Jump to content

pythagorean Therom!!!1


megachirops

Recommended Posts

The Theorum states that the hypotenuse of a triangle with a right angle is equal to the measure of each side squared and added together:

 

a2+b2=C2

So, for example, if you have a right triangle with two sides that measure 4 and 5, then you'd plug them into the formula above.

 

a2+b2=c2

42 + 52=c2

Which would simplify out to:

16 + 25 = C2

41=c2

You'd then take the square of 41, and that would be the measure of the hypotenuse.

So, what's your question?

Link to comment
Share on other sites

The Theorum states that the hypotenuse of a triangle with a right angle is equal to the measure of each side squared and added together:

 

a2+b2=C2

So, for example, if you have a right triangle with two sides that measure 4 and 5, then you'd plug them into the formula above.

 

a2+b2=c2

42 + 52=c2

Which would simplify out to:

16 + 25 = C2

 

41=c2

You'd then take the square of 41, and that would be the measure of the hypotenuse.

 

So, what's your question?

 

 

 

Thanks that answers my question, i was confused on the steps of the problem, my teacher just rushed through this therom and i was left stranded! Thanks!

Link to comment
Share on other sites

I think it's worth noting that the geometrical proof has nothing to do with numbers per se. Its about areas. There are several proofs in the link. The one I used in school was proof 12 for example. http://www.cut-the-k...org/pythagoras/

 

The Pythagorean theorem is only periphally about area.

 

What distinguishes Euclidean geometry from non-Euclidean (planar) geometries is the parallel postulate: Given a line and a point not on the line there is a unique line through that point parallel to the given line.

 

It is not obvious, but the Pythagorean theorem is equivalent to the parallel postulate. It is quite proper to say that the Pythagorean theorem characterizes Eucllidean geometry in the plane. One can extend that to the fact that it is the Pythagorean theorem that gives us the usual notion of distance and it is that notion of distance that characterizes Euclidean geometries in higher dimensions -- and that is what distinguishes the Euclidean model that we learn in high school from the model that seems to actually describe the universe as one lerns in general relativity.

 

So, no matter what specific proof and bag of tricks you use to prove the Pythagorean theorem, you can be sure that in any valid proof the parallel postulate will play a role somewhere, either directly or in the proof of a result that is in turn used to prove the theorem.

Link to comment
Share on other sites

The Pythagorean theorem is only periphally about area.

 

What distinguishes Euclidean geometry from non-Euclidean (planar) geometries is the parallel postulate: Given a line and a point not on the line there is a unique line through that point parallel to the given line.

 

It is not obvious, but the Pythagorean theorem is equivalent to the parallel postulate. It is quite proper to say that the Pythagorean theorem characterizes Eucllidean geometry in the plane. One can extend that to the fact that it is the Pythagorean theorem that gives us the usual notion of distance and it is that notion of distance that characterizes Euclidean geometries in higher dimensions -- and that is what distinguishes the Euclidean model that we learn in high school from the model that seems to actually describe the universe as one lerns in general relativity.

 

So, no matter what specific proof and bag of tricks you use to prove the Pythagorean theorem, you can be sure that in any valid proof the parallel postulate will play a role somewhere, either directly or in the proof of a result that is in turn used to prove the theorem.

 

I left school at 16 so my maths is pretty basic. I believe that the parallel postulate is only applicable to Euclidean geometry (i.e. applicable to a plane).

It seems to me that if you draw a right angled triangle and place three dots at different distances from each of the sides and draw lines parallel to the sides you will draw another right angled triangle. The basic proof of the addition of areas will apply to the triangle you have just drawn. I'm probably missing something, if so I apologise.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.