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  • Level: GCSE
  • Subject: Maths
  • Word count: 1438

The Fencing Problem.

Extracts from this document...

Introduction

The Fencing Problem A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the largest size. I will investigate different shapes the fencing can make to achieve the largest area. Firstly I am going to investigate: Squares I am investigate the use of a square with a maximum area and a 1000m perimeter. The general formula to work out the area for this square is: AREA= xy NOT TO SCALE As the square has four equal sides there can only be one length of each side and one overall area. The length of this side must be: 1000m � 4 = 250m NOT TO SCALE Therefore the maximum area of the square is 62500m�. As I know this, I do not need to display my results in a table or graph as there is only one possible result of a 250m length. RECTANGLES To begin with I am going to investigate numerous rectangles that all have a perimeter of 1000 meters, to find out if there is any relevant pattern in my results. Firstly, the basic formula for quadrilateral shapes such as the rectangle and square is AREA = LENGTH x WIDTH In this case it is 498m x 2m = 994m� NOT TO SCALE The generalized formula to find out the area of this rectangle is: 2x+2y =1000 x +y=500 Area =xy =x (500-x) ...read more.

Middle

sin89.99 52499.48 200 300 sin89.99 59999.4 250 250 sin89.99 62499.38 300 200 sin89.99 59999.4 350 150 sin89.99 52499.48 400 100 sin89.99 39999.6 450 50 sin89.99 22499.78 490 10 sin89.99 4899.951 From my results I have found out the maximum area that can be achieved from a parallelogram is approximately 62499m� however this is when going by all the known rules of the parallelogram. If the maximum angle of a side can be 90� and both sides can become 250m then the maximum area of a parallelogram would be 62500m�. This shows the relation between all quadrilaterals in which that they are very much closely linked. Trapezium I am now going to investigate the use of trapeziums for the fence. The formula to work out the area of a trapezium is: AREA = (AB + CD � 2) x h When we look at the trapezium it is easy to see that it is made up of a rectangle and two right-angle triangles. This makes it a lot easier to work out the area of a trapezium. To work out the area of this trapezium I must first find out the area s of the triangles, so I then have the width of the rectangle. h� + 110�= 200� h= V200� -110� = V40,000 - 12,100 = 167.03m So now I have the height of the triangle, I also have the width of the rectangle and I can now find out the total area of this trapezium. ...read more.

Conclusion

Side (m) Height (m) Area (m ) 0 500.0 500.000 0.000 50 475.0 474.342 11858.541 100 450.0 447.214 22360.680 150 425.0 418.330 31374.751 200 400.0 387.298 38729.833 250 375.0 353.553 44194.174 333.3 333.3 288.675 48112.522 350 325.0 273.861 47925.724 400 300.0 223.607 44721.360 450 275.0 158.114 35575.624 500 250.0 0.000 0.000 Scalene I am now going to investigate the scalene triangle. With all triangles the area is worked out with the general formula: AREA = 1/2 x b x h However in order to work out the area of these triangles we must first work out the height. We can do this by splitting this triangle into two. Now the height of this triangle can be worked out. h� + 50� = 150� h� = 150� - 50� h� = V22500-2500 h = 141.421m AREA= 1/2 x 400 x 141.421 = 1/2 56568.4m� = 28284.271m� I am now going to work out the areas of other scalene triangles. AREA= 1/2 x 450 x 143.614 = 1/2 64626.3m� = 32313.165m� BASE SIDE 1 SIDE 2 HEIGHT AREA 400 150 450 141.421 28284.271 450 175 475 143.614 32313.165 500 200 300 132.288 33071.891 From this table of results it is easy to see that the scalene triangle with the largest area is the one with the 500m base and a 132.288m height. This triangle has an area of 33071.891m� AREA= 1/2 x 500 x 132.288 = 1/2 66144m� = 33071.891m� ALEX YORKSTON ...read more.

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