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Setsuna-Yuregeshi

  • one year ago

How do you find the area of this triangle?

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  1. Setsuna-Yuregeshi
    • one year ago
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    |dw:1438116778704:dw|I need to find the area of this. How do I do that?

  2. Setsuna-Yuregeshi
    • one year ago
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    |dw:1438116845340:dw| I forgot the right angle

  3. phi
    • one year ago
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    area is 1/2 base * height

  4. Setsuna-Yuregeshi
    • one year ago
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    Really? That's it?

  5. phi
    • one year ago
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    yes

  6. Setsuna-Yuregeshi
    • one year ago
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    5x3/2?

  7. phi
    • one year ago
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    yes

  8. Setsuna-Yuregeshi
    • one year ago
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    But what about:

  9. Setsuna-Yuregeshi
    • one year ago
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    |dw:1438116970196:dw| What about that part? WHen they ask find the area which triangle am i finding it for exactly?

  10. phi
    • one year ago
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    You found the area of the triangle. we can prove it

  11. ChillOut
    • one year ago
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    You're finding the are for the full lines one.

  12. Setsuna-Yuregeshi
    • one year ago
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    I found the area of which triangle?

  13. ChillOut
    • one year ago
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    area*

  14. ChillOut
    • one year ago
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    |dw:1438117079817:dw|

  15. phi
    • one year ago
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    |dw:1438117108451:dw|

  16. Setsuna-Yuregeshi
    • one year ago
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    Ooh, so the |dw:1438117132170:dw| is giving me the height for the triangle?

  17. ChillOut
    • one year ago
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    Absolutely.

  18. Setsuna-Yuregeshi
    • one year ago
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    Ooh, Brainfart!

  19. Setsuna-Yuregeshi
    • one year ago
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    Since I am taking geometry during summerschool I expected it to be some complicated formula. At first I ruled out that it could be that...

  20. ChillOut
    • one year ago
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    There are other formulas involving angles and stuff. But leave that for later :)

  21. phi
    • one year ago
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    here is a proof. We assume we know the area of a rectangle is width * height |dw:1438117298679:dw|

  22. Setsuna-Yuregeshi
    • one year ago
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    So if I answer using 1/2BxH on a graded assessment it will be correct?

  23. phi
    • one year ago
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    now put in a diagonal. the two triangles are congruent, so the area of each is 1/2 of the rectangle's|dw:1438117369207:dw|

  24. Setsuna-Yuregeshi
    • one year ago
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    So if I answer using 1/2BxH on a graded assessment it will be correct?

  25. Setsuna-Yuregeshi
    • one year ago
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    They're not expecting a different method?

  26. phi
    • one year ago
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    now make an obtuse triangle |dw:1438117452871:dw|

  27. Setsuna-Yuregeshi
    • one year ago
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    @phi Ooh, I see. That makes sense...

  28. Setsuna-Yuregeshi
    • one year ago
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    ooh. C is the height of |dw:1438117583062:dw|

  29. phi
    • one year ago
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    if we subtract of the small triangle on the left we are left with the obtuse triangle that is \[ \frac{1}{2} (a+b)\cdot c - \frac{1}{2} b c \\\frac{1}{2} ac + \frac{1}{2} bc - \frac{1}{2} bc \\ =\frac{1}{2} ac \]

  30. phi
    • one year ago
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    hopefully you can follow the logic

  31. Setsuna-Yuregeshi
    • one year ago
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    I can, so you just proved 1/2BxH right?

  32. phi
    • one year ago
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    yes, that the formula works for obtuse triangles.

  33. Setsuna-Yuregeshi
    • one year ago
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    Ok thanks. Can I ask another quick question?

  34. Setsuna-Yuregeshi
    • one year ago
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    How do you find the area of a kite?

  35. phi
    • one year ago
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    1/2 the product of the diagonals http://www.mathopenref.com/kitearea.html

  36. Setsuna-Yuregeshi
    • one year ago
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    Thanks!!!!

  37. Setsuna-Yuregeshi
    • one year ago
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    Would the diagonals be these? |dw:1438117923899:dw|

  38. Setsuna-Yuregeshi
    • one year ago
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    @phi

  39. phi
    • one year ago
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    the "crossbars" |dw:1438118022805:dw|

  40. Setsuna-Yuregeshi
    • one year ago
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    Ooh, ok sorry but: |dw:1438118151545:dw| It would be 1+1 x 2 + 3/2 right? @phi

  41. Setsuna-Yuregeshi
    • one year ago
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    OOps I mean 1+2 x 3+3/2

  42. phi
    • one year ago
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    it would be except that would not be a kite (the diagonals of a kite bisect each other) see http://www.coolmath.com/reference/kites

  43. phi
    • one year ago
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    correction: the longer diagonal bisects the shorter diagonal

  44. Setsuna-Yuregeshi
    • one year ago
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    Ooh, so what i said was right? I have a problem asking for the area where they give me the length of half of both lines.

  45. Setsuna-Yuregeshi
    • one year ago
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    Like they give me this...

  46. Setsuna-Yuregeshi
    • one year ago
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    |dw:1438118486114:dw|

  47. Setsuna-Yuregeshi
    • one year ago
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    I just add the 2's and 1's and use the 1/2bxh formula right?

  48. phi
    • one year ago
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    yes

  49. Setsuna-Yuregeshi
    • one year ago
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    Ooh ok. Thanks.

  50. phi
    • one year ago
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    1/2 d1 * d2 (though you can think of the diagonals as the width and height)

  51. Setsuna-Yuregeshi
    • one year ago
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    Ooh, thank you so much!!!!!!!!!!!!!!

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