An open box at the top has its outer dimensions \[10 \times 9 \times 2.5{\rm{cm}}\] and its thickness is \[0.5{\rm{cm}}\]. Find the volume of the metal used.

Hint:
Here, we will find the outer volume of this cuboidal box using the given information. Then, subtracting the thickness from each dimension, we will be able to find the inner dimensions and multiplying them together, we will find the inner volume of this box. Now, if we subtract the inner volume from the outer one, then, we will be able to find the required volume of the metal used.

Formula Used:
Volume of a cuboid \[ = l \times b \times h\], where \[l,b,h\] are length, breadth and height respectively.

Complete step by step solution:
Given outer dimensions of the open box are:
Length, \[L = 10{\rm{cm}}\]
Breadth, \[B = 9{\rm{cm}}\]
Height, \[H = 2.5{\rm{cm}}\]
Now, outer volume of this cuboidal box \[ = L \times B \times H\]
This is because Volume of a cuboid \[ = l \times b \times h\], where \[l,b,h\] are length, breadth and height respectively.
\[ \Rightarrow \]Outer Volume\[ = 10 \times 9 \times 2.5 = 9 \times 25 = 225{\rm{c}}{{\rm{m}}^2}\]
Now, it is given that the thickness of the open box is \[0.5{\rm{cm}}\]
Hence, the inner dimensions of this box will be \[0.5{\rm{cm}}\] less than the outer dimensions.
Thus, the inner dimensions:
Length, \[l = \left( {L - 2 \times 0.5} \right) = \left( {10 - 1} \right) = 9{\rm{cm}}\]
Breadth, \[b = \left( {B - 2 \times 0.5} \right) = \left( {9 - 1} \right) = 8{\rm{cm}}\]
Height, \[h = \left( {H - 0.5} \right) = \left( {2.5 - 0.5} \right) = 2{\rm{cm}}\]
Now, here, we have subtracted the thickness twice in the length as well as the breadth because in a cuboidal box, we have its surface area as \[2\left( {lb + bh + hl} \right)\]
Hence, we will subtract the thickness from both the lengths and both the breadths of the cuboid.
But, since the cuboidal box is open at the top.
Hence, we will not consider that and subtract the thickness only once from the height.
Now, using these inner dimensions, we will find the inner volume of this cuboidal box.
Thus, Inner Volume \[ = l \times b \times h\]
\[ \Rightarrow \] Inner Volume \[ = 9 \times 8 \times 2 = 9 \times 16 = 144{\rm{c}}{{\rm{m}}^2}\]
Therefore, in order to find the volume of the metal used, we will simply subtract the inner volume from the outer volume to find the required volume.
Thus, the Volume of metal used \[ = \left( {225 - 144} \right) = 81{\rm{c}}{{\rm{m}}^2}\]

Therefore, the volume of metal used is \[81{\rm{c}}{{\rm{m}}^2}\]
Hence, this is the required answer.

Note:
Here the shape of the box is not given but we will assume it to be a cuboid because the dimensions of the box are given as length, breadth and height. A cuboid is a three dimensional shape whose sides are rectangular. In real life, rooms are usually in the shape of cuboid but if in case all the 4 walls and the top and bottom of a room have the same length, breadth, and height, this would mean that it is cubical in shape as a cube has all the sides equal and square-shaped. Some other examples of cuboidal figures which we see in our day to day life are matchboxes, shoeboxes, books, etc. All of these have a length, breadth and height respectively