Volume of a Orthohedron: How many times does the shape fit inside of another shape?

To solve this problem, we'll calculate the volumes of both cuboids:

• Step 1: Determine the volume of the larger cuboid
• Step 2: Determine the volume of the smaller cuboid
• Step 3: Calculate how many smaller cuboids fit inside the larger cuboid

Now, let's work through each step:
Step 1: Let the dimensions of the larger cuboid be $(L, W, H)$. The volume is $V_{large} = L \times W \times H$.
Step 2: The dimensions of the smaller cuboid are $\left(\frac{L}{3}, \frac{W}{3}, \frac{H}{3}\right)$. Thus, the volume is $V_{small} = \frac{L}{3} \times \frac{W}{3} \times \frac{H}{3} = \frac{L \times W \times H}{27}$.
Step 3: To find the number of smaller cuboids that fit inside the larger cuboid, divide the larger volume by the smaller volume:
$\frac{V_{large}}{V_{small}} = \frac{L \times W \times H}{\frac{L \times W \times H}{27}} = 27$

Therefore, the number of times the smaller cuboid fits into the larger cuboid is 27.

To solve this problem, we need to determine how many smaller cuboids fit into the larger cuboid given that each side of the smaller cuboid is one-eighth the length of the corresponding side of the larger cuboid.

First, let's calculate the scaling effect on volume:

• The length $AB$ of the smaller cuboid is $\frac{1}{8}$ of length $AC$ of the larger cuboid.
• The width $BF$ of the smaller cuboid is $\frac{1}{8}$ of width $CG$ of the larger cuboid.
• The height $AD$ of the smaller cuboid is $\frac{1}{8}$ of height $AE$ of the larger cuboid.

The volume of a cuboid is given by multiplying its three dimensions (length, width, and height). Thus, the volume of the smaller cuboid is:

$\left(\frac{1}{8} \times AC\right) \times \left(\frac{1}{8} \times CG\right) \times \left(\frac{1}{8} \times AE\right)$

$= \frac{1}{8} \times \frac{1}{8} \times \frac{1}{8} \times (AC \times CG \times AE)$

$= \frac{1}{8^3} \times (\text{volume of the larger cuboid})$

$= \frac{1}{512} \times (\text{volume of the larger cuboid})$

Therefore, the volume of the smaller cuboid is $\frac{1}{512}$ of the larger cuboid's volume. This indicates that:

$512$ smaller cuboids fit into the larger cuboid.

Therefore, the number of times the small cuboid fits inside the larger cuboid is 512.

To solve this problem, the approach involves determining the volume of each cuboid and assessing how many times the volume of the small cuboid fits into the volume of the large cuboid.

Step-by-step solution:

• First, note the relationships: $AB = DF$, $BE = FG$, and $AC = \frac{1}{4} AD$. These indicate equivalent heights and widths but differing lengths.
• The sides of the larger cuboid can be assumed as follows for simplification: If $AD = 4x$, then $AC = x$. Since $AB = DF$ and $BE = FG$, widths and heights remain the same for both cuboids.
• The volume of the larger cuboid can be given as: $V_{\text{large}} = (4x) \times w \times h$.
• The volume of the smaller cuboid can be given as: $V_{\text{small}} = x \times w \times h$.
• The ratio of volumes then is $\frac{V_{\text{large}}}{V_{\text{small}}} = \frac{(4x) \times w \times h}{x \times w \times h} = 4$.

Therefore, the smaller cuboid can fit into the larger cuboid 4 times.

The correct answer is 4.