Let's solve the question step-by-step to determine the volume of one of the square pyramids formed by drawing diagonals in a cube.
1. Volume of the Cube:
- A cube has side length \( b \).
- The volume of the cube is given by:
[tex]\[
V_\text{cube} = b^3
\][/tex]
2. Volume of One Pyramid:
- The cube is divided into 6 pyramids of equal volume.
- Therefore, the volume of one pyramid is:
[tex]\[
V_\text{pyramid} = \frac{1}{6} V_\text{cube} = \frac{1}{6} b^3
\][/tex]
3. Volume Formula for a Pyramid:
- The volume of a pyramid is given by:
[tex]\[
V_\text{pyramid} = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\][/tex]
- The base of each pyramid is one of the faces of the cube, which is a square with side length \( b \), so the base area is:
[tex]\[
\text{Base Area} = b^2
\][/tex]
- The height of each pyramid is given as \( h \).
4. Relate Volume Expressions:
- We need to equate the volume we calculated from dividing the cube to the formula for the volume of a pyramid:
[tex]\[
\frac{1}{6} b^3 = \frac{1}{3} \times b^2 \times h
\][/tex]
5. Finding the Correct Option:
- Let’s simplify and check the given options:
Option 1:
[tex]\[
\frac{1}{6} b^2 \times 2h = \frac{1}{3} b^2 \times h
\][/tex]
Which simplifies to:
[tex]\[
\frac{1}{3} b^2 h
\][/tex]
This does not match \(\frac{1}{6} b^3\).
Option 2:
[tex]\[
\frac{1}{6} b^2 \times 6h = b^2 \times h
\][/tex]
Which simplifies to:
[tex]\[
b^2 h
\][/tex]
This matches \(\frac{1}{6} b^3\) considering the height \(h\) fits correctly in the context of cube.
Option 3:
[tex]\[
\frac{1}{3} b^2 \times 6h = 2b^2 \times h
\][/tex]
This simplifies to:
[tex]\[
2b^2 h
\][/tex]
This does not match \(\frac{1}{6} b^3\).
Option 4:
[tex]\[
\frac{1}{3} b^2 \times 2h = \frac{2}{3} b^2 \times h
\][/tex]
This simplifies to:
[tex]\[
\frac{2}{3} b^2 h
\][/tex]
This does not match \(\frac{1}{6} b^3\).
After evaluating all the options, Option 2: \(\frac{1}{6}(b)(b)(6 h)\) or \( b^2 h \) matches our derived volume for one pyramid.
Therefore, the correct answer is:
[tex]\[
\boxed{2}
\][/tex]
