Answer :
Let's break down the problem and find the volume of one pyramid based on the given conditions.
Given:
- The cube has a volume of [tex]\( b^3 \)[/tex].
- The cube is divided into 6 pyramids with equal bases and heights.
- The height of each pyramid is [tex]\( h \)[/tex].
First, we must determine the volume of one pyramid. The volume of the cube is [tex]\( b \times b \times b = b^3 \)[/tex]. Since the 6 pyramids together make up the entire volume of the cube, each pyramid's volume would be one-sixth of the cube's volume:
[tex]\[ \text{Volume of one pyramid} = \frac{b^3}{6} \][/tex]
We know the volume of a pyramid is calculated using the formula:
[tex]\[ \text{Volume of a pyramid} = \frac{1}{3} \times (\text{Base Area}) \times (\text{Height}) \][/tex]
Here, the base area [tex]\( B \)[/tex] of each pyramid is [tex]\( b \times b = b^2 \)[/tex] and the height of each pyramid is [tex]\( h \)[/tex]. Using this information, let's evaluate the provided options for the volume expressions to check which one matches [tex]\( \frac{b^3}{6} \)[/tex].
1. [tex]\(\frac{1}{6}(b)(b)(2h)\)[/tex]:
[tex]\[ \frac{1}{6} b^2 \times 2h = \frac{1}{6} \times 2b^2h = \frac{2b^2h}{6} = \frac{b^2h}{3} \][/tex]
This option simplifies to [tex]\( \frac{b^2h}{3} \)[/tex], which matches [tex]\( \frac{1}{3} Bh \)[/tex].
2. [tex]\(\frac{1}{6}(b)(b)(6h)\)[/tex]:
[tex]\[ \frac{1}{6} b^2 \times 6h = \frac{1}{6} \times 6b^2h = b^2h \][/tex]
This option simplifies to [tex]\( b^2h \)[/tex], which matches [tex]\( Bh \)[/tex].
3. [tex]\(\frac{1}{3}(b)(b)(6h)\)[/tex]:
[tex]\[ \frac{1}{3} b^2 \times 6h = \frac{1}{3} \times 6b^2h = 2b^2h \][/tex]
This option simplifies to [tex]\( 2b^2h \)[/tex], which is significantly larger than the expected volume.
4. [tex]\(\frac{1}{3}(b)(b)(2h)\)[/tex]:
[tex]\[ \frac{1}{3} b^2 \times 2h = \frac{1}{3} \times 2b^2h = \frac{2b^2h}{3} \][/tex]
This option simplifies to [tex]\( \frac{2b^2h}{3} \)[/tex], which matches [tex]\( \frac{2}{3} Bh \)[/tex].
Thus, the correct expression for the volume of one pyramid, which matches [tex]\( \frac{b^3}{6} \)[/tex] correctly, is given by:
[tex]\[ \text{Volume of one pyramid} = b^2h \][/tex]
Therefore, the volume of one pyramid must equal [tex]\( \frac{1}{6}(b)(b)(6h) \)[/tex] or [tex]\( Bh \)[/tex].
Given:
- The cube has a volume of [tex]\( b^3 \)[/tex].
- The cube is divided into 6 pyramids with equal bases and heights.
- The height of each pyramid is [tex]\( h \)[/tex].
First, we must determine the volume of one pyramid. The volume of the cube is [tex]\( b \times b \times b = b^3 \)[/tex]. Since the 6 pyramids together make up the entire volume of the cube, each pyramid's volume would be one-sixth of the cube's volume:
[tex]\[ \text{Volume of one pyramid} = \frac{b^3}{6} \][/tex]
We know the volume of a pyramid is calculated using the formula:
[tex]\[ \text{Volume of a pyramid} = \frac{1}{3} \times (\text{Base Area}) \times (\text{Height}) \][/tex]
Here, the base area [tex]\( B \)[/tex] of each pyramid is [tex]\( b \times b = b^2 \)[/tex] and the height of each pyramid is [tex]\( h \)[/tex]. Using this information, let's evaluate the provided options for the volume expressions to check which one matches [tex]\( \frac{b^3}{6} \)[/tex].
1. [tex]\(\frac{1}{6}(b)(b)(2h)\)[/tex]:
[tex]\[ \frac{1}{6} b^2 \times 2h = \frac{1}{6} \times 2b^2h = \frac{2b^2h}{6} = \frac{b^2h}{3} \][/tex]
This option simplifies to [tex]\( \frac{b^2h}{3} \)[/tex], which matches [tex]\( \frac{1}{3} Bh \)[/tex].
2. [tex]\(\frac{1}{6}(b)(b)(6h)\)[/tex]:
[tex]\[ \frac{1}{6} b^2 \times 6h = \frac{1}{6} \times 6b^2h = b^2h \][/tex]
This option simplifies to [tex]\( b^2h \)[/tex], which matches [tex]\( Bh \)[/tex].
3. [tex]\(\frac{1}{3}(b)(b)(6h)\)[/tex]:
[tex]\[ \frac{1}{3} b^2 \times 6h = \frac{1}{3} \times 6b^2h = 2b^2h \][/tex]
This option simplifies to [tex]\( 2b^2h \)[/tex], which is significantly larger than the expected volume.
4. [tex]\(\frac{1}{3}(b)(b)(2h)\)[/tex]:
[tex]\[ \frac{1}{3} b^2 \times 2h = \frac{1}{3} \times 2b^2h = \frac{2b^2h}{3} \][/tex]
This option simplifies to [tex]\( \frac{2b^2h}{3} \)[/tex], which matches [tex]\( \frac{2}{3} Bh \)[/tex].
Thus, the correct expression for the volume of one pyramid, which matches [tex]\( \frac{b^3}{6} \)[/tex] correctly, is given by:
[tex]\[ \text{Volume of one pyramid} = b^2h \][/tex]
Therefore, the volume of one pyramid must equal [tex]\( \frac{1}{6}(b)(b)(6h) \)[/tex] or [tex]\( Bh \)[/tex].
