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Thread: Area of triangle inscribed in a rectangular prism

  1. #1
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    Area of triangle inscribed in a rectangular prism

    Area of triangle inscribed in a rectangular prism-potw-6-2.pngi dont know where to start

    *Finding the volume of the rectangular prism
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    Re: Area of triangle inscribed in a rectangular prism

    Area of triangle inscribed in a rectangular prism-clipboard01.jpg

    Use the Pythagorean theorem to develop a system of equations for $a,~b,~c$ given the lengths you are given.

    Solve that system.

    for example

    $HJ^2 = a^2+b^2 = 4^2 = 16$

    2 other similar equations can be developed.
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    Re: Area of triangle inscribed in a rectangular prism

    Quote Originally Posted by Ilikebugs View Post
    Click image for larger version. 

Name:	potw 6 2.png 
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ID:	37385i dont know where to start
    *Finding the volume of the rectangular prism
    Area of triangle inscribed in a rectangular prism-iiii.gif
    The coordinates are: $O: (0,0,0),~A: (0,0 ,2a ),~B: (0,2b ,2a ),~C: (0,2b ,0 ),~D: (-2c, 2b,0 ),~E: (-2c,2b ,2a )$
    Now the coordinates of $ H,~J,~K$ are the midpoints of the diagonals of their faces.
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    Re: Area of triangle inscribed in a rectangular prism

    b^2+c^2=(JK)^2=36
    a^2+c^2=HK^2=25
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  5. #5
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    Re: Area of triangle inscribed in a rectangular prism



    ok ... you've got three equations, and the volume of the prism is $V=8abc$

    $a^2+b^2=4^2$

    $a^2+c^2=5^2$

    $b^2+c^2=6^2$

    ... what next?
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  6. #6
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    Re: Area of triangle inscribed in a rectangular prism

    2-1=c^2-b^2=9
    3-2=b^2-a^2=11
    3-1=c^2-a^2=20

    c=sqr(20+a^2)
    b=sqr(11+a^2)

    V=8*a*sqr(20+a^2)*sqr(11+a^2)

    a^2+11+a^2=16
    a=sqr(5/2)

    V=8*5/2*sqr(20+5/2)*sqr(11+5/2)?

    (I think I messed up
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    Re: Area of triangle inscribed in a rectangular prism

    Quote Originally Posted by skeeter View Post
    [IMG]http://mathhelpforum.com/geometry/

    ok ... you've got three equations, and the volume of the prism is $V=8abc$

    $a^2+b^2=4^2$

    $a^2+c^2=5^2$

    $b^2+c^2=6^2$

    ... what next?
    subtract (1) from (3)

    $c^2 - a^2 = 20$

    add this to (2)

    $2c^2 = 45$

    $c = \sqrt{\dfrac{45}{2}} = 3\sqrt{\dfrac 5 2}$

    plug this into (2)

    $a^2 + \dfrac{45}{2} = 25$

    $a^2 = \dfrac{5}{2}$

    $a = \sqrt{\dfrac{5}{2}}$

    plug this into (1)

    $\dfrac 5 2 + b^2 = 16$

    $b^2 = \dfrac{27}{2}$

    $b = \sqrt{\dfrac{27}{2}}=3\sqrt{\dfrac 3 2}$

    $V = 8abc = 8\left(\sqrt{\dfrac{5}{2}}\right)\left(3\sqrt{ \dfrac 3 2}\right)\left(3\sqrt{ \dfrac 5 2}\right)=90\sqrt{6}$
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