anonymous
  • anonymous
I need help!! Find the volume for the regular pyramid. V=
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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Michele_Laino
  • Michele_Laino
are all sides 4 units long? right?
anonymous
  • anonymous
yes

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Michele_Laino
  • Michele_Laino
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?
Michele_Laino
  • Michele_Laino
I think that P is the common point of the three bisectors of the triangle
anonymous
  • anonymous
i have no clue
Michele_Laino
  • Michele_Laino
|dw:1438359407990:dw| it is the base triangle
Michele_Laino
  • Michele_Laino
|dw:1438359464896:dw|
anonymous
  • anonymous
oh.....
anonymous
  • anonymous
Still clueless lol
Michele_Laino
  • Michele_Laino
we have to find the measure of the height of your pyramid
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
would it be 8
Michele_Laino
  • Michele_Laino
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
anonymous
  • anonymous
oh ummmm is it 4
anonymous
  • anonymous
because it looks like it may be 4
Michele_Laino
  • Michele_Laino
we have to compute it first
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
I'm pondering...
anonymous
  • anonymous
lol okay me too
Michele_Laino
  • Michele_Laino
There exists a theorem which states that the distance x is given by the subsequent formula: \[\Large x = \frac{{4\sqrt 3 }}{3}\]
anonymous
  • anonymous
thats okay i finished the lesson
Michele_Laino
  • Michele_Laino
so, in order to compute the height of your pyramid we can apply the theorem of Pitagora: |dw:1438360380159:dw|
Michele_Laino
  • Michele_Laino
there fore we can write this: \[\Large h = \sqrt {{4^2} - {x^2}} = \sqrt {16 - \frac{{16 \times 3}}{9}} = ...?\]
Michele_Laino
  • Michele_Laino
oops..therefore*
Michele_Laino
  • Michele_Laino
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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