Calculate Cuboid Volume: 5 Orthohedra with AB=5, BC=4 and 1/3 Junction Ratio

Let's solve the problem by calculating the volume of the large cuboid step-by-step:

Step 1: Determine the dimensions of each small orthohedron

• Given $AB = 5$ and $BC = 4$, we understand these are the sides of a triangle segment within the cuboid's formation.
• The problem states $DB = \frac{1}{3}$ of the total sum of $AB$ and $CB$, which implies $CB$ is as large as $BC$.
• Hence, $DB = \frac{1}{3} (5 + 4) = \frac{1}{3} \times 9 = 3$.
• This deduction allows us to assume the height of each small orthohedron $h = 3$.

Step 2: Calculating the volume of one small orthohedron

• Each orthohedron has dimensions: $AB = 5$, $BC = 4$, and the height $h = 3$.
• Therefore, the volume $V_{\text{small}}$ is calculated as:
• $V_{\text{small}} = AB \times BC \times h = 5 \times 4 \times 3 = 60$ cm³

Step 3: Calculate the total volume of the large cuboid

• The large cuboid is composed of 5 such small orthohedra, so:
• $V_{\text{large}} = 5 \times V_{\text{small}} = 5 \times 60 = 300$ cm³

Thus, the volume of the large cuboid is $300$ cm³.