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anonymous

  • one year ago

I need help!! Find the volume for the regular pyramid. V=

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  1. anonymous
    • one year ago
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  2. Michele_Laino
    • one year ago
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    are all sides 4 units long? right?

  3. anonymous
    • one year ago
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    yes

  4. Michele_Laino
    • one year ago
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    we have to understand where is the intersection point P|dw:1438359138429:dw| between the height of your pyramid and its base. Have you any idea about that?

  5. Michele_Laino
    • one year ago
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    I think that P is the common point of the three bisectors of the triangle

  6. anonymous
    • one year ago
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    i have no clue

  7. Michele_Laino
    • one year ago
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    |dw:1438359407990:dw| it is the base triangle

  8. Michele_Laino
    • one year ago
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    |dw:1438359464896:dw|

  9. anonymous
    • one year ago
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    oh.....

  10. anonymous
    • one year ago
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    Still clueless lol

  11. Michele_Laino
    • one year ago
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    we have to find the measure of the height of your pyramid

  12. anonymous
    • one year ago
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    oh ok

  13. anonymous
    • one year ago
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    would it be 8

  14. Michele_Laino
    • one year ago
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    I don't know, in order to find the height of yopur pyramid I have to find this length: |dw:1438359730025:dw| namely the distance between each vertex of the triangle and the point P

  15. anonymous
    • one year ago
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    oh ummmm is it 4

  16. anonymous
    • one year ago
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    because it looks like it may be 4

  17. Michele_Laino
    • one year ago
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    we have to compute it first

  18. anonymous
    • one year ago
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    ok

  19. Michele_Laino
    • one year ago
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    I'm pondering...

  20. anonymous
    • one year ago
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    lol okay me too

  21. Michele_Laino
    • one year ago
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    There exists a theorem which states that the distance x is given by the subsequent formula: \[\Large x = \frac{{4\sqrt 3 }}{3}\]

  22. anonymous
    • one year ago
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    thats okay i finished the lesson

  23. Michele_Laino
    • one year ago
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    so, in order to compute the height of your pyramid we can apply the theorem of Pitagora: |dw:1438360380159:dw|

  24. Michele_Laino
    • one year ago
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    there fore we can write this: \[\Large h = \sqrt {{4^2} - {x^2}} = \sqrt {16 - \frac{{16 \times 3}}{9}} = ...?\]

  25. Michele_Laino
    • one year ago
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    oops..therefore*

  26. Michele_Laino
    • one year ago
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    hint: \[\Large h = \sqrt {{4^2} - {x^2}} = \sqrt {16 - \frac{{16 \times 3}}{9}} = 4\sqrt {\frac{2}{3}} \] so being the base area A equal to: \[\Large A = \frac{{4 \times 2\sqrt 3 }}{2} = 4\sqrt 3 \] the requested volume V is: \[\Large V = \frac{{A \times h}}{3} = \frac{1}{3} \times 4\sqrt 3 \times 4\sqrt {\frac{2}{3}} = ...?\]

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