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  • Level: GCSE
  • Subject: Maths
  • Word count: 4850

isopometric quotients

Extracts from this document...

Introduction

Cara Foley 10i

Isoperimetric Quotients

Isoperimetric Quotients of plane shapes are calculated using the formula:

I.Q. = 4π x Area of shape

(Perimeter of shape)²

I am going to investigate isoperimetric quotients of plane shapes and interpret my findings.

Firstly, I am going to look at flat shapes. Using the formula, I will calculate the isoperimetric quotients of the shapes.  Starting with the smallest 2D shape- a triangle- I will calculate the I.Q s of right-angled triangles. I will also do this with isosceles and equilateral triangles. I will move on to quadrilaterals and look at the I.Q s. Maybe there will be something about the results that will help me with further plane shapes; pentagon, hexagon, heptagon, nonagon, decagon and possibly a circle. With comparison, the results might show something about the shapes, such as a pattern.

Triangles

I am now going to study right-angled triangles.

Right-angled triangles

I will first look at the 3, 4, 5 right-angled triangles and then enlargements of it.

1.     image00.png

Perimeter= 3+4+5= 12 cm

Area= ½ x 4 x 3= 6cm²

I.Q. = 4 x π x 6

                    12²

I.Q. = 0.5236

I will look at similar enlarged right-angled triangles, based on the 3, 4, 5 triangle.

2.  image00.png

Perimeter= 6+8+10= 24 cm

Area= ½ x 6 x 8= 24 cm ²

I.Q= 4 x π x 24

                   24²

I.Q. = 0.5236

3.image00.png

Perimeter= 16+20+12=48 cm

Area= ½ x 16 x 12= 96 cm²

I.Q. = 4 x πx 96

48²

=0.5236

Similar 345 right-angled triangles have the same I.Q. The answers are all the same.  I predict that similar enlarged shapes make the same I.Q. answer.

I am now going to look at other right-angled triangles that are not versions of the 3, 4, 5.

Other right angled triangles

Firstly, in these triangles, I will need to find the hypotenuse length so I will use Pythagoras’ theorem.

4. image00.png

Hypotenuse = 4² + 6² = 52

                           =    √52

                           = 7. 2111 cm

Perimeter = 4+6+7.2111

                     = 17.2111 cm

Area= ½ x 4 x 6

           = 12 cm ²

I.Q. = 4 x π x 12

              17.2111²

I.Q. = 0.5090

5.  image00.png

H = 8²+ 12² = 208

=208

       = 14.422 cm

P = 12+8+14.422

     = 34.422

A = ½ x 8 x 12

      = 48

I.Q. = 4 x π x 48

               34.422²

I.Q. = 0.5090

6. image00.png

H = 3² + 2² = 13

      = 13

      = 3.6

P = 3+2+3.6

    = 8.6

A= ½ x 2 x 3

      = 3

I.Q. = 4 x π x 3

                  8.6²

I.Q. = 0.5097

7.image00.png

...read more.

Middle

Parallelogram

I will now look at parallelograms. I think that the I.Q. s will be less than that of a square but enlarged versions will have equal I.Q. s. Different parallelograms, however, will have different I.Q. s.

1. image07.png

C² = a² + b²

C² = 5² + 3²

C² = 34

C    = 34

C    = 5.830951895

P = 5.830951895 + 5.830951895 + 7 + 7

     = 25.66190379

A = 7 x 5 = 35

I.Q. = 4 x π x 35

              25.66190379²

I.Q. = 0.667882652

I will now look at an enlargement of the above parallelogram.

2. image07.png

C² = a² + b²

C² = 10² + 6²

C² = 136

C    = 136

C    = 11.66190379

P = 11.66190379 + 11.66190379 + 14 + 14

    = 51.32380758

A = 14 x 10 = 140

I.Q. = 4 x π x 140

            51.32380758²

I.Q. = 0.667882652

The I.Q. s were equal. I will now look at different parallelograms.

3.   image07.png

C² = a² + b²

C² = 6² + 4²

C² = 52

C    = 52

C    = 7.211102551

P= 7.211102551 + 7.211102551 + 9 +9

   = 32.4222051

A = 9 x 6 = 54

I.Q. = 4 x π x 54

             32.4222051²

I.Q. = 0.645533115

Here is an enlargement. I predict that it will have the same I.Q. as the above shape.

4. image07.png

C² = a² + b²

C² = 12² + 8²

C² = 208

C    = 208

C    = 14.4222051

P = 14.4222051 + 14.4222051 + 18 +18

    = 64.8444102

A = 18 x 12 = 216

I.Q. = 4 x π x 216

             64.8444102²

I.Q.= 0.645533115

The I.Q. s of the enlarged parallelograms are equal. My prediction was correct; the I.Q. s of parallelograms are less than that of a square.

I will now construct a table comparing the I.Q. s of my quadrilaterals.

Table of results for I.Q. s of Quadrilaterals

Square

I.Q

1

0.78539163

2

0.78539163

3

0.78539163

4

0.78539163

Rectangle

I.Q

1

0.736310778

2

0.736310778

3

0.502654824

4

0.502654824

1

0.120637157

2

0.384684814

3

0.623125815

4

0.753982236

Parallelogram

I.Q.

1

0.667882652

2

0.667882652

3

0.645533115

4

0.645533115

Enlarged similar quadrilaterals have the same I.Q. s Squares, a regular plane shape, have the largest I.Q. s of quadrilaterals. All square I.Q. s are the same.

Regular shapes have the largest I.Q. s out of their ‘family’ of shapes.

I can now just study regular shapes.

Regular pentagons

I will now look at regular pentagons.

image08.png

  To find the sum of the interior angles of a polygon, I can use the formula:

180(n-2) where n represents the number of sides.

For a pentagon

180(5-2) = 180 x 3 = 540°

Because I am looking at a regular pentagon each angle equals:

...read more.

Conclusion

°

40

50

0.99868

π Tan 86.4°

50

60

0.999

π Tan 87°

60

75

0.9994

π Tan 87.6°

75

90

0.99959

π Tan 88°

90

100

0.99967

π Tan 88.2°

100

My attached graph shows all of my results.

The following is a computer-generated version of my graph.

image15.png

As my graph is based on my table, it shows similar results. However, I can conclude that as the number of sides of a regular polygon increases, so does the I.Q. The I.Q. tends towards the number 1, but it does not actually reach the limit of 1.

I am now going to look at a circle. I predict that it will have an I.Q. of 1 because it has an infinite number of sides and it’s a regular shape.

I will now look at a circle with a radius of 6.

image11.png

1.

I.Q. = 4π x Area of shape

(Perimeter of shape)²

A = πr²

A = π x6² = 113.09

Perimeter/ Circumference = πD

                                                       = 37.69

I.Q. = 4 x π x113.09 = 1.000

                  (37.69)²

A general formula for a circle

image11.png

Let the radius be A.

A== πr²

A = π xA²

C= πD

     = π2A

I.Q. = 4 x π x π A²

                  (π2A)²

I.Q. = 4 x π² A²

π 2A x π 2A

I.Q. = 4 x π² A²

             4 x π² A²

  I.Q. of any circle= 1

This has shown that a shape with an infinite number of sides i.e circle, has the highest Isoperimetric Quotient, which is 1.

I can conclude that the I.Q. of any regular polygon is constant and the more sides a regulars polygon has, the bigger the I.Q. However, irregular polygons, or shapes with different sized sides, have differing isoperimetric quotients.

I have found a general formula to work out the I.Q. for any regular polygon.

I.Q. =

From my table of results and the graphs I have shown that as the number of sides of a regular polygon increases, the I.Q. approaches 1 but never actually reaches it. However, the circle, a regular polygon with an infinite number of sides, had an I.Q. of exactly 1.

This shows that the circle with its infinite number of sides has the highest I.Q. of 1 and that the limit of a plane shape is 1.

GCSE Mathematics Coursework

...read more.

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