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• Level: GCSE
• Subject: Maths
• Word count: 1818

# Fencing Prblem

Extracts from this document...

Introduction

Introduction

A Farmer has a plot of land. She has only 1000m of fencing. She wants to fence off a plot of land. She is not concerned about the shape but the perimeter has to be exactly 1000m. The fenced off area has to be the maximum contained area. I am going to investigate the shape or shapes that could be used to fence in the maximum area using exactly 1000 metres of fencing each time.

Firstly, I am going to investigate 3 - sided shapes - triangles, 4 – sided shapes-quadrilaterals and 5 sided shapes-pentagons. From there on I am going to gather the results and see if I need to investigate any further.

TRIANGLES

In first part of my coursework I am going to investigate different types of triangles.

Firstly I will look at Isosceles triangles.

The formula for triangles:

There are two ways of working out the area. You will get the same answer with either formula.

• Area = ½ x base x height
• Area = base x height

2

Triangle 1

450m                        450m

100m

I divide the Isosceles triangle

450 m2 - 50m2 = Height m2

202,500 – 2,500 = 200,000

Height = square root of 200,000

Height = 447.21 to two decimal places.

To calculate the area of this Isosceles triangle we calculate:

Area = ½ x base x height

Area = ½ x 100 x 447.21

Area = 22,361 m2 (meter squared) to the nearest whole number.

Triangle 2

I chose a different size isosceles triangle to continue my investigation

425m                        425m

150m

...read more.

Middle

354

44,250

Triangle 5

293

293

42,924.5

In the above results, I have found out that the equilateral Triangle had the largest area. I also noticed that there is a pattern. The pattern is that as the base increases and the height decreases, the area also increases.

QUADRILATERALS

I am going to investigate different quadrilaterals making sure that the perimeter is 1000 metres at all times.

The formula for the quadrilaterals:

• Perimeter = length + width + length + width.
• Area = length x width.

My first shape is going to be a square:

250m

A square has 4 sides. The 4 sides of the square are all equal. Therefore that means that all the edges will have the length of 250m due to the perimeter being 1000m.

Area = length x width.

Area = 250 meters x 250 meters

Area = 62,5002 (meters squared)

The next 5 shapes are going to be different types of rectangles. They will be numbered for further investigation:

(1)                                450m

50m

(2)                400m

100m

(3)        300m

200m

(4)        350m

150m

(5)        280m

220m

Results of the Quadrilaterals.

 Shape Length(m) Width(m) Area(m2) Square 250 250 62500 Rectangle 1 450 50 22250 Rectangle 2 400 100 40000 Rectangle 3 300 200 60000 Rectangle 4 350 150 52500 Rectangle 5 280 220 61600

In the above results, I have found out that the Square had the largest area.

200 cm

To find out the area of a pentagon, I am going to divide the pentagon into 5 isosceles triangles.

Each length is equal to 200 cm as 1000

5

= 200

hyp                  hyp

h

200cm

## To find out the hypotenuse, I am going to use the tangent formula

Tan = Opp

Adj

Tan 36 =100

...read more.

Conclusion

 Shape Base(m) Height(m) Area(m2) Triangle 1 100 447 22,350 Triangle 2 150 418 31,350 Triangle 3 200 387 38,700 Triangle 4 250 354 44,250 Triangle6 300 316 47,400 Equilateral triangle 333.33 289 48,166 Triangle7 350 274 47,950 Triangle8 400 224 44,800

I then put these values in a graph, shown below. I realised I had reached the peak of the area as I reached the area of the equilateral triangle.

The following table consists of all the measurements to the quadrilaterals above.

 Shape Length(m) Width(m) Area(m2) Square 250 250 62500 Rectangle 1 450 50 22250 Rectangle 2 400 100 40000 Rectangle 3 300 200 60000 Rectangle 4 350 150 52500 Rectangle 5 280 220 61600

Bearing the calculations in mind, I then put the quadrilaterals in order and compared the base of the quadrilaterals with the Area. I than calculated the area for a few more quadrilaterals and tabulated them below in order and compared the base of the quadrilaterals with the Area.

 Base (M) Height(M) Area(M) 100 400 40000 150 350 52500 200 300 60000 220 280 61600 250 250 62500 280 220 61600 300 200 60000 350 150 52500 400 100 40000

I then put these values in a graph, shown below. I realised I had reached the peak of the area as I reached the area of the Square.

I put the maximum results together to compare them.

The Circle gave me a coverage area of 79,382.34 metres squared. The Square gave me a maximum coverage area of 62,5002 (metres squared). The Triangle gave me a maximum coverage area of 48,1662 (metres squared).

Comparing all the values investigated so far in comparison the Circle covers the largest farm area.  Therefore I would recommend the Farmer to build her fence in a circle, as she will cover the area of 79,382.342 metres squared.

...read more.

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

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