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Math Help - Solid Geometry

  1. #1
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    Solid Geometry

    A pyramid whose altitude is 5 feet weighs 800lbs . At what distance from its vertex must it be cut by a plane parallel to its base so that the two solids of equal weight will be formed?

    i have no idea how to solve this
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  2. #2
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    Re: Solid Geometry

    Quote Originally Posted by TechnicianEngineer View Post
    A pyramid whose altitude is 5 feet weighs 800lbs . At what distance from its vertex must it be cut by a plane parallel to its base so that the two solids of equal weight will be formed?

    i have no idea how to solve this
    m=\rho \cdot V

    m_1=m_2 \Rightarrow \rho \cdot V_1=\rho \cdot V_2 \Rightarrow V_1=V_2 \Rightarrow V_1=V-V_1 \Rightarrow 2\cdot V_1=V \Rightarrow

    \Rightarrow 2\cdot\frac{B_1\cdot H_1}{3}=\frac{B\cdot H}{3} \Rightarrow H_1=\frac{B\cdot H}{2\cdot B_1}

    where H_1 is distance from vertex , B_1 is base of the small pyramid , B is the base of the large pyramid and H is altitude.
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  3. #3
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    Re: Solid Geometry

    Quote Originally Posted by TechnicianEngineer View Post
    A pyramid whose altitude is 5 feet weighs 800lbs . At what distance from its vertex must it be cut by a plane parallel to its base so that the two solids of equal weight will be formed?

    i have no idea how to solve this
    1. I assume that the material of the pyramid is homogenous. Then the weight correspond directly to the volume:

    w = k \cdot V where k is a real constant.

    2. b_0 denotes the base area of the complete pyramide, b the base area of the smaller pyramide. Both areas are similar. Then

    V = \frac13 \cdot b_0 \cdot 5

    with w = k \cdot \frac53 \cdot b_0 = 800

    3. h denotes the height of the smaller pyramid. Then the volume is calculated by:

    V_{pyr.\ small} = \frac13 \cdot b \cdot h

    with w = k \cdot \frac13 \cdot b \cdot h = 400

    4. Since b and b_0 are similar you can use the proportion:

    \frac b{b_0} = \left(\frac h5 \right)^2~\implies~b=\frac{h^2}{25} \cdot b_0

    5. Plug in this term into the equation at #3:

    k \cdot \frac13 \cdot \frac{h^2}{25} \cdot b_0 \cdot h = 400

    \underbrace{\left(k \cdot \frac13 \cdot b_0 \cdot 5\right)}_{= 800} \cdot \frac{h^3}{5 \cdot 25}  = 400

    \frac{h^3}{125}=\frac12

    Solve for h.
    Last edited by earboth; December 3rd 2011 at 11:00 AM.
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