To solve this problem effectively, we need to understand the relationship between the volume of a cube and the volume of pyramids that fit inside it.
Let's start with the basics:
1. Volume of a Cube: A cube's volume is calculated as [tex]\( V = h^3 \)[/tex], where [tex]\( h \)[/tex] is the height (which is equal to the width and length since all sides of a cube are equal).
2. Volume of a Pyramid: The volume of a square pyramid can be calculated using the formula [tex]\( V = \frac{1}{3} \text{Base Area} \times \text{Height} \)[/tex].
Given that six identical pyramids can fill up a cube, we infer that the sum of the volumes of these six pyramids is equal to the volume of the cube.
Let:
- [tex]\( H \)[/tex] = height of the cube,
- [tex]\( b \)[/tex] = side length of the base of the pyramids (which is also the side length of the cube since the base of the pyramid matches the face of the cube),
- [tex]\( h \)[/tex] = height of each pyramid.
The volume of the cube is:
[tex]\[ V_{\text{cube}} = H^3 \][/tex]
The volume of one pyramid is:
[tex]\[ V_{\text{pyramid}} = \frac{1}{3} b^2 h \][/tex]
Since the base of each pyramid is identical to the face of the cube:
[tex]\[ b = H \][/tex]
Thus, the volume of one pyramid becomes:
[tex]\[ V_{\text{pyramid}} = \frac{1}{3} H^2 h \][/tex]
The total volume of the six pyramids must equal the volume of the cube:
[tex]\[ 6 \times \frac{1}{3} H^2 h = H^3 \][/tex]
Simplify the equation step-by-step:
[tex]\[ 2H^2 h = H^3 \][/tex]
Divide both sides by [tex]\(H^2\)[/tex] (considering [tex]\(H > 0\)[/tex]):
[tex]\[ 2h = H \][/tex]
Thus:
[tex]\[ h = \frac{H}{2} \][/tex]
Therefore, the height of each pyramid is [tex]\(\frac{1}{2} h\)[/tex].
So, the correct statement:
[tex]\[ \text{The height of each pyramid is} \ \frac{1}{2} h \ \text{units}. \][/tex]
