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  • Level: GCSE
  • Subject: Maths
  • Word count: 4280

For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician.

Extracts from this document...

Introduction

        --

Sarah Williams

BEYOND PYTHAGORAS

Introduction

For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician. Pythagoras lived on the island of Samos and was born around 569BC. He did not write anything but he is regarded as one of the world’s most important characters in maths. His most famous theorem is named after him and is called the Pythagoras Theorem. It is basically a²+b²=c². This is what the coursework is based on.image05.png

I am going to look at the patterns, which surround this theorem and look at the different sequences that can be formed.

The coursework

The numbers 3, 4 and 5 satisfy the condition 3²+4²=5² because

3²=3x3=9

        4²=4x4=16

        5²=5x5=25

And so

        3²+4²=9+16=25=5²

I will now check that the following sets of numbers also satisfy the similar condition of (smallest number) ²+(middle number) ²=(largest number) ²

a) 5, 12, 13

5²=5x5=25

        12²=12x12=144

        25+144=169

        √169 = 13

This satisfies the condition as

        5²+12²=25+144=169=13²

b) 7, 24, 25

        7²=7x7=49

        24²=24x24=576

        49+576=625

        √625=25

This satisfies the condition as

        7²+24²=49+576=625=25²

The numbers 3,4 and 5 can be the lengths – in appropriate units – of the side of a right-angled triangle.

image00.png

            5                        

3                                        

image01.png

The perimeter and area of this triangle are:

Perimeter = 3+4+5=12 units

        Area = ½x3x4=6 units ²

image02.png

The numbers 5,12,13 can also be the lengths – in appropriate units – of a right-angled triangle.         

image01.png

Perimeter = 5+12+13=30

Area=½x5x12=30

                                                        This is also true for the numbers 7,24,25image03.png

image01.png

Perimeter = 7+24+25=56

Area=½x7x24=84                        

I have put these results into a table to see if I can work out any patterns.

Length of shortest side

Length of middle side

Length of longest side

Perimeter

Area

3

4

5

12

6

5

12

13

30

30

7

24

25

56

84

I have noticed that there seems to be a recurring pattern between the length of the middle side and the length of the longest side. They are always consecutive numbers. I shall now investigate this.

I will assume that the hypotenuse has length b + 1 where b is the length of the middle side.

...read more.

Middle

2256

25944

24

49

1200

1201

2450

29400

25

51

1300

1301

2652

33150

Another pattern that I have noticed is that there must be a connection between the sides of the triangle because the middle side and the longest side have a difference of one. I will now try and work out the formula. I will basically use trial and error and try out different formulae that I think of.

(Shortest side) ²+(middle side) ²=(longest side) ²

I already know that this is correct because it is Pythagoras theorem. However I will still check it using the Pythagorean triple 3,4,5.

3²+4²=5²

I know that this is correct because

3²=9

4²=16

9 + 16 = 25, which is also 5²

(Middle side) ²+(longest side) ²=(shortest side) ²

To see if this statement is true I shall use the Pythagorean triple 9,40,41.

40²+41²=9²

This is not true because

40² = 1600

41² = 1681

which add together to make 3281

and

9²=81

This means that this is not a correct formula.

(Longest side) ²/(shortest side) ² = (middle side)

To check this I will use the Pythagorean triple 15,112,113.

113²/15²=112²

This is wrong as

113²=12769

15²=225

and

12769/225 = 56.751111

and as

The middle side = 113 this formula has to be wrong.

Shortest side x longest side = (middle side) ²

To check this I will use the Pythagorean triple 7,24,25

7 x 25 = 175

This is wrong because

24²= 576 and this is very different

(Longest side) ²-(middle side) ²=shortest side

To check this formula I will use the Pythagorean triple 5,12,13

13²-12²=5²

This is right as

13²=169

12²=144

169-144=25 which is also 5².

So this formula is correct

(Middle side) ²+Smallest side=Largest side

To check this I will use the Pythagorean triple 5, 12,13.

12²+5=13

This is silly and I shouldn’t really have tried it, as I should have been able to work out that this formula is incorrect just by looking at the numbers involved. 144 is bigger than 13 without adding the 5.

Largest side + Middle side = (shortest side) ²

To check this I will use the Pythagorean triple 3,4,5

5+4=3²

This is correct because 5 + 4 = 9 which is also 9.

Just to check that this formula works with all triples I shall pick 3 random numbers from the N column of the table and use the formula with them. I shall use (i) 23 (ii) 10 (iii) 16

  1. 1105+1104=47² which is also 2209. This is correct
  2. 221+220=21² which is also 441. This is correct
  3. 545+544=33² which is also 1089. This is correct

This seems to be a correct formula.

So the formulae  I have worked  out in this section are:

  • (Shortest side)²+(middle side)²=(longest side)²
  • (Longest side) ²-(middle side) ²=shortest side
  • Largest side + Middle side = (shortest side) ²

All these are based on the assumption that the difference between the longest side and the middle side is 1.

However, I am interested about what happens when the difference between the longest side and the middle side is different from 1. I shall investigate this going up to c=b+10.

In the next few investigations I was required to work out the square roots of numbers. However, although it is possible to work out a square root of every number I chose only to investigate the perfect squares. Those numbers which are whole numbers. This is because I am only investigating Pythagorean triples with positive integers.

Investigating c=b+2

I will now investigate when longest side – middle side= 2. (c-b=2)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 2(2b+2)

   =4b+4

I can now produce a table for different values of b. Perfect squares are coloured red.

B

a²=4b+4

√a²

Triples (a,b,c)

1

8

2

12

3

16

4

This becomes 4,3,5

4

20

5

24

6

28

7

32

8

36

6

This becomes 6,8,10

9

40

10

44

11

48

12

52

13

56

14

60

15

64

8

This becomes 8,15,17

16

68

17

72

18

76

19

80

20

84

21

88

22

92

23

96

24

100

10

This becomes 10, 24, 26, which is a multiple of 5,12,13.

25

104

Investigating c=b+3

I will now investigate when longest side – middle side=3. (c-b=3)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 3(2b+3)

   =6b+9

I can now produce a table for different values of b. Perfect squares are coloured red.

B

a²=6b+9

√a²

Triples (a,b,c)

1

15

2

21

3

27

4

33

5

39

6

45

7

51

8

57

9

63

10

69

11

75

12

81

9

This becomes 9,12,15, which is a multiple of 3,4,5

13

87

14

93

15

99

16

105

17

111

18

117

19

123

20

129

21

135

22

141

23

147

24

153

25

159

Investigating c=b+4

I will now investigate when longest side – middle side=4. (c-b=4)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 4(2b+4)

   =8b+16

I can now produce a table for different values of b. Perfect squares are coloured red.

B

a²=8b+16

√a²

Triples (a,b,c)

1

24

2

32

3

40

4

48

5

56

6

64

8

This becomes 8,6,10 which is a multiple of 4,3,5

7

72

8

80

9

88

10

96

11

104

12

112

13

120

14

128

15

136

16

144

12

This becomes 12,16,20 which is a multiple of 3,4,5

17

152

18

160

19

168

20

176

21

184

22

192

23

200

24

208

25

216

Investigating c=b+5

I will now investigate when longest side – middle side=5. (c-b=5)

a²+b²=c²

therefore

a²=c²-b²

factorising

a²=(c-b)(c+b)

   = 5(2b+5)

   =10b+25

I can now produce a table for different values of b. Perfect squares are coloured red.

B

a²=10b+25

√a²

Triples (a,b,c)

1

35

2

45

3

55

4

65

5

75

6

85

7

95

8

105

9

115

10

125

11

135

12

145

13

155

14

165

15

175

16

185

17

195

18

205

19

215

20

225

15

This becomes 15,20.25 which is a multiple of 3,4,5

21

235

22

245

23

255

24

265

25

275

...read more.

Conclusion

Internet

An interesting thing I found on the Internet, which I had never heard of before, was Pythagorean Quads (quadruples).

Pythagorean quads are the 3D version of the triples and can all be generated by the following formula, where a, b, c and d are whole numbers.
If p = a
2 + b2 - c2 - d2

   q = 2(ad - bc)  

   r = 2(ac + bd)

and s = a2 + b2 + c2 + d2


then p
2 + q2 + r2 = s2.


I shall now try and work out a few of these to check if this formula is correct

I shall choose the numbers a=4, b=1, c=1 and d=1.

This means that p=16+1-1-1 = 15

                      q=2(4-1)=6

                       r=2(4+1)=10             

                       s=16+1+1+1=19

From looking further on the Internet for confirmation I conclude that this formula does work because the quad that I worked out above is correct.

Conclusion

 I started this investigation by noticing that there was a difference of 1 between the middle side and the longest side. I then went on to work out formulae relating to this. All these formulae were based on the assumption that c-b+1. however after investigating further I realised that the difference between the middle side and the longest side could be any number. So I investigated all the differences up to a difference of 5. I decided that I would stop each table at the value of b=25. I realised that this investigation could go on forever and you could investigate unlimited values of b and  there are unlimited differences that I could have investigated but I had to set a limit. I also decided that I would only investigate up to c=b+10 as I felt that was a sufficient number of investigations. I wrote the extension because I thought it was interesting to see what else other people had discovered

...read more.

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

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