# Solid Geometry

• December 3rd 2011, 10:11 AM
TechnicianEngineer
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
• December 3rd 2011, 11:02 AM
princeps
Re: Solid Geometry
Quote:

Originally Posted by TechnicianEngineer
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.
• December 3rd 2011, 11:49 AM
earboth
Re: Solid Geometry
Quote:

Originally Posted by TechnicianEngineer
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.