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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.

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

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

5² + 12² = 13² (25+144=169)

Area = 0.5 x 5 x 12 = 30cm²

Perimeter= 5cm+12cm+13cm=30cm²

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)

Area = 0.5 x 9 x 40 = 180cm²

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

11²+60² = 61² (121+3600 = 3721)

Area = 0.5 x 11 x 60 = 330cm²

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

13²+84² = 85² (169+7056 = 7225)

Area = 0.5 x 13 x 84 = 546cm²

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

15²+112² = 113² (225+12544 = 12769)

Area = 0.5 x 15 x 112 = 840cm²

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

17²+144² = 145² (289+20736 = 21025)

Area = 0.5 x 17 x 144 = 1224cm²

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

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

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

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.

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