To find the height of each pyramid given that six identical square pyramids can fill the same volume as a cube with the same base, let's derive this step-by-step.
1. Volume of the Cube:
Let the height of the cube be [tex]\( h \)[/tex] units.
The volume of the cube, [tex]\( V_{\text{cube}} \)[/tex], is given by:
[tex]\[
V_{\text{cube}} = h^3
\][/tex]
2. Geometry of Pyramids:
Each of the six pyramids has the same square base as the cube. If the side length of the base of the cube is [tex]\( h \)[/tex] units, then the area of the base of each pyramid is:
[tex]\[
\text{Base Area of Pyramid} = h^2
\][/tex]
3. Volume of a Square Pyramid:
The volume [tex]\( V_{\text{pyramid}} \)[/tex] of a pyramid with a base area [tex]\( B \)[/tex] and height [tex]\( H \)[/tex] is given by:
[tex]\[
V_{\text{pyramid}} = \frac{1}{3} B H
\][/tex]
Substituting [tex]\( B = h^2 \)[/tex] and [tex]\( H \)[/tex] as the height of the pyramid:
[tex]\[
V_{\text{pyramid}} = \frac{1}{3} h^2 H
\][/tex]
4. Relationship Between Volumes:
Six such pyramids fill the same volume as the cube. Therefore:
[tex]\[
6 \times V_{\text{pyramid}} = V_{\text{cube}}
\][/tex]
Substituting the volumes:
[tex]\[
6 \times \frac{1}{3} h^2 H = h^3
\][/tex]
5. Solve for [tex]\( H \)[/tex]:
[tex]\[
2 h^2 H = h^3
\][/tex]
Divide both sides by [tex]\( 2 h^2 \)[/tex]:
[tex]\[
H = \frac{h^3}{2 h^2} = \frac{h}{2}
\][/tex]
Thus, the height of each pyramid is [tex]\( \frac{1}{2} h \)[/tex] units.
Therefore, the correct answer is:
[tex]\[
\text{The height of each pyramid is } \frac{1}{2} h \text{ units.}
\][/tex]
