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Three cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are used to form a single cube. Find its edge. Also, find the surface area of the new cube.

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Hint: First of all, find the volume of the 3 cubes of edges 3 cm, 4 cm, and 5 cm. Then equal the total volume of three cubes to the volume of the new cube to get the edge of the new cube. Then find the surface area of the cube, that is, \[S=6{{\left( \text{edge} \right)}^{2}}\].

Complete step-by-step answer:

Here, we are given 3 cubes whose edges measure 3 cm, 4 cm, and 5 cm are used to form a single cube. We have to find the edge as well as the surface area of the new cube.
Before proceeding with the question, we must know that when one object of a particular shape is converted into other objects/objects of the same or different shape, then volume always remains constant. That is, the volume of the object before and after the conversion would be the same, while the other quantities such as surface area, length, breadth, radius, etc. could change.
Now, we are given 3 cubes of edges 3 cm, 4 cm, and 5 cm. We will first find the volume of these cubes.
We know that the volume of the cube \[={{a}^{3}}....\left( i \right)\] where ‘a’ is the edge of the cube.
So, we get,
The volume of the cube of edge 5 cm \[={{\left( 5 \right)}^{3}}c{{m}^{3}}=125c{{m}^{3}}....\left( ii \right)\]
The volume of the cube of edge 4 cm \[={{\left( 4 \right)}^{3}}c{{m}^{3}}=64c{{m}^{3}}....\left( iii \right)\]
The volume of the cube of edge 3 cm \[={{\left( 3 \right)}^{3}}c{{m}^{3}}=27c{{m}^{3}}....\left( iv \right)\]
Now, we are given that these 3 cubes are used to form a new cube. Let us consider the edge of the new cube as A. So, we get,
The volume of the new cube \[={{\left( A \right)}^{3}}c{{m}^{3}}....\left( v \right)\]
As we already know that when one or more object is converted into another, then volume remains constant, therefore we get,
(Volume of the cube of edge 5 cm) + (Volume of the cube of edge 4 cm) + (Volume of the cube of edge 3 cm) = (Volume of the cube of edge A cm)
By substituting the values of LHS and RHS from the equation (ii), (iii), (iv) and (v), we get,
\[125+64+27={{A}^{3}}\]
Or, \[{{A}^{3}}=216\]
By taking cube root on both the sides, we get,
\[A=\sqrt[3]{216}\]
\[A=6\text{ cm}\]
Hence, the edge of the new cube formed is 6 cm.
Now, we know that the surface area of the cube \[=6{{\left( edge \right)}^{2}}\] or \[6{{a}^{2}}\]
By substituting a = A = 6 cm, we get,
The surface area of the cube \[=6{{\left( 6 \right)}^{2}}\]
\[=216\text{ c}{{\text{m}}^{2}}\]

Note: Students must remember that whenever a 3-dimensional object is converted into other objects, then volume always remains constant. Also, students must take care of the units like the volume of the object is always in cubic meters while the area of the object is always in square meters and so on in any question.