Surface Area of a Cuboid: Identifying and defining elements

Every cuboid is made up of rectangles. These rectangles are the faces of the cuboid.

As we know that in a rectangle the parallel faces are equal to each other, we can conclude that for each face found there will be two rectangles.

Let's first look at the face painted orange,

It has width and height, 5 and 3, so we already know that they are two rectangles of size 5x6

Now let's look at the side faces, they also have a height of 3, but their width is 6,

And then we understand that there are two more rectangles of 3x6

Now let's look at the top and bottom faces, we see that their dimensions are 5 and 6,

Therefore, there are two more rectangles that are size 5x6

That is, there are
2 rectangles 5X6

2 rectangles 3X5

2 rectangles 6X3

Let's go through the options:

A - In this option, we can observe that there are two flaps on the same side.

If we try to turn this net into a box, we should obtain a box where on one side there are two faces one on top of the other while the other side is "open",
meaning this net cannot be turned into a complete and full box.

B - This net looks valid at first glance, but we need to verify that it matches the box we want to draw.

In the original box, we see that we have four flaps of size 9*4, and only two flaps of size 4*4,
if we look at the net we can see that the situation is reversed, there are four flaps of size 4*4 and two flaps of size 9*4,
therefore we can conclude that this net is not suitable.

C - This net at first glance looks valid, it has flaps on both sides so it will close into a box.

Additionally, it matches our drawing - it has four flaps of size 9*4 and two flaps of size 4*4.

Therefore, we can conclude that this net is indeed the correct net.

D - In this net we can see that there are two flaps on the same side, therefore this net will not succeed in becoming a box if we try to create it.

To solve this problem, we'll follow these steps:

• Step 1: Identify the dimensions of each pair of rectangles given the orthohedron's dimensions.
• Step 2: Each face of a cuboid corresponds to a rectangle, with three distinct pairs of dimensions.
• Step 3: Identify all pairs of dimensions and count the pairs that form sets of rectangles.

Now, let's work through each step:
Step 1: The orthohedron's dimensions are given as $4$, $7$, and $10$.
Step 2: A cuboid (orthohedron) has three pairs of opposite rectangular faces:
- Pair 1: Two rectangles with dimensions $4 \times 7$.
- Pair 2: Two rectangles with dimensions $4 \times 10$.
- Pair 3: Two rectangles with dimensions $7 \times 10$.
Step 3: Count each of the pairs to verify the total number of rectangles formed.
We find there are 6 rectangles in total, with the dimensions specified above fulfilling the conditions for each face of the cuboid.

The solution to the problem is that the orthohedron is formed of:
2 Rectangles $4 \times 7$,
2 Rectangles $4 \times 10$,
2 Rectangles $7 \times 10$.

These dimensions and quantities match choice #3 in the answer options provided.

To determine the feasability of a cuboid composed of two 4x3 rectangles and four 4x4 squares, we start by calculating the total surface area these would provide:

The total surface area contributes as follows:
- Two 4x3 rectangles: $2 \times 4 \times 3 = 24$
- Four 4x4 squares: $4 \times 4 \times 4 = 64$

The total surface area is $24 + 64 = 88$.

When forming a cuboid with dimensions $l \times w \times h$, the surface area should satisfy:
$2(lw + lh + wh) = 88$.

Now, let us examine possible dimensions that can result from the given face dimensions:

• Dimension 1: 4 (from the squares).
  
• Dimension 2: 3 (from the rectangles).
• Dimension 3 needs consideration from remaining panels.

Since using the given two 4x3 rectangles and four 4x4 squares in a valid arrangement providing 6 surface faces does not meet the criteria without repeating or extending beyond six faces, the random assembly of these square and rectangular panels cannot result in a valid orthogonal shape (cuboid).

Conclusively, this orthohedron is not possible.

Thus, the solution is that 'This orthohedron is not possible.'