Answer:
[tex]3456\; \rm cm^{2}[/tex].
Step-by-step explanation:
The diagram suggests that the length of a row of five cubes is [tex]40\; \rm cm[/tex]. Hence, the length of the edge of each cube would be:
[tex]\displaystyle \frac{40\; \rm cm }{5} = 8\; \rm cm[/tex].
That would correspond to an area of [tex](8\; \rm cm)^{2} = 64\; \rm cm^{2}[/tex] for each face of the cube.
Refer to the diagram attached.
Viewing from the front and from the rear each shows [tex]11[/tex] faces of the cubes.
Viewing from the top and from the bottom each shows [tex]10[/tex] faces of the cubes.
Viewing from the left and from the right each shows [tex]6[/tex] faces of the cubes.
When viewed from the six perspectives, [tex]2 \times 11 + 2 \times 10 + 2 \times 6 = 54[/tex] faces of the cubes are visible in total.
For this particular construction, all faces that need to be painted are visible when viewed from exactly one of the six perspectives: front, rear, top, right, left, right. Hence, Sam would need to paint [tex]54[/tex] of such [tex]64\; \rm cm^{2}[/tex] squares.
The area that Sam needs to paint would be [tex]54 \times (64\; \rm cm^{2}) = 3456\; \rm cm^{2}[/tex].
