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Math Help - Area of a triangular pyramid

  1. #1
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    Area of a triangular pyramid

    TA= .5Pl+B

    The base of the triangle is an equilateral triangle with length of 6

    The height of the pyramid is 9

    How do I find the slant height of the pyramid?
    I have attached a picture to show the pyramid.

    I know that I have to do the Pyth. Thm. but am having trouble finding the length on the base of the triangle. I think that it is a 30-60-90[IMG]file:///Users/schinb64/Desktop/Picture%202.png[/IMG]
    Attached Thumbnails Attached Thumbnails Area of a triangular pyramid-picture-2.png  
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  2. #2
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    Quote Originally Posted by IDontunderstand View Post
    TA= .5Pl+B

    The base of the triangle is an equilateral triangle with length of 6

    The height of the pyramid is 9

    How do I find the slant height of the pyramid?
    I have attached a picture to show the pyramid.

    I know that I have to do the Pyth. Thm. but am having trouble finding the length on the base of the triangle. I think that it is a 30-60-90[IMG]file:///Users/schinb64/Desktop/Picture%202.png[/IMG]
    Yes, take the distance from any corner of the base triangle to the
    base of the perpendicular height line as

    3cos30^o

    then s=\sqrt{9^2+3^2cos^230^o} is the length of an edge.

    To get the length of the line from the midpoint of a side of the base to the top, use

    x=3tan30^o

    then apply Pythagoras' theorem.

    You can also calculate the volume, if needed, by equating the base area to a circle area,
    then calculating the volume of a cone of same height and base areas.

    (0.5)6^2sin60^o={\pi}r^2

    V=\frac{{\pi}r^2h}{3}=\frac{18sin60^o(9)}{3}=54sin  60^o
    Last edited by Archie Meade; April 18th 2010 at 12:28 PM. Reason: added data
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  3. #3
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    Hello, IDontunderstand!

    If you don't know Trig, we can still work it out.


    The base of the pyramid is an equilateral triangle with length of 6

    The height of the pyramid is 9

    How do I find the slant height of the pyramid?

    The base is an equilateral triangle of side 6.

    Code:
                  A
                  *
                 /|\
                / | \
               /  |  \
            6 /   |h  \ 6
             /    |    \
            /     |     \
           /      |      \
        B * - - - + - - - * C
              3   D   3

    In right triangle ADC\!:\;\;h^2 +3^2 \:=\:6^2 \quad\Rightarrow\quad h \:=\:3\sqrt{3}

    We want the "center" of the triangle (the centroid G).
    It divides an altitude in the ratio 2:1


    Code:
                  A
                  *
                 /|\
                / | \
               /  |  \
              /   |   \ 
             /    |  _ \
            /     |2√3  \
           /      |      \
          /      G*       \
         /        | _      \
        /         |√3       \
       /          |          \ 
    B * - - - - - - - - - - - * C

    This is the point directly below the top vertex of the pyramid (V).


    Now consider the side view of the pyramid.

    Code:
          V
          *
          |\
          | \
          |  \
        9 |   \ s
          |    \
          |     \
          |      \
        G * - -_- * A
             2√3

    We have: . s^2 \:=\:9^2 + (2\sqrt{3})^2 \:=\:81 + 12 \:=\:93

    Therefore, the slant height is: . s \:=\:\sqrt{93}

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