The problem involves a cube and the creation of six square pyramids within the cube. Here is a step-by-step breakdown to solve the given problem:
1. Volume of the Cube:
The volume of a cube with side length [tex]\( b \)[/tex] is given by [tex]\( b \cdot b \cdot b = b^3 \)[/tex].
2. Understanding the Structure and Height:
When the four diagonals of a cube are drawn, they intersect at the center of the cube, forming six identical square pyramids. The height [tex]\( h \)[/tex] of each pyramid is from the center of the cube to one of its faces.
3. Volume of the Cube Split into Pyramids:
Since the entire cube is divided into 6 pyramids of equal volume, the volume of one pyramid is:
[tex]\[
\text{Volume of one pyramid} = \frac{1}{6} \times (\text{Volume of the cube}) = \frac{1}{6} \times b^3
\][/tex]
4. Volume Formula of a Pyramid:
The general formula for the volume [tex]\( V \)[/tex] of a pyramid with a base area [tex]\( B \)[/tex] and height [tex]\( h \)[/tex] is:
[tex]\[
V = \frac{1}{3} B h
\][/tex]
In this case, the base [tex]\( B \)[/tex] of each pyramid is a square with side length [tex]\( b \)[/tex], so:
[tex]\[
B = b \cdot b = b^2
\][/tex]
5. Equating the Pyramid’s Volume:
Setting the volume formula equal to the calculated volume:
[tex]\[
\frac{1}{3} B h = \frac{1}{3} b^2 h
\][/tex]
Recall from the setup that each pyramid’s volume is also equal to:
[tex]\[
\frac{1}{6} b^3
\][/tex]
6. Combining the Information:
From the problem, we have:
[tex]\[
\frac{1}{6} b^3 = \frac{1}{3} b^2 h
\][/tex]
Simplifying this expression:
[tex]\[
\frac{1}{6} b^3 = \frac{1}{3} b^2 h
\][/tex]
7. Given Options Analysis:
Simplifying [tex]\( \frac{1}{3} B h \)[/tex]:
[tex]\[
\frac{1}{3} b^2 h = \frac{1}{3} b b (2 h)
\][/tex]
- The provided solution for the options yields:
[tex]\[
\frac{1}{6} b b (2 h) = \frac{1}{6} b^2 \cdot 2h = \frac{1}{3} b^2 h
\][/tex]
From the provided results, the correct expression for the volume per pyramid is:
\]
\frac{1}{6} (b)(b)(2h)
\]
Thus, identifying the matching option:
Option: [tex]\(\frac{1}{6} (b)(b)(2 h)\)[/tex].
