I.Q = 4π x area of shape
(Perimeter of shape) ²
= 4π x χ
(4χ) ²
= 4π x 1
16
= π
4
I.Q = 0.785398163
= 0.79 to 2d.p
Therefore, the I.Q of a square is always 0.79 to 2d.p.
Now I will try the formula on pentagons (5 sides) to see if it has the same effects:
Height of one triangle = tan54 x 1
Area of one triangle = tan54 x 1
Area of pentagon = tan54 x 5
I.Q = 4 π x tan54 x 5
100
= 4π x Tan54
20
= π x tan54
5
= 0.864806260
I.Q = 0.86 to 2d.p
Height of one tri = Tan 54 x 5
Area of one tri = Tan54 x 5 x 5
= Tan54 x 25
Area of Pentagon =Tan54 x 25 x 5
= Tan54 x 125
I.Q. = 4π x Tan54 x 125
2500
= π x Tan54 x 125
625
= π x Tan54
5
= 0.865806260
I.Q. = 0.86 to 2d.p
Using Algebra:
After using the formula: 4π x area of shape on two different sized
(Perimeter of shape) ²
regular pentagons, I will now use it on a pentagon of side χ.
Height of one triangle = Tan54 x ½χ
Area of one triangle = Tan54 x 0.25χ
Area of pentagon = Tan54 x 1.25χ²
I.Q. = 4π x Tan54 x 1.25χ2
25χ²
= 4π x Tan54
20
= π x Tan54
5
= 0.864806206
I.Q. = 0.86 to 2d.p
Therefore, the I.Q of a regular pentagon is always 0.86 to 2d.p.
Now I will find the I.Q of a regular 6-sided shape and see if it has the same effect as it does on 1, 3, 4, and 5-sided shapes:
Height of one triangle = Tan60 x 1
Area of one triangle = Tan60 x 1
Area of hexagon = Tan60 x 6
I.Q = 4π x Tan60 x 6
144
= 4π x Tan60
24
= π x Tan60
6
= 0.906899682
I.Q = 0.90 to 2d.p
Height of one triangle = Tan60 x 5
Area of one triangle = Tan60 x 25
Area of hexagon = Tan60 x 150
I.Q = 4π x Tan60 x 180
3600
= 4π x Tan60
24
= π x Tan60
6
= 0.906899682
I.Q = 0.91 to 2d.p
Using Algebra:
Now I will use algebra to see if the I.Q of a hexagon is always the same:
Height of one triangle = Tan60 x ½χ
Area of one triangle = Tan60 x 0.25χ
Area of hexagon = Tan60 x 1.5χ²
I.Q = 4π x area of shape
(Perimeter of shape) ²
= 4π x Tan60 x 1.5χ²
36χ²
= 4π x Tan60
24
= π x tan60
6
I.Q = 0.906899682
= 0.90 to 2d.p
Therefore the I.Q of a regular hexagon is always 0.90 to 2d.p
I will now try the I.Q formula on a 7-sided shape because
Height of one triangle = Tan64.28… x 1
Area of one triangle = Tan64.28… x 1
Area of Heptagon =Tan64.28… x 7
I.Q = 4π x Tan64.28… x 7
196
= 4π x Tan64.28…
28
= π x Tan64.28…
7
= 0.931940623
I.Q = 0.93 to 2d.p
Height of one tri = Tan64.28… x 175
Area of one tri = Tan64.28… x 25
Area of Heptagon = Tan64.28… x 175
I.Q = 4π x Tan64.28… x 175
4900
= 4π x Tan64.28…
28
= π x Tan64.28…
7
= 0.931940623
I.Q = 0.93 to 2d.p
Using Algebra:
I will now use algebra to find out if the I.Q of a heptagon is always the same:
Height of one triangle = Tan64.28… x ½ χ
Area of one triangle = Tan64.28… x 0.25χ
Area of heptagon = Tan64.28… x 1.75χ²
I.Q = 4π x area of shape
(Perimeter of shape) ²
= 4π x Tan64.28… x 1.75χ²
49χ²
= 4π x Tan64.28…
28
= π x Tan64.28
7
= 0.931940623
I.Q. = 0.93 to 2d.p
Therefore, the I.Q of a regular heptagon is always 0.93to 2d.p.
I will now carry on increasing the number of sides of plane shapes and find the I.Q of octagons:
Height of one triangle = Tan67.5 x 1
Area of one triangle = Tan67.5 x 1
Area of octagon = Tan67.5 x 8
I.Q = 4π x Tan67.5 x 8
256
= 4π x Tan67.5
32
= π x Tan67.5
8
= 0.948059449
I.Q. = 0.95 to 2d.p
Height of one triangle = Tan67.5 x 5
Area of one triangle = Tan67.5 x 25
Area of Octagon = Tan67.5 x 200
I.Q = 4π x Tan67.5 x 200
6400
= 4π x Tan67.5
32
= π x Tan67.5
8
= 0.948059449
I.Q.= 0.95 to 2d.p
Using Algebra:
I will now use algebra to try and prove that the I.Q of a regular octagon is always the same:
Height of one triangle = Tan67.5 x ½χ
Area of one triangle = Tan67.5 x 0.25χ
Area of octagon = Tan67.5 x 2χ²
I.Q = 4π x area of shape
(Perimeter of shape) ²
= 4π x Tan67.5 x 4χ²
64χ²
= 4π x Tan67.5
32
= π x Tan67.5
8
= 0.948059449
I.Q = 0.95 to 2d.p
Therefore, the I.Q of a regular octagon is always 0.95 to 2d.p.
Now I know what the I.Q’s of 1,3,4,5,6,7, and 8-sided shapes are, I will try and find out the I.Q of a decagon (10 sides).
Height of one triangle = Tan72 x 1
Area of one triangle = Tan72 x 1
Area of decagon = Tan72 x 10
I.Q. = 4π x Tan72 x 10
400
= 4π x Tan72
40
= π x Tan72
10
= 0.966882799
I.Q = 0.97 to 2d.p
Height of one triangle = Tan72 x 250
Area of one triangle = Tan72 x 25
Area of Decagon = Tan72 x 250
I.Q = 4π x Tan72 x 250
10000
= 4π x Tan72
40
= π x Tan72
10
= 0.966882799
I.Q = 0.97 to 2d.p
Now I will try and prove using algebra that the I.Q of a regular decagon is always the same:
Height of one triangle = Tan 72 x ½χ
Area of one triangle = Tan72 x 0.25χ²
Area of decagon = Tan72 x 2.5χ²
I.Q = 4π x Tan72 x 2.5χ²
= 4π x Tan72
40
= π x Tan72
10
= 0.966882799
I.Q = 0.97 to 2d.p
Therefore, the IQ of a regular decagon is always 0.97 to 2d.p.
Table of Results
From the table, I can see that the possible rule is:
I.Q = π x Tan (½ interior angles)
No of sides
However, I worked out the I.Q of a square without using Tan, so now I will break it up into triangles to find the I.Q, and see if it follows the above rule.
Height of one triangle = Tan45 x 3
Area of one triangle = Tan45 x 3 x 3
= Tan45 x 9
Area of square = Tan45 x 9 x 4
= Tan45 x 36
I.Q = 4π x tan45 x 36
576
= 4π x Tan 45
16
= π x Tan45
4
Therefore this square fits the rule. I will prove this using algebra:
Height of one triangle = Tan45 x ½χ
Area of one triangle = Tan45 x 0.25χ
Area of square = Tan45 x χ²
I.Q = 4π x Tan45 x χ²
16χ²
= 4π x Tan45
16
= π x Tan45
4
This now shows that squares do follow the rule:
I.Q = π x Tan (½ interior angles)
No of sides
In making my answers, I also used ‘sin’ instead of ‘tan’ when working out the I.Q of an equilateral triangle. I will try it again using ‘tan’, and by dividing the triangle into 3 because it has three sides. I followed this pattern with my other shapes.
Height of one triangle = Tan30 x 2
Area of one triangle = Tan30 x 4
Area of whole triangle = Tan30 x 4 x 3
= Tan30 x 12
I.Q = 4π x Tan30 x 12
144
= 4π x Tan30
12
= π x Tan30
3
This triangle follows the rule: I.Q = π x Tan (½ interior angles)
No of sides
I will now prove using algebra that this formula works on all equilateral triangles:
Height of one triangle = Tan30 x ½χ
Area of one triangle = Tan30 x 0.25χ
Area of whole triangle = Tan30 x 0.75χ²
I.Q = 4π x Tan 30 x 0.75χ²
9χ²
= 4π x Tan30
12
= π x Tan30
3
This now shows that the rule: I.Q = π x Tan (½ interior angles)
No of sides
will work on any equilateral triangle.
By putting my new results into my table:
I can see that the formula I.Q = π x Tan (½ interior angles) works on any
No of sides
size regular plane shape.
Evaluation
From the results in my table, I found that the I.Q for any regular plane shape was π x Tan (½ interior angles) divided by the number of sides that shape had. However to get to this formula, I had to make sure that however many sides a shape has, that’s how many triangles it was divided into to find the I.Q of that shape.
I can also see that the more sides a shape has, the nearer it’s I.Q is to the I.Q of a circle, which has the highest I.Q of 1.
This could be because the more sides there are to a shape; the more it will resemble a circle. So if we carried on adding sides to a shape, eventually we would get to a circle, which has an infinite number of sides.
So you can see that when an equilateral triangle is fitted into a circle, there is a lot of space where the shapes don’t meet.
But when we put a decagon inside a circle, there is not much space. Therefore, if we kept putting shapes with more sides inside circles, the space would gradually decrease until it became non-existent, and it would’ve become a circle, with an I.Q of one.
Further investigation – irregular shapes
As I have used the formula for finding the IQ of a shape only on regular shapes, I will now try it on rectangles, irregular shapes.
I.Q = 4π x area of shape
(Perimeter of shape) ²
= 4π x 8
144
= 4π
18
=0.69813170079773183076947630739545
=0.70 to 2d.p
I.Q = 4π x 12
256
= 4π
21.3333
=0.589048622548086232211745634365
=0.59 to 2d.p
I.Q = 4π x 16
400
= 4π
25
I.Q = 0.50265482457436691815402294132472
= 0.50 to 2d.p
I have noticed that as a rectangle becomes more stretched, its IQ decreases. I have shown this by finding the IQ of rectangles length = 2 times the width, length = 3 times the width, and length = 4 times the width.
