A solid cube of side length $4 \mathrm{~cm}$ is cut into two pieces by a plane that passed through the midpoints of six edges, as shown. To the nearest square centimetre, the surface area of each half cube created is
(A) 69
(B) 48
(C) 32
(D) 65
(E) 58

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-113.jpg?height=320&width=293&top_left_y=214&top_left_x=1293)
When the cube is sliced in half in this manner, each half cube will have 7 faces: one hexagonal face from the slice, and 6 faces each of which is part of a face of the original cube.

By symmetry, the total area of these last 6 faces will be half of the surface area of the original cube, or $\frac{1}{2} \times 6 \times 4^{2}=48 \mathrm{~cm}^{2}$.

Lastly, we must calculate the area of the hexagonal face. By symmetry, the face is a regular hexagon. The side length is equal to the length of the line segment joining the midpoints of adjacent sides of a square of side length 4 , or $\sqrt{2^{2}+2^{2}}=2 \sqrt{2}$.

So we need to calculate the area of a regular hexagon of side length $2 \sqrt{2}$.

Consider a regular hexagon $A B C D E F$. Each interior angle is $120^{\circ}$.

Draw diagonals $A D, B E$ and $C F$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-223.jpg?height=312&width=350&top_left_y=2094&top_left_x=947)

By symmetry, these diagonals divide the hexagon into 6 equilateral triangles, each of side length $2 \sqrt{2}$.

Consider one of these equilateral triangles.

Draw altitude from vertex $O$ to $P$ on side $D E$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-224.jpg?height=244&width=263&top_left_y=301&top_left_x=991)

Since $\triangle O D E$ is equilateral, $P$ is the midpoint of $D E$, so $E P=\sqrt{2}$.

But $\triangle O P E$ is $30^{\circ}-60^{\circ}-90^{\circ}$, so $O P=\sqrt{3} E P=\sqrt{6}$.

Therefore, the area of $\triangle O D E=\frac{1}{2}(E D)(O P)=\frac{1}{2}(2 \sqrt{2})(\sqrt{6})=\sqrt{12}=2 \sqrt{3} \mathrm{~cm}^{2}$.

Therefore, the area of hexagon $A B C D E F$ is $6(2 \sqrt{3})=12 \sqrt{3} \mathrm{~cm}^{2}$.

Therefore, the total surface area of each half cube is $48+12 \sqrt{3} \approx 69 \mathrm{~cm}^{2}$.

ANSWER: (A)