4
The general formula for squares = IQ = ∏ tan 45
4
IQ = 0.785398163
Now I have to prove my square formulas for area perimeter and IQ are correct. To do this I will use my formulas and the original formulas and then I will take them away from each other, if they are correct the answer to my formula subtracted by the original formula should be exactly 0.
We will now investigate squares substituting numbers for sides.
1 cm sided square
1 cm
1cm 1 cm
1 cm
Area of the square.
Using my own formula ; x2 * tan 45
Substitute 1 in for x
= 12 *tan 45
= 1 * tan 45
= 1 cm2
Using the original formula ; base * height
Substitute 1 in for base and height
= 1cm * 1cm
= 1cm2
I will now take the two formulas away from each other to see if my formula is correct.
1cm2 - 1cm2
= 0
My formula for the area of a square is correct.
I am now going to do the same as I just did but I am going to do it the perimeter instead.
1 cm sided square
1 cm
1cm 1 cm
1 cm
Perimeter of a square.
Using my formula = 4x
Substitute 1 in for x
= 4 * 1
= 4cm
using original formula = x + x + x + x
substitute 1 in for x
= 1 + 1 + 1 + 1
= 4 cm
I will now take the two formulas away from each other to see if my formula is correct.
4cm - 4cm
= 0
My formula for the perimeter of a square is correct.
Now I know my area and perimeter formulas were correct my IQ formula should also be correct because the IQ formula was made up of the area and perimeter formulas. But I am going to make sure any way.
I am now going to do the same as I just did but I am going to do it with the IQ instead.
1 cm sided square
1 cm
1cm 1 cm
1 cm
Using my formulas;-
Perimeter = 4x
Substitute 1 in for x
= 4 * 1
= 4cm
Area = x2 * tan 45
Substitute 1 in for x
= 12 *tan 45
= 1 * tan 45
= 1 cm2
IQ = ∏ * tan 45
4
= 0.785398163
Using the original formulas;-
Using the original formula ; base * height
Substitute 1 in for base and height
= 1cm * 1cm
= 1cm2
using original formula = x + x + x + x
substitute 1 in for x
= 1 + 1 + 1 + 1
= 4 cm
IQ = 4∏ * area of the shape
(Perimeter of the shape)2
= 4∏ * 1
42
=12.56637061
16
= 0.785398163
I will now take the two formulas away from each other to see if my formula is correct.
0.785398163 – 0.785398163
= 0
All my formulas for the square are correct.
Prediction
We can see that four is the denominator and it’s a four-sided shape, therefore we will investigate further to see if a pattern emerges. I am not sure if my IQ formula is correct because if it is correct it means every square must have the same IQ (0.785398163) however large its sides are.
5 cm sided square
5 cm
5 cm 5cm
5 cm
Area
Using my formula = x2 tan 45
= 52 tan 45
= 25 tan 45
= 25 cm2
Perimeter
Using my own formula = 4 x
= 4 * 5
= 20 cm
Formula given
IQ = 4∏ * area of the square
(perimeter of the square)2
IQ = 4∏ * 25
(20)2
IQ = 4 ∏ *25
400
IQ = 4 ∏
16
IQ = 1∏
4 _
IQ = 0.785398163
Using my formula
IQ = ∏ tan 45
4
= 0.785398163
I will now take the two formulas away from each other to see if my formula is correct.
0.785398163 – 0.785398163
= 0
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.
We now have a right angle triangle, but we don’t yet have another angle in it, so I still cant perform trigonometry.
The pentagon is now made up of 10 right angle triangles, the center of the pentagon = 360
To find the angle of 1 of these 10 right-angled triangles we divide 360 by 10
Which = 36
Exterior angle = 360/n
= 360/5
= 72
Interior angle = 180 – 72
= 108/2
= 54
To check if the angles are correct I am going to add them all together to see if they make 180
90 + 36 + 54 = 180. The angles are correct.
We now have a right angled triangle with another angle so we can now perform trigonometry.
I will now enlarge one of the ten segments to make it easier to find the area of the square, which I need to find the IQ.
54
x/2
The base = x/2 because the side of the pentagon = x and the segments side is half the size of the pentagons side therefore x is divided by 2, so base = x/2.
Now I will find the area of the segment using trigonometry to find the height and then multiplying the height by the base and dividing it all by 2, I do this because the formula for an are of a triangle = ½ (base * height). When I find the area of the segment to find the area of the pentagon all I will have to do is times the area of the segment by ten because the segment is 1/10 of the pentagon.
O H
h
90 54
A x/2
H= Hypotenuse
O=Opposite
A=Adjacent
h=height
Using trigonometry.
Height =using tan 54=O/A
= tan 45 = h/x/2
h = x/2 * tan 54
Area of 2 triangular segments = ½ * (base * height)
= ½ (x* tan 54 * x/2)
=½ * x * x/2*tan 54
= x2/4 * tan 54
Area of a pentagon = 5* x2/4 * tan 45
=5* x2 * tan 45
4
total area of the pentagon =5/4* x2 * tan 45
Now I’ve got the area of the pentagon I need the perimeter of the pentagon so I can complete the IQ, because you need both the area and the perimeter to complete the formula.
To find the perimeter all I need to do is add the five sides of the pentagon together.
Perimeter = x + x + x + x +x = 5x
Now I have the area and the perimeter for a pentagon I can substitute them into the IQ formula that has been given to us to find a general IQ formula for a pentagon.
Formula given
IQ = 4 ∏ * area of the square
(perimeter of the square)2
Substitute 5/4 *x2 * tan 54 in for area of the shape, and substitute (5x) in for perimeter of the shape.
IQ = 4∏ * 5/4 * x2 * tan 54
(5x)2
IQ = 4∏ * 5/4 * x2 * tan 54
25x2
the 2 x2 ‘s cancel each other out to leave
IQ = 4∏ *5/4 * tan 54
25
5/4 goes into 25 so they cancel down.
IQ = 4∏ * tan 54
20
4 and 20 go into each other so they cancel down.
IQ = 1∏ * tan 54
5
The general formula for pentagons = IQ = ∏ tan 54
5
IQ = 0.864806266
My formulas are:-
Perimeter = 5x
Area = 5/4 * x2 tan 54
IQ= ∏*tan 54
5
Proving my pentagon formulas correct.
Now I have to prove my pentagon formulas for area perimeter and IQ are correct. To do this I will use my formulas and the original formulas and then I will take them away from each other, if they are correct the answer to my formula subtracted by the original formula should be exactly 0.
We will now investigate squares substituting numbers for sides.
1 cm sided pentagon
1 cm 1 cm
1cm 1 cm
1 cm
Area of the pentagon.
Using my own formula ; 5/4 * x2 tan 54
Substitute 1 in for x
= 5/4*12 *tan 54
= 5/4 *1 * tan 54
= 1.25 * 1
= 1.25 cm2
Using the original formula ;
Exterior angle = 360/n
= 360/5
= 72
Interior angle = 180 – 72
= 108/2
= 54
O H
h
90 54
A .5
H= Hypotenuse
O=Opposite
A=Adjacent
h=height
Using trigonometry.
Height =using tan 54=O/A
= tan 54 = h/.5
h = ½ * tan 54
Area of 2 triangular segments = ½ * (base * height)
= ½ (1* tan 54 * ½ )
=½ * 1 * ½ *tan 54
= ¼ * tan 54
Area of a pentagon = 5* ¼ * tan 45
=5* 1 * tan 54
4
total area of the square =5/4* tan 54
= 1.25 cm2
I will now take the two formulas away from each other to see if my formula is correct.
1.25cm2 – 1.25cm2
= 0
My formula for the area of a pentagon is correct.
I am now going to do the same as I just did but I am going to do it the perimeter instead.
1 cm sided pentagon
1 cm 1cm
1cm 1 cm
1 cm
Perimeter of a pentagon.
Using my formula = 5x
Substitute 1 in for x
= 5 * 1
= 5cm
using original formula = x + x + x + x +x
substitute 1 in for x
= 1 + 1 + 1 + 1 + 1
= 5 cm
I will now take the two formulas away from each other to see if my formula is correct.
5cm - 5cm
= 0
My formula for the perimeter of a pentagon is correct.
Now I know my area and perimeter formulas were correct my IQ formula should also be correct because the IQ formula was made up of the area and perimeter formulas. But I am going to make sure any way.
I am now going to do the same as I just did but I am going to do it with the IQ instead.
1 cm sided pentagon
1 cm 1cm
1cm 1 cm
1 cm
Using my formulas;-
Perimeter = 5x
Substitute 1 in for x
= 5 * 1
= 5cm
Area = 5/4 * x2 tan 54
Substitute 1 in for x
= 5/4 * 12 *tan 54
= 5/4 * 1 * tan 54
= 1.25 cm2
IQ = ∏ * tan 54
5
= 0.864806266
Using the original formula ;
Exterior angle = 360/n
= 360/5
= 72
Interior angle = 180 – 72
= 108/2
= 54
O H
h
90 54
A .5
H= Hypotenuse
O=Opposite
A=Adjacent
h=height
Using trigonometry.
Height =using tan 54=O/A
= tan 54 = h/.5
h = ½ * tan 54
Area of 2 triangular segments = ½ * (base * height)
= ½ (1* tan 54 * ½ )
=½ * 1 * ½ *tan 54
= ¼ * tan 54
Area of a pentagon = 5* ¼ * tan 45
=5* 1 * tan 54
4
total area of the pentagon =5/4* tan 54
= 1.25 cm2
IQ = 4∏ * area of shape
(perimeter of shape)2
= 4∏ * 1.25 * tan 54
52
= 4∏ * 1.25 * tan 54
25
= ∏ * 5 * tan 54
25
= ∏ * 1 * tan 54
5
= ∏ * tan 54
5
= 0.864806266
I will now take the two formulas away from each other to see if my formula is correct.
= 0.864806266– = 0.864806266
= 0
All my formulas for the pentagon are correct.
Substitute 10 for x.
We are going to use 10 as the sides, and all the interior angles are 108.
tan 54 = o/a
tan 54 = h/5
h=tan 54 * 5
Area of triangular segment = ½ (b*h)
= ½ (5 * tan 54 *5)
= ( 25 * tan 54)
2
Area of pentagon = 10( 25 * tan 54)
2
= 25 * 10 * tan 54
2
= 250 * tan 54
2
= 125 * tan 54
IQ = 4 ∏ * 125 * tan 54
502
IQ = 4∏ * 125 * tan 54
2500
IQ = ∏ * 500 * tan 54
2500
IQ = ∏ * tan 54
5
Regular hexagons
A regular hexagon is a six, sided shape with all its sides and angles the same.
Exterior angle = 360/6
= 60
Interior angle = (180 – 60)/2
= 120/2
= 60
We are going to try some simple examples of regular hexagons and try and find a pattern. By using the formula that has been given to us.
We are going to use x as the sides, and all the interior angles are 120.
Using tan 60 = o/a
Tan 60 = h/x
2
Perimeter of hexagon = 6x
IQ= 4∏ * area of hexagon
(perimeter)2
IQ= 4 ∏ * 3/2 X2 * tan 60
(6x ) 2
IQ= 4 ∏ * 3/2 X2 * tan 60
36x2
IQ= ∏ * tan 60
6
IQ = 0.906899682
General rule so far.
We can see 2 patterns emerging, the first I noticed was that the denominator is n, n being the number of sides. The second is that the top part is
∏ * tan interior angle/2, because the 4 from 4∏ gets cancelled out to leave ∏, and the area of the shape is tan interior angle/2.
Original rule:
IQ = 4∏ * area of shape
(perimeter)2
So the rule so far is:
IQ = ∏ * tan interior angle/2
Number of sides.
Lets cancel it down further.
IQ = ∏ * tan (180 – 360/n)/2
N
IQ = ∏ * tan (90 – 180/n)
N
IQ = ∏ * tan (90n – 180)/n
N
IQ = ∏ * tan 90 (n-2)/n
N
We will now investigate regular hexagons substituting numbers for sides.
2 cm sided hexagon
Exterior angle = 360/6
= 60
Interior angle = (180 – 60)/2
= 120/2
= 60
tan 60 = h/1
1* tan 60 = h * 12
total area = 12(1*(1* tan 60)/2
= (12 tan 60)/2
IQ = 4 ∏ * (12 tan 60)/2
122
= 2 ∏ * 12 tan 60
144
= ∏ * tan 60
144/24
= ∏ * tan 60
6
Lets see if my formula is correct.
IQ = ∏ * tan 90 (n-2)/n
N
IQ = ∏ * tan 90 (6-2)/6
6
IQ = ∏ * tan 90 (4)/6
6
IQ = ∏ * tan 90 (0.666)
6
IQ = ∏ * tan 60
6
My formula so far is correct because the answer came out the same as with the original way of working it out.
My formula enables me to find the IQ of a shape without working out the area, perimeter and exterior and interior angles of the shape.
3 cm sided hexagon
Exterior angle = 360/6
= 60
Interior angle = (180 – 60)/2
= 120/2
= 60
tan 60 = h/1.5
1.5* tan 60 = h * 12
total area =12(1.5*(1.5 * tan 60)/2)
=(1.5 tan 60 * 1.5 * 12)/2
=(27 tan 60)/2
IQ = 4 ∏ * (27 tan 60)/2
182
= 2 ∏ * 27 tan 60
324
= ∏ * 54 * tan 60
324
= ∏ * tan 60
324/54
=∏ * tan 60
6
Regular octagons.
I am going to study the regular octagon next, a regular octagon is a eight sided regular polygon, all its sides and angles equal. A regular octagon is not a simple shape like the square.
Exterior angle =360/8
=45
Interior angle =(180- 45)/2
=135/2
=67.5
We are going to try some simple examples of regular octagons
and find a pattern. By using the formula that I have made.
I found out the exterior and interior angles just to check my answer was correct, they don’t serve any other purpose.
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
IQ = 0.948059449
Now we have this formula it doesn’t matter how large the sides are because we don’t have to use them in our workings.
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
IQ = 0.966882799
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
IQ = 0.977048616
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
IQ = 0.999999837
Circle
IQ = 4∏ * ∏ r*r
(2∏r)2
= 4∏ * ∏ x*x
2∏x * 2∏x
= 4∏ * ∏
2∏ * 2∏
=∏
∏
= 1
1
= 1
Results table
This table shows the higher the number of sides a shape has the higher the IQ will be.
