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eroshea

  • 2 years ago

A regular hexagon A has the midpoints of its edges joined to form a smaller hexagon B. This process is repeated by joining the midpoints of the edges of the hexagon B to get a third hexagon C. What is the ratio of the area of hexagon C to the area of hexagon A?

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  1. hartnn
    • 2 years ago
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    Each smaller hexagon formed will have side length equal to 0.866(half times square root of 3)times the length of its previous hexagon

  2. eroshea
    • 2 years ago
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    how did you get that?

  3. hartnn
    • 2 years ago
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    |dw:1350381371686:dw|

  4. hartnn
    • 2 years ago
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    sorry for bad drawing, but could u get the side of smaller hex from that ?

  5. eroshea
    • 2 years ago
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    it's fine.. i'll analyze it for a while

  6. hartnn
    • 2 years ago
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    |dw:1350381577682:dw|

  7. eroshea
    • 2 years ago
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    what was that d/2 ??

  8. hartnn
    • 2 years ago
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    yes, d/2 or u asked , why was that d/2 ?

  9. eroshea
    • 2 years ago
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    i meas WHAT is that d/2? where did you get it? sorry but i find solid mensuration and geometry hard ..

  10. hartnn
    • 2 years ago
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    d is the side of smaller hexagon , so the length i have marked here will be d/2 |dw:1350381918942:dw|

  11. eroshea
    • 2 years ago
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    ahh.. got it.. i was confused from that equation :D

  12. hartnn
    • 2 years ago
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    so u will have ratio of sides, know how to find ratio of areas ?

  13. eroshea
    • 2 years ago
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    no, i don't

  14. hartnn
    • 2 years ago
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    (ratio of sides)^2 = ratio of areas ALWAYS.

  15. eroshea
    • 2 years ago
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    owh.. so like the equation on how i got the corresponding sides of the 2 triangles? ok.. got it.. i'll just now solve for the sides of the hexagon C

  16. hartnn
    • 2 years ago
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    yes, find the ratio of sides of hex A and hex C.

  17. eroshea
    • 2 years ago
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    9/16 ?? ahm.. is it??

  18. hartnn
    • 2 years ago
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    yes!

  19. hartnn
    • 2 years ago
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    9/16 is ratio of areas

  20. eroshea
    • 2 years ago
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    i thought i was wrong :D that problem is complicated for me. thanks for the help man! :))

  21. hartnn
    • 2 years ago
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    welcome ^_^

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