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anonymous
 4 years ago
How do you come to the conclusion that when considering a equilateral triangle inside a circumscribed circle, the height of the triangle is 2/3 of the radius of the circle?
anonymous
 4 years ago
How do you come to the conclusion that when considering a equilateral triangle inside a circumscribed circle, the height of the triangle is 2/3 of the radius of the circle?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327064516603:dw the height is actually greater than the radius sin(30) = h/r = 1/2 > h = r/2 therefore the height of triangle is 3/2 the radius

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327064957563:dw I've tried to replicate your diagram, with names to the vertexes.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So you're considering triangle CEP or ACP?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0CEP, sorry i can't draw :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327065261653:dw thats where the "h" comes in

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0naah, you draw just fine. but isnt the height of the triangle AP, not EP?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0correct i added the length AE which is just the radius r + r/2 = 3/2r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OHHH! oops. my bad. Thanks a bunch!
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