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  • Level: GCSE
  • Subject: Maths
  • Word count: 3187

Pythagorian Triples

Extracts from this document...

Introduction

Introduction

I am going to investigate Pythagorean Triples.  In this investigation, I shall explore right-angled triangles as used in Pythagoras’s Theorem.  Pythagoras’s Theorem is ‘A²+B²=C².  Some examples of Pythagorean Triples are (3,4,5), (5,12,13) & (7,24,25).  These could be used in the triangle below.

image00.png

In right-angled triangles like this one, the area can be found by using this equation -         0.5 x Base x Height.

The perimeter in these triangles is always found using this equation – Side A + Side B + Side C.

To begin my investigation, I am going to investigate the family of Pythagorean Triples where all three sides are positive integers and the shortest side is an odd number.  I will start with the shortest side being 3cm and progress until I have 9 readings.

‘B + 1’

 3² + 4² = 5² (9+16=25)
Area = 0.5 x 3 x 4 =6cm²
image01.png

Perimeter = 3cm + 4cm + 5cm = 12cm

image02.png

5² + 12² = 13² (25+144=169)

Area = 0.5 x 5 x 12 = 30cm²

Perimeter= 5cm+12cm+13cm=30cm²

image03.png

7²+24² = 25² (49+576 = 625)

Area = 0.5 x 7 x 24 = 84cm²

Perimeter= 7cm+24cm+25cm=56cm²

        9²+40² = 41² (81+1600 = 1681)image04.png

Area = 0.5 x 9 x 40 = 180cm²

Perimeter = 9cm+40cm+41cm = 90cm

11²+60² = 61² (121+3600 = 3721)image05.png

Area = 0.5 x 11 x 60 = 330cm²

Perimeter = 11cm+60cm+61cm = 132cm

        13²+84² = 85² (169+7056 = 7225)image06.png

Area = 0.5 x 13 x 84 = 546cm²

Perimeter = 13cm+84cm+85cm = 182cm

        15²+112² = 113² (225+12544 = 12769)image07.png

Area = 0.5 x 15 x 112 = 840cm²

Perimeter = 15cm+112cm+113cm = 240cm

        17²+144² = 145² (289+20736 = 21025)image08.png

Area = 0.5 x 17 x 144 = 1224cm²

Perimeter = 17cm+144cm+145cm = 306cm

image09.png

19²+180² = 181² (360+32400 = 32760)

        Area = 0.5 x 19 x 180 = 1710cm²

Perimeter = 19cm+180cm+181cm = 380cm

I will now add all of my results into a table so I can see them more clearly and see if there are any trends and rules.

Term Number

Side A

Side B

Side C

Area (cm²)

Perimeter (cm)

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

380

...read more.

Middle

64

510

514

16320

1088

15

68

576

580

19584

1224

16

72

646

650

23256

1368

17

76

720

724

27360

1520

18

80

798

802

31920

1680

19

84

880

884

36960

1848

20

88

966

970

42504

2024

From this table, I have found a rule to find Side A, a rule to find Side B and rule to find Side C.

Rules For B+4

        Side A -        4n+8.  This is explained by four multiplied by the term number then plus eight.  For example, if the term number is four, then Side A would be twenty-four.

        Side B -        2(n²+4n+3).  This is explained by squaring the term number, adding 4 multiplied by the term number, adding 3 and then multiplying that number by 2.

Side C -         B+4.  This is explained by Side B plus 4.  For example, if Side B is 4, Side C would be 8.

‘B+5’

        Here is my results table for ‘B+5’.

Term Number

Side A

Side B

Side C

Area (cm²)

Perimeter (cm)

1

15

20

25

150

60

2

25

60

65

750

150

3

35

120

125

2100

280

4

45

200

205

4500

450

5

55

300

305

8250

660

6

65

420

425

13650

910

7

75

560

565

21000

1200

8

85

720

725

30600

1530

9

95

900

905

42750

1900

10

105

1100

1105

57750

2310

From this table, I have found a rule to find Side A, a rule to find Side B and rule to find Side C.

Rules For B+5

        Side A -        10n+5.  This is explained by ten multiplied by the term number then plus five.  For example, if the term number is one, then Side A would be fifteen.

        Side B -        an+5n.  This is explained by Side A multiplied by the term number then plus five times the term number.  For example, if the term number was 2 and side A was 25, Side B would be 60.

Side C -         B+5.  This is explained by Side B plus 5.  For example, if Side B is 10, Side C would be 15.

‘B+6’

        Here is my results table for ‘B+6’.

Term Number

Side A

Side B

Side C

Area (cm²)

Perimeter (cm)

1

18

24

30

216

72

2

24

32

38

384

94

3

30

72

78

1080

180

4

36

105

111

1890

252

5

42

142

148

2982

332

6

48

189

195

4536

432

7

54

240

246

6480

540

8

60

297

303

8910

660

9

66

360

366

11880

792

10

72

429

435

15444

936

11

78

504

510

19656

1092

12

84

585

591

24570

1260

13

90

672

678

30240

1440

14

96

765

771

36720

1632

15

102

864

870

44064

1836

16

108

969

975

52326

2052

17

114

1080

1086

61560

2280

18

120

1197

1203

71820

2520

19

126

1320

1326

83160

2772

From this table, I have found a rule to find Side A, a rule to find Side B and rule to find Side C.

Rules For B+6

        Side A -        6n+6.  This is explained by six multiplied by the term number then plus six.  For example, if the term number is one, then Side A would be fifteen.

        Side B -        3(n²+4n+3) This is explained by squaring the term number, adding 4 multiplied by the term number, adding 3 and then multiplying that number by 3.

Side C -         B+6.  This is explained by Side B plus 6.  For example, if Side B is 10, Side C would be 16.

‘B+7’

        Here is my results table for ‘B+7’.

Term Number

Side A

Side B

Side C

Area (cm²)

Perimeter (cm)

1

21

28

35

294

84

2

35

84

91

1470

210

3

49

168

175

4116

392

4

63

280

287

8820

630

5

77

420

427

16170

924

6

91

588

595

26754

1274

7

105

784

791

41160

1680

8

119

1008

1015

59976

2142

9

133

1260

1267

83790

2660

10

147

1540

1547

113190

3234

...read more.

Conclusion

8

8n+16

4(n²+4n+3)

B+8

0.5xAxB

A+B+C

9

18n+9

an+9n

B+9

0.5xAxB

A+B+C

10

10n+20

5(n²+4n+3)

B+10

0.5xAxB

A+B+C

I have noticed that in the above table there are two different patterns.  The two patterns are one when ‘B+…’ is even and when ‘B+…’ is an odd number.  Because of this, I will now draw two separate tables.

Odds

B+…

Side A

Side B

Side C

Area (cm²)

Perimeter (cm)

1

2n+1

an+n

B+10

0.5xAxB

A+B+C

3

6n+3

an+3n

B+3

0.5xAxB

A+B+C

5

10n+5

an+5n

B+5

0.5xAxB

A+B+C

7

14n+7

an+7n

B+7

0.5xAxB

A+B+C

9

18n+9

an+9n

B+9

0.5xAxB

A+B+C

Evens

B+…

Side A

Side B

Side C

Area (cm²)

Perimeter (cm)

2

2n+4

(n²+4n+3)

B+2

0.5xAxB

A+B+C

4

4n+8

2(n²+4n+3)

B+4

0.5xAxB

A+B+C

6

6n+6

3(n²+4n+3)

B+6

0.5xAxB

A+B+C

8

8n+16

4(n²+4n+3)

B+8

0.5xAxB

A+B+C

10

10n+20

5(n²+4n+3)

B+10

0.5xAxB

A+B+C

Overall Rules

Key – X = B+… (whatever side c is)

Odds

        Side A – 2X + X.          This is explained by 2 times by Side C plus Side C.

        Side B – AN + XN.  This is explained by Side A times the term number, plus Side C times by the term number.

Area – ½ A X B.  This is explained by 0.5 times Side A times Side B.

        Perimeter – A+B+C.  This is explained by Side A plus Side B plus Side C.

Evens

Side A – XN + 2X.  This is explained by Side C times the term number, plus two time Side C.

        Side B – (0.5 X)(n²+4n+3).  This is explained by 0.5 times Side C, times by the term number squared plus four times the term number plus 3.

        Area – ½ A X B.  This is explained by 0.5 times Side A times Side B.

        Perimeter - A+B+C.  This is explained by Side A plus Side B plus Side C.

...read more.

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