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# 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

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

^ 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

^ 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

^ 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

^ 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 |

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

n² 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.

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

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