4 N 2 B example essay topic

Pythagoras Theorem is a 2 + b 2 = c 2.? a? being the shortest side, ? b? being the middle side and? c? being the longest side (hypotenuse) of a right angled triangle. The numbers 3, 4 and 5 satisfy this condition? 32 + 42 = 52 because? 32 = 3 x 3 = 9? 42 = 4 x 4 = 16?

52 = 5 x 5 = 25 and so? 32 + 42 = 9 + 16 = 25 = 52 The numbers 5, 12, 13 and 7, 24, 25 also work for this theorem? 52 + 122 = 132 because? 52 = 5 x 5 = 25?

122 = 12 x 12 = 144? 132 = 13 x 13 = 169 and so? 52 + 122 = 25 + 144 = 169 = 132? 72 + 242 = 252 because? 72 = 7 x 7 = 49?

242 = 24 x 24 = 576? 252 = 25 x 25 = 625 and so? From these observations I have worked out the next two terms. I will now put the first five terms in a table format. Term Number? n? Shortest Side? a?

Middle Side? b? Longest Side? c? How to get? b? from? n? How to get? c? from? n? How to get the perimeter from? n?

How to get the area from? n? From these numbers you can see that the formula is 2 n + 1 because these are consecutive odd numbers (2 n + 1 is the general formula for consecutive odd numbers) You may be able to see the formula if you draw a graph From looking at my table of results, I noticed that? an + n? = b? So I took my formula for? a? (2 n + 1) multiplied it by? n? to get? 2 n 2 + n? I then added my other? n? to get?

2 n 2 + 2 n? This is a parabola as you can see from the equation and also the graph Side? c? is just the formula for side? b? +1 The perimeter = a + b + c. Therefore I took my formula for? a? (2 n + 1), my formula for? b? (2 n 2 + 2 n) and my formula for? c?

(2 n 2 + 2 n + 1). 4 n 2 + 6 n + 2 = perimeter The area = (a x b) divided by 2. (2 n + 1) and my formula for? b? (2 n 2 + 2 n).

I then did the following: – (2 n + 1) (2 n 2 + 2 n) = area? 2 Multiply this out to get 4 n 3 + 6 n 2 + 2 n = area? From my table of results I know that the perimeter = area at term number 2. Therefore (n-2) is my factor I would like to find out at what other places (if any) the perimeter is equal to the are aTo find this out I have been to the library and looked at some A-level textbooks and learnt?

Polynomials? 2 n 2 + 3 n + 1? b n? 2) 2 n 3? n 2? 5 n? 2? 2 n 3?

4 n 2? b? 3 n 2? 5 n? 2? 3 n 2? 6 n? b? n?

2? n? 2? b? 0 This tells us that the only term where the perimeter = area is term number 2 Therefore when f (x) = 2 n 3? n 2? 5 n? 2 is divided by n? 2 there is no remainder and a quotient 2 n 2 + 3 n + 1.

The result can be written asf (x) = 2 n 3? n 2? 5 n? 2 = (n? 2) (2 n 2 + 3 n + 1) If 2 was substituted for the x in this identity so that n? 2 = 0, the quotient is eliminated giving f (2) = + 2 Now I will complete the square on?

4 n 2 + 6 n + 2? to see what the solution to this is. 4 n 2 + 6 n + 2 4 (n +3) 2? 9 + 2 4 (n +3) 2? To find this out I have been to the library and looked at some A-level textbooks and learnt? Arithmatic Progression? 3d 2?

2 ad = 0? Change a to xx 2? 3d 2? 2 dx = 0 Factor ise this equation to get (x + d) (x? 3d) Therefore x = -d x = 3d? x = -d is impossible as you cannot have a negative dimension a, a+d, a + 2d Is the same as 3d, 4d, 5 this tells us that the only Pythagorean triples are 3, 4, 5 or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc.