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Austin_Rain

  • 2 years ago

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

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  1. Austin_Rain
    • 2 years ago
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  2. KingGeorge
    • 2 years ago
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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\[2\cdot(24\sqrt3)=48\sqrt3\]

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

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

  5. Austin_Rain
    • 2 years ago
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    The height of the triangle is 3. Yes.

  6. KingGeorge
    • 2 years ago
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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 \[24\sqrt3+6\cdot6=24\sqrt3+36\]

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

  8. Austin_Rain
    • 2 years ago
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    That is a possible answer. :D

  9. Austin_Rain
    • 2 years ago
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    Is that answer squared?

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

  11. Austin_Rain
    • 2 years ago
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    Oh, sweet. :D Thanks man!

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

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