• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16
17. 17
17
18. 18
18
19. 19
19
20. 20
20
21. 21
21
22. 22
22
23. 23
23
24. 24
24
25. 25
25
26. 26
26
27. 27
27
28. 28
28
29. 29
29
30. 30
30
31. 31
31
32. 32
32
33. 33
33
34. 34
34
35. 35
35
36. 36
36
37. 37
37
38. 38
38
39. 39
39
40. 40
40
41. 41
41
42. 42
42
43. 43
43
44. 44
44
45. 45
45
• 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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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:

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.

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.

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

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

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

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE IQ Correlation essays

1. ## Mathematics Statistics Coursework

the relevant data for my hypotheses and questions that I will investigate. My tables do have clear headings as they are typed out in bold. All my headings are relevant to what I need. I added another heading, 'Average SATS Score', as when a graph, chart or a diagram is shown, there are only two axes, not four.

2. ## GCSE Statistics Coursework

-61 240 29 370 19 240 29 349 2 243 32 302 49 243 32 392 -41 244 33 428 -77 245 34 300 -51 250 39 400 -49 250 39 400 -49 250 39 500 149 260 49 421 -70 266 55 411 -60 266 55 453 -102 267

1. ## Mayfield Maths Coursework

leaf diagram for the correlation between sick day and weight, to find the mode, mean and median now should be easier Average of age :( B2:B51) = 36.98 Average of weight: (E2:E51) = 73.18 That is one way of finding an average to support the correlation between age and weight.

2. ## An Investigation into Gender-Based Stereotyping Using IQ Estimates

the simple observable fact that s*x roles, have changed and are in the process of changing as time passes. Woman who are considered to be unable to hold positions of power or careers involving intelligence are at last being allowed to serve in such capacities.

1. ## Maths Coursework: investigation into the correlation between IQ and KS2 results

4 4 100 4 67 9 Smith Anjelina F 4 4 100 4 68 9 Hardman Rachael F 4 4 103 4 69 9 Mohammed Kiran F 3 3 74 3 70 10 Bolt Jim M 5 5 106 5 71 10 Urfon Homeed M 3 3 88 3 72

2. ## The 3 statements I am going to investigate are: -Does the gender of the ...

The frequency table has two extra columns, as shown in the example below: Below is the table needed for standard deviation using a frequency table, which I made by taking the levels achieved by girls in both years for English and subtracting them from the mean (3.78), then squaring this answer.

1. ## HYPOTHESIS Blonde girls are more intelligent than non blonde girls. Blonde girls that ...

90-99 11 18 Hall 93 10 Grimshaw 95 10 100-109 15 33 Hen 94 17 Calg 97 14 110-119 13 46 Yo 94 15 McFadden 97 35 Earnshaw 96 20 Lloyd 98 10 Jones 97 10 Iilyas 98 6 UNUSED DATA Green 97 42 Bron 99 50 Hunt 98 18

2. ## This experiment will show that there is a significant positive correlation between males and ...

esteem initially which assists in gaining academic success, rather than the other way around which I have been investigating. Expectations are another aspect of a persons psychological state that I have not been able to include in my estimations or predictions of my results.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to