• 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
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  • Level: GCSE
  • Subject: Maths
  • Word count: 2086

The aim of this coursework is to investigate which shape gives the largest enclosed area for a fixed perimeter of 1000m. In the coursework I will be investigate different shapes with different number of sides to see which encloses the largest area.

Extracts from this document...

Introduction

GCES Mathematics Coursework                                              Jugdeesh Singh 10 Blue        

Jugdeesh Singh

Maths Coursework 2003

Mrs Phull  

10 Blue

Aim

The aim of this coursework is to investigate which shape gives the largest enclosed area for a fixed perimeter of 1000m. In the coursework I will be investigate different shapes with different number of sides to see which encloses the largest area.

Three sided shape Triangle

The only three sided shapes are triangle. There area depends on the length of each side. For the triangle I will be investigating which triangle with has the largest area by changing the lengths of each side and eventually getting the triangle with the largest area.

Prediction:

I predict that the triangle with equal sides will have the largest area. This is the equilateral triangle.  

First I will start by changing the length of the base. To calculate the area of the triangle I will be using the following formula

½ Base x Perpendicular Height

E.g.image00.png

image05.pngimage09.png

½ Base x Perpendicular Height

(½ x 10m) x 15mimage20.pngimage24.png

=        5m x 15m

=        75 m2

For some triangles the perpendicular height is not given, therefore we have to work the height out our selves. We will do this by applying Pythagoras’ Theorem, which is. The square on the hypotenuse is equal to the sum of the squares on the other two sides.

a2 + b2 = c2image45.pngimage08.png

image01.png

image02.pngimage03.png

E.g.image04.png

...read more.

Middle

424.264

29698.485

150

425.0

418.330

31374.751

160

420.0

412.311

32984.845

170

415.0

406.202

34527.163

180

410.0

400.000

36000.000

190

405.0

393.700

37401.537

200

400.0

387.298

38729.833

210

395.0

380.789

39982.809

220

390.0

374.166

41158.231

230

385.0

367.423

42253.698

240

380.0

360.555

43266.615

250

375.0

353.553

44194.174

260

370.0

346.410

45033.321

270

365.0

339.116

45780.727

280

360.0

331.662

46432.747

290

355.0

324.037

46985.370

300

350.0

316.228

47434.165

310

345.0

308.221

47774.209

320

340.0

300.000

48000.000

330

335.0

291.548

48105.353

333.3

333.3

288.675

48112.522

340

330.0

282.843

48083.261

350

325.0

273.861

47925.724

360

320.0

264.575

47623.524

370

315.0

254.951

47165.931

380

310.0

244.949

46540.305

390

305.0

234.521

45731.554

400

300.0

223.607

44721.360

410

295.0

212.132

43487.067

420

290.0

200.000

42000.000

430

285.0

187.083

40222.817

440

280.0

173.205

38105.118

450

275.0

158.114

35575.624

460

270.0

141.421

32526.912

470

265.0

122.474

28781.504

480

260.0

100.000

24000.000

490

255.0

70.711

17324.116

500

250.0

0.000

0.000

As you can see from the table above the maximum area occurs when all the sides are the same length. The length is 333 1/3m Hence the triangle with the largest area is an equilateral triangle.

Four sided shape

Again for the four sided shapes I will start with a base of a 10m and increase keep on increasing it. I predict that the largest area again will occur when all the sides will be equal. I will first start with a rectangle. Below will show a quadratic curve which will show what lengths give the maximum area for a four sided shape.

image29.png

image30.png

Perimeter = 2(500–x) + 2(x)

Area of rectangle = x(500-x)

X(500 – x) is a quadratic curve and looks like this.

y = x(500-x)

image36.pngimage37.pngimage38.pngimage39.pngimage40.pngimage32.pngimage41.pngimage33.pngimage31.pngimage34.pngimage35.png

 The formula for working out the area of a rectangle is.

Area = Length x Width

E.g.

image42.pngimage43.png

image44.png

image43.png

Area = Length x Width

A = 100 x 400

A = 4000m2

A = 10 x 490

A = 4900m2

A = 20 x 480

A = 9600m2

A = 30 x 470

A= 14100m2

A = 40 x 460

A = 18400m2

A = 50 x 450

A = 22500m2

...read more.

Conclusion

A

When ‘n’ tends to infinity

n                ∞                        n = number of sides

The length of each side tends to 0

l 0        l = length

and the area tend a fixed value

A              a fixed value        A = area

As the number of sides increase the shape becomes more circular, and a circle can be considered to be a regular polygon with infinite sides.

Below is a graph showing as the maximum area for a shape with a certain number of sides.

Circle

Area of circle = ∏r2

Since we do not know the radius we have to work it out by using the perimeter formula which is

Circumference = 2∏r

1000 = 2 x ∏ x r

500 = ∏ x r

500 = r

r = 159.155m (2dp)

Area = ∏r2

A = ∏ (159.1552)

A = 79577.472m2 (3dp)

Conclusion

In conclusion we can say that as the number of sides increase so does the area of a regular polygon. This is shown in the table above. We can also say that there is a general formula to find out the maximum area for a regular polygon and that is;

Area for any regular                        1000   2

polygon =              n                                                

                                                           4Tan   360

where n =number of sides                        2n

The diagram below shows how a circle achieves the largest area.

As the number of sides increase it takes up a larger area, this eventually leads to the circle which has the greatest area.

...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

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Investigating different shapes of gutters.

    I can show on a graph the maximum area of the rectangle Crosses x-axis when A=0 x(L - 2x) = 0 x=0 L-2x=0 2x=L x=L/2 Therefore it crosses the x-axis at 0 and L/2. I know the graph is a quadratic n-shaped graph because the Area= x(L-2x)

  2. Investigate different shapes of guttering for newly built houses.

    Triangle conclusion After investigating different angles for the triangular guttering, I have found that the triangle with the sides of 15cm and an angle of 90� gives the maximum cross-sectional area for the triangle, which is 112.5cm� Rectangle The angles in the rectangle cannot be varied, however the only restrictions

  1. Medicine and mathematics

    Thus we can say that the time of interval is proportional to the Amount of pencillin remaining in the body. Now if we compare the remaining amount of penicillin after 48 hours from administration with a interval of 6 hours the figure was: to 0.00149mg, and for this it was 0.031914 mg.

  2. Investigation to find out the number of matchsticks on the perimeter in a matchstick ...

    2 2 1st difference 2nd difference When the 1st difference is not the same but the 2nd difference is, the formula will follow the quadratic pattern: t = an2 + bn + c But I am going to use different letters.

  1. To investigate the isoperimetric quotient (IQ) of plane shapes using the calculation shown below.

    =4sin45 Perimeter = = = 8 IQ = = = =0.948059448 Algebraic formula for decagon: Area = 0.5sin36 =0.5sin36 =5 sin36 =5sin36 Perimeter = = = 8 IQ = = = = 0.966882799 I can now see that a pattern has emerged a general formula can be used to find the IQ of a shape with sides.

  2. To investigate the effects of a parachutes shape and surface area, on it time ...

    Another problem I had was cutting the shapes out along the lines I had draw. Because the plastic keeps slipping in between the blades of the scissors and not cutting the plastic or it would cut of the lines. This was overcome by simply getting a sharper pair of scissors and stretch the plastic out as I cut it.

  1. t shape t toal

    + (T+2) + (T+6) + (T+1) = T-total 5T + 7 = T-total Does this formula work for other t-shapes in the grid. 5T + 7 = T-total 5 ? 9 + 7 = T-total 52 = 52 Data from 5 by 5 grid 5 + 10 + 15 + 9 + 8 = 47 T + (T-3)

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

    80*420=33600 33600 9 410 90 90*410=36900 36900 Square 25 250 250 250*250=62500 62500 Analysis From this graph I can tell that there is an increase in Area as I decrease the gap between the length and width of the square.

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