In the cuboid $A B C D E F G H$, a solid $P Q R S T U V X$ is placed, whose vertices are the midpoints of the edges of the cuboid, see the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_3f052d529d7f6d768bd1g-4.jpg?height=709&width=1196&top_left_y=1116&top_left_x=453)

Calculate the volume and surface area of the solid, given: $|A B|=8 \mathrm{~cm},|B C|=6 \mathrm{~cm},|B F|=4 \mathrm{~cm}$.
Quadrilateral $P Q R S$ in rectangle $A D E H$ is a rhombus, whose area is $S_{p}=\frac{1}{2}|A D| \cdot|A E|=\frac{6 \cdot 4}{2}=12\left(\mathrm{~cm}^{2}\right)$. The body $P Q R S T U V X$ is a prism with a rhombus base and height $|A B|$, so its volume is equal to $S_{p} \cdot|A B|=12 \cdot 8=96\left(\mathrm{~cm}^{3}\right)$.

To calculate the surface area, we need to know the length of the side of the rhombus. Side $R S$ is the hypotenuse in the right-angled triangle $R H S$, so $|R S|=\sqrt{3^{2}+2^{2}}=\sqrt{13}(\mathrm{~cm})$. The lateral surface area of the prism $P Q R S T U V X$ is $S_{p l}=4 \cdot|R S| \cdot|A B|=4 \cdot \sqrt{13} \cdot 8 \doteq 115.4\left(\mathrm{~cm}^{2}\right)$. The surface area of the prism is therefore $2 \cdot S_{p}+S_{p l} \doteq 2 \cdot 12+115.4 \doteq 139.4\left(\mathrm{~cm}^{2}\right)$.