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  • Level: GCSE
  • Subject: Maths
  • Word count: 1346

I have been set the task of finding equable shapes and eventually find a possible formula for them. I will intend to do this by first using the rule of an equable shape, where the perimeter and the area have the same numerical value

Extracts from this document...

Introduction

Maths Investigation Introduction - I have been set the task of finding equable shapes and eventually find a possible formula for them. I will intend to do this by first using the rule of an equable shape, where the perimeter and the area have the same numerical value and extending it to regular polygons. E.g. Taking the sides as X Area = Perimeter X� = 4X X� - 4X = 0 X(X-4) = 0 X = 0 X = 4 I intend to set out the results, eventually in a table and possibly in a graph depending on the type of results I get. I hope to see a pattern in the results so I can predict the results in more complicated shapes. I intend to extend the task to surface area and volume and find a formula for them. I am only going to do regular polygons because irregular polygons could get very complicated and I doubt a general formula could be found for them. The first shape I am going to take is an equilateral triangle because it is a regular polygon and it has the least amount of sides a shape can have. The area of a triangle is 1/2 base * height and I am going to use L for the sides E.g. ...read more.

Middle

The final regular shape I will be testing is the decagon, Area = Perimeter 10 * area of triangle = 10L 10 * 1/2 L * 1/2 L Tan18� = 10L 1/2 ' Tan18� = 2 1/2 = 2Tan18� L = 4Tan18� Here I am doing the triangle again, I am going to split it up into 3 triangles and use the same method on this as the other shapes. Area = Perimeter 3 * area of triangle = 3L 3 * 1/2 L * 1/2 L Tan60� = 3L 1/2 ' Tan60� = 2 1/2 = 2Tan60� L = 4Tan60� As you can see the result I got fits a pattern, I am now going to put these results into a table and hopefully produce a general formula for any regular shape with any amount of sides. Number of sides Formula 3 4Tan60� 4 4Tan45� 5 4Tan36� 6 4Tan30� 8 4Tan22.5� 10 4Tan18� A table of results showing how sides relate to the formula As you can probably see, the greater the number of sides the less the angle is. I predict if you take the number of sides as N, then you start with 4tan and divide 180, because that's what all the angles of a triangle add up to and divide it by then you get your answer. ...read more.

Conclusion

Surface area = volume 5 times area of triangle = surface area of face 2 * 5 * 1/2 L * 1/2 L + 5L� = 5 * 1/2 * L * 1/2 L * L Tan45� Tan36� 1/2 L 1/4 L Tan36� + 1 = Tan36� 1/2 + Tan36� = 1/4 L 4(1/2 + Tan36�) = L 2 + 4 Tan36� = L Already a visible pattern has emerged. All the 3D shapes seem to have only one equable answer. Sides (N) Formula 3 2+4tan60 4 2+4tan45 5 2+4tan36 I predict the general case formula for any number of sides for a regular 3D shape will be 2+4Tan (180�) N Conclusion - I have discovered some key formulas for finding equable lengths for regular shapes and regular 3D shapes. There is a link between the 2D and 3D shapes, both 3 sided shapes had tan of 60 and 4 sided shapes had a tan of 45 and so on. If I continued with my investigation I would of liked to have done something with cones and shapes with a slant. The investigation went quite well although I found it quite difficult at the beginning to work out and understand a method that could be used for a shape with any number of sides. Once I had the method of breaking the shapes down into triangles it was fairly easy. My method could have been improved if it was set out neater then it would be more easy to understand. ?? ?? ?? ?? 1 ...read more.

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