Beyond Pythagoras

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Beyond Pythagoras

Pythagoras’ Theorem for a right-angled triangle states that the square of the hypotenuse is equal to the sum of the squares of the other two sides, which can be written as a formula for the above triangle:

 

C2 = A2 + B2

A Pythagorean Triple is a set of these numbers, such as 3, 4, 5, where the square of the largest number is equal to the squares of the other two numbers.  My task is to investigate different relationships between the numbers in families of Pythagorean Triples.  To begin, I will investigate the family where the shortest side of the triangle is an odd number, all 3 sides are positive integers and side C is always 1 more than side B.  I am going to start with the shortest side being 1cm and work out 10 more triangles after that.

1cm2 + 2cm2 ≠ 3cm2, so there is no relation between the sides when the shortest side of a right-angled triangle is 1cm.

As this triangle doesn’t follow the rules, I will ignore it and start again with 10 more triangles.

32 + 42 = 9 + 16 = 25

√25 = 5

Area = (4×3)/2 = 6cm2

Perimeter = 4 + 3 + 5 = 12cm

52 + 122 = 25 + 144 = 169

√169 = 13

Area = (12×5)/2 = 15cm2

Perimeter = 13 + 12 + 5 = 30cm

72 + 242 = 49 + 576 = 625

√625 = 25

Area = (7×24)/2 = 84cm2

Perimeter = 7 + 24 + 25 = 56cm

92 + 402 = 81 + 1600 = 1681

√1681 = 41

Area = (9×40)/2 = 180cm2

Perimeter = 9 + 40 + 41 = 90cm

112 + 602 = 121 + 3600 = 3721

√3721 = 61

Area = (11×60)/2 = 330cm2

Perimeter = 11 + 60 + 61 = 132cm

132 + 842 = 169 + 7056 = 7225

√7225 = 85

Area = (13×84)/2 = 546cm2

Perimeter = 13 + 84 + 85 = 182cm

152 + 1122 = 225 + 12544 = 12769

√12769 = 113

Area = (15×112)/2 = 840cm2

Perimeter = 15 + 112 + 113 = 240cm

172 + 1442 = 289 + 20736 = 21025

√21025 = 145

Area = (17×144)/2 = 1224cm2

Perimeter = 17 + 144 + 145 = 306cm

192 + 1802 = 361 + 32400 = 32761

√32761 = 181

Area = (19×180)/2 = 1710cm2

Perimeter = 19 + 180 + 181 = 380cm

212 + 2202 = 441 + 12544 = 48400

√48400 = 221

Area = (21×220)/2 = 2310cm2

Perimeter = 21 + 220 + 221 = 462cm


Now I am going to put all the information that I have found out into a table to make it easier to spot any relationships or patterns between the numbers.

Now that the results are in a clear table, I can see that there are various patterns emerging around the numbers.  I am going to try to find a formula for each of the sides, the area and the perimeter of the triangles.

Shortest side (a)

The length of this side increases by 2 for each triangle.  It is also 1 more than double the triangle number (n).  This means that the formula is  (2×n) + 1, which is more commonly written as 2n + 1

Middle side (b)

The length of this side goes up in multiples of 4 each time.

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