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Sagot :
To determine the volume of one of the six square pyramids formed within a cube, let's break down the problem step by step.
1. Volume of the Cube:
The volume [tex]\( V \)[/tex] of a cube with side length [tex]\( b \)[/tex] is given by:
[tex]\[ V_{\text{cube}} = b^3 \][/tex]
2. Understanding the Pyramids:
Inside the cube, when the four diagonals are drawn, it creates six square pyramids. Each pyramid has its base as one of the square faces of the cube and a height equal to half the body diagonal of the cube.
3. Height of the Pyramids:
For each of these pyramids, the height can be considered as the body diagonal of the cube divided by 2. Given that the height [tex]\( h \)[/tex] provided corresponds with the cube's side, we have:
[tex]\[ h = \frac{\sqrt{3}}{2}b \][/tex]
4. Volume of a Pyramid:
The volume [tex]\( V \)[/tex] of a pyramid with a square base is given by:
[tex]\[ V_{\text{pyramid}} = \frac{1}{3} \times [\text{Base Area}] \times [\text{Height}] \][/tex]
The base area of the pyramid is:
[tex]\[ \text{Base Area} = b^2 \][/tex]
Substituting the height [tex]\( h \)[/tex] as:
[tex]\[ V_{\text{pyramid}} = \frac{1}{3} \times b^2 \times 2h \][/tex]
Simplifying:
[tex]\[ V_{\text{pyramid}} = \frac{1}{3} \times b^2 \times 2 \times \frac{1}{2} = \frac{1}{6} \times b^2 \times b \times 2 = \frac{1}{6}(b^2)(2h) \][/tex]
5. Comparing the Options:
We need to determine which expression from the options matches our derived formula:
[tex]\[ \frac{1}{6}(b^2)(2h) = \frac{1}{3}b^2h \][/tex]
We compare it to the options given:
- [tex]\(\frac{1}{6}(b)(b)(2h)\)[/tex] or [tex]\(\frac{1}{3}bh\)[/tex]
- [tex]\(\frac{1}{6}(b)(b)(6h)\)[/tex] or [tex]\(bh\)[/tex]
- [tex]\(\frac{1}{3}(b)(b)(6h)\)[/tex] or [tex]\(2bh\)[/tex]
- [tex]\(\frac{1}{3}(b)(b)(2h)\)[/tex] or [tex]\(\frac{2}{3}bh\)[/tex]
The correct one that matches our derived volume is:
[tex]\[ \frac{1}{6}(b^2)(2h) \][/tex]
or equivalently:
[tex]\[ \frac{1}{3}b^2h \][/tex]
Therefore, the correct choice is:
[tex]\[ \frac{1}{6}(b)(b)(2h) \text{ or } \frac{1}{3}bh. \][/tex]
1. Volume of the Cube:
The volume [tex]\( V \)[/tex] of a cube with side length [tex]\( b \)[/tex] is given by:
[tex]\[ V_{\text{cube}} = b^3 \][/tex]
2. Understanding the Pyramids:
Inside the cube, when the four diagonals are drawn, it creates six square pyramids. Each pyramid has its base as one of the square faces of the cube and a height equal to half the body diagonal of the cube.
3. Height of the Pyramids:
For each of these pyramids, the height can be considered as the body diagonal of the cube divided by 2. Given that the height [tex]\( h \)[/tex] provided corresponds with the cube's side, we have:
[tex]\[ h = \frac{\sqrt{3}}{2}b \][/tex]
4. Volume of a Pyramid:
The volume [tex]\( V \)[/tex] of a pyramid with a square base is given by:
[tex]\[ V_{\text{pyramid}} = \frac{1}{3} \times [\text{Base Area}] \times [\text{Height}] \][/tex]
The base area of the pyramid is:
[tex]\[ \text{Base Area} = b^2 \][/tex]
Substituting the height [tex]\( h \)[/tex] as:
[tex]\[ V_{\text{pyramid}} = \frac{1}{3} \times b^2 \times 2h \][/tex]
Simplifying:
[tex]\[ V_{\text{pyramid}} = \frac{1}{3} \times b^2 \times 2 \times \frac{1}{2} = \frac{1}{6} \times b^2 \times b \times 2 = \frac{1}{6}(b^2)(2h) \][/tex]
5. Comparing the Options:
We need to determine which expression from the options matches our derived formula:
[tex]\[ \frac{1}{6}(b^2)(2h) = \frac{1}{3}b^2h \][/tex]
We compare it to the options given:
- [tex]\(\frac{1}{6}(b)(b)(2h)\)[/tex] or [tex]\(\frac{1}{3}bh\)[/tex]
- [tex]\(\frac{1}{6}(b)(b)(6h)\)[/tex] or [tex]\(bh\)[/tex]
- [tex]\(\frac{1}{3}(b)(b)(6h)\)[/tex] or [tex]\(2bh\)[/tex]
- [tex]\(\frac{1}{3}(b)(b)(2h)\)[/tex] or [tex]\(\frac{2}{3}bh\)[/tex]
The correct one that matches our derived volume is:
[tex]\[ \frac{1}{6}(b^2)(2h) \][/tex]
or equivalently:
[tex]\[ \frac{1}{3}b^2h \][/tex]
Therefore, the correct choice is:
[tex]\[ \frac{1}{6}(b)(b)(2h) \text{ or } \frac{1}{3}bh. \][/tex]
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