# The Fencing Problem.

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Introduction

Prediction

Now that I have looked at the different triangles and quadrilaterals, I can predict that the shapes with the greater number of sides, will have the maximum area as the number of sides, will affect the calculation of the area of the shape of the plot of land. Therefore, the larger the number of sides in the shape the greater the area of land. In turn this will benefit the farmer as the maximum area of land is achieved with the minimum amount of fencing which in the long run will save the farmer money.

Further to my prediction, I think that a circle would give the greatest area of land because a circle is technically made

Middle

c²= 333⅓²- 166.67²

c²=83 333⅓

c= 288.68 metres

Area = ½bh

=½ x333⅓ x 288.68

=48 122.52 metre²

OR

Area = ½bc SinA

=½ x 333⅓ x 333⅓ x sin 60º

=48 122.52

For the isosceles triangle I will take an example of different bases and heights to find out which combination gives the maximum area.

Number 1

a=100 b=c=450

height a²=b²+c²

50²=450²+c²

c²=450²-50²

c²= 200 00

c+ 447.21 meters

Area=½bh

=½x 100x 447.21

= 22 360.68 metres²

Number 2

a=200 b=c=400

height a²=b²+c²

100²=400²+c²

c²=400²-100²

c²=150 000

c=387.29 meters

Area=½bh

=½x200x387.29

=38729.83 metres²

Number 3

a=300 b=c=350

Conclusion

Area(m²)

1

100

447.21

22 360.68

2

200

387.29

38 729.83

3

300

316.23

47 434.16

4

400

223.61

44 722

The isosceles triangle which has the maximum area out of the ones which I have investigated is number 3. Number 3 has a base of 300 meters and a height of 316.23 meters. It also has an area of 47 434.16 metres²

I am now going to investigate a right angle triangle.

Ratio of length of sides

a : b : c

3 : 4 : 5

1=1000= 83½m

12

a = 3 x 83½= 250m

b = 4 x 83½= 333⅓m

c = 5 x 83½= 416⅔m

Angle A

a = c

SinA SinC

250 = 416⅔

SinA = Sin 90º

SinA x 416⅔= Sin 90º x 250

Sin A = Sin90º x 250

416⅔

SinA = 0.6

A=36.87º

Area= ½bc SinA

=½x333⅓x416⅔x Sin36.87º

=41 666.77 m²

The last type of triangle I am going to invest is the scalene.

Number 1

a=400, b= 450 & c=150

CosA= b²+c²-a²

2bc

CosA=450²+150²-400²

2x450x150

CosA= 65 000

135 000

CosA=0.48

A=61.22º

Area= ½bcSinA

=½x450x150xSin61.22º

=29 581.02

Number 2

a=350 b=450 c=200

CosA=b²+c²-a²

2bc

CosA=450²+200²-350²

2x450x200

CosA=120 000

180 000

CosA=0.67

A=48.19º

Area= ½bcSinA

=½x450x200xSin48.19º

=33 541.18m²

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

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