Beyond Pythagoras.
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Introduction
Mathematics Coursework  Beyond Pythagoras
Mathematics Coursework BEYOND PYTHAGORAS
By Asif Azam
1)
The numbers 3, 4, and 5 satisfy the condition
3² + 4² = 5²
because 3² = 3x3 =9
4² = 4x4 = 16
5² = 5x5 = 25
and so
3² + 4² = 9 + 16 = 25 = 5²
I will now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) ² + (middle number) ² = (largest number) ².
a) 5, 12, 13
5² + 12² = 25 + 144 = 169 = 13²
b) 7, 24, 25
7² + 24² = 49 + 576 = 625 = 25²
2) Perimeter
 
 

 
 

wwfe fewstfefeud efe fentcfe enfetral fecofe uk:
b)
Nth term  Length of shortest side  Length of middle sidewwdd ddwstddddud edd ddntcdd enddtral ddcodd uk.  Length of longest side  Perimeter  Area  
1 wwbf bfwstbfbfud ebf bfntcbf enbftral bfcobf uk.  3  4  5  12  6  
2  5  12  13  30  30  
3  7  24  25  56  84  
4 wwef efwstefefud eef efntcef eneftral efcoef uk;  9  40  41  90  180  
5  11  60  61  132  330  

Middle
182
546
7
15
112
113
240
840

8
17
144
145
306
1224
9
19
180
181
380
1710 wwdd ddwstddddud edd ddntcdd enddtral ddcodd uk.
10
21
220
221
462
2310
I looked at the table and noticed that there was only 1 difference between the length of the middle side and the length of the longest side. And also if you can see in the shortest side column, it goes up by 2. I have also noticed that the area is
½ (shortest side) x (middle side).
3)
In this section I will be working out and finding out the formulas for:
 Shortest side
 Middle side
 Longest side
In finding out the formula for the shortest side I predict that the formula will be something to do with the differences between the lengths (which is 2). But I don’t know the formula so I will have to work that out. wwfe fewstfefeud efe fentcfe enfetral fecofe uk.
Firstly I will be finding out the formula for the shortest side.
3 5 7 9 11
2 2 2 2
The differences between the lengths of the shortest side are 2. This means the equation must be something to do with 2n.
Let’s see…
Nth term  Length of shortest side 
1  3 
2n
2 x 1 = 2 (wrong)
There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1. To see if I’m correct I will now test this formula.
2n+1
Conclusion
= 2n + 1 + 2n² + 2n + 2n² + 2n + 1
= 2n² + 2n² + 2n + 2n + 2n + 1 + 1
= 4n² + 6n + 2
If I have done my calculations properly then I should have the right answer. To check this I am going use the 4th, 5th and 6th terms.
4th term:
4n² + 6n² + 2 = perimeter
4 x 4² + 6 x 4 + 2 = 9 + 40 + 41
64 + 24 + 2 = 90
90 = 90
It works for the 4th term
Let’s see if it works for the 5th term:
4n² + 6n² + 2 = perimeter
4 x 5² + 6 x 2 = 11 + 60 + 61
100 + 30 + 2 = 132
132 = 132
And it works for the 5th term
And finally the 6th term:
4n² + 6n² + 2 = perimeter
4 x 6² + 6 x 6 + 2 = 13 + 84 + 85
144 + 36 + 2 = 182
182 = 182
It works for all the terms so:
Perimeter = 4n² + 6n + 2 wwed edwstededud eed edntced enedtral edcoed uk!
Like the area I know that the area of a triangle is found by:
Area = ½ (b x h)
b = base
h = height
Depending on which way the right angled triangle is the shortest or middle side can either be the base or height because it doesn’t really matter which way round they go, as I’ll get the same answer either way.
Area = ½ (shortest side) X (middle side)
= ½ (2n + 1) x (2n² + 2n)
= (2n + 1)(2n² + 2n)
I will check this formula on the first two terms:
(2n + 1)(2n² + 2n) = ½ (b x h)
(2 x 1 + 1)(2 x 1² + 2 x 1) = ½ x 3 x 4
3 x 4 = ½ x 12
12 = 6
6 = 6
2nd term:
(2n + 1)(2n² + 2n) = ½ b h
(2 x 2 + 1)(2 x 2² + 2 x 2) = ½ x 5 x 12
5 x 12 = ½ x 60
60 = 30
30 = 30
It works for both of the terms. This means:
Area = (2n + 1)(2n² + 2n)
This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras section.
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