• 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
• Level: GCSE
• Subject: Maths
• Word count: 2237

# The Fencing problem.

Extracts from this document...

Introduction

The Fencing problem A farmer has exactly 1000 metres of fencing to fence of a piece of land. She wishes to fence off a plot of land, which contains the maximum area. I will investigate which shape will provide the maximum amount of area, whilst still maintaining a perimeter of 1000m. I shall start by looking at the shape with the least number of sides, the triangle The equilateral triangle The equilateral triangle is a set of special triangles in which all of the sides have are the same length, and the angles are equal. The isosceles triangle These, as well as the aforementioned equilateral triangle are the only triangles I shall be investigating. This is due to scalene and other irregular triangles have more than one variable, and so it is very difficult to calculate the area for them, as there are infinite combinations. With the isosceles triangle, if I know the base length, then I can work out the other two lengths, because they will be equal. If the base is 300m long then I can subtract that from 1000 and divide it by 2. ...read more.

Middle

The area of a parallelogram is equal to the base x vertical height. The vertical height can be worked out using 'side length sin ?'. I realised that the maximum value of ? was 90, as the graph for sine peaks at this point. I then worked out that this produced the result 250. The largest possible outcome, using a parallelogram is therefore the same as a square. The Rhombus The rhombus, like all quadrilaterals must have fours sides, but unlike others, these must all be equal, although the angles can change. Therefore, as I have found that regular shapes produce the greatest area, there can only be one value for any side, 250, and so the same outcome is produced as that of a square, producing an area of 62500m2. The Trapezium The trapezium can be worked out in almost the same way the same way as the parallelogram using the 'vertical height = side sin 90' formula. Using the area formula 'area = 1/2 x (a + b) x h' I realised that, as in the parallelogram, the largest value of sin could be 90. ...read more.

Conclusion

Using this formula, I worked out the area to be; 68819.1 Whilst, previously I found it to be; 68820 (when using a manual method to a lesser decimal place) Using this formula, I also worked out the area of shapes with large numbers of sides. No. of sides Area (m ) 50 79472.724 100 79551.29 200 79570.926 500 79476.424 1000 79577.21 And its graph; I have now come thoroughly proved that as the number of sides increases, so will the area. The results seem to be reaching a limit though, as they are getting closer and closer to one point. This point, or limiting factor, is the case of the circle, as it has an infinite number of sides. Hence, it will have a larger area, due to there being shorter vertices in a circle than any other shape, hence there is more area enclosed within the fence. To work out the area of a circle, I have to use a different formula, as I cannot put the number "infinity" into my formula. Instead, I shall use the formulae, A= ?r2, and C= ?d; C= ?d 1000 = ?d d = 1000 / ? d=318.31 (5 s.f) Hence, I have proved that the most efficient shape which the farmer should use with her fencing is a circle of perimeter 1000m. Will Witt The Fencing problem ...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

48056.97 326 337 337 1000 500 48078.09 328 336 336 1000 500 48094.24 330 335 335 1000 500 48105.35 332 334 334 1000 500 48111.37 334 333 333 1000 500 48112.23 336 332 332 1000 500 48107.88 338 331 331 1000 500 48098.24 340 330 330 1000 500 48083.26 After

2. ## Fencing problem.

= 1/2 � 250 m � 333.33 m Area of the right angled triangle = 41666 m2 Hence the area of this right-angled triangle is equal to 41666 m2 Scalene triangle The final triangle that I shall be investigating is a scalene triangle.

1. ## The Fencing Problem.

h = 2502-2502 h = 0m Area = 250 x 0 / 2 Area = 0m2 Triangle B: b = 400m s = 300m b/2 = 200m h = 3002-1002 h = ?50000m h = 223.6068m Area = 400 x 223.6068 / 2 Area = 44721.35955m2 Triangle C: b =

2. ## The Fencing Problem

30 70 20493.90 100 460 440 1000 500 400 40 60 21908.90 100 450 450 1000 500 400 50 50 22360.68 100 440 460 1000 500 400 60 40 21908.90 100 430 470 1000 500 400 70 30 20493.90 100 420 480 1000 500 400 80 20 17888.54 100 410

1. ## Fencing Problem

I then used hero's formula which is needed to find the area of the scalene triangle. The graph shows the side lengths in relation to the area of the scalene. The graph shows that as the side lengths get closer together, the area increases.

2. ## Maths:Fencing Problem

This table shows that 300 is the maximum an isosceles triangle will go up to because 350 will not make a triangle. So I will try to find out if 300 base is the real maximum. I will do this by uinvestigating around the numbers 300 to 349.

1. ## The Fencing Problem

380 310 244.949 46540.305 390 305 234.521 45731.554 400 300 223.607 44721.360 410 295 212.132 43487.067 420 290 200.000 42000.000 430 285 187.083 40222.817 440 280 173.205 38105.118 450 275 158.114 35575.624 460 270 141.421 32526.912 470 265 122.474 28781.504 480 260 100.000 24000.000 490 255 70.711 17324.116 Using this

2. ## The Fencing Problem

polygon is made up of n (n being the number of sides in that shape) number of equilateral triangles.

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