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
$\displaystyle m=\rho \cdot V$
$\displaystyle m_1=m_2 \Rightarrow \rho \cdot V_1=\rho \cdot V_2 \Rightarrow V_1=V_2 \Rightarrow V_1=V-V_1 \Rightarrow 2\cdot V_1=V \Rightarrow$
$\displaystyle \Rightarrow 2\cdot\frac{B_1\cdot H_1}{3}=\frac{B\cdot H}{3} \Rightarrow H_1=\frac{B\cdot H}{2\cdot B_1}$
where $\displaystyle H_1$ is distance from vertex , $\displaystyle B_1$ is base of the small pyramid , $\displaystyle B$ is the base of the large pyramid and $\displaystyle H$ is altitude.
1. I assume that the material of the pyramid is homogenous. Then the weight correspond directly to the volume:
$\displaystyle w = k \cdot V$ where k is a real constant.
2. $\displaystyle b_0$ denotes the base area of the complete pyramide, b the base area of the smaller pyramide. Both areas are similar. Then
$\displaystyle V = \frac13 \cdot b_0 \cdot 5$
with $\displaystyle w = k \cdot \frac53 \cdot b_0 = 800$
3. h denotes the height of the smaller pyramid. Then the volume is calculated by:
$\displaystyle V_{pyr.\ small} = \frac13 \cdot b \cdot h$
with $\displaystyle w = k \cdot \frac13 \cdot b \cdot h = 400$
4. Since $\displaystyle b $ and $\displaystyle b_0$ are similar you can use the proportion:
$\displaystyle \frac b{b_0} = \left(\frac h5 \right)^2~\implies~b=\frac{h^2}{25} \cdot b_0$
5. Plug in this term into the equation at #3:
$\displaystyle k \cdot \frac13 \cdot \frac{h^2}{25} \cdot b_0 \cdot h = 400$
$\displaystyle \underbrace{\left(k \cdot \frac13 \cdot b_0 \cdot 5\right)}_{= 800} \cdot \frac{h^3}{5 \cdot 25} = 400$
$\displaystyle \frac{h^3}{125}=\frac12$
Solve for h.