Question
The four diagonals of a cube are drawn to create 6 square pyramids with the same base and height. The volume of the cube is (b)(b)(b). The height of each pyramid is h.
Four diagonals of a cube are drawn to create 6 square pyramids inside of the cube with the same base and height. The sides of the cube have lengths b. The height of each pyramid is h.
Therefore, the volume of one pyramid must equal one-sixth the volume of the cube, or
One-sixth (b)(b)(2h) or One-thirdBh.
One-sixth (b)(b)(6h) or Bh.
One-third (b)(b)(6h) or One-thirdBh.
One-third (b)(b)(2h) or Two-thirdsBh.
Four diagonals of a cube are drawn to create 6 square pyramids inside of the cube with the same base and height. The sides of the cube have lengths b. The height of each pyramid is h.
Therefore, the volume of one pyramid must equal one-sixth the volume of the cube, or
One-sixth (b)(b)(2h) or One-thirdBh.
One-sixth (b)(b)(6h) or Bh.
One-third (b)(b)(6h) or One-thirdBh.
One-third (b)(b)(2h) or Two-thirdsBh.
Answers
GPT-4o mini
Let's start by analyzing the situation described in the question.
1. **Volume of the Cube**: The volume of a cube with side length \( b \) is given by:
\[
V_{\text{cube}} = b^3
\]
2. **Pyramids Inside The Cube**: The cube is divided into 6 square pyramids with the same base and height. The base of each pyramid is a square that forms part of the base of the cube, and the height is defined as \( h \).
3. **Volume of Each Pyramid**: The volume \( V \) of a pyramid is given by the formula:
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]
Since each pyramid has a square base, let's denote the side length of the base square as \( B \). The base area \( A \) will then be:
\[
A = B^2
\]
Therefore, the volume of one pyramid can be expressed as:
\[
V_{\text{pyramid}} = \frac{1}{3} B^2 h
\]
4. **Total Volume of 6 Pyramids**: Since there are 6 pyramids, the total volume of the pyramids is:
\[
V_{\text{total pyramids}} = 6 \times V_{\text{pyramid}} = 6 \times \frac{1}{3} B^2 h = 2 B^2 h
\]
5. **Setting Volumes Equal**: We know that the total volume of all 6 pyramids must equal the volume of the cube:
\[
2 B^2 h = b^3
\]
From this, we can derive a relationship between the base area \( B^2 \), height \( h \), and side length \( b \) of the cube.
To answer the initial question regarding the volume of one pyramid as one-sixth the volume of the cube:
Since we derived that the total volume of the pyramids \( 2 B^2 h = b^3 \), it follows that:
\[
\frac{1}{6} b^3 = \frac{1}{2} B^2 h \Rightarrow B^2 h = \frac{1}{3} b^3
\]
Therefore, the volume of one pyramid is:
\[
V_{\text{pyramid}} = \frac{1}{3} B^2 h = \frac{1}{3} \times \frac{1}{3} b^3 = \frac{1}{9} b^3
\]
However, this shows that the answer you provided does not match the working above exactly, as it discusses several formulas without clear definitions of \( h \) or \( B \). In light of the question and derivation, it seems that the valid conclusion is:
The volume of one pyramid is \( \frac{1}{6} b^3 \).
Thus, it's more about ensuring coherent relationships rather than matching specific other statements provided, as the computation leads to the understanding of relationships between base area, height, and cube volume \(\mathbf{\frac{1}{3} B^2 h = \frac{1}{6} b^3}\).
1. **Volume of the Cube**: The volume of a cube with side length \( b \) is given by:
\[
V_{\text{cube}} = b^3
\]
2. **Pyramids Inside The Cube**: The cube is divided into 6 square pyramids with the same base and height. The base of each pyramid is a square that forms part of the base of the cube, and the height is defined as \( h \).
3. **Volume of Each Pyramid**: The volume \( V \) of a pyramid is given by the formula:
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]
Since each pyramid has a square base, let's denote the side length of the base square as \( B \). The base area \( A \) will then be:
\[
A = B^2
\]
Therefore, the volume of one pyramid can be expressed as:
\[
V_{\text{pyramid}} = \frac{1}{3} B^2 h
\]
4. **Total Volume of 6 Pyramids**: Since there are 6 pyramids, the total volume of the pyramids is:
\[
V_{\text{total pyramids}} = 6 \times V_{\text{pyramid}} = 6 \times \frac{1}{3} B^2 h = 2 B^2 h
\]
5. **Setting Volumes Equal**: We know that the total volume of all 6 pyramids must equal the volume of the cube:
\[
2 B^2 h = b^3
\]
From this, we can derive a relationship between the base area \( B^2 \), height \( h \), and side length \( b \) of the cube.
To answer the initial question regarding the volume of one pyramid as one-sixth the volume of the cube:
Since we derived that the total volume of the pyramids \( 2 B^2 h = b^3 \), it follows that:
\[
\frac{1}{6} b^3 = \frac{1}{2} B^2 h \Rightarrow B^2 h = \frac{1}{3} b^3
\]
Therefore, the volume of one pyramid is:
\[
V_{\text{pyramid}} = \frac{1}{3} B^2 h = \frac{1}{3} \times \frac{1}{3} b^3 = \frac{1}{9} b^3
\]
However, this shows that the answer you provided does not match the working above exactly, as it discusses several formulas without clear definitions of \( h \) or \( B \). In light of the question and derivation, it seems that the valid conclusion is:
The volume of one pyramid is \( \frac{1}{6} b^3 \).
Thus, it's more about ensuring coherent relationships rather than matching specific other statements provided, as the computation leads to the understanding of relationships between base area, height, and cube volume \(\mathbf{\frac{1}{3} B^2 h = \frac{1}{6} b^3}\).