anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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dumbcow
  • dumbcow
|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
anonymous
  • anonymous
|dw:1327064957563:dw| I've tried to replicate your diagram, with names to the vertexes.
anonymous
  • anonymous
So you're considering triangle CEP or ACP?

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dumbcow
  • dumbcow
CEP, sorry i can't draw :)
dumbcow
  • dumbcow
|dw:1327065261653:dw| thats where the "h" comes in
anonymous
  • anonymous
naah, you draw just fine. but isnt the height of the triangle AP, not EP?
dumbcow
  • dumbcow
correct i added the length AE which is just the radius r + r/2 = 3/2r
anonymous
  • anonymous
OHHH! oops. my bad. Thanks a bunch!

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