• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

The fencing problem - maths

Extracts from this document...

Introduction

The Fencing Problem In this piece of coursework, I will try and find out how I can get the biggest area of fencing, using only 1000m. I will do this by investigating different shapes I could make the fencing. I will start with rectangles: Rectangles Width (m) Length (m) Area (m2) 50 450 22500 100 400 40000 150 350 52500 200 300 60000 250 250 62500 300 200 60000 350 150 52500 400 100 40000 450 50 22500 Triangles Base(cm) Hypotenuse(cm) Height(cm) Area(cm2) 100 450 447 22350 150 425 418 31350 200 400 387 38730 250 375 354 44250 300 350 316 47400 350 325 274 47950 400 300 224 44800 450 275 158 35550 I found that an equilateral triangle is also classed as an isosceles triangle. So I decided to write down its data: Base(cm) Hypotenuse(cm) ...read more.

Middle

166.6666667 60 83.33333333 1.047197551 1.732050808 173.2050808 14433.75673 86603 7 51.42857143 142.8571429 64.28571429 71.42857143 1.121997376 2.076521397 207.6521397 14832.29569 103826 8 45 125 67.5 62.5 1.178097245 2.414213562 241.4213562 15088.83476 120711 9 40 111.1111111 70 55.55555556 1.221730476 2.747477419 274.7477419 15263.76344 137374 10 36 100 72 50 1.256637061 3.077683537 307.7683537 15388.41769 153884 11 32.72727273 90.90909091 73.63636364 45.45454545 1.285196995 3.405687239 340.5687239 15480.39654 170284 12 30 83.33333333 75 41.66666667 1.308996939 3.732050808 373.2050808 15550.2117 186603 13 27.69230769 76.92307692 76.15384615 38.46153846 1.329135353 4.057159486 405.7159486 15604.45956 202858 14 25.71428571 71.42857143 77.14285714 35.71428571 1.346396852 4.381286268 438.1286268 15647.45096 219064 15 24 66.66666667 78 33.33333333 1.361356817 4.704630109 470.4630109 15682.10036 235232 16 22.5 62.5 78.75 31.25 1.374446786 5.027339492 502.7339492 15710.43591 251367 17 21.17647059 58.82352941 79.41176471 29.41176471 1.385996759 5.349527506 534.9527506 15733.90443 267476 18 20 55.55555556 80 27.77777778 1.396263402 5.67128182 567.128182 15753.56061 283564 19 18.94736842 52.63157895 80.52631579 26.31578947 1.405449345 5.992671459 599.2671459 15770.18805 299634 20 ...read more.

Conclusion

This theory, shown and proven by the graph, can be applied to as many sides as possible until it reaches the point of a true circle. A circle has an infinite number of sides, therefore making the area as big as it could possibly be. The graph at this point wouldn't ascend any further, it would simply flatten out to show the lack of increase. Therefore, in conclusion, I have found in this piece of coursework, that the more sides the fence has the more its area will increase. Therefore I believe that if I carried on this pattern that the shape with the biggest area would be a circle as it has an infinite amount of sides, so its area couldn't get any bigger. ?? ?? ?? ?? Martin Sales ...read more.

The above preview is unformatted text

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. GCSE Maths Coursework Growing Shapes

    Odd Pattern no. (n) No. of lines No. of lines - 9n2- 9n 1 1 3 +3 3 2 21 +3 5 3 57 +3 7 4 111 +3 9 5 183 +3 Formula for working out the number of lines in an even pattern number = 9n2 -9n+3.

  2. The Fencing Problem

    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)

  1. Math Coursework Fencing

    So I decided to investigate a general formula for all polygon shapes after pentagon. This will allow me to calculate large numbers of polygons very quickly and accurately. As doing it step by step, as shown above, will raise the chances of doing mistakes in my calculations.

  2. The Fencing Problem.

    The area of one isosceles triangle will be calculated using the formula =(B3*F3)/2 where B3 is one equal side and F3 is the height. Finally the total area will be calculated by multiplying the area of one isosceles triangle by the number of sides.

  1. Fencing problem.

    111.1m The angles of the triangle from the centre of the nonagon can be found by dividing 3600 by the number of sides of the shape that has been shown above: Exterior angles = 3600 � Number of sides Exterior angles = 3600 � 9 Exterior angles = 400 I shall now find the interior angles of the above shape.

  2. Fencing Problem

    my rectangle I can multiply it by the height to find me the area of the rectangle. * 70.71067812 x 329.2893219 = 23284.27125 m� * Now that I have the area of my 2 Triangles and my Rectangle I can add the areas I have, to give me the overall area of the parallelogram.

  1. The Fencing Problem

    However as the base is increasing, the height is decreasing. This makes the area decrease back again. The area is largest somewhere around the 300m-400m so I'm going to zoom in around that point and do exactly the same as I did in the table above except this time I am going to go up by 10m.

  2. The Fencing Problem

    Area = (250000 x Tan (90 - (180/10))) / 10 = 76,942 I am now definitely sure that my formula is right. Finding the shape with the biggest area Now that I have a general formula, I can now check which shape has the biggest are.

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