can find the area by using the formula: Area of triangle = 1/2 base length x height. = (1/2 x 1000/5) x 137.64 = 13,764m� To find the area of the whole pentagon, we times this number (13,764) by 5, as it was split equally into 5 isosceles triangles, and at the moment this number is only the area of one of the triangles. 13,764 x 5 = 68,820m� (to 5 s.f) I have decided that I'm not going to investigate what would happen to the area of a pentagon if I were to change the lengths of the sides, still keeping the perimeter to 1000m The reason for this choice is because firstly it would become too complicated (the pentagon couldn't be split equally into triangles), and secondly I have seen a pattern for the triangles and rectangles. Their regular states (all sides are of the same length), have the biggest area compared to when the lengths are changed. So for triangles the equilateral triangle, and for the rectangles, the square have the biggest area. Thirdly I have noticed another pattern. It seems like the more sides the shape has, the bigger the area it has. I am going to put all the regular shapes areas that I have done so far into a table, to show that the more sides, the bigger the area. Number of Sides Area (m�) 3 48,113 4 62,500 5 68,820 To support this fact, that the more sides the shape has the bigger the area, I will investigate what the area of a regular hexagon is. 6-Sided Shapes Again we can use the formula of: Exterior angle of a polygon = 360/number of sides = 360/6 From this we can find the interior angle: 180 - exterior angle, so in this case it is 180 - (360/6) ...read more.

Conclusion

So by the time we get to 1,000,000 sides the area stays the same no matter how many more sides you try. However the area only stays the same on the calculator, in theory it still increases when you increase the number of sides, but a calculator can only do a certain amount of digits, and it rounds the answers up. There is a major shape that I haven't yet tried, and that is the circle. The reason why I have left it till last is because I have always believed that a circle has an infinite amount of sides (straight sides), and therefore according to my theory that the more sides the bigger the area, the circle has the biggest area. So I am going to find the area of a circle and see if it is bigger than the amount we have got so far for the number of sides. The area of a circle = ? x r� The circumference of a circle = ? x D Therefore if we know the circumference of the circle (1000m), and we know ?, then we can find D. D = circumference/ ? = 1000/ ? Now that we know the diameter, we can find the radius by halving it. We do this because we need to find 'r' so that we can find the area of the circle. Therefore: r = 1/2 x 1000/ ? Area of circle = ? x ((1/2 x 1000/ ?)�) = 79,577.47155m� (to 5 d.p) This answer is the same as my answers to all the sides above 1,000,000 sides. The reason for them being the same is because of the calculators rounding up. Therefore the only way that I can prove that the circle has the biggest area is because of its theory of having an infinite amount of sides, and I have proved that the more sides there are the bigger the area. Therefore I conclude that the farmer should choose a circle fence of 1000m to fence of the plot of land, as it is the maximum amount of possible area. ...read more.

The above preview is unformatted text

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. ## Fencing Problem

of the right angled triangle by 10, (number of sides x 2), to give me the area of the pentagon instead of multiplying it by 2 to give me the area of the isosceles triangle and then multiplying the answer of that by the amount of sides in the shape.

2. ## t shape t toal

- (7 x 6) In this there is a common trend of 30. Both 45 and 30 are multiples of 5. I worked out that if you translate the T-shape up one square you time's the grid size (g) by 5 For a movement upwards you use this formula (-5g)

1. ## Beyond Pythagoras

How to get shortest side from 'n' 2. How to get middle side from 'n' 3. How to get longest side from 'n' To start with, now I am going to find an expression which gives the length of the shortest side, by using the expression for the term number as 'n'.

2. ## Find the biggest value of the ratio area / perimeter for some triangles.

= square root of 90x(90-80)x(90-40)x(90-60) = square root of 90x10x50x30 = square root of 1350000 = 1161.9 Ratio = a/p = 1161.9 � 180 = 6.45 I will now try to find the ratio of the triangle "a". The ratio is impossible to find using the formulas that are mention at the beginning.

1. ## I have been given 3 tasks in this piece of work. These tasks are ...

square rooted. So this meant I could do one of my statements above: If I have the middle side can I work out the short and long side? Yes. Long side= m+1 and Short side=m + (m+1) square rooted. Now that I had one of my statements proved I thought

2. ## Pythagoras Theorem.

+ 6n2 + 2n)/2 = 2n3 + 3n2 + n 2n3 +3n2 + n = n(2n2 +3n + 1) OR nth term Orig. seq. 1st diff. 2nd diff. 3rd diff 2n3 New seq. 1st diff. 2nd diff. 3n2 New seq.

1. ## The Fencing Problem

To find the area I must first find the height of the triangle. To do this I will need to split the triangle into two right-angled triangles. I will use triangle A to find the height, but I do not have enough information on the right-angled triangle to do so yet.

2. ## The Fencing Problem. My aim is to determine which shape will give me ...

I can also prove this with shape 1 + 2. In shape 1, which had a length of 10 and a width of 490, the area was 4900, while with shape 2, which had a length of 20 and width of 480, the area was 9600.

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