A right triangle is one in which ( leg1)^2 + (leg 2)^2 = (hypothenuse)^2
1^2 + 1^2 = rad2^2
hypothenuse =rad2
Any equal legged right rt triangle has a hypothenuse= to leg* rad2
bjh
See the drawing at the attached file, the sides of the triangle are 1,1,√2
Ιwant a geometric relation for √2, i.e. something like that:
a/1=b/a=>b=a^2
and thus
√2=1+a+b+c+...=1+a+a^2+a^4+a^8+a^16+...
But a/1=b/ais probably wrong. Can you proove it if it’s correct?
Any other similar relation I could get?
Hello Thodorisk,
I read your largest triangle as a right triangle with sides 1 and 1 and hyoothenuse =rad 2
1^2 + 1^2 =rad2^2 proving that it is a right triangle If the two legs of a right triangle are equal then the hypothenuse = leg times rad2 Call a leg a then the hypot is atimes rad2
bjh
It's not clear exactly what you are trying to do. I can see that you have a right triangle with legs of length 1 and so hypotenuse of length . Then, I think, you have marked of a length of 1 on the hypotenuse and drawn a line from the right angle to that point. But how is the second line, separting the segments marked "a" and "b" determined. Is it parallel to the first line from the right angle?