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The fencing problem.
Member rating: No Rating | Words: | Submitted: Mon Jul 12 2004
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THE FENCING PROBLEM INTRODUCTION In this investigation, I have to find out a farmers problem who needs to build a fence that is 1000m long. I am investigating which shape would give the maximum area with a 1000m perimeter. I will be investigating the properties of a 1000m perimeter fence. I am going to start by drawing several regular and irregular rectangles, all with a perimeter of 1000m. All drawings are not to scale. m = Metres 4 SIDED SHAPES Square To find out an area of a square = base x length In this case it will be = 250m x 250m = 62500m. Rectangles 1. Area = 300 x 200 = = 60000m 2. Area = 350 x 150 =52500m 3. Area = 400 x 100 =40,000m 4. Area = 450 x 50 =22,500m In a rectangle, any two different length sides will add up to 500m, because each side has an opposite with the same length. That's why when we look at the triangles above...
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