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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 [tex]$(b)(b)(b)$[/tex]. The height of each pyramid is [tex]$h$[/tex].

Therefore, the volume of one pyramid must equal one-sixth the volume of the cube, or

A. [tex]$\frac{1}{6}(b)(b)(2h)$[/tex] or [tex][tex]$\frac{1}{3} Bh$[/tex][/tex]
B. [tex]$\frac{1}{6}(b)(b)(6h)$[/tex] or [tex]$Bh$[/tex]
C. [tex]$\frac{1}{3}(b)(b)(6h)$[/tex] or [tex][tex]$\frac{1}{3} Bh$[/tex][/tex]
D. [tex]$\frac{1}{3}(b)(b)(2h)$[/tex] or [tex]$\frac{2}{3} Bh$[/tex]


Sagot :

Let's tackle this problem step-by-step.

1. Understand the given information:
- Volume of the cube: [tex]\( V_{cube} = b \times b \times b = b^3 \)[/tex]
- The height of each pyramid is denoted as [tex]\( h \)[/tex].

2. Divide the cube:
- The cube is divided into 6 pyramids, each with the same base and height.
- The volume of one pyramid, therefore, must be one-sixth (since there are 6 pyramids) of the total volume of the cube.

3. Volume of one pyramid:
- Since the volume of the cube is evenly divided among the 6 pyramids, the volume of one pyramid is:
[tex]\[ V_{pyramid} = \frac{1}{6} V_{cube} = \frac{1}{6} b^3 \][/tex]

4. Formula for the volume of a pyramid:
- The general formula for the volume of a pyramid is:
[tex]\[ V_{pyramid} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \][/tex]
- Here, the Base Area [tex]\( B \)[/tex] of each pyramid is [tex]\( b \times b = b^2 \)[/tex].

5. Height of the pyramid:
- The height of each pyramid in this context is not the same as [tex]\( h \)[/tex]; it's given as [tex]\( 2h \)[/tex] to consider the proportional relationship with the cube.

6. Equating the volumes:
- Using the given height [tex]\( 2h \)[/tex] and the base area formula, we write:
[tex]\[ V_{pyramid} = \frac{1}{3} \times b^2 \times 2h = \frac{2}{3} b^2 h \][/tex]

7. Verify the equality:
- We know the previously calculated volume of one pyramid is [tex]\( \frac{1}{6} b^3 \)[/tex].
- By computing the derived formula:
[tex]\[ \frac{2}{3} b^2 h \][/tex]

8. Compare both volumes:
- Notice that:
[tex]\[ \frac{2}{3} b^2 h = \frac{1}{6} b^3 \][/tex]

- This confirms that the derived formula [tex]\( \frac{2}{3} b^2 h \)[/tex] matches the volume requirement for one-sixth of the cube's volume.

Thus, the correct volume formula for one of the pyramids, considering the height [tex]\( h \)[/tex], is:
[tex]\[ \frac{2}{3} b^2 h \][/tex]

In the options provided, the correct choice is therefore:
[tex]\[ \frac{1}{3}(b)(b)(2 h) \text{ or } \frac{2}{3} B h \][/tex]