 Posers and Puzzles

1. 20 Jan '10 08:382 edits
in a triangle, if a2 + b2=c2, then find c if

a=1
b=1
2. 20 Jan '10 11:54
Originally posted by uzless
in a triangle, if a2 + b2=c2, then find c if

a=1
b=1
Don't be irrational
3. 20 Jan '10 12:12
Originally posted by uzless
in a triangle, if a2 + b2=c2, then find c if

a=1
b=1
c = 2

1x2 + 1x2 = 2c
4. 20 Jan '10 16:10
Originally posted by AThousandYoung
c = 2

[hidden]1x2 + 1x2 = 2c[/hidden]
Thats a funny triangle with lengths 1,1,2 !!!
5. 20 Jan '10 16:26
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 🙂
6. 20 Jan '10 16:311 edit
Originally posted by TheMaster37
Technically a line, but it IS the correct solution to the posed problem 🙂
meh, nvm
7. 23 Jan '10 22:28
I don't understand this entire thread. 😕
By a2 do you mean a^2?
If so, why isn't c = sqrt(2) ?
8. 24 Jan '10 03:48
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
9. 24 Jan '10 22:40
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
10. 25 Jan '10 01:051 edit
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"
11. 25 Jan '10 19:58
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.
12. 25 Jan '10 20:00
Originally posted by David113
Don't be irrational
no moves , yet you've figured out the forums already?

what does 3a + 3a =?
13. 02 Feb '10 01:03
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.
14. 03 Feb '10 11:42
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.
15. 03 Feb '10 11:55
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...