To solve the problem, we need to compare the volumes of the cube and the pyramids with a common base. Let's begin by identifying the essential properties and calculating the volumes step-by-step.
1. Volume of the Cube:
- The cube has a height [tex]\( h \)[/tex], and since it is a cube, all sides are equal to [tex]\( h \)[/tex].
- The volume [tex]\( V_{cube} \)[/tex] of the cube is given by:
[tex]\[
V_{cube} = h^3
\][/tex]
2. Volume of a Square Pyramid:
- The volume [tex]\( V_{pyramid} \)[/tex] of a square pyramid with a base area [tex]\( A \)[/tex] and height [tex]\( H \)[/tex] is given by:
[tex]\[
V_{pyramid} = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\][/tex]
- Since the base of each pyramid is the same as the base of the cube and assuming the base area (A) is [tex]\( h^2 \)[/tex], we have:
[tex]\[
V_{pyramid} = \frac{1}{3} \times h^2 \times H_{pyramid}
\][/tex]
3. Equating Volumes:
- We are given that six identical pyramids fill the same volume as the cube. Thus:
[tex]\[
6 \times V_{pyramid} = V_{cube}
\][/tex]
- Substituting the volumes, we get:
[tex]\[
6 \times \left(\frac{1}{3} \times h^2 \times H_{pyramid}\right) = h^3
\][/tex]
- Simplifying:
[tex]\[
2 \times h^2 \times H_{pyramid} = h^3
\][/tex]
- Solving for [tex]\( H_{pyramid} \)[/tex]:
[tex]\[
2 \times h^2 \times H_{pyramid} = h^3
\][/tex]
[tex]\[
H_{pyramid} = \frac{h^3}{2 \times h^2}
\][/tex]
[tex]\[
H_{pyramid} = \frac{h}{2}
\][/tex]
4. Choosing Correct Options:
- The given options are:
[tex]\[
\text{The height of each pyramid is } \frac{1}{2} h \text{ units.}
\][/tex]
[tex]\[
The height of each pyramid is \frac{1}{3} h \text{ units.}
\][/tex]
[tex]\[
The height of each pyramid is \frac{1}{6} h \text{ units.}
\][/tex]
[tex]\[
The height of each pyramid is h \text{ units.}
\][/tex]
5. Conclusion:
- From the calculations, the height of each pyramid is found to be [tex]\( \frac{1}{3} h \)[/tex] units.
- Therefore, the correct answer is:
[tex]\[
The height of each pyramid is \frac{1}{3} h \text{ units.}
\][/tex]
