To find the length, breadth, and height of a cuboidal box when the areas of three adjacent faces are given, let's denote the dimensions as:

• $l$ = length
• $b$ = breadth
• $h$ = height

The areas of the three adjacent faces are given as:

• $l \times b = 120 \, \text{cm}^2$ (Area of the face with length and breadth)
• $b \times h = 72 \, \text{cm}^2$ (Area of the face with breadth and height)
• $h \times l = 60 \, \text{cm}^2$ (Area of the face with height and length)

To find $l$, $b$, and $h$, we can follow these steps:

1. Multiply all three equations together: $(l \times b) \times (b \times h) \times (h \times l) = 120 \times 72 \times 60$

   This simplifies to: $(l \times b \times h)^2 = 120 \times 72 \times 60$

2. Take the square root of both sides: $l \times b \times h = \sqrt{120 \times 72 \times 60}$

3. Calculate the square root: $l \times b \times h = \sqrt{518400} = 720 \, \text{cm}^3$

   So, the volume of the cuboid $l \times b \times h = 720 \, \text{cm}^3$.

4. Now, using the given areas and the volume, you can find the individual dimensions:

   • $l = \frac{720}{b \times h}$
   • $b = \frac{720}{l \times h}$
   • $h = \frac{720}{l \times b}$

   Substitute the values: $l = \frac{720}{72} = 10 \, \text{cm}, \quad b = \frac{720}{60} = 12 \, \text{cm}, \quad h = \frac{720}{120} = 6 \, \text{cm}$

5. Thus, the dimensions of the box are: $\text{Length } l = 10 \, \text{cm}, \quad \text{Breadth } b = 12 \, \text{cm}, \quad \text{Height } h = 6 \, \text{cm}$

Would you like more details or have any questions?

Here are some related questions you might find useful:

1. How can you verify the correctness of the dimensions calculated?
2. What is the total surface area of the cuboidal box?
3. What is the diagonal length of the cuboid?
4. How does the volume change if the height is doubled?
5. How do the dimensions change if the area of one face increases by 50%?

Tip: When dealing with cuboids, remember that multiplying the areas of adjacent faces and taking the square root gives you the product of all three dimensions, which simplifies solving for individual lengths.