# [SOLVED] Pythagoras theorem

• Jun 13th 2006, 12:24 AM
[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?!?! :mad:

can any1 help with that? like now.. lol i have to give it in 2moro at skool.
• Jun 13th 2006, 01:10 AM
JakeD
The page has 64 proofs of the Theorem. Is yours proof number 4?
http://www.cut-the-knot.org/pythagoras/proof31.gif
• Jun 13th 2006, 03:23 AM
ticbol
Quote:

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?!?! :mad:

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.
• Jun 13th 2006, 04:11 AM
CaptainBlack
Don't post the same question multiple times, it wastes the time of the helper
and if ThePerfectHacker catches you he will ban you :eek:

RonL
• Jun 13th 2006, 10:27 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
and if ThePerfectHacker catches you he will ban you :eek:

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.
• Jun 13th 2006, 01:41 PM
ThePerfectHacker
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, $BD\cdot EB=BC^2$.
19)Thus, $BD\cdot (EA+AD+BD)=BC^2$
23)Thus, $BD\cdot (AC+AC+BD)=BC^2$
29)Thus, $BD\cdot (2AC+BD)=BC^2$
31)Thus, $2AC\cdot BD+BD^2=BC^2$
37)Thus, $AC^2+2AC\cdot BD+BD^2=AC^2+BC^2$
39)Thus, $(AC+BD)^2=AC^2+BC^2$
41)Thus, $(AD+BD)^2=AC^2+BC^2$
43)Thus, $AB^2=AC^2+BC^2$

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 $EB=EA+AD+DE$ since it is a line.
23) $EA=AD$ since it is radius of circle.