## Task 5 - 010725

If you place a cube on the table, only 5 of its 6 faces are still visible. Now, three cubes with edge lengths $a_{1}=20 \mathrm{~cm}, a_{2}=10 \mathrm{~cm}$, and $a_{3}=4 \mathrm{~cm}$ are to be stacked on top of each other in size order. The largest cube stands at the bottom on the tabletop. The centers of the cubes are exactly aligned with each other.

How large is the total visible area of all three cubes?
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0316.jpg?height=391&width=303&top_left_y=1598&top_left_x=248)

(see figure) Of each cube, five side faces are visible (all except the bottom side), minus the covered top sides of the two lower cubes (which are as large as the bottom sides of the cubes placed on top of them).

The visible area is thus

$$
5 a_{1}^{2} + 5 a_{2}^{2} + 5 a_{3}^{2} - a_{2}^{2} - a_{3}^{2} = 5 a_{1}^{2} + 4 a_{2}^{2} + 4 a_{3}^{2} = 2464 \, \text{cm}^{2}
$$

Solutions of the 2nd Round 1961 adopted from [5]

### 4.3.3 3rd Round 1961, Class 7