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# Equible Shapes

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Introduction

Jamie Watson 11HC                                            Equable Shapes Coursework

Maths Coursework - Equable Shapes

A 2 dimensional equable shape is a shape whose area and perimeter are equal.

Example:

Area         =  4 * 4             = 16

4cm                              Perimeter =  4 + 4 + 4+ 4 = 16

4cm

The formula for all two dimensional equable shapes must be:

Perimeter = Area

I tried to work out the formula of any regular polygon. I realised that all regular polygons could be broken down into isoceles triangles:

Using only the side, length, x, it is possible to dicover the height of each of these triangles. Using the height I can then calculate the area of each triangle. By multiplying the number of triangles, of which there are the same amount as of sides,  by the area of each, I can discover the total area of the shape.

Middle

= tan[(180(n-2))/2n] * (x/2)

Using the height, I can work out the area of the individual triangle:

a ∆ = (h * x)/2

Which, with the formula for height added, looks like this:

a ∆ = tan[(180(n-2))/2n] *(x/2) * (x/2)

To find the area of the total shape, I will need to multiply the above formula by the number of triangles there are, which is also the number of sides, n:

a = tan[(180(n-2))/2n] *(x/2) * (x/2) * n

The perimeter of the total shape can be worked out by multiplying the length of the sides by the number of them:

p = nx

To work out the side length of an equable shape, therefore, I need to put the two formulas together and rearange it so I am able to calculate x:

nx = tan[(180(n-2))/2n]  * (x/2)

Conclusion

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x = 4/tan[(90(n-2))/n]

x = 4/tan[(90(3-2))/3]

x = 4/tan[90/3]

x = 4/tan30

x = 4/0.58…

x = 6.90cm (2dp)

## Square

x = 4/tan[(90(n-2))/n]

x = 4/tan[(90(4-2))/4]

x = 4/tan[180/4]

x = 4/tan45

x = 4/1

x = 4cm

## Octagon

x = 4/tan[(90(n-2))/n]

x = 4/tan[(90(8-2))/8]

x = 4/tan[540/8]

x = 4/tan67.5

x = 4/2.31…

x = 1.66cm (2dp)

### 673 Sided shape

x = 4/tan[(90(n-2))/n]

x = 4/tan[(90(673-2))/673]

x = 4/tan[60390/673]

x = 4/tan89.73…

x = 4/214.22…

x = 0.019cm (2sf)

To conclude, I have discovered a formula that will let me discover the side length required for an equible shape of any amount of sides. I think that my formulas will be reliable for any amount of sides, this is demonstrated by my example of a random large number of sides. However, there is a problem it will give an answer for shapes with 1 side, saying the length should be -7, which is impossible.

of

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