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• Level: GCSE
• Subject: Maths
• Word count: 1272

# Pythagorean Theorem Coursework

Extracts from this document...

Introduction

Jaymin Patel 10Q                                                               Page  of

Pythagoras’ theorem states that a²+b²=c²

As you can see, a is the longest side, b is the middle side and c is the longest side (hypotenuse). The point of this coursework is to find b when a is an odd number and all of the sides are positive integers. Then after that I will go looking when a is a positive number.

The numbers 3, 4 and 5 work in Pythagoras’ theorem,

3²+4²=5²

because        3²=3×3=9

4²=4×4=16

5²=5×5=25

and so                3²+4²=9+16=25=5²

The numbers 5, 12 and 13 also work,

5²+12²=13²

because        5²=5×5=25

12²=12×12=144

13²=13×13=169

and so                5²+12²=25+144=169=13²

The numbers 7, 24 and 25 also work,

7²+24²=25²

because        7²=7×7=49

24²=24×24=576

25²=25×25=625

and so                7²+24²=49+576=625=25²

3, 4 and 5

Perimeter= 3+4+5=12

Area= ½ ×3×4=6

5, 12 and 13

Perimeter= 5+12+13=30

Area= ½ ×5×12=30

7, 24 and 25

Perimeter= 7+24+25=56

Area= ½ ×7×24=84

From the first three terms I have realised that: -

• a increases by 2 each time
• a is equal to the formula ×2+1 from n
• b is always even
• c is always odd
• c is always +1 of b
• b=(a×n) + n

I have also added 2 more terms using what I think are my formulae.

 ‘n’ ‘a’ ‘b’ ‘c’ Perimeter Area 1 3 4 5 12 6 2 5 12 13 30 30 3 7 24 25 56 84 4 9 40 41 90 180 5 11 60 61 132 330

Here are my formulas.

Middle

From looking at my table I could see that n×a+n=b. so then I substituted a for its formula and I got 2n²+2n.Side c is +1 of b so you just put the formula of b and then +1. So 2n²+2n+1.
 ‘n’ ‘a’ ‘b’ ‘c’ area perimeter 1 3 4 5 6 12 2 5 12 13 30 30 3 7 24 25 84 56 4 9 40 41 180 90 5 11 60 61 330 132 6 13 84 85 546 182 7 15 112 113 840 240 8 17 144 145 1224 306 9 19 180 181 1710 360 10 21 220 221 2310 462

Now I have to prove that my formulae for a, b and c work. To do this I must incorporate a²+b²+c²: -

a²+b²=c²

(2n+1)²+(2n²+2n) ²=(2n²+2n+1) ²

(2n+1)(2n+1) + (2n²+2n)(2n²+2n) = (2n²+2n+1)(2n²+2n+1)

4n²+2n+2n+1+4n⁴+4n²+4n³+4n³ = 4n⁴+8n³+8n²+4n+1

4n⁴+8n³+8n²+4n+1 = 4n⁴+8n³+8n²+4n+1

This proves that my formulae for a, b and c are all correct!

Now for part 2, I will try to solve the problem of Pythagorean triples with even numbers for a.

The numbers 6, 8 and 10 work for Pythagoras’ theorem,

6²+8²=10²

because        6²=6×6=36

8²=8×8=64

10²=10×10=100

and so                6²+8²=36+64=100=10²

Conclusion

So                2n+n²-1+n²+1

=                2n+2n²

=                2n²+2n

Here is how I got to my b and c formulae.

Getting b

a²+b²=c²

=c²= (b+2) ²

=(2n) ²+b²=(b+2) ²

=4n²+b²=b²+4b+4

=4n²-4=4b

=n²-1=b

=b=n²-1

Getting c

By looking at b you can see that c is always +2 than b.

So c is n²-1+2

=n²+1

Here is a table with all of the results for even numbers.

 ‘n’ ‘a’ ‘b’ ‘c’ Perimeter Area 1 2 0 2 4 0 2 4 3 5 12 6 3 6 8 10 24 24 4 8 15 17 40 60 5 10 24 26 60 120 6 12 35 37 84 210 7 14 48 50 112 336 8 16 63 65 144 504 9 18 80 82 180 720 10 20 99 101 220 990

Now I have to prove my formulae for a, b and c. To do this I have to incorporate a²+b²=c²

a²+b²=c²

=        (2n)²+(n²-1) ²=(n²+1) ²

=        (2n)(2n)=4n²

(n²-1)(n²-1)=n⁴-n²-n²+1

(n²+1)(n²+1)=n⁴+n²+n²+1

=        4n²+n⁴-n²-n²+1=n⁴+n²+n²+1

=        n⁴+2n³+1=n⁴+2n³+1

This proves that my formulas for a, b and c are all correct.

This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras section.

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