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Attachment 28375
I wanted to see if I could take the top half of a unit circle and create a right triangle such that one side is equal to the circle's diameter and the other equal to half that. I think I've shown that you can't create such a triangle using the contraint of a circle.
Sorry, can't read it.
if you click on it twice it should bring up a full screen (it does for me). Sorry about that, I'll redo it if anything.
Attachment 28380the problem is one of creating a right triangle using the diameter as the horizontal side and letting the other two sides (hypotenuse and the vertical side equal half the circumference = Pi if radius = 1). The question is: is that possible? I think the answer is no.
I calculated that you can't create a right triangle with
radius=1,
base = 2r,
height = r
and have the hypotenuse and the vertical side sum to pi. I would like to know if I was wrong or not.
You are very right it is just not possible. the reason is if one side is of right triangle is of length 2r ( diameter ) and other r 9 radius ) the hypotenuse will definitely be greater than the diameter and so will take the third vertex outside the circle.
