• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
• Level: GCSE
• Subject: Maths
• Word count: 3111

# 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 ofShort 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

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## Beyond Pythagoras

84 + 85 = 182cm 15� = 225 C=113 112� = 12544 A=15 113� = 12769 225 + 12544 = 12769 B=112 Area = 15 x 112 = 1680 2 = 840 cm� Perimeter = 15 + 112 + 113 = 240cm 17� = 289 144� = 20736 C=145 145�

2. ## Beyond pythagoras - First Number is odd.

2n� + 2n +1 = 13 2 x 2� + 2 x 2 + 1 = 13 8 + 4 + 1 = 13 13 = 13 The formula also works for the 2nd term. 2n� + 2n +1 = 25 2 x 3� + 2 x 3 + 1

1. ## Beyond Pythagoras

* Area formula is 1/2 x base x height = 1/2 x b x a * Perimeter formula is a + b + c ( I am going to find three more Pythagorean triples. The shortest side has been given to me as 3, 5, 7, and I am supposed to find out the middle side and the longest side.

2. ## Beyond Pythagoras

� 2n + 1 � 2n2 + 2n P = 4n2 + 6n + 2 Now that I have found all the relationships and formulas for the family of Pythagorean Triples where the shortest side of the triangle is an odd number, all 3 sides are positive integers and c

1. ## Beyond Pythagoras.

+ 8n3 + 8n2 + 4n + 1 4n4 + 8n3 + 8n2 + 4n + 1 = 4n4 + 8n3 + 8n2 + 4n + 1 This proves that my 'a', 'b' and 'c' formulas are correct Extension Polynomials Here are the formulas for the perimeter and the area

2. ## Beyond Pythagoras.

I will then be left with "a" the co-efficient of "n�". 2n�=4 I am now able to fine "c" by calculating the output for T0. If I subtract 4 from the first difference between the 1st and 2nd term and then subtract the output of the first term from my resulting number, I am given "c".

1. ## Beyond Pythagoras.

Shortest side 'a' Middle side 'b' Longest side (Hypotenuse) 'h' Perimeter Area 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 85 154 546 Side 'a' For side 'a' is could easily see that the formula would be 2n+1 as this is the general formula for odd numbers.

2. ## Pythagoras Theorem is a2 + b2 = c2. 'a' being the shortest side, 'b' ...

2n2 + 2n + 1 4. 4n2 + 6n + 2 5. 2n3 + 3n2 + n To get these formulas I did the following 1. Take side 'a' for the first five terms 3, 5, 7, 9, 11. From these numbers you can see that the formula is 2n + 1 because these are consecutive

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to