4. The areas of three faces of a wooden cuboid are $6 d m^{2} ; 8 d m^{2}$ and $12 d m^{2}$. At each of the vertices, a cube is cut out and removed such that one of its vertices coincides with the vertex of the cuboid, with the side of the cube being $25 \%$ of the smallest side of the cuboid. Calculate:

a) the number of vertices, edges, and faces of the resulting body,

b) the area and volume of the cuboid and the body.

c) The percentage of the area and the percentage of the volume of the body relative to the corresponding measure of the cuboid.
Solution. a) The obtained body has 56 vertices; 84 edges and 30 faces. Let $a, b, c, (a<b<c)$ be the dimensions of the cuboid expressed in $d m$. From $a b=6, a c=8, b c=12$ we get $(a b c)^{2}=576$, i.e., $a b c=24$. Thus, $a=2 d m, b=3 d m, c=4 d m$. Let $x$ be the edge of the cube (in $d m$). Then $x=\frac{25 \cdot 2}{100}=0.5 d m$.

b) Now it is easy to see that

$$
P_{kv}=52 d m^{2}, V_{kv}=24 d m^{3}, P_{t}=52 d m^{2} \text { and } V_{t}=24-8 \cdot 0.5^{3}=23 d m^{3}
$$

c) Clearly,

$$
p_{P}=\frac{100 P_{t}}{P_{kv}}=100 \% \text { and } p_{V}=\frac{100 V_{t}}{V_{kv}}=\frac{2300}{24} \%
$$

## VIII Department