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• Level: GCSE
• Subject: Maths
• Word count: 4266

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

Extracts from this document...

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;

1. Trigonometry
2. Algebra
3. 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

My formula is correct I know this because I subtracted them from each other and they were equal.

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

## 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

## 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

## 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

## 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 4500a circle 0.9999998371

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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