## Task 3 - 240613

If you place a cube on a table so that it does not protrude over the tabletop on any side, exactly five of its six faces are visible.

Similarly, a smaller cube can be placed on a larger one so that exactly five of the smaller cube's six faces are visible, while the sixth lies completely on the larger cube, without protruding over it on any side.

In this way, three cubes with edge lengths $a_{1}=20 \mathrm{~cm}, a_{2}=10 \mathrm{~cm}, a_{3}=4 \mathrm{~cm}$ are to be stacked on top of each other in size order, with the largest cube at the bottom on the tabletop. What is the sum of the areas of all visible face parts of the three cubes?
Visible are, for each of the three dice, first the four side faces (the mantle). They have the areas

$$
A_{1}=4 \cdot a_{1}^{2}=1600 \mathrm{~cm}^{2}, \quad A_{2}=4 \cdot a_{2}^{2}=400 \mathrm{~cm}^{2}, \quad A_{3}=4 \cdot a_{3}^{2}=64 \mathrm{~cm}^{2}
$$

Visible parts are secondly parts of the top faces of the three dice.

The areas of these parts together amount to the area of the top face of the largest die, i.e., $A_{D}=a_{1}^{2}=400 \mathrm{~cm}^{2}$. No further visible parts occur.

Therefore, for the sum of the areas of all visible parts, we have

$$
A=A_{1}+A_{2}+A_{3}+A_{D}=2464 \mathrm{~cm}^{2}
$$