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Austin_Rain

How would I solve this question? All of the triangles in the figure below are congruent. What is the area of the figure? Note that all measurements are in centimeters. Note that the apothem shown is equal to 2 (sqrt)3

  • one year ago
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  1. Austin_Rain
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    • one year ago
  2. KingGeorge
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    A nice easy trick to this problem, is to notice that the area of the equilateral triangles is the same as the area of the hexagon. So you just have to find one to find the total area. The area of the hexagon is \[A={1 \over 2}ap\]where a is the apothem and p is the perimeter. We know the apothem to be \(2\sqrt3\) and the perimeter to be \(6\cdot4=24\) So the area of the hexagon is\[{1 \over 2}48\sqrt3 ={24\sqrt3}\]Therefore, the total area is\[2\cdot(24\sqrt3)=48\sqrt3\]

    • one year ago
  3. KingGeorge
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    Wait, is the height of the triangle 3? Was I misreading the diagram?

    • one year ago
  4. Austin_Rain
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    But, that's not an answer. D: However, 24 (sqrt)3 ^2 is. WOuld that be it?

    • one year ago
  5. Austin_Rain
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    The height of the triangle is 3. Yes.

    • one year ago
  6. KingGeorge
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    If the height of the triangle is 3, then the area of each triangle is \[{1\over2}\cdot3\cdot4=6\]So then the answer should be \[24\sqrt3+6\cdot6=24\sqrt3+36\]

    • one year ago
  7. KingGeorge
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    There are 6 triangles, so that's why I multiplied the 6 by 6.

    • one year ago
  8. Austin_Rain
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    That is a possible answer. :D

    • one year ago
  9. Austin_Rain
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    Is that answer squared?

    • one year ago
  10. KingGeorge
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    No. That's the entire answer. \[24\sqrt3+36 \;\;\;\text{cm}^2\]

    • one year ago
  11. Austin_Rain
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    Oh, sweet. :D Thanks man!

    • one year ago
  12. KingGeorge
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    You're welcome.

    • one year ago
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