# Equable shapes Maths Investigation

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Introduction

GCSE Maths Coursework: Strand 1 - Making and monitoring decisions Equable shapes: The task set is to investigate equable rectangles and further shapes. An equable shape is a shape whose perimeter is 'numerically equal to its area'. Equable Rectangles To start off my investigation I decided to experiment by drawing several rectangles with random lengths and widths. My drawings are included on graph paper. (see page ) I then decided using a table I would investigate every rectangle between that had a width or length between one and ten (whole numbers only). I worked out each shapes area and perimeter and if they were equal I highlighted them: Length Base Area Perimeter 1 1 1 4 1 2 2 6 1 3 3 8 1 4 4 10 1 5 5 12 1 6 6 14 1 7 7 16 1 8 8 18 1 9 9 20 1 10 10 22 2 1 2 6 2 2 4 8 2 3 6 10 2 4 8 12 2 5 10 14 2 6 12 16 2 7 14 18 2 8 16 20 2 9 18 22 2 10 20 24 3 1 3 8 3 2 6 10 3 3 9 12 3 4 12 14 3 5 15 16 3 6 18 18 3 7 21 20 3 8 24 22 ...read more.

Middle

the perimeter also equals 19.6 so it is an equable shape Before testing this formula on other numbers I needed an algebraic formula so replaced the known length or width with z Area = zx Perimeter = 2z + 2x Equation: zx = 2x + 2z I then solved this equation to have a final formula by moving the 2x to the other side of the equal sign but in brackets and as (-2) 2z = zx (-2x) So z = 2x x-2 This is the formula for working out equable rectangles but I tested it to see if it was accurate using trial and error. But the shape cannot exist because if 1 represents x, 2 multiplied between one is two, so two divided between (1-2 = -1) -1 = -2 and a length or width of a shape cannot be below zero. The same rule applies for 2. I already know that 3 works as the rectangle with a length and width of 3 and 6 was given as an example. I also know that 4 works as we already discovered the equable square/rectangle 4 by 4. Length Width Area Perimeter 1 -2 -2 -2 2 0 0 4 3 6 18 18 4 4 16 16 5 3.3 16.6 16.6 6 3 18 18 7 2.8 19.6 19.6 8 2.6 21.3 21.3 ...read more.

Conclusion

+ (b ) x b x 1/2 Regular Polygons I decided at this point as I could not find a formula for equilateral triangle on their own I could find a formula for a regular n-sided polygon. ( As equilateral triangle are regular polygons) To do this I needed to know at what angle does each edge subtend into the centre. To explain how to do this I would need to use tan to find the length. When drawing a regular polygon it can always be divided into triangles this is how I worked out the angle. The angle in my equation is represented by 90-180 divided by n. The equation I came up with which should work on any regular polygon is this: (x is length and n is number of sides) X = 4 Tan (90-180) N The formula works with any regular polygon such as a square X = 4 Tan 45? So x = 4 and already from my investigations I know a square with a side equalling 4 is an equable shape. Regular Polygons include: equilateral triangle, square, pentagon, hexagon, heptagon, octagon So I tried and tested my formula on hexagons. Hexagons have six sides so n would equal 6. X = 4 = 4 Tan 60? 3 The formula can be adapted to work for any regular polygon. ...read more.

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