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

Fencing Problem

Extracts from this document...


Fencing Problem In this piece of coursework I shall try to obtain the biggest possible area enclosed by 1800m of fencing. I shall look at different types of sizes and shapes of fences to obtain the biggest area. First I shall look at rectangles as they are easiest to calculate. P=2L+2w 1800=2L+2w 900=L+w 900-w=L I shall use this formula to work out different lengths by substituting a different number for w. A=L x w A=(900-w) x w w l=900-w Area=w(900-w) 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900-50=850 900-100=800 900-150=750 900-200=700 900-250=650 900-300=600 900-350=550 900-400=500 900-450=450 900-500=400 900-550=350 900-600=300 900-650=250 900-700=200 900-750=150 900-800=100 900-850=50 50(900-50)=42500 100(900-100)=80000 150(900-150)=112500 200(900-200)=140000 250(900-250)=162500 300(900-300)=180000 350(900-350)=192500 400(900-400)=200000 450(900-450)=202500 500(900-500)=200000 550(900-550)=192500 600(900-600)=180000 650(900-650)=162500 700(900-700)=140000 750(900-750)=112500 800(900-800)=80000 850(900-850)=42500 I tabulated the results because it allows you to see the different areas with the different lengths. This table shows that a rectangle of both lengths and widths of 450 (square) gives the biggest area in terms of the rectangle. ...read more.


I have found that each time I have found the area the sides and angles are all the same. Use Trigonometry to find the height Use O to help you when you are trying to find A T A A=O/T A=(900/n)/tan(180/n) 1/2x1800/n x ((900/n)/tan(180/n)) 1x1800x900/n 2x n tan(180/n) (810000/n) ntan(180/n) nx (810000/n) ntan(180/n) Therefore, after all of my calculations my formula for an n-sided shape has come to 810000 ntan(180/n) I shall check it with a pentagon 810000 =222973.8711 5tan(180/5) Centre angle=72 Each side =360 72/2=36 (T) 360/2=180 (O) A =O/T We need to find the height A=180/tan36 A=247.7487457 Now to find the area of the triangle. (247.7487457x360)/2=44594.77422 Times the answer by 5 because there are 5 triangles in a pentagon 44594.77422x5=222973.8711 I shall now use my new formula to calculate 3-50 sided shapes instead of working everything out like a pentagon. I have also used excel to do this. N-sided shape 810,000/(n(tan(180/n))) 3 155885 4 202500 5 222974 6 233827 7 240283 8 244439 9 247273 10 249292 11 250782 12 251913 13 252792 14 253489 15 254050 16 254509 17 254889 18 255208 19 255477 20 255707 21 255905 ...read more.


With this vital piece of information I came to estimate that the area will be biggest in an equilateral triangle with all sides 600. After a couple of workings out of some triangles I came to discover another faster and easer way to calculate a triangle and this formula is called the semi perimeter. After a lot of working out I found that my estimate was right with an area of 155884.5727. I then moved onto other regular polygons and saw that each time the area increased however the amount it decreased by got smaller each time. I also noticed that each time you add a side onto a shape it looks more and more like a circle. After I had finished the regular polygons I guessed that because it was the only shape I had not done the shape with the most maximum area will be a circle. I was right in guessing that the shape with the largest areas was a circle with 257,831. ?? ?? ?? ?? ...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. Fencing Problem

    34641.016 80 10 100 17.36481777 98.4807753 382.6351822 1000 39392.31 89 1 100 1.745240644 99.98476952 398.2547594 1000 39993.908 89.5 0.5 100 0.872653549 99.99619231 399.1273465 1000 39998.48 By analysing my table of results I can see that the larger "angle a" Is the larger the area of the parallelogram is.

  2. Fencing problem.

    I shall now substitute the height into the formula below: Area of a triangle = 1/2 � Base � Height Area of a triangle = 1/2 � 166.7 � 144.4 = 12035.7m2 I have now found the area of one triangle.

  1. 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.

  2. Maths Coursework - The Fencing Problem

    When I multiply this by five to get the total area of the pentagon I get the result, 68800 m� Conclusion Using the area of a right angled triangle, I was able to find the total area of a pentagon with a 1000m perimeter, which is 68800 m� The Hexagon The hexagon is very similar, only with one extra side.

  1. The Fencing Problem

    Base (m) Sloping Height (m) Perpendicular Height (m) Perimeter (m) Area (m�) 50 =(1000-A4)/2 =SQRT((B4*B4)-((0.5*A4)*(0.5*A4))) =A4+(2*B4) =(A4*C4)/2 =A4+50 =(1000-A5)/2 =SQRT((B5*B5)-((0.5*A5)*(0.5*A5))) =A5+(2*B5) =(A5*C5)/2 =A5+50 =(1000-A6)/2 =SQRT((B6*B6)-((0.5*A6)*(0.5*A6))) =A6+(2*B6) =(A6*C6)/2 =A6+50 =(1000-A7)/2 =SQRT((B7*B7)-((0.5*A7)*(0.5*A7))) =A7+(2*B7) =(A7*C7)/2 =A7+50 =(1000-A8)/2 =SQRT((B8*B8)-((0.5*A8)*(0.5*A8))) =A8+(2*B8) =(A8*C8)/2 =A8+50 =(1000-A9)/2 =SQRT((B9*B9)-((0.5*A9)*(0.5*A9))) =A9+(2*B9) =(A9*C9)/2 =A9+50 =(1000-A10)/2 =SQRT((B10*B10)-((0.5*A10)*(0.5*A10))) =A10+(2*B10) =(A10*C10)/2 =A10+50 =(1000-A11)/2 =SQRT((B11*B11)-((0.5*A11)*(0.5*A11))) =A11+(2*B11) =(A11*C11)/2 =A11+50 =(1000-A12)/2 =SQRT((B12*B12)-((0.5*A12)*(0.5*A12)))

  2. The Fencing Problem

    with sides of 333.3m long and a perimeter of 1000m.The area of this triangle is 48112.522m�. The triangle is an equilateral triangle. Quadrilaterals I will investigate if and how the interior angle of a parallelogram effects the area of the parallelogram and see if the parallelogram has a bigger area than a rectangle with the same perimeter of 1000m.

  1. t shape t toal

    72 73 74 75 76 77 78 79 80 81 If we start with the 1st T-shape drawn on the grid we get a T-number of 50 and a T-total of 187. If we than translate the T-shape up to have the T-number as 41 and then the T-total is 142.

  2. The Fencing Problem

    This shape is a circle and I will try to work out the area of a circle with a circumference of 1000m. However, I will need to know the radius of the circle before I can calculate its area.

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