Maths Coursework
Beyond Pythagoras. Area. Perimeter. Shortest. Middle. Longest

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Maths Coursework
Beyond Pythagoras
32+42= 52
5cm 9+16= 25
3cm
Area= b x h = 3 x 4 = 6cm2
4cm 2 2
Perimeter= 3+4+5= 12cm
52+122=132
5cm 13cm 25+144+=169
Area= (12 x 5) : 2 = 30cm2
12cm Perimeter= 13+5+12= 30cm
72+242=252
49+576=625
7cm 25cm
Area= (7 x 24) :2= 84cm2
Perimeter= 7+24+25= 56cm
24cm
| No | Shortest A | Middle B | Longest C | Area | Perimeter |
| 1 | 3cm | 4cm | 5cm | 6cm2 | 12cm |
| 2 | 5cm | 12cm | 13cm | 30cm2 | 30cm |
| 3 | 7cm | 24cm | 25cm | 84cm2 | 56cm |
| 4 | 9cm | 40cm | 41cm | 180cm2 | 90cm |
| 5 | 11cm | 60cm | 61cm | 330cm2 | 132cm |
| 6 | 13cm | 84cm | 85cm | 546cm2 | 182cm |
| 7 | 15cm | 112cm | 113cm | 840cm2 | 240cm |
| nth | 2n+1 | 2n2+2n | 2n2+2n+1 | 4n2+6n+2 | (4n3+6n2+2n) 2 |
Nth term:
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an2+bn+c
Middle side
12= 8+2b+c
4=2b+c 2
as the second difference = 4 then a=2
sequence= 2n2+bn+c
When n=1 b=4
N=2 b=12
4= 2 x 1 x 1 + b x 1 + c
4=2+b+c 2=b+c 1
--- 4= 2b+c
2=b+c
= 2=b
C=0
Fomula for middle side= 2n2+2n
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Large side
A=2 because the second difference is 4
Formula= 2n2+2n+1
Pythagoras rule
(2n+1)2 + (2n2+2n)2 = (2n2+2n +1)x(2n2+2n+1)
(4n2+2n+2n+1+4n4+4n3+4n3+4n2) = (4n4+4n3+2n2+4n3+4n2+2n+2n2+2n+1)
8n2+4n+4n4+8n3+1= 4n4+8n3+8n2+4n+1
Nth term of perimeter
A=4; sequence is 4n2+bn+c
When n is 1 b=12cm
When n is 2 b=30cm
12= 4+b+c
8=b+c
30=16+2b+c
14=2b+c
Formula for the perimeter= 4n2+6n+2
Area
(2n+1)(2n2+2n) = 4n3+4n2+2n2+2n
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2
4n3+6n2+2n
2