Let's walk through the problem step-by-step to understand the relationship between the height of the cube and the height of the pyramids that fill it.
1. Volume of the Cube:
- Given the height [tex]\( h \)[/tex] of the cube, we know that the side length of the cube is also [tex]\( h \)[/tex] because all sides of a cube are equal.
- The volume [tex]\( V_{\text{cube}} \)[/tex] of the cube is calculated as:
[tex]\[
V_{\text{cube}} = h^3
\][/tex]
2. Volume of a Square Pyramid:
- The base area [tex]\( A_{\text{base}} \)[/tex] of the square pyramid is the same as one face of the cube, so:
[tex]\[
A_{\text{base}} = h^2
\][/tex]
- Let the height of the pyramid be [tex]\( h_{\text{pyramid}} \)[/tex].
- The volume [tex]\( V_{\text{pyramid}} \)[/tex] of a square pyramid is given by:
[tex]\[
V_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height}
= \frac{1}{3} \times h^2 \times h_{\text{pyramid}}
\][/tex]
3. Relationship Between the Cubic and Pyramidal Volumes:
- We are given that six identical square pyramids perfectly fill the volume of the cube. Therefore:
[tex]\[
6 \times V_{\text{pyramid}} = V_{\text{cube}}
\][/tex]
- Substitute the volumes into this equation:
[tex]\[
6 \times \left( \frac{1}{3} \times h^2 \times h_{\text{pyramid}} \right) = h^3
\][/tex]
- Simplify the equation:
[tex]\[
2 \times h^2 \times h_{\text{pyramid}} = h^3
\][/tex]
4. Solve for the Height of Each Pyramid:
- To find [tex]\( h_{\text{pyramid}} \)[/tex], divide both sides of the equation by [tex]\( 2h^2 \)[/tex]:
[tex]\[
h_{\text{pyramid}} = \frac{h^3}{2h^2} = \frac{h}{2}
\][/tex]
Therefore, the height of each pyramid is:
[tex]\[
h_{\text{pyramid}} = \frac{1}{2} h
\][/tex]
The correct answer is:
- The height of each pyramid is [tex]\(\frac{1}{2} h\)[/tex] units.
