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anonymous

  • one year ago

If the circle with center O has area 9 pi, what is the area of equilateral triangle ABC?

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  1. anonymous
    • one year ago
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    |dw:1441314908068:dw|

  2. misty1212
    • one year ago
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    HI!!

  3. anonymous
    • one year ago
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    hey misty!

  4. misty1212
    • one year ago
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    area is \(9\pi \)r right?

  5. anonymous
    • one year ago
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    yea so the diameter is 6

  6. misty1212
    • one year ago
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    ok so you got that the height is 6

  7. anonymous
    • one year ago
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    |dw:1441315064479:dw|

  8. anonymous
    • one year ago
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    ^is that for the height of the triangle? Like we're converting the 6 to an apropriate side that corresponds to the 30-60-90 triangle right?

  9. misty1212
    • one year ago
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    you are way ahead of me

  10. anonymous
    • one year ago
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    |dw:1441315150836:dw|

  11. misty1212
    • one year ago
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    yeah each side of your triangle is \[\frac{2\times 6}{\sqrt3}\]

  12. misty1212
    • one year ago
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    you lost me there height of the triangle is 6

  13. misty1212
    • one year ago
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    |dw:1441315234092:dw|

  14. misty1212
    • one year ago
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    the short leg of either right triangle on the left and right is \(\frac{6}{\sqrt{3}}\)

  15. misty1212
    • one year ago
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    area of each right triangle is base times height

  16. misty1212
    • one year ago
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    rather one half base times height but together it is base times height so \[6\times \frac{6}{\sqrt3}\]

  17. misty1212
    • one year ago
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    |dw:1441315367905:dw|

  18. anonymous
    • one year ago
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    wait what? so \[4\sqrt{3}\] is for the bottom left and right of the traingles?

  19. anonymous
    • one year ago
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    how why?

  20. misty1212
    • one year ago
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    |dw:1441315446382:dw|

  21. misty1212
    • one year ago
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    the short leg is the long leg divided by root 3

  22. misty1212
    • one year ago
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    so the area of the two right triangles taken together is base times height

  23. misty1212
    • one year ago
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    ok i see \[\frac{6}{\sqrt{3}}=2\sqrt3\]

  24. misty1212
    • one year ago
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    so total area is \[6\times 2\sqrt3 = 12\sqrt 3\]

  25. anonymous
    • one year ago
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    fml i've been doin it all wrong this whole time

  26. misty1212
    • one year ago
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    you got the height right away then i think you confused yourself and turned the height in to something else first you said it was 6 (correct) then you said it was \(4\sqrt3\)

  27. anonymous
    • one year ago
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    So |dw:1441315715658:dw|

  28. misty1212
    • one year ago
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    yeah that looks good to me

  29. anonymous
    • one year ago
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    i dont get why it's 2 root 3 is the problem like why does it convert that way and the process

  30. anonymous
    • one year ago
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    i thought in a 30-60-90 triangle the root three would replace the 6

  31. misty1212
    • one year ago
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    k lets go slow you said you had a 30 - 60 -90 triangle right?

  32. anonymous
    • one year ago
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    yea

  33. misty1212
    • one year ago
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    the ratios are \[1:\sqrt3:2\] for short leg ; long leg ; hypotenuse

  34. anonymous
    • one year ago
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    yea

  35. misty1212
    • one year ago
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    the long leg here is 6

  36. misty1212
    • one year ago
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    so the short leg is \[\frac{6}{\sqrt3}=2\sqrt3\]

  37. anonymous
    • one year ago
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    how?

  38. misty1212
    • one year ago
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    you multiply the short leg by root 3 to get the long leg right?

  39. misty1212
    • one year ago
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    so to go the other way, from the long leg to the short leg, you divide

  40. anonymous
    • one year ago
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    yea but like why is it that way? shouldn't the short leg not contain the root 3?

  41. misty1212
    • one year ago
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    lol no it does here

  42. anonymous
    • one year ago
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    that so weird

  43. misty1212
    • one year ago
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    if the short leg was a whole number the long leg would contain root 3

  44. misty1212
    • one year ago
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    but the long leg is a whole number, so the short leg as the root 3 in it

  45. misty1212
    • one year ago
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    it is not weird, only seems weird because you usually see it with the short leg a whole number

  46. anonymous
    • one year ago
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    hmmmm yea i guess so

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