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  • Level: GCSE
  • Subject: Maths
  • Word count: 6538

The Fencing Problem

Extracts from this document...

Introduction

Fencing Coursework        Haaziq Farook  11o

Yr 10 GCSE Maths Coursework:

The Fencing Problem

In this investigation I plan to explore various polygons and deduce which type would yield the greatest area while complying with a perimeter of 1000m. I am only investigating regular polygons, because the otherwise would prove to be far too complex and involve an immeasurable amount of data. I will expand on this decision later in the investigation.

I intend to find a general formula for the area of any polygon and prove it by applying it to each polygon I explore.

The polygons I plan to investigate include:

  • Triangle (Isosceles, Scalene and Equilateral)
  • Rectangle
  • Parallelogram
  • Pentagon
  • Hexagon
  • Octagon
  • Decagon
  • A polygon with 15 sides (Pentadecagon)
  • A polygon with 20 sides (Icosagon)
  • A polygon with n number of sides (N-Gon)
  • Circle

As you can see, I am working my way up chronologically in aspects of the number of sides on each polygon. For each one I will produce a table of data displaying the progressive areas for the shape; the areas will increase/decrease respectively according to the varying unit (e.g. 50) used for each instance of the shape.

For every polygon I will produce a line graph presenting their progressive areas from the table of data. I will indicate the highest value on the graph (i.e. the highest area for the shape) and evaluate it accordingly.

The area formulas for each shape cited are;

  • Triangle (Isosceles) => A = ½(b x h)
  • Triangle (Scalene) => A = √s(s - a)(s - b)(s - c), where s = ½(a+b+c)
  • Triangle (Equilateral) => A = ½(b x h)
  • Rectangle => A = l x w
  • Parallelogram => A = b x perpendicular height
  • Pentagon => A = {½[(1000 ÷ 5) x h]} x 5
  • Hexagon => A = {½[(1000 ÷ 6) x h]} x 6
  • Octagon => A = {½[(1000 ÷ 8) x h]} x 8
  • Decagon => A = {½[(1000 ÷ 10) x h]} x 10
  • A polygon with 15 sides (Pentadecagon) => A = {½[(1000 ÷ 15) x h]} x 15
  • A polygon with 20 sides (Icosagon) => A = {½[(1000 ÷ 20) x h]} x 20
  • A polygon with n number of sides => Remains to be seen…mWaHaHa!
  • Circle => A = πr²
...read more.

Middle

50

475

474.34

1000

11858.54

100

450

447.21

1000

22360.68

150

425

418.33

1000

31374.75

200

400

387.30

1000

38729.83

250

375

353.55

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. As we can see, the highest value shown is 47925.72, before the area begins to decline to 44721.36. We can evaluate that an isosceles triangle with base = 350m, both sloping sides = 325m each accommodates the highest area whilst adhering to the set perimeter of 1000m.

One other thing we notice in the table is the last result which has an area of zero, although the perimeter is intact. This is an anomaly due to the fact that this triangle is physically impossible to draw; the sloping sides are too short to meet each other as the base is not shorter than both their lengths combined. i.e.

250250  

        500

Nevertheless, I have noticed that the difference between the highest value and the values either side of it is quite large. Therefore, I can further develop this analysis by producing another set of data, this time increasing the base values by a much smaller-scale number such as 10.

On the following pages, you will see the drawings and written method used to find the areas – I have implemented Pythagoras’ Theorem to find out the height of each triangle, and have substituted the gathered information into the formula accordingly. A graph which corresponds to the areas that I have just examined will also be shown following the drawings; the page which

succeeds that will continue my investigation.

1)

  475        475

50

     h                475

   25        

2)

     h                450

   50        

3)

     h                425

   75        

4)

     h                400

   100

5)

     h                375

   125        

6)

     h                350

   150        

7)

     h                325

   175        

8)

     h                300

   200        


image08.png


^ The graph has a smooth curve,; however, unlike the scalene triangles, the isosceles triangles results aren’t “symmetrical”; they ascend to a point, and then decline with greater magnitude. The explanation for this is because the sum of the two sloping sides must be greater than that of the base. As the base values are increasing, the sloping lengths must increase in order to comply with this standard.

Base (m)

Sloping Height (m)

Perpendicular Height (m)

Perimeter (m)

Area (m²)

300

350

316.23

1000

47434.16

310

345

308.22

1000

47774.21

320

340

300.00

1000

48000.00

330

335

291.55

1000

48105.35

340

330

282.84

1000

48083.26

350

325

273.86

1000

47925.72

360

320

264.58

1000

47623.52

370

315

254.95

1000

47165.93

Using the same formulas, but only changing the base values in scale of 10, I have produced this table of data.

Evidently, we see that the data has revealed a higher value that the previous table. As there is still a gap around the largest triangle, I will produce another table of data, this time increasing the base values by an even smaller number; 2. Again, a graph depicting the areas will be displayed after the diagrams.

1)

     h                350

   150        

2)

     h                345

   155        

3)

     h                340

   160        

4)

     h                335

   165        

5)

     h                330

   170        

6)

     h                325

   175        

7)

     h                320

   180        

8)

     h                315

   185        


image09.png


^ The graph resembles an “n” to a much greater extent than before; though it still, obviously, remains “unsymmetrical”.

Base (m)

Sloping Height (m)

Perpendicular Height (m)

Perimeter (m)

Area (m²)

320

340

300.00

1000

48000.00

322

339

298.33

1000

48030.92

324

338

296.65

1000

48056.97

326

337

294.96

1000

48078.09

328

336

293.26

1000

48094.24

330

335

291.55

1000

48105.35

332

334

289.83

1000

48111.37

334

333

288.10

1000

48112.23

336

332

286.36

1000

48107.88

338

331

284.60

1000

48098.24

340

330

282.84

1000

48083.26

342

329

281.07

1000

48062.87

344

328

279.28

1000

48036.99

346

327

277.49

1000

48005.55

The highest value has changed yet again; and, to make no room for error next, I will create another table of data with the base values being increased by 0.5. A final graph will be shown after the diagrams.

1)

     h                340

   160        

2)

     h                339

   161        

3)

     h                338

   162        

4)

     h                337

   163        

5)

     h                336

   164        

6)

     h                335

   165        

7)

     h                334

   166        

8)

     h                333

   167        

9)

     h                332

   168        

10)

     h                331

   169        

11)

     h                330

   170        

12)

     h                329

   171        

13)

     h                328

   172        

14)

     h                327

   173        


image10.png


^ We can see that the shape of the graph is very similar to an “n” shape now. We know that it cannot be symmetrical – you can notice that there is no exact middle point in the graph.  I have showed this on the graph by putting a line through the highest point (highest value).

Base (m)

Sloping Height (m)

Perpendicular Height (m)

Perimeter (m)

Area (m²)

330

335

291.55

1000

48105.35

330.5

334.75

291.12

1000

48107.34

331

334.5

290.69

1000

48109.00

331.5

334.25

290.26

1000

48110.35

332

334

289.83

1000

48111.37

332.5

333.75

289.40

1000

48112.07

333

333.5

288.96

1000

48112.45

333.5

333.25

288.53

1000

48112.50

334

333

288.10

1000

48112.23

334.5

332.75

287.66

1000

48111.64

335

332.5

287.23

1000

48110.71

335.5

332.25

286.79

1000

48109.46

336

332

286.36

1000

48107.88

336.5

331.75

285.92

1000

48105.97

Once again, the highest area has changed.

1)

     h                335

   165        

2)

     h                334.75

 165.25        

3)

     h                334.5

  165.5        

4)

     h                334.25

 165.75        

5)

     h                334

   166        

6)

     h                333.75

 166.25        

7)

     h                333.5

 166.5        

8)

     h                333.25

 166.75        

9)

     h                333

   167        

10)

     h                332.75

 167.25        

11)

     h                332.5

  167.5        

12)

     h                332.25

 167.75        

13)

     h                332

   168        

14)

     h                331.75

 168.25        


image11.png


^ Once more, we have discovered a higher area between the gaps. Judging by the measurements, I can predict that the largest area would be that of an equilateral triangle.  

Base (m)

Sloping Height (m)

Perpendicular Height (m)

Perimeter (m)

Area (m²)

333.33

333.33

288.68

1000

48112.52

This proves that my prediction is correct.

Triangles - Equilateral

Bearing in mind that the perimeter = 1000m, I have divided it by 3 to give us the one equilateral triangle.

Area        =  ½(b x h)

           Because 1000 ÷ 3 =333.33 (2d.p.)

333.33m333.33m

333.33m

333.33m        

166.67m

Since there is only one instance of the equilateral triangle, I surmise that there is no need for a graph to plot one point on. We can clearly see that the only area that an equilateral triangle will produce is 48112.63m²; this is the maximum area any triangle can hold while observing the 1000m perimeter.

Nevertheless, I can prove logically that neither triangle will hold the maximum potential area while abiding by the 1000m perimeter. This is because the shape of the triangle (regardless of whether it is isosceles, equilateral or even scalene) hinders it, as shown in various examples below. I indicate the “wasted space” which, if filled, not only gives us a bigger potential area, but tells us that quadrilaterals (specifically rectangles) have the ability to hold a higher maximum potential area (with the perimeter of 1000m).

        Isosceles

        Equilateral

“Wasted Space”

        Scalene


Quadrilaterals – Parallelograms

Continuing on with my investigation, I will now examine the extent of which parallelograms will go to hold its maximum area, again, whilst abiding by the decided perimeter of 1000m. The following table I created in Excel show the areas of parallelograms, with each interior angle increased by 10.

Area             = b x perpendicular height

Side B is set at 400, and the base (which has to comply in order to fulfil the 1000m perimeter.

a

b

a

Here is the table of data in formula view.

Base [a] (m)

Side [b] (m)

Angle (°)

Height [h] (m)

Perimeter (m)

Area (m²)

100

400

10

=SIN(RADIANS(C4))*400

=(2*A4)+(2*B4)

=D4*A4

100

400

20

=SIN(RADIANS(C5))*400

=(2*A5)+(2*B5)

=D5*A5

100

400

30

=SIN(RADIANS(C6))*400

=(2*A6)+(2*B6)

=D6*A6

100

400

40

=SIN(RADIANS(C7))*400

=(2*A7)+(2*B7)

=D7*A7

100

400

50

=SIN(RADIANS(C8))*400

=(2*A8)+(2*B8)

=D8*A8

100

400

60

=SIN(RADIANS(C9))*400

=(2*A9)+(2*B9)

=D9*A9

100

400

70

=SIN(RADIANS(C10))*400

=(2*A10)+(2*B10)

=D10*A10

100

400

80

=SIN(RADIANS(C11))*400

=(2*A11)+(2*B11)

=D11*A11

100

400

90

=SIN(RADIANS(C12))*400

=(2*A12)+(2*B12)

=D12*A12

100

400

100

=SIN(RADIANS(C13))*400

=(2*A13)+(2*B13)

=D13*A13

100

400

110

=SIN(RADIANS(C14))*400

=(2*A14)+(2*B14)

=D14*A14

100

400

120

=SIN(RADIANS(C15))*400

=(2*A15)+(2*B15)

=D15*A15

100

400

130

=SIN(RADIANS(C16))*400

=(2*A16)+(2*B16)

=D16*A16

100

400

140

=SIN(RADIANS(C17))*400

=(2*A17)+(2*B17)

=D17*A17

100

400

150

=SIN(RADIANS(C18))*400

=(2*A18)+(2*B18)

=D18*A18

100

400

160

=SIN(RADIANS(C19))*400

=(2*A19)+(2*B19)

=D19*A19

100

400

170

=SIN(RADIANS(C20))*400

=(2*A20)+(2*B20)

=D20*A20

...read more.

Conclusion

                                                                            A                               B

          1000 ÷ n

2)

To find out length AC;

AC = AB ÷ 2 = 0.5 x (1000 ÷ n) = 500 ÷ n

3)

To find out the interior angle AÔC

AÔB = 360 ÷ n         AÔC = 0.5 x (360 ÷ n)                A                            B

         = 180 ÷ n                                                1000 ÷ n

         2

4)

To acquire the perpendicular height;

        O

Tan AÔC =  AC ÷ h                                                                  Tan θ        A

∴ h Tan AÔC = AC

∴ h = AC ÷ Tan AÔC

        = 500 ÷ n

         Tan (180 ÷ n)

        = 500

          n Tan (180 ÷ n)

5)

Step 4) must be multiplied by 1000 ÷ n (base of triangle), as the area formula for a triangle is ½bh.

[ Continued overleaf ]

Continued…

= ½  1000    x   500

n        n Tan (180 ÷ n)

=        500

n² Tan(180 ÷ n)

6)

Now that I have found the general formula for one of the triangles in the polygon, all I must do is multiply the area of that triangle by the total number of triangles/sides the polygon has.

=   n       x         500²

 Tan (180 ÷ n)

Final simplification;

=        250,000

         n Tan (180 ÷ n)

I will now proceed to test (authenticate) the formula by testing it on a hexagon (n = 6) and a 1000-sided-shape (n = 1000).

1) Hexagon:

=        250,000

     6 x Tan (180 ÷ 6)

=        250,000        = 72168.78 (2d.p.)

        6 x Tan 30

2) 1000-sided-shape:

=               250,000

     1000 x Tan (180 ÷ 1000)

=               250,000        = 79577.21 (2d.p.)

            1000 x Tan 0.18

Now that I have verified that the formula works I am able to use it to find the formula of any regular polygon by substituting n with the number of sides the polygon has.

In Conclusion;

To conclude, I will summarise what I discovered during this investigation.

I found that as n (number of sides) increases, so does the area; this, in turn, led me to investigate the circle which, as I predicted, held the largest area whilst keeping to the set rule of the perimeter/circumference being 1000m.

I have also found out and showed in stages a general formula in which the area of any regular polygon can be found, just by substituting n for the number of sides the polygon has.

I have also shown a graph which shows clearly the maximum areas for each polygon, and this also reinforces the idea that as the number of sides increase, so does the maximum potential area.

...read more.

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