1. ## [SOLVED] Pythagoras theorem

okay, ive got a big square, with a square inside it diagonally. leaving 4 right angled triangles. one at each cornor of the large square.

sides of the internal squares are 'z' and the 3 sides of triangle are x y and z.. z is the hypotenuse...

i have shown that the large square's area can be written as (x+y) squared and also as z squared + 2xy.

now i have to use these results to prove pythagoras theorum?!?!

can any1 help with that? like now.. lol i have to give it in 2moro at skool.

2. The page has 64 proofs of the Theorem. Is yours proof number 4?

3. Originally Posted by adam gerrett
okay, ive got a big square, with a square inside it diagonally. leaving 4 right angled triangles. one at each cornor of the large square.

sides of the internal squares are 'z' and the 3 sides of triangle are x y and z.. z is the hypotenuse...

i have shown that the large square's area can be written as (x+y) squared and also as z squared + 2xy.

now i have to use these results to prove pythagoras theorum?!?!

can any1 help with that? like now.. lol i have to give it in 2moro at skool.
Ummm, you were able to show that the area, A, of the large square is:
A = (x+y)^2 -----------(1)

Or,
A = z^2 +2xy -------------(2) ?

Very good. In fact, those two equations are both correct.

Each side of the large square is (x+y), so, A = (side)^2 = (x+y)^2.

Each of the 4 right triangles has an area of (1/2)(x)(y) = (1/2)xy,
so the 4 rigth triangles have a collective area of 4*(1/2)xy = 2xy.

The area of the inner, smaller square of side z is z^2, hence the areaof the large square is
A = 2xy +z^2
or, A = z^2 +2xy.

Now, since A = A, then,
(x+y)^2 = z^2 +2xy
Simplifying,
x^2 +2xy +y^2 = z^2 +2xy
The 2xy cancells out,
x^2 +y^2 = z^2 ----------on any of the 4 right tgriangle. The Pythagorean Theorem!

On any of the 4 right triangles,
---z is the hypotenuse
---x and y are the two legs
z^2 = x^2 +y^2
Phythagorean Theorem.

4. Don't post the same question multiple times, it wastes the time of the helper
and if ThePerfectHacker catches you he will ban you

RonL

5. Originally Posted by CaptainBlack
and if ThePerfectHacker catches you he will ban you
First he is going to really mad.
-----------
I made up my own proof to the pythagorean theorem using circles maybe I will post it. I am just warning you that I might not because it is difficult to draw geometrical shapes through the internet.

6. As promised, here is my proof.
I searched the 69 proofs and did not seem to be there.

2)Triangle ABC with right angle at C.
3)Using AC draw a circle with that radius at C.
5)The circle in 2) intersect hypotenuse AB at D because it is longer than the radius.
7)Extend AD so that it intersect this circle at E.
11)EB is a secant by definition.
13)BC is a tangent at C because C is a right angle.
17)Thus, $\displaystyle BD\cdot EB=BC^2$.
19)Thus, $\displaystyle BD\cdot (EA+AD+BD)=BC^2$
23)Thus, $\displaystyle BD\cdot (AC+AC+BD)=BC^2$
29)Thus, $\displaystyle BD\cdot (2AC+BD)=BC^2$
31)Thus, $\displaystyle 2AC\cdot BD+BD^2=BC^2$
37)Thus, $\displaystyle AC^2+2AC\cdot BD+BD^2=AC^2+BC^2$
39)Thus, $\displaystyle (AC+BD)^2=AC^2+BC^2$
41)Thus, $\displaystyle (AD+BD)^2=AC^2+BC^2$
43)Thus, $\displaystyle AB^2=AC^2+BC^2$

Notes,
7)Postulate #2
13)A line perpendicular to the radius on the circumfurence is the tangent.
17)The theorem of tangent and secants.
19)Because $\displaystyle EB=EA+AD+DE$ since it is a line.
23)$\displaystyle EA=AD$ since it is radius of circle.