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:

'Maths Coursework'
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

'Maths Coursework'

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

  • 2

4n3+6n2+2n

2

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