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enya.gold

  • 3 years ago

[Work Included] In ABC, centroid D is on median AM. AD = x + 5 and DM = 2x – 1 Find AM.

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  1. enya.gold
    • 3 years ago
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    |dw:1357236456605:dw|

  2. enya.gold
    • 3 years ago
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    It looks more like an orthocenter, but for the sake of it, let's say it is a centroid.

  3. enya.gold
    • 3 years ago
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    \[ x + 5 = 2x - 1\]\[5 + x - 2x = 2x - 2x - 1\]\[5 - x = -1\]

  4. enya.gold
    • 3 years ago
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    Subtract five from both sides. \[-x = -1 - 5\]\[-x = -6\]

  5. enya.gold
    • 3 years ago
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    \[x = 6\]

  6. enya.gold
    • 3 years ago
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    Now I can plug in the missing numbers. \[6 + 5 = 2(6) - 1\]\[11 = 12 - 1\]\[= 11\]

  7. KingGeorge
    • 3 years ago
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    I think you're incorrect in assuming that \(x+5\) must equal \(2x-1\). The centroid of a triangle is located at a point 2/3 of the way along the median, and not 1/2 of the way along the median.

  8. enya.gold
    • 3 years ago
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    In that case, I'll tweak a few things here. \[x + 5 = 2(2x - 1)\]\[x + 5 = 4x - 2\] Subtracting 4 from both sides.

  9. enya.gold
    • 3 years ago
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    \[5 - 3x = -2\]\[-3x = -2 - 5\]\[-3x = -7\]

  10. enya.gold
    • 3 years ago
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    \[x = \frac{ -7 }{ -3 }\]

  11. KingGeorge
    • 3 years ago
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    That looks good to me. You can simplify that to \(\dfrac{7}{3}\) if you want.

  12. KingGeorge
    • 3 years ago
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    You're welcome.

  13. enya.gold
    • 3 years ago
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    @KingGeorge The answer was really 11. Guess I should have trusted my gut, ha.

  14. KingGeorge
    • 3 years ago
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    Oops :( I guess that would be my fault.

  15. KingGeorge
    • 3 years ago
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    However, I think we just forgot the last step in the problem. We found \(x=7/3\), but AM is equal to \((x+5)+(2x-1)\). If we plug in our value for \(x\) into this, we get\[\frac{7}{3}+4+\frac{14}{3}=\frac{21}{3}+4=7+4=11\]

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