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anonymous

  • 5 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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  1. dumbcow
    • 5 years ago
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    |dw: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

  2. anonymous
    • 5 years ago
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    |dw:1327064957563:dw| I've tried to replicate your diagram, with names to the vertexes.

  3. anonymous
    • 5 years ago
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    So you're considering triangle CEP or ACP?

  4. dumbcow
    • 5 years ago
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    CEP, sorry i can't draw :)

  5. dumbcow
    • 5 years ago
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    |dw:1327065261653:dw| thats where the "h" comes in

  6. anonymous
    • 5 years ago
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    naah, you draw just fine. but isnt the height of the triangle AP, not EP?

  7. dumbcow
    • 5 years ago
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    correct i added the length AE which is just the radius r + r/2 = 3/2r

  8. anonymous
    • 5 years ago
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    OHHH! oops. my bad. Thanks a bunch!

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