Table of Results for the I.Q. s of Triangles
All the I.Q. s are under 1; they are 0. numbers.
From the table, equilateral triangles have the largest I.Q. s. This could possibly be because it is a regular plane shape with equal sides and angles.
Quadrilaterals
I am now going to study quadrilaterals and their enlargements. I think that enlargements will give the same I.Q. s. I think also that a square will give the largest I.Q. because it is a regular plane shape; this is my prediction. I will look at the following:
- Square
- Rectangle
- Parallelogram
Square
Squares are all enlargements of each other. I predict, from this, that all squares will have the same I.Q. and the largest I.Q. of all the quadrilaterals because it is a regular four-sided shape.
1.
P=2+2+2+2= 8
A= 2² =4
I.Q. = 4 x π x 4
8²
I.Q. =0.785398163
2.
P = 4+4+4+4 = 16
A= 4²=16
I.Q. = 4 x π x 16
16²
I.Q. = 0.785398153
3.
P=5+5+5+5 = 20
A= 5² = 25
I.Q.= 4 x π x 20
25²
I.Q. =0.78539153
4.
P = 6+6+6+6= 24
A= 6² = 36
I.Q.= 4 x π x 36
24²
I.Q.= 0.785398163
My prediction was correct; the I.Q. s were all the same. The square’s I.Q. was also larger compared to the I.Q. of an equilateral triangle.
Rectangles
I will now look at rectangles. I think that enlarged similar rectangles will have equal I.Q. s but have different I.Q. s to other rectangles. I predict that the I.Q. s will be smaller than that of a square.
1.
P= 5+5+3+3= 16
A= 5x3 = 15
I. Q. = 4 x π x 15
16²
I.Q. =0.736310778
Here is an enlargement of the above triangle.
2.
P= 6+6+10+10 = 32
A = 6x10 = 60
I.Q. = 4 x π x 60
32²
I.Q. = 0.736310778
I will now look at a different rectangle.
3.
P = 8+8+2+2 = 20
A =8x2 = 16
I.Q. =4 x π x 16
20²
I.Q.= 0.502654824
This is an enlargement of the previous rectangle.
4.
P = 16+16+4+4 = 40
A = 4x16 = 64
I.Q. = 4 x π x 64
40²
I.Q. = 0.502654824
The I.Q. s of rectangles are less than that of a square. Enlarged rectangles have equal I. Q. s but different I.Q. s to different triangles.
I will now investigate rectangles that have the same area (24cm² but different perimeters. I will look what happens to the size of the I.Q. s. I predict that the more ‘square’ a rectangle is, the bigger its I.Q.
1.
P= 24+24+1+1= 50
A = 24
I.Q. = 4 x π x 24
50²
I.Q. = 0.120637157
2.
P = 12+12+2+2 = 28
A = 24
I.Q. = 4 x π x 24
28²
I.Q.= 0.384684814
3.
P = 8+8+3+3 = 22
A = 24
I.Q. = 4 x π x 24
22²
I.Q. = 0.623125815
4.
P = 6+4+6+4= 20
A= 24
I.Q.= 4 x π x 24
20²
I.Q.= 0.753982236
My prediction was correct; the more ‘square’ the rectangle, the larger the I.Q.
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
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:
Each angle = 540 = 108°
5
I think that the I.Q. s of all regular pentagons will be the same because they are all enlargements of another.
1. There are 5 triangles inside the pentagon, they are the same as each other.
SOH CAH TOA
Tan A = O
A
Tan 54°= O
2.5
2.5 x Tan 54 = O
3.440 cm = O
P= 5+5+5+5+5 = 25
A of 1 triangle =½ x b x h
= ½ x 5 x 3.440
=8.6
Total area of pentagon = 8.6 x 5
= 43
I.Q. = 4 x π x 43
25²
I.Q. = 0.864566298
The regular pentagon’s I.Q. is bigger than the I.,Q. of a square. The square’s I.Q. is larger than that of an equilateral triangle.
It seems regular shapes with more sides have a larger I.Q.
I will now look at another regular pentagon; the I.Q. should equal 0.864566298.
2.
Tan A = O
A
Tan 54°= O
3.5
3.5 x Tan 54 = O
4.8173 cm = O
P = 7+7+7+7+7 = 35
A of I triangle = ½ x 7 x 4.8173
=16.86055
Total area of pentagon= 16.86055 x 5
= 84.30275
I.Q. = 4 x π x 84.30275
35²
I.Q. = 0.864799673
3.
Tan A = O
A
Tan 54°= O
4.5
4.5 x Tan 54 = O
6.19371 cm = O
P = 9+9+9+9+9 = 45
A of 1 triangle =½ x 9 x 6.19371
= 27.871695
Total area of pentagon = 27.871695 x 5
= 139.358475
I.Q.= 4 x π x 139.358475
45²
I.Q.= 0.864805059
All regular pentagons have the same I.Q. s
A general formula for a regular pentagon
Sum of interior angles = 180(n-2)
=180(5-2)
=180 x 3
= 540°
Each angle = 540 = 108°
5
Let the length of each side of a pentagon by 2A.
For the area, we need to calculate the perpendicular height (h) of each of the triangles.
Tan 54° = O
A
A x Tan 54° = O
Area of 1 triangle=½ x b x h
= ½ x 2A X A x Tan 54°
= A²Tan 54°
Area of 5 triangles= 5A²Tan 54°
(whole pentagon)
P = 2A+2A+2A+2A+2A = 10 A
I.Q. = 4π x Area of shape
(Perimeter of shape)²
I.Q.= 4x π x 5A²Tan 54°
(10A) ²
I.Q.= 4x π x 5A²Tan 54°
100A²
I.Q. = 20 π Tan54°
100
I.Q. for any regular pentagon = π Tan54°
5
I.Q. =0.864806266, which is the I.Q. for all my regular pentagons.
Regular Hexagons
I think the regular hexagons will have an I.Q. larger than that of the shapes I have already looked at. I think that all regular hexagons will have the same I.Q. s.
Sum of interior angles = 180(n-2)
=180(6-2)
=180 x 4
= 720
Each angle = 720 = 120°
6
Tan A = O
A
Tan 60°= O
3
3 x Tan 60 = O
5.19615 cm = O
A of 1 triangle = ½ x 6 x 5.19615
= 15.58845
A of 6 triangles = 6 x 15.58845
= 93.5307
P = 16+16+16+16+16+16 = 96
I.Q. = 4 x π x 93/5307
96²
I.Q. = 0.906899259
2.
Tan A = O
A
Tan 60°= O
2.5
2.5 x Tan 60 = O
4.3301 cm = O
A of 1 triangle =½ x 5 x 4.3301
= 10.82525
A of 6 triangles = 6 x 10.82525
= 64.9515
P = 5+5+5+5+5+5 = 30
I.Q. = 4 x π x 64.9515
30²
I.Q. =o.90894023
I will now look at a final regular hexagon.
3.
Tan A = O
A
Tan 60°= O
8
8 x Tan 60 = O
13.8564 cm = O
Area of 1 triangle =½ x 16 x 13.8564
=110.8512
A of 6 triangles = 110.8512 x 6
= 665.1072
P =16+16+16+16+16+16 = 96
I.Q. = 4 x π x 665.1072
96²
I.Q. = 0.906899259
The I.Q. s of regular hexagons are equal.
A general formula for a regular hexagon
Sum of interior angles = 180(n-2)
= 180(6-2)
=180 x 4
= 720
Each angle = 720 = 120
6
Tan 60° = O
A
A x Tan 60° = O
Area of 1 triangle=½ x b x h
= ½ x 2A X A x Tan 60°
= A²Tan 60°
Area of 6 triangles= 6A²Tan 60°
(whole hexagon)
P = 2A+2A+2A+2A+2A+2A = 12 A
I.Q. = 4π x Area of shape
(Perimeter of shape)²
I.Q.= 4x π x 6A²Tan 60°
(12A) ²
I.Q.= 4x π x 6A²Tan 60°
144A²
I.Q. = 24 π Tan60°
144
I.Q. for any regular hexagon = π Tan60°
6
I.Q. =0.906899682, which is the I.Q. for all my regular hexagons.
Regular octagons
I think that regular octagons will have an I.Q. larger than any other regular polygon I have studied. I predict that they will all have the same I.Q.
1.
Sum of interior angles = 180(n-2)
= 180 (8-2)
= 180 x 6
=1080
Each angle = 1080 = 135°
8
Tan A = O
A
Tan 67.5°= O
2
2 x Tan 67.5 = O
4.8284 cm = O
Area of 1 triangle =½ x 4 x 4.8284
= 9.6568
A of 8 triangles = 8 x 9.6568
= 77.2544
P =4+4+4+4+4+4+4+4 = 32
I.Q. = 4 x π x 77.2544
32²
I.Q. = 0.948054123
2.
Tan A = O
A
Tan 67.5°= O
2.5
2.5 x Tan 67.5 = O
6.0355cm = O
Area of 1 triangle =½ x 5 x 6.0355
= 15.0875
A of 8 triangles = 8 x 15.0875
= 120.7
P = 5+5+5+5+5+5+5+5 = 40
I.Q. = 4 x π x 120.7
40²
I.Q. = 0.947975583
My prediction was correct; the I.Q. s are the same.
A general formula for a regular octagon
Sum of interior angles = 180(n-2)
= 180 (8-2)
= 180 x 6
=1080
Each angle= 1080 = 135
8
Tan 67.5° = O
A
A x Tan 67.5° = O
Area of 1 triangle=½ x b x h
= ½ x 2A X A x Tan 67.5°
= A²Tan 67.5°
Area of 8 triangles= 8A²Tan 67.5°
(whole octagon)
P = 2A+2A+2A+2A+2A+2A+2A +2A= 16 A
I.Q. = 4π x Area of shape
(Perimeter of shape)²
I.Q.= 4x π x 8A²Tan 67.5°
(16A) ²
I.Q.= 4x π x 8A²Tan 67.5°
256A²
I.Q. = 32 π Tan67.5°
256
I.Q. for any regular octagon = π Tan67.5°
8
I.Q. =0.948059449 which is the I.Q. for all my regular octagons.
Regular Nonagons
I’ve noticed that the general formula always has a π, tan, ½ interior angle and then÷ by the number of sides.
I think that nonagons will have a larger I.Q. than that of an octagon.
Sum of interior angles = 180(n-2)
=180(9-2)
=180 x 7
=1260
Each angle = 1260 = 140
9
Tan A = O
A
Tan 70°= O
3
3 x Tan 70 = O
8.24243 cm = O
Area of 1 triangle =½ x 6 x 8.24243
= 24.72729
A of 9 triangles = 9 x 24.72729
= 222.54561
P =6+6+6+6+6+6+6+6+6 = 54
I.Q. = 4 x π x 222.54561
54²
I.Q. = 0.959050279
2.
Tan A = O
A
Tan 70°= O
4.5
4.5 x Tan 70 = O
12.36364 cm = O
Area of 1 triangle =½ x 9 x 12.36364
= 55.63638
A of 9 triangles = 9 x 55.63638
= 500.72742
P =9+9+9+9+9+9+9+9+9 = 81
I.Q. = 4 x π x 500.72742
81²
I.Q. = 0.959049891
The I.Q. s of regular nonagons are equal; just like my prediction.
A general formula for a regular nonagon
Sum of interior angles = 180(n-2)
=180(9-2)
=180 x 7
=1260
Each angle = 1260 = 140
9
Tan 70° = O
A
A x Tan 70° = O
Area of 1 triangle=½ x b x h
= ½ x 2A X A x Tan 70°
= A²Tan 70°
Area of 9 triangles= 9A²Tan 70°
(whole nonagon)
P = 2A+2A+2A+2A+2A+2A+2A +2A+2A= 18 A
I.Q. = 4π x Area of shape
(Perimeter of shape)²
I.Q.= 4x π x 9A²Tan 70°
(18A) ²
I.Q.= 4x π x 9A²Tan 70°
324 A²
I.Q. = 36 π Tan70°
324
I.Q. for any regular nonagon = π Tan 70°
9
I.Q. =0.959059541 which is the I.Q. for all my regular nonagons.
Regular Decagons
I will now look at a regular 10-sided shape. I predict that all the I.Q. s will be the same and they will be larger than the I.Q. s I have already studied.
1.
Sum of interior angles= 180(n-2)
= 180(10-2)
=180 x 8
=1440
Each angle = 1440 = 144
10
Tan A = O
A
Tan 72°= O
2
2 x Tan 72 = O
6.15536 cm = O
Area of 1 triangle =½ x 4 x 6.15536
= 12.31072
A of 10 triangles = 10 x 12.31072
= 123.1072
P =4+4+4+4+4+4+4+4+4+4= 40
I.Q. = 4 x π x 123.1072
40²
I.Q. = 0.966881687
2.
Tan A = O
A
Tan 72°= O
4
4 x Tan 72 = O
12.3107 cm = O
Area of 1 triangle =½ x 8 x 12.3107
= 49.2428
A of 10 triangles = 10 x 49.2428
= 492.428
P =8+8+8+8+8+8+8+8+8+8= 80
I.Q. = 4 x π x 492.428
80²
I.Q. = 0.966880117
The I.Q. s of both my regular decagons were the same and larger than that of other plane shapes I have studied.
A general formula for a regular decagon.
Sum of interior angles= 180(n-2)
= 180(10-2)
=180 x 8
= 1440
Each angle = 1440 = 144
10
Tan 72° = O
A
A x Tan 72° = O
Area of 1 triangle=½ x b x h
= ½ x 2A X A x Tan 72°
= A²Tan 72°
Area of 10 triangles= 10A²Tan 72°
(whole decagon)
P = 2A+2A+2A+2A+2A+2A+2A +2A+2A+2A= 20 A
I.Q. = 4π x Area of shape
(Perimeter of shape)²
I.Q.= 4x π x 10A²Tan 72°
(20A) ²
I.Q.= 4x π x 10A²Tan 72°
400 A²
I.Q. = 40 π Tan72°
400
I.Q. for any regular decagon = π Tan 72°
10
I.Q. =0.966882799 which is the I.Q. for all my regular decagons.
Table comparing regular plane shapes
Each general formula is divided by the number of sided. The angles are all half of the interior angles. The I.Q. s increase, as the number of sides increase. And Tan are in every formula. From my observations, I predict that the general formula of an equilateral triangle is:
π Tan 30°
3
And the general formula for a square is:
π Tan 45°
4
I will check my formula
I.Q. = π Tan 30° for equilateral triangle number 1 that I constructed on page 9.
3
The I.Q. for this triangle was 0.6045.
I.Q. = π Tan 30° = 0.6045
3
I will check my formula
I.Q. = π Tan 45° for square number 1 that I constructed on page 12.
4
The I.Q. for this square was 0.7854
I.Q. = π Tan 45° = 0.7854
4
Overall general formula for any 2D regular plane shape.
I.Q. = π Tan ½ interior angle°
Number of sides
Let = interior angle
n= number of sides
I.Q. =
I will now test my formula by constructing a 12-sided shape (dodecagon).
Each interior angle = 180(n-2) ÷ 12
= 180 x 10 ÷ 12
= 1800 ÷ 12
= 150 °
I predict what the I.Q. will be using my formula:
I.Q. =
I.Q. = π Tan 75 = 0.9770
12
Now I will do full calculations to prove my prediction and formula to be correct.
Tan 75= h
2
H = 2 x Tan 75 = 7.46
A of triangle= 2 x 7.46 = 14.92
A of 12 triangles = 14.92 x 12 =179.04
(whole shape)
Perimeter= 48
I.Q. 4 x π x 179.04
48²
I.Q. = 0.977
Due to the fact my answers are the same, my formula is correct.
So far the 12-sided regular shape has the largest I.Q. As the shapes get bigger (i.e. more sides), the I.Q. gets bigger. The I.Q. is going towards 1.
I will continue the table of I.Q.s for other regular polygons to show that they are tending towards a limit of 1. I will then draw a graph to display all of my results.
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 x 6² = 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
Let the radius be A.
A== πr²
A = π x A²
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.