The Length of The Longest Side (c)
The length of the longest side (c) can be calculated using a, n and b.
The relationship between c, n and a is that: -
c = an + n + 1
For example, if a = 9 and n = 4
c = an + a + 1
c = [(9 x 4) + 4] +1
c = [36 + 4] + 1
∴c = 40 + 1 = 41
The relationship between c and b is that: -
c = b + 1
For example, if b = 40
c = b + 1
c = 40 + 1
∴c = 41
The Perimeter
The perimeter can be calculated using a, b and c.
The relationship between the perimeter, a, b and c is that: -
perimeter = a + b + c
For example, if a = 17, b = 144 and c = 145
perimeter = a + b + c
perimeter = 17 + 144 + 145
∴perimeter = 306
The Area
The area can be calculated using a and b.
The relationship between the area, a and b is that: -
area = ½ab
For example, if a = 15 and b = 112
area = ½ab
area = ½(15 x 112)
area = ½ x 480
∴area = 240
Generalisations
Odd numbers for Length of Shortest Side (a).
Nth Term for Shortest Side (a)
Linear equation = an + b
a = 1st difference = 2
b = 1st no. in sequence - 1st difference = 3 – 2 = 1
nth term for shortest side (a) =
2n + 1
Nth Term for Middle Side (b)
New sequence = Original sequence – 2n2
Quadratic equation = an2 + bn + c
a = 2nd difference/2 = 4/2 = 2
b = 1st difference = 2
c = 1st no. in New sequence – 1st difference = 2 – 2 = 0
∴nth term for middle side (b) =
2n2 + 2n + 0 =
2n2 + 2n
2n2 + 2n =
2(n2 + n)
Nth Term for Longest Side (c)
New sequence = Original sequence – 2n2
Quadratic equation = an2 + bn + c
a = 2nd difference/2 = 4/2 = 2
b = 1st difference = 2
c = 1st no. in New sequence – 1st difference = 3 – 2 = 1
∴nth term for longest side (c) =
2n2 + 2n + 1
Nth Term for Perimeter
New sequence = Original sequence – 4n2
Quadratic equation = an2 + bn + c
a = 2nd difference/2 = 8/2 = 4
b = 1st difference = 6
c = 1st no. in New sequence – 1st difference = 8 – 6 = 2
∴nth term for perimeter =
4n2 + 6n + 2
4n2 + 6n + 2 =
2(2n2 + 3n + 1)
Nth Term for Area
area = ab/2
a = 2n + 1
b = 2n2 + 2n
area = (2n + 1)(2n2 + 2n)/2
(2n + 1)(2n2 + 2n) =
2n(2n2 + 2n)+1(2n2 + 2n) =
4n3 + 4n2 + 2n2 + 2n =
4n3 + 6n2 + 2n
(4n3 + 6n2 + 2n)/2 =
2n3 + 3n2 + n
2n3 +3n2 + n =
n(2n2 +3n + 1)
OR
New sequence = Original sequence – 2n3
New sequence = New sequence – 3n2
Cubic equation = an3 + bn2 + cn + d
a = 3rd difference/6 = 12/6 = 2
b = 2nd difference/2 = 6/2 = 3
c = 1st difference = 1
d = 1st no. in New sequence – 1st difference = 1 – 1 = 0
∴nth term for area =
2n3 + 3n2 + 1n + 0 =
2n3 +3n2 + n
2n3 +3n2 + n =
n(2n2 +3n + 1)
Results Table
Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).
Fig.2
Relationships
Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).
The Length of The Shortest Side (a)
The length of the shortest side (a) can be calculated using n.
The relationship between a and n is that: -
a = 4n + 2
For example, if n = 6
a = 4n +2
a = (4 x 6) + 2
a = 24 + 2
∴a = 26
The Length of The Middle Side (b)
The length of the middle side (b) can be calculated using n and a.
The relationship between b, a and n is that: -
b = an + 2n
For example, if a = 18 and n = 4
b = an + 2n
b = (18 x 4) + (2 x 4)
b = 72 + (2 x 4)
b = 72 + 8
∴b = 80
The Length of The Longest Side (c)
The length of the longest side (c) can be calculated using a, n and b.
The relationship between c, n and a is that: -
c = an + 2n + 2
For example, if a = 18 and n = 4
c = an + 2n + 2
c = [(18 x 4) + (2 x 4] +2 = [72 + (2 x 4] + 2 = [72 + 8] + 2
∴c = 80 + 2 = 82
The relationship between c and b is that: -
c = b + 2
For example, if b = 80
c = b + 2
c = 80 + 2
∴c = 82
The Perimeter
The perimeter can be calculated using a, b and c.
The relationship between the perimeter, a, b and c is that: -
perimeter = a + b + c
For example, if a = 34, b = 288 and c = 290
perimeter = a + b + c
perimeter = 34 + 288 + 290
∴perimeter = 612
The Area
The area can be calculated using a and b.
The relationship between the area, a and b is that: -
area = ½ab
For example, if a = 30 and b = 224
area = ½ab
area = ½(30 x 224)
area = ½ x 6720
∴area = 3360
Generalisations
Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).
Nth Term for Shortest Side (a)
Linear equation = an + b
a = 1st difference = 4
b = 1st no. in sequence - 1st difference = 6 – 4 = 2
∴nth term for shortest side (a) =
4n + 2
4n + 2 =
2(2n +1)
Nth Term for Middle Side (b)
New sequence = Original sequence – 4n2
Quadratic equation = an2 + bn + c
a = 2nd difference/2 = 8/2 = 4
b = 1st difference = 4
c = 1st no. in New sequence – 1st difference = 4 – 4 = 0
∴nth term for middle side (b) =
4n2 + 4n + 0 =
4n2 + 4n
4n2 + 4n =
4(n2 + n)
Nth Term for Longest Side (c)
New sequence = Original sequence – 4n2
Quadratic equation = an2 + bn + c
a = 2nd difference/2 = 8/2 = 4
b = 1st difference = 4
c = 1st no. in New sequence – 1st difference = 6 – 4 = 2
∴nth term for longest side (c) =
4n2 + 4n + 2
4n2 + 4n + 2 =
2(2n2 + 2n + 1)
Nth Term for Perimeter
New sequence = Original sequence – 8n2
Quadratic equation = an2 + bn + c
a = 2nd difference/2 = 16/2 = 8
b = 1st difference = 12
c = 1st no. in New sequence – 1st difference = 16 – 12 = 4
∴nth term for perimeter =
8n2 + 12n + 4
8n2 + 12n + 4 =
4(2n2 + 3n + 1)
Nth Term for Area
area = ab/2
a = 4n + 2
b = 4n2 + 4n
area = (4n + 2)(4n2 + 4n)/2
(4n + 2)(4n2 + 4n) =
4n(4n2 + 4n)+2(4n2 + 4n) =
16n3 + 16n2 + 8n2 + 8n =
16n3 + 24n2 + 8n
(16n3 + 24n2 + 8n)/2 =
8n3 + 12n2 + 4n
8n3 +12n2 + 4n =
4n(2n2 + 3n + 1)
OR
New sequence = Original sequence – 8n3
New sequence = New sequence – 12n2
Cubic equation = an3 + bn2 + cn + d
a = 3rd difference/6 = 48/6 = 8
b = 2nd difference/2 = 24/2 = 12
c = 1st difference = 4
d = 1st no. in New sequence – 1st difference = 4 – 4 = 0
∴nth term for area =
8n3 + 12n2 + 4n + 0 =
8n3 +12n2 + 4n
8n3 +12n2 + 4n =
4n(2n2 + 3n + 1)
Conclusion
To conclude this investigation, I have made relationships between the lengths of right-angled triangles, the perimeter and areas of these triangles. I have made predictions about the relationships in both cases where the shortest side ‘a’ is odd and even. I have also made generalisations about the lengths of the lengths of right-angled triangles, the perimeter and areas of these triangles. Also, I have made predictions about the nth terms in both cases where the shortest side ‘a’ is odd and even.
Below is the proof that the nth terms I have produced satisfy the condition a2 + b2 = c2.
Odd numbers for Length of Shortest Side (a).
nth term for a = 2n + 1
nth term for b = 2n2 + 2n
nth term for c = 2n2 + 2n + 1
a2 + b2 = c2
(2n + 1) 2 + (2n2 + 2n) 2 + (2n2 + 2n + 1) 2
a = (2n + 1) 2 =
(2n + 1)(2n + 1) =
2n(2n + 1) + 1(2n + 1) =
4n2 + 2n + 2n + 1=
4n2 + 4n + 1
b = (2n2 + 2n) 2 =
(2n2 + 2n)(2n2 + 2n) =
2n2 (2n2 + 2n) + 2n(2n2 + 2n) =
4n4 + 4n3 + 4n3 + 4n2 =
4n4 + 8n3 + 4n2
a2 + b2 =
(4n2 + 4n + 1) + (4n4 + 8n3 + 4n2) =
4n4 + 8n3 + 8n2 + 4n + 1
c = (2n2 + 2n + 1) 2 =
(2n2 + 2n + 1)(2n2 + 2n + 1) =
2n2 (2n2 + 2n + 1) + 2n(2n2 + 2n + 1) + 1(2n2 + 2n + 1) =
4n4 + 4n3 + 2n2 + 4n3 + 4n2 + 2n +2n2 + 2n + 1 =
4n4 + 8n3 + 8n2 + 4n + 1 = a2 + b2 = c2
Even numbers for Length of Shortest Side (a) (previous odd numbers doubled).
nth term for a = 4n + 2
nth term for b = 4n2 + 4n
nth term for c = 4n2 + 4n + 2
a2 + b2 = c2
(4n + 2) 2 + (4n2 + 4n) 2 + (4n2 + 4n + 2) 2
a = (4n + 2) 2 =
(4n + 2)(4n + 2) =
4n(4n + 2) + 2(4n + 2) =
16n2 + 8n + 8n + 4=
16n2 + 16n + 4
b = (4n2 + 4n) 2 =
(4n2 + 4n)(4n2 + 4n) =
4n2 (4n2 + 4n) + 4n(4n2 + 4n) =
16n4 + 16n3 + 16n3 + 16n2 =
16n4 + 32n3 + 16n2
a2 + b2 =
(16n2 + 16n + 4) + (16n4 + 32n3 + 16n2) =
16n4 + 32n3 + 32n2 + 16n + 4
c = (4n2 + 4n + 2) 2 =
(4n2 + 4n + 2)(4n2 + 4n + 2) =
4n2 (4n2 + 4n + 2) + 4n(4n2 + 4n + 2) + 2(4n2 + 4n + 2) =
16n4 + 16n3 + 8n2 + 16n3 + 16n2 + 8n + 8n2 + 8n + 4 =
16n4 + 32n3 + 32n2 + 16n + 4 = a2 + b2 = c2