Perimeter = 3 + 4 + 5 = 12
Finally to find the areas we times the smallest and middle side and then divide by to.
Area = ½ x 3 x 4 = 6
I will now put the first term in to a table and try to find the next terms up to 10.
- s increases by +2 each term
- s is equal to the term number times 2 then add 1
- the square root of (m + l) = a
- (s x n) + n = m
I have worked out formulas for
- How to get s from n
- How to get m from n
- How to get l from n
- How to get the p from n
- How to get the a from n
To find the Smallest side (s) you use the formula 2n + 1
To find the Middle side (m) you use the formula 2n2 + 2n or (s x n) + n
To find the Longest side (l) you use the formula 2n2 + 2n + 1 or (n x s) + n + 1
To find the Perimeter (p) you use the formula 4n2 + 6n +2 or s + m + l
To find the Area (a) you use the formula 2n3 + 3n2 + n or s x m divided by 2
To get these formulas I did the following
- Take side ‘s’ 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 odd numbers (2n + 1 is the general formula for consecutive odd numbers) You may be able to see the formula if you draw a graph.
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From looking at my table of results, I noticed that ‘(s x n) + n = m’. So I took my formula for ‘s’ (2n + 1) multiplied it by ‘n’ to get ‘2n2 + n’. I then added my other ‘n’ to get ‘2n2 + 2n’.q=A graph will help explain this. (Also you can use the longer formula (n x s) + n)
- Side ‘l’ is just the formulas for side ‘m’ +1
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The perimeter = s + m + l. Therefore I took my formula for ‘s’ (2n + 1), my formula for ‘m’ (2n2 + 2n) and my formula for ‘l’ (2n2 + 2n + 1). I then did the following: -
2n + 1 + 2n2 + 2n + 2n2 + 2n + 1 = perimeter
Rearranges to equal
4n2 + 6n + 2 = perimeter (Also you can use the formula s + m + l)
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The area = (s x m) divided by 2. Therefore I took my formula for ‘s’ (2n + 1) and my formula for ‘m’ (2n2 + 2n). I then did the following: -
(2n + 1)(2n2 + 2n) = area 2
Multiply this out to get
4n3 + 6n2 + 2n = area
2
Then divide 4n3 + 6n2 + 2n by 2 to get
2n3 + 3n2 + n = area (Also you can use the formula s x m divided by 2)
To prove my formulas for ‘s’, ‘m’ and ‘l’ are correct. I decided incorporate my formulas into s2 + m2 = l2
s2 + m2= l2
(2n + 1)2+ (2n2 + 2n)2= (2n2 + 2n + 1)2 (2n + 1)(2n + 1) + (2n2 + 2n)(2n2 + 2n) = (2n2 + 2n + 1)(2n2 + 2n + 1)
4n2 + 2n + 2n + 1 + 4n4 + 4n2 + 4n3 + 4n3= 4n4 + 8n3 + 8n2 + 4n + 1
4n2 + 4n + 1 + 4n4 + 8n3 + 4n2= 4n4 + 8n3 + 8n2 + 4n + 1
4n4 + 8n3 + 8n2 + 4n + 1 = 4n4 + 8n3 + 8n2 + 4n + 1
This proves that my ‘s’, ‘m’ and ‘l’ formulas are correct