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• Level: GCSE
• Subject: Maths
• Word count: 3537

# Maths GCSE coursework: Beyond Pythagoras

Extracts from this document...

Introduction

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:

 Short side (1st difference 2) Middle side(2nd difference 4) Long side(2nd difference 4) Perimeter(2nd difference 8) Area (units²)(3rd difference 12) 3 4 5 12 6 5 12 13 30 30 7 24 25 56 84 9 40 41 90 180 11 60 61 132 330 13 84 85 182 546

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)

Middle

8        8          8            8                2nd difference

The 2nd difference is 8 so we must find the formula for the ‘perimeter’.

8 / 2 = 4                                        [8 is halved due to the use of n²

So we must look at 4n²

Sequence                       12         30          56          90          132        182

Value of 4n²                4        16        36        64        100        144

Difference                8        14        20        26        32        38

Difference in differences 6            6           6          6          6

Difference                 8        14        20        26        32        38

+2

Value of 6n                6        12        18        24        30        36

So the formula is:

4n² + 6n + 2

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

n = 3 the sequence number for this is 56.

4n² + 6n + 2

(4 x 9) + (6 x 3) +2

36 + 18 + 2

= 56

It is correct!

We can now look at the area to complete the table of formulas,

6          30          84           180        330        546

24             56             96             150        216                  1st difference

30          42          54           76        2nd difference

12             12             12                         3rd difference

This is hard to do due to the 3rd difference rather than the first or second differences in the previous formulas, so I had to teach myself how to 3rd differences with the use of ‘n’.

I will not worry about the how to get this formula in this way, but first a shall use the original formula to work out area with the use of the other numbers, small number x middle number x a half, I will use this to work out, simply my formula. I will use the formulas for the small number and multiply it by the middle number formula then divide it by 2.

Formula for small number x Formula for middle number x a half

2n + 1                             x 2n² + 2n                               x 0.5

I will work out the first part of multiplying the formulas together:

2n x 2n² = 4n³

(2n + 1)(2n² + 2n)

1 x 2n = 2n

Conclusion

The final topic I will mention is a circle drawn to join all of the interior sides of a triangle and show you the relations between the length of sides of the triangle and the radius of the circle.

The “r” stands for radius of the circle inscribed in the triangle, “a” stands for the short side of the right-angled triangle and “b” stands for the medium side. I will now let you think about the diagrams for a little bit and see if you can see the relationships between the radius and sides.

Note: this can also prove that a² + b² = c²

The conclusion:

My conclusion for the task set is that Pythagorean triples can be worked out by the use of the first odd numbered Pythagorean triples (3 – 13) these can be multiplied to any Pythagorean triple and there for the formulae must be changed, to the other Pythagorean triples needs by multiplying formulae by the multiple of the original triples apart from the area which needs to be multiplied again by 2 so the different measurement (units³) can be accounted for. I have also noticed that the shortest and medium sides of the triangle are made up of the radius of an interior circle, either by the medium side minus the radius to for the shortest side or the shortest side minus the radius added to the medium side minus the radius to form the longest side.

The investigation has shown that Pythagorean triples can be predicted but there may be some that I have not accounted for in my investigation.

Lok Man Lee

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

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