• 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
14. 14
14
• Level: GCSE
• Subject: Maths
• Word count: 1405

# The fencing problem

Extracts from this document...

Introduction

Guy McALL        Page         5/7/2007

Maths coursework

The fencing problem

## Aim

My aim is to find out the largest area that can be put into a field. That has a perimeter of fencing that is 1000meters long. I am going to find out the biggest area I can get using different shapes that only have a perimeter of 1000meters.

## Investigating rectangles

I have started with rectangles because they are the simplest to find the area of (H*B=area)

The perimeter=1000 because 250 + 250 + 250 + 250 the area is 62500 meters² I used the equation h*b.

### The perimeter=1000 because 400 + 400 +100 + 100 the area is 40000 meters² I used the same equation again h*b.

The perimeter=1000 because 300 + 300 + 200 + 200 the area is 60000 meters² I used the equation h*b

Conclusion for rectangles

### I found that the square had the biggest area than all the other rectangles. I found that the smaller the width the smaller the area (the width 100 and length 400 the are equals 40000.) I also found that the smaller the length the smaller the area (length 100 and width 400 the area equals 40000.)

...read more.

Middle

250

375

44194.17382

300

350

47434.1649

333⅓

333⅓

48112.52243

350

325

47925.72378

450

275

35575.62368

Investigating a Pentagon

Aim

I have to decided to find out the area of shapes with more sides to see if the area will increase compared with the square and triangle. For this irregular pentagon I am first going to divide the square and triangle up and work out the area of them then add them together to get the area.

Irregular pentagon

The perimeter is 1000 meters because 250 + 250 + 100 + 100 + 300=1000

The area=45000 meters²

Because: -

I split the pentagon in to a triangle and a square then I found the area of the two and add them up: -

Square: -

(B*H)=area

100*300=30000

Square + triangle = area

30000 + 15000= 45000

Regular Pentagon

For this sided shape and the other shapes like this I am going to first divide the pentagon in to triangles. Then work out the area of the triangle and then times by the number triangles I have split.

The perimeter of the pentagon is 1000 the area= 68819.09602meters² because

I first divided the pentagon in to 5 triangles then worked out the angles of the triangle. Then I worked out the height (o) Tan54=O/100

...read more.

Conclusion

## Investigating a 100 sided shape

The perimeter is 1000 because 10*100=1000

The area=79551.28988meters²

Because

Tan88.2=31.82051595

Tan88.2=O/A

31.82051595*5=

159.1025798

Height=159.1025798

## Investigating a circle

First I have to use a equation to find out what the radius is once I found out the radius is I can find the area.

First I am going to take the equation 2πr

Then I take the equation 1000=2πr

From this I work out the radius: -

1000=2πr  /2

500=πr     /π (3.141592654)

159.1549431=r

Now I can work out the area of the circle using the equation πr²

πr²=π*159.1549431²= 79577.47155

## The area is larger because the more side you have to a shape the larger the area will become this is only when the perimeter equals the same on the shapes. I have worked out a formula to find out all the areas of shapes

Now I can draw a graph that will show how the areas increase as you get more sides.

I would suggest after these results that the farmer should make a circle out of the 1000 meters of fencing. This would give him the largest area for his cows to graze.

...read more.

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. ## Investigating different shapes to see which gives the biggest perimeter

I have the angle, 36�, the opposite value, 100m, and I need to work out the height on the adjacent to the angle. So I can now apply the tan rule which is adjacent = opposite/tan 36�.

2. ## Fencing Problem

Rectangle 62499.99 1000 metres 249.1 x 250.1 2nd largest quadrilateral. The more narrowed down the closer dimension were to a square. Parallelogram (Trapezium) 39998.48 1000 metres On table of parallelograms. The smallest quadrilateral amongst the other 2. I stopped testing the values, as the area was nowhere near the area of a square.

1. ## Fencing problem.

This can be found by excluding 1800 from the exterior angle. Interior angle = 1800 - Exterior angle Interior angle = 1800 - 51.4 = 128.60 Now by using the help of the interior angle I have calculated I shall find of the angles within one of the five triangles that can be made in a regular heptagon.

2. ## The Fencing Problem.

of 1 Triangle (rad.) (rad.) (m) (square m) (square m) 5 =1000/A3 =B3*A3 =2*PI()/A3 =D3/2 =(B3/2)/TAN(E3) =(B3*F3)/2 =G3*A3 Having entered the correct information I will be able to calculate the areas of many regular polygons with different numbers of sides and with a perimeter of 1000m.

1. ## The Fencing Problem

1000 44194.17 300 350 316.23 1000 47434.16 350 325 273.86 1000 47925.72 400 300 223.61 1000 44721.36 450 275 158.11 1000 35575.62 Naturally, the perimeter must always equal 1000, and the measurements of the triangles have all met this criterion.

2. ## The Fencing Problem

For example, if the length of a parallelogram was 300m and the width of a parallelogram was 200m, I wouldn't be able to use the formula shown above. The formula to work out the perpendicular height is: h = w * sinx?

1. ## Math Coursework Fencing

Angle 30� 60� 90� Length m 450 450 450 Width m 50 50 50 Height m 25 43.3 50 Area m� 11250 19485 22500 Angle 30� 60� 90� Length m 400 400 400 Width m 100 100 100 Height m 50 86.6 100 Area m� 20000 34640 40000 Angle 30�

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

350 300 350 47430 Scalene Triangles 4 250 350 400 43301.27 5 300 300 400 44721.36 6 350 250 400 43301.27 7 400 150 450 29580.399 Equilateral Triangle 8 333.33 333.33 333.33 48,126.3 In order for me to analysis these results I will have to remove the results I got

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