# Maths GCSE coursework: Beyond Pythagoras

Lok Man Lee

10 MWY

Maths GCSE coursework:  Beyond Pythagoras

Within this investigation I will look at the relationships between the lengths, perimeters and areas of right-angled triangles. This looks at Pythagorean Triples, three numbers that satisfy the condition of:

(smallest number)2 + (middle number)2 = (largest number)2

This can also be expressed as:

a2 + b2 = c2

I will look first at an odd number as the ‘smallest number’ then continue to even numbers.

The objective of this investigation is to be able to:

Make generalisations about the lengths of sides.

Make generalisations about the perimeter and area of the corresponding triangles.

To prove a Pythagorean triple you must check that the three numbers satisfy the condition.

So…

5, 12, 13

52 + 122 = 25 + 144 = 169 = 132

It is a Pythagorean triple because 5² + 12² = 13².

To try a second one:

7, 24, 25

7² + 24² = 49 + 576 = 625 = 25²

This is also a Pythagorean triple 7² + 24² = 25²

Pythagorean triples also can be used in right-angled triangles

Perimeter: 5 + 12 + 13 = 30

Area: 0.5(5 x 12) = 30

13

5

12

To work out the perimeter of these triangles you add together all the sides, and to find out the area of the triangle you have to multiply the shortest sides and then you half the answer.

A table showing the perimeter, area of Pythagorean triples:

Within this table there are many patterns that can be seen, the first is simple which is that the numbers on the middle and longest side are consecutive (b + 1 = c), but this is a part of another pattern which is that the square of the shortest side is the same as the middle and longest sides added together,

So…

a² = b + c

The theory works because:

12 +13 = 25 = 5²

5² = 12 + 13                        This now allows to predict, odd starting, Pythagorean triples.

It is correct!

We have to, now, look at the smallest number.

3        5            7          9        11       13

2              2              2              2           2                  1st difference

The 1st difference is 2 so we must find the formula for the ‘smallest number’.

So we must look at 2n

Sequence                3          5          7          9          11         13

Value of 2n                2        4        6        8        10        12

Difference                1        1        1        1         1         1

So the formula is:

2n + 1

To check that the formula is correct we can try and put it into the sequence:

n = 3 the sequence number for this is 7 .

2n + 1

2 x 3 + 1

6 + 1

= 7

It is correct!

To see if there are any more formulae in the table we will first look at the ‘middle number’ sequence:

4         12       24        40        60       84

8             12             16            20        24                  1st difference

4            4            4            4                2nd difference

The 2nd difference is 4 so we must find the formula for the ‘middle number’.

4 / 2 = 2                                        [4 is halved due to the use of n²

So we must look at 2n²

Sequence                4        12        24        40        60        84

Value of 2n²                2        8        18        32        50        72

Difference                2        4        6        8        10        12

Difference in differences  2          2        2            2         2

Difference                 2        4        6        8        10        12

0

Value of 2n                2        4        6        8        10        12

So the formula is:

2n² + 2n

To check that the formula is correct we can try and put it into the sequence:

n = 3 the sequence number for this is 24.

2n² + 2n

2(3²) + (2 x 3)

18 + 6

= 24

It is correct!

We will now look at the long side and look for a pattern:

5          13       25        41        61        85

8             12             16             20          24                  1st difference

4         4         4         4                2nd difference

The 2nd difference is 4 so we must find the formula for the ‘largest number’.

4 / 2 = 2                                        [4 is halved due to the use of n²

So we must look at 2n²

Sequence                5        13        25        41        61        85

Value of 2n²                2        8        18        32        50        72

Difference                3        5        7        9        11        13

Difference in differences    2       2             2         2          2

Difference                 3        5        7        9        11        13

+  1

Value of 2n                2        4        6        8        10        12

So the formula is:

2n² + 2n + 1

To check that the formula is correct we can try and put it into the sequence:

n = 3 the sequence number for this is 25.

2n² + 2n + 1

2(3²) + (2 x 3) + 1

18 + 6 + 1

= 25

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