# An Investigation into the Varying Isoperimetric Quotients of Differing Shapes.

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Introduction

An Investigation into the Varying Isoperimetric Quotients of Differing Shapes

The purpose of this investigation is to find.

1: To find a general formula for the IQ of regular shapes.

2: To find a maximum and minimum IQ of shapes.

3: Try to find a pattern linking between the IQ and the dimensions of the shape.

To do this the investigation will be conducted with varying techniques to gain a general formula. Some of these techniques are;

- Trigonometry
- Algebra
- Area and perimeter of all shapes

A developing pattern in results.

When studying the results for each shape, one fact becomes apparent the higher the number of sides the shape has, the higher the IQ.

A circle, which has an infinite amount of sides, has the highest IQ of all shapes.

Therefore, shapes with a high number of sides also have a high IQ due to their similarity with a circle. It also stands to reason the triangle had the lowest IQ because it has the least number of sides of all shapes.

## Polygons

A convex polygon is one in which all of its corners point outwards.

The sum of the exterior angles of any convex polygon equals 360.

A

b

e

c

d

a+b+c+d+e=360

if a regular polygon with n sides has an exterior angle of ø then n ø= 360

it follows that : ø = 360/n and n = 360/ø

At each vertex of a convex polygon, the interior angle and exterior angle adds up to 180.

The sum of the interior angles of any convex polygon with n sides is (n-2) 180.

For example, the sum of the interior angle of a pentagon is (5-2)180 = 3 * 180 = 540

For any triangle, the exterior angle is equal to the sum of the interior opposite angles.

a

b c

c=a + b

## Squares

A square is a basic four-sided shape, all sides and angles are equal.

Middle

10 cm sided square

I will now just work with the original IQ formula and use tan 45 in the area.

Interior angle = 360/4

= 90

Exterior angle = 180 – 90

2

= 45

Sides o=5 /2 a=5/2

Area of triangular segment = tan 45 o/a

o = 10/2 * tan 45

o = 5 * tan 45

½ * (base * height)/2

= ½ (5 * tan 45 * 5)

=½ * 10 * 5 *tan 45

= 102/4 * tan 45

Total area = 4* 102/4 * tan 45

=4* 102 * tan 45

4

total area of the square = 102 * tan 45

= 100* tan 45

Perimeter = 10+ 10 + 10 + 10

= 40

## Formula given

IQ = 4∏ * area of the square

(perimeter of the square)2

IQ = 4∏ * 100 * tan 45

(40)2

IQ = 4∏ * 100* tan 45

402

IQ = 4∏ * 100 * tan 45

1600

IQ = 4∏ * tan 45

16

IQ = 1∏ * tan 45

4

IQ =∏ tan 45

4

This proves our prediction correct for squares.

I have spotted that the result of canceling the equation down as far as possible, results with the denominator of n, n being the number of sides.

## Regular pentagons

I am going to study the regular pentagon next, a regular pentagon is a five sided regular polygon, all its sides and angles equal. A regular pentagon is not a simple shape like the square even though it only has one more side than it, it’s the third smallest amount of sides a polygon can have (5). To work out the IQ of any regular polygon I need to find the area and perimeter of the shape. I can see this when I look at the isoperimetric formula

IQ = 4∏ * Area of the shape

(perimeter of the shape)2

I have labeled the length of the sides . To find the area of the pentagon I am going to use trigonometry. To use trigonometry I need a right angled triangle with another angle. At the moment I don’t have this. So I am going to cut the square into 10 equal segments.

Conclusion

We are going to use x as the sides.

Lets see if my formula is correct.

IQ = ∏ * tan 90 (n-2)/n

N

IQ = ∏ * tan 90 (8 – 2)/8

8

IQ = ∏ * tan 90 (6)/8

8

IQ = ∏ * tan 90 (0.75)

8

IQ = ∏ * tan 67.5

8

## IQ = 0.948059449

We can see this is definitely correct because it =

IQ = ∏ * tan interior angle/2

N

We will now investigate regular octagons substituting numbers for sides.

25 km sided octagon

Lets use my formula to find the IQ.

IQ = ∏ * tan 90 (n-2)/n

N

IQ = ∏ * tan 90 (8 – 2)/8

8

IQ = ∏ * tan 90 (6)/8

8

IQ = ∏ * tan 90 (0.75)

8

IQ = ∏ * tan 67.5

8

## IQ = 0.948059449

## Now we have this formula it doesn’t matter how large the sides are because we don’t have to use them in our workings.

## Regular decagons.

A regular decagons is a ten sided, shape with all its sides and angles the same.

We are going to try some simple examples of regular decagons. By using the formula that I have made.

We are going to use x as the sides.

Lets use my formula to find the IQ.

IQ = ∏ * tan 90 (n-2)/n

N

IQ = ∏ * tan 90 (10 – 2)/10

10

IQ = ∏ * tan 90 (8)/10

10

IQ = ∏ * tan 90 (0.8)

10

IQ = ∏ * tan 72

10

### IQ = 0.966882799

## Regular dodecahedrons.

A regular dodecahedrons is a 12 sided, shape with all its sides and angles the same.

We are going to try a example of regular dodecahedrons. By using the formula that I have made.

Lets use my formula to find the IQ.

IQ = ∏ * tan 90 (n-2)/n

N

IQ = ∏ * tan 90 (12 – 2)/12

12

IQ = ∏ * tan 90 (10)/12

12

IQ = ∏ * tan 90 (0.8333)

12

IQ = ∏ * tan 75

12

## IQ = 0.977048616

## A regular 4500 sided shape.

A regular 4500 sided shape has 4500 sides, with all its sides and angles the same.

A regular 4500 sided shape is basically a circle.

We are going to try a example of regular 4500 sided shape.. By using the formula that I have made.

Lets use my formula to find the IQ.

IQ = ∏ * tan 90 (n-2)/n

N

IQ =∏ * tan 90(4500-2)/4500

4500

IQ = ∏ * tan 90(4498)/4500

4500

IQ = ∏ * tan 90(0.999555555)

4500

IQ = ∏ * tan 89.96

4500

## IQ = 0.999999837

## Circle

IQ = 4∏ * ∏ r*r

(2∏r)2

= 4∏ * ∏ x*x

2∏x * 2∏x

= 4∏ * ∏

2∏ * 2∏

=∏

∏

= 1

1

= 1

Number of sides of shape | IQ |

4 | 0.785398163 |

5 | 0.864806266 |

6 | 0.906899682 |

8 | 0.948059449 |

10 | 0.966882799 |

12 | 0.977048616 |

4500 a circle | 0.999999837 1 |

Results table

This table shows the higher the number of sides a shape has the higher the IQ will be.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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