To find the Perimeter I simply had to find the total of all three sides. For the area I had to
P = a + b + h
find multiply side ‘a’ with side ‘b’ and half the total
A = ab
2
Bellow is have shown in detail how I found the perimeter and area for each triple.
2)
(5,12,13)
Perimeter = 5 + 12 + 13
= 30
Area = 5 × 12
2
= 60
2
= 30
(7,24,25)
Perimeter = 7 + 24 + 25
= 56
Area = 7 × 24
2
= 168
2
= 84
I then constructed an extended table by investigating the formulas for each side which are recorded under the table.
Side ‘a’
For side ‘a’ is could easily see that the formula would be 2n+1 as this is the general formula for odd numbers.
I made a sequence table to prove this:
Term Number Sequence
1 (×2 + 1) 3
2 (×2 + 1) 5
3 (×2 + 1) 7
I then tested this formula with the equation I found:
Term 2
‘a’ = 2n + 1
= 2 × 2 + 1
= 4 + 1
= 5
Term 3
‘a’ = 2n + 1
= 2 × 3 + 1
= 6 + 1
= 7
Side ‘b’
I used quadratic rule to find the formula for this side:
1 diff = 8 The first difference would be: difference of 1st sequence and 2nd sequence
2 diff = 4 The second difference would be: 2nd sequence – 1st sequence
1 term = 4 The first term is just the first number in the sequence
an² + bn + c = 0 I will find the values of a, b, c and use this formula to find the nth term
for side ‘b’
Bellow are the rules that I will use to find a, b and c
2a = 2nd diff
3a + b = 1st diff
a + b + c = 1st term
2a = 2nd diff
= 2a = 4
= a = 4
2
= a = 2
= 3a + b = 1st diff
= 3 × 2 + b = 8
= 6 + b = 8
= b = 8 – 6
= b = 2
term:
= a + b + c = 1st term
= 2 + 2 + c = 4
= 4 + c = 4
= c = 4 – 4
= c = 0
Bellow are the figures I fount for a, b and c:
a = 2
b = 2
c = 0
I then used this to replace the letters in the formula to find the nth term
an² + bn + c = 0
When replaced
2n² + 2n + 0
So the nth term formula would be : 2n² + 2n
I then tested this formula for 2 randomly selected terms.
Term 4:
‘b’ = 2n2 + 2n
= 2 × 4² + 2 × 4
= 2 × 16 + 2 × 4
= 32 + 8
= 40
Term 5:
‘b’ = 2n2 + 2n
= 2 × 5² + 2 × 5
= 2 × 25 + 2 × 5
= 50 + 10
= 60
Side ‘h’
It is obviouse that the set of results found for side ‘h’ are just one plus to the results of side ‘b’
So I took the formula for side ‘b’ and simply added one onto it:
2n2 + 2n + 1
I tested this formula by randomly selecting 2 terms.
Term 5:
‘h’ = 2n2 + 2n + 1
= 2 × 5² + 2 × 5 + 1
= 2 × 25 + 10 + 1
= 50 + 11
= 61
Term 4:
‘h’ = 2n2 + 2n + 1
= 2 × 4² + 2 × 4 + 1
= 2 × 16 + 8 + 1
= 32 + 9
= 41
Perimeter
To find the perimeter for any shape you must find the total of all the sides so I would do
a + b + c. So I took the formulas that I fount for all 3 sides and added them together to make the formula for the perimeter.
2n+1 + 2n2 + 2n + 2n2 + 2n + 1
4n2 + 6n +2
I then tested this formula
Term 2:
‘p’ = 4n2 + 6n + 2
= 4×22 + 6 × 2 + 2
= 4×4 + 12 + 2
= 16 + 14
= 30
Term 5:
‘p’ = 4n2 + 6n + 2
= 4×52 + 6 × 5 + 2
= 4×25 + 30 + 2
= 100 + 32
= 132
Area
To find the area of a right angled triangle I would have to multiply the base and the height and half the answer as one would to find the area for a simple right angled triangles. So I took both formulas from the base and height (‘a’’b’) and put them into the following formula:
½ (2n+1) (2n2 + 2n)
Then multiplied it out to get:
4n3 + 6n2 + 2n
2
Then divide 4n3 + 6n2 + 2n by 2 to get
2n3 + 3n2 + n
I then looked at the relationship between the perimeter and area.
P = 4n2 + 6n + 2
A = 2n3 + 3n2 + n
It can be easily seen that the half of the perimeter is equal to the area so the area would be
A = ½Pn
The Pythagoras thorium shows that to find the area you must use a²+b²=h² so I used this formula to find the area from side ‘h’.
h = b + 1
a² + b² = h²
a² + b² = (b+1)²
a² + b² = (b+1) (b+1)
a² + b² = b² + 2b+1
a² =2b + 1
a² - 1= 2b + 1
a²-1
2
So from this I can see that when I have the relationship between side ‘b’ and ‘h’ I can use this to find a formula for the area by replacing the letters with the correct formula.
So the formula would be:
Area = a²-1
2
Bellow I have made a table listing all the formulas that I fount
I then double each number in the extended table to make an even set of results. Then I investigated the formulas for even.
I could see that when I doubled the table the whole of the area was not correct and the last perimeter was not correct. Bellow I have made the corrections for the table by using the simple formulas for perimeter and area.
1th row
Area = 6 × 8
2
= 48
2
= 24
2nd row
Area = 10 × 24
2
= 240
2
= 120
3rd row
Area = 14 × 48
2
= 672
2
= 336
4th row
area = 18× 80
2
= 1440
2
= 720
5th row
area = 22 × 120
2
= 2640
2
= 1320
6th row
perimeter = 26 + 168 + 170
= 364
area = 26 × 168
2
= 4368
2
= 2184
I then reconstructed the table with the right perimeter and area filled in.
Side ‘a’ even
As the numbers are doubled in this table all I will have to do is double the formula.
2n +1
4n + 2
The term table will also double
Term Number Sequence
1 (×4 + 2) 6
2 (×4 + 2) 10
3 (×4 + 2) 14
Side ‘b’
For this side I will also double the formula:
2n2 + 2n
4n2 + 2n
I will use the quadratic rule to check this and see what changes acquire in that
1 diff = 16 The first difference would be: difference of 1st sequence and 2nd sequence
2 diff = 8 The second difference would be: 2nd sequence – 1st sequence
1 term = 8 The first term is just the first number in the sequence
an² + bn + c = 0 I will find the values of a, b, c and use this formula to find the nth term
for side ‘b’
Bellow are the rules that I will use to find a, b and c
2a = 2nd diff
3a + b = 1st diff
a + b + c = 1st term
2a = 2nd diff
= 2a = 8
= a = 8
2
= a = 4
= 3a + b = 1st diff
= 3 × 4 + b = 16
= 12 + b = 16
= b = 16 – 12
= b = 4
= a + b + c = 1st term
= 4 + 4 + c = 8
= 8 + c = 8
= c = 8 – 8
= c = 0
Bellow are the figures I fount for a, b and c:
a = 4
b = 4
c = 0
I then used this to replace the letters in the formula to find the nth term
an² + bn + c = 0
When replaced
4n² + 4n + 0
So the nth term formula would be : 2n² + 2n
It could see that everything double and I ended up with a double nth term equation
I then tested this formula for 2 randomly selected terms.
Term 4:
‘b’ = 4n2 + 4n
= 4 × 4² + 4 × 4
= 4 × 16 + 4 × 4
= 64 + 16
= 80
Term 5:
‘b’ = 4n2 + 4n
= 4 × 5² + 4 × 5
= 4 × 25 + 4 × 5
= 100 + 20
= 120
Side ‘h’ even
I only had to double the formula for for the hypotenuse
2n2 + 2n + 2
4n2 + 4n + 2
But as the results have doubled, instead of there being a difference of 1 between side ‘a’ and ‘h’ there is a difference of 2
I then tested this formula
Term 5:
‘h’ = 4n2 + 4n + 2
= 4 × 5² + 4 × 5 + 2
= 4 × 25 + 20 + 2
= 100 + 22
= 122
Term 4:
‘h’ = 4n2 + 4n + 2
= 4 × 4² + 4 × 4 + 2
= 4 × 16 + 16 + 2
= 64 + 18
= 82
Perimeter even
There are 2 ways to find the perimeter for even. I can add all the formulas I have found together or I can multiply the whole equation by 2. Either way this is the formula I got:
4n+2 + 4n2 + 4n + 4n2 + 4n + 2
8n2 + 12n +4
I then tested this formula
Term 2:
‘p’ = 8n2 + 12n + 4
= 8×22 + 12 × 2 + 4
= 8×4 + 24 + 4
= 32 + 28
= 60
Term 5:
‘p’ = 8n2 + 12n + 4
= 8×52 + 12 × 5 + 4
= 8×25 + 60 + 4
= 200 + 64
= 264
Area even
The area for the even results would be
A = pn
If I multiplied the perimeter by n I should get the area.
P = 8n2 + 12n + 4
I will now see if this is correct by using the formulas for side a and b (simple area formula)
Area = ½ (4n + 1) (4n2 + 4n)
= 16n³ + 10n + 8n2 + 8n
2
= 16n³ + 12n2 + 8n
= 8n + 12n2 + 4n
Area = p × n
This shows that if the perimeter is multiplied by n it well equal to the answer.
