# Beyond Pythagoras.

Extracts from this document...

Introduction

The numbers 3,4 and 5 satisfy the condition because

3 + 4 = 5

= 9 + 16 = 25 3 5

And so 3 + 4 = 5

4

The perimeter and area of this triangle are:

Perimeter = 3 + 4 + 5= 12 units

Area= ½ base x height = ½ x 4 x 3 = 6 units

I am going to investigate to see if the following numbers satisfy the similar condition of (smallest number) + (middle number) = (largest number)

I am keeping the smallest side an odd number. I will draw diagrams to help me solve the problem more easily.

5, 12, 13

5 + 12 = 13 5 13

25 + 144 = 169

And so 5 + 12 = 13

12

The perimeter and area of this triangle are:

Perimeter = 5 + 12 + 13 = 30 units

Area = ½ base x height = ½ x 12 x 5 = 30 square units

I am now going to investigate with the numbers 7, 24 and 25

7 + 24 = 25

49 + 576 = 625

and so 7 + 24 = 25 7 25

24

The perimeter and area of this triangle are :

Perimeter = 7 + 24 +25 = 56

Area = ½ base x height = ½ x 24 x 7 = 84 square units

Here is a table to show you my results

n | Length of Short Side | Length of Middle Side | Length of Longest Side | Perimeter (Units) | Area (Units Squared) |

1 | 3 | 4 | 5 | 12 | 6 |

2 | 5 | 12 | 13 | 30 | 30 |

3 | 7 | 24 | 25 | 56 | 84 |

4 | 9 | 40 | 41 | 90 | 180 |

5 | 11 | 60 | 61 | 132 | 330 |

6 | 13 | 84 | 85 | 142 | 546 |

7 | 15 | 112 | 113 | 240 | 840 |

8 | 17 | 144 | 145 | 306 | 1224 |

I only investigated on triangles 1, 2 and 3. I tried to see if I could find a pattern going down each length to fill in the rest of the table, I did. As I am investigating on odd numbers I would make the length of the shortest side odd i.e. 3,5,7,9,11 etc.

To find the middle length I found that it was going down by the 4 times table.

4 12 24 40 60 84 112 144

8 12 16 20 24 28 32

So if I add the answer from before to the next multiplication of 4 it would give me the next answer so for example.

Middle

To work out the formula for the longest length

I know that the sequence goes like this

5 13 25 41 61

8 12 16 20

4 4 4

The second difference is 4. I am now going to use the following formula to help me find a formula fort he longest length.

an + bn + c = 0

I know that a = 2 because the second difference is 4 so if I divide 4 by 2 I would get 2.

a = 2 2n + bn + v

n = 1 2 + b + c = 5

B + c = 3 (1)

n = 2 8 + 2b + c = 13

2b + c = 5 (2)

(2) – (1) b + c = 3

b = 2

so c = 1

I now know that a = 2, b = 2 and c = 1. So if I put this in the formula of an + bn + c

I should get 2n + 2n + 1

I am going to see if this works. I am going to use the first length of the longest length to se if this formula works. By looking at my results table it tells me that the answer should be 5. Lets see.

2n + 2n + 1

2 x 1 + 2 x 1 + 1

2 + 2 + 1 = 5 correct

I am now going to see if this formula would work for another long length. I am now going to use length 2 to see if 2n + 2n + 1 is the right formula. It should be 13. Lets see.

2n + 2n + 1

2 x 2 + 2 x 2 + 1

8 + 4 + 1 = 13 correct

If it didn’t equal to 13 I would need to work on my formula again.

I have just work out the formulas and double checked to see if they were right. They all were correct.

Formula for the shortest side 2n + 1

Formula for the middle side 2n + 2n

Conclusion

= 8n + 26n + n + 40n +25

a + b = c so (2n + 4) + (n + 4n + 3) = (n + 4n + 5)

Lets see

a + b (2n + 4) + (n + 4n + 3)

4n + 16 + 16n + 16n + 24n + 6n + 9 + 8n + n

= n + 40n + 25 + 26n + 8n

This should equal to c = (n + 4n + 5) which is 8n + 26n + n + 40n + 25

And so from his I can see that a + b = c . So yes my formulas fit into Pythagoras theorem.

I now need to find a formula to work out the area. I know that the formula for an area of the triangle is ½ base x height. So I would apply the formulas I have found to find a formula for the area of each triangle.

Small (2n + 4) Long

2 (n + 2)

Middle (n + 4n + 3)

½ x 2 (n + 2) x (n + 4n + 3)

(n + 2) (n 4n + 3)

I now need to see if this is right. So I am going to use triangle 1 to see if this formula is right. It should be 24.

(n + 2) (n + 4n + 3)

(1 +2) (1x1 + 4x1 + 3)

(3) (8) = 24 correct

I am now going to check again and see if this formula is right. I am going to use triangle 2. The answer should be 60.

(n + 2) (n + 4n + 3)

(2+2) (2 x 2 + 4 x 2 + 3)

(4) (15) = 60 correct

I now need to find a formula to find the perimeter of each triangle. To find the perimeter of any shapes you just need to add up all the sides. So I am now going to add up all the formulas.

2n + 4 + n + 4n + 3 + n + 4n + 5

= 2n + 10n + 12

Lets see if this formula is right. I am going to use triangle 1 to see if this perimeter formula is right.

2n + 10n + 12

= 2 x 1 + 10 x 1 + 12

= 2 + 10 + 12

= 24 correct

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

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month