surface area of a solid

The surface area of the "half" cuboid is the area of the hexagon plus half the area of the original cuboid. To calculate the area of the hexagon, we first assign the center of the cuboid as the origin $O$ of a vector space.

Then the vectors to the vertices of the hexagon are $\begin{cases} {\bf A} = -5 {\bf i} + 0 {\bf j} + 4 {\bf k} \\ {\bf B} = 0 {\bf i} + 6 {\bf j} + 4 {\bf k} \\ {\bf C} = 5 {\bf i} + 6 {\bf j} + 0 {\bf k} \\ {\bf D} = 5 {\bf i} + 0 {\bf j} - 4 {\bf k} \\ {\bf E} = 0 {\bf i} - 6 {\bf j} - 4 {\bf k} \\ {\bf F} = - 5 {\bf i} - 6 {\bf j} + 0 {\bf k} \end{cases}$

Consider the cross products, we note that ${\bf A \times B} = {\bf B \times C} = {\bf C \times D} = {\bf D \times E} = {\bf E \times F} = {\bf F \times A} = -24{\bf i} + 20 {\bf j} -30 {\bf k}$ . $\implies |{\bf A \times B}| = \sqrt{24^2+20^2+30^2}=2\sqrt{469}$ . Note that $\dfrac {|{\bf A\times B}|}2=\sqrt{469}$ is the area of the $\triangle AOB$ . This means that the hexagon is made up of 6 triangle of equal area and the area of the hexagon is $6\sqrt{469}$ . Therefore, the surface area of the "half" cuboid is $6\sqrt{469} + 8 \times 10+ 10 \times 12 + 12 \times 8 \approx \boxed{426}$