poor pythagorus....

in a triangle, if a2 + b2=c2, then find c if

a=1
b=1

Originally posted by uzless
in a triangle, if a2 + b2=c2, then find c if

a=1
b=1

Don't be irrational

Originally posted by uzless
in a triangle, if a2 + b2=c2, then find c if

a=1
b=1

c = 2

Reveal Hidden Content

Originally posted by AThousandYoung
c = 2

[hidden]1x2 + 1x2 = 2c[/hidden]

Thats a funny triangle with lengths 1,1,2 !!!

Originally posted by wolfgang59
Thats a funny triangle with lengths 1,1,2 !!!

Technically a line, but it IS the correct solution to the posed problem 🙂

Originally posted by TheMaster37
Technically a line, but it IS the correct solution to the posed problem 🙂

meh, nvm

I don't understand this entire thread. 😕
By a2 do you mean a^2?
If so, why isn't c = sqrt(2) ?

Originally posted by crazyblue
I don't understand this entire thread. 😕
By a2 do you mean a^2?
If so, why isn't c = sqrt(2) ?

that's what aty is making fun of

Originally posted by crazyblue
I don't understand this entire thread. 😕
By a2 do you mean a^2?
If so, why isn't c = sqrt(2) ?

poor crazyblue

Actually, I think the theorem goes....." in any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides"

Originally posted by missal
Actually, I think the theorem goes....." in any [b]right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides"[/b]

I was giving you guys credit to know that already. I'll remember not to do that next time.

Originally posted by David113
Don't be irrational

no moves , yet you've figured out the forums already?


what does 3a + 3a =?

The full formula from which the Pythagorus Theorem is based is as follows (as I recall)

C^2 = A^2 + B^2 - 2*A*B*cos(o) (where o = the angle between sides A and B)

At 90 degrees, cos(o) = 0, thus eliminating the third term.

At 180 degrees, cos(o) = -1, making the formula C^2 = A^2 + B^2 + 2*A*B, which simplifies into C^2 = (A+B)^2. Given lengths are generally considered positive, C is thus A+B.

Originally posted by geepamoogle
The full formula from which the Pythagorus Theorem is based is as follows (as I recall)

C^2 = A^2 + B^2 - 2*A*B*cos(o) (where o = the angle between sides A and B)

At 90 degrees, cos(o) = 0, thus eliminating the third term.

At 180 degrees, cos(o) = -1, making the formula C^2 = A^2 + B^2 + 2*A*B, which simplifies into C^2 = (A+B)^2. Given lengths are generally considered positive, C is thus A+B.

Pythagoras pretty much predates cos.

Originally posted by TheMaster37
Pythagoras pretty much predates cos.

This is the Pythagorean Theorem:

http://en.wikipedia.org/wiki/Pythagorean_theorem

Some resemblance to the Fermat's Last Theorem...